VECTOR CALCULUS

Solved exercises

Equacions Diferencials

Grau en Enginyeria en Tecnologies Industrials (GETI)

ETSEIB – UPC

Departament de MatematiquesETS d’Enginyeria Industrial de Barcelona (ETSEIB)Universitat Politecnica de Catalunya (UPC)

Version: May 2021

https://mat-web.upc.edu/etseib/ed/

Note: All the underlined text provides links to websites which may help you better understand each problem, or

obtain further information.

Line integrals

1. Length of the arc of the cycloid σ(t) = (R(t− sin t), R(1− cos t)), 0 ≤ t ≤ 2π.

Resolution: A cycloid is the curve traced by a point on the rim of a circular wheel as the wheelrolls along a straight line without slipping. To get the parameterization, we first parameterize the rimby (R cos(3π/2− t), R sin(3π/2− t)), 0 ≤ t ≤ 2π, since the wheel goes clockwise and we measure theangle from the first contact point whose angle is 3π/2. As the center of the wheel travels like (Rt,R),the parameterization of the cycloid turns out to be

σ(t) = (R(cos(3π/2− t) +Rt, sin(3π/2− t) +R) = R(t− sin t, 1− cos t) , 0 ≤ t ≤ 2π.

Therefore σ′(t) = R(1− cos t, sin t), so that∥∥σ′(t)∥∥ = R

√2(1− cos t) = 2R

∣∣∣∣sin t

2

∣∣∣∣ and

L(cycloid) = 2R

ˆ 2π

0

∣∣∣∣sin t

2

∣∣∣∣ dt = 2R

ˆ 2π

0sin

t

2dt = 2R

[−2 cos

t

2

]t=2π

t=0

= 8R.

2. Length of the epicycloid with k cusps parameterized by

σ(t) = (R(k + 1) cos t−R cos((k + 1)t), R(k + 1) sin t−R sin((k + 1)t)), 0 ≤ t ≤ 2π.

Resolution: An epicycloid or hypercycloid is a plane curve produced by tracing the path of a chosenpoint on the circumference of a circlecalled an epicyclewhich rolls without slipping around a fixedcircle. It is a particular kind of roulette. To get the length, as we have the parameterization, we firsthave to calculate its first derivative which is

σ′(t) = (−R(k + 1) sin(t) +R sin((k + 1)t)(k + 1), R(k + 1) cos(t)−R cos((k + 1)t)(k + 1))

Now let’s simplify this expression saying that a = k + 1 :

σ′(t) = R(−a sin(t) + sin(at)a, a cos(t)− cos(at)a)

The norm is∥∥σ′(t)∥∥ =R√

(−a sin(t) + sin(at) a)2 + (a cos(t)− cos(at) a)2 =

=R

√a2(sin2(t)− 2 sin(t) sin(at) + sin2(at) + cos2(t)− 2 cos(t) cos(at) + cos2(at)) =

=aR

√sin2(t)− 2 sin(t) sin(at) + sin2(at) + cos2(t)− 2 cos(t) cos(at) + cos2(at) =

=aR√

2− 2 sin(t) sin(at)− 2 cos(t) cos(at) =

=aR√√√√2− 2 (sin(t) sin(at)︸ ︷︷ ︸

12(cos(t−at)−

cos(t+at))

+ cos(t) cos(at))︸ ︷︷ ︸12(cos(t−at)+

cos(t+at))

=

=aR√

2− 2 cos(t− at) = aR√

2√

1− cos(t− at) =

=aR√

2√

1− cos((1− a)t) = (cosx is an even function so) = aR√

2√

1− cos((a− 1)t) =

(k + 1)R√

2√

1− cos(kt) (where we have changed again k = a− 1)

Finally let us compute the length

L(epicycloid) =

ˆ 2π

0(k + 1)R

√2√

1− cos(kt) dt

= (k + 1)R√

2

ˆ 2π

0

√1− cos(kt)︸ ︷︷ ︸

|sin( kt2 )|=√

12− cos(kt)

2=

√1−cos(kt)√

2

dt

= (k + 1)R√

2

ˆ 2π

0

√2

∣∣∣∣sin(kt2)∣∣∣∣ dt = 2(k + 1)R

ˆ 2π

0

∣∣∣∣sin(kt2)∣∣∣∣ dt.

Now we apply the change u = kt2 , du = k

2 dt ⇐⇒ dt = 2k du and use that |sinu| is π-periodic and

that |sinu| = sinu ≥ 0 for 0 ≤ u ≤ π,

L(epicycloid) = 4k + 1

kR

ˆ πk

0|sinu| du = 4(k + 1)R

ˆ π

0|sinu| du = 4(k + 1)R [− cosu]u=πu=0

= 8(k + 1)R.

An epicycloid with k cusps has k archs, each one with length 8(k+1)Rk . You can see a figure for some

k just clicking on it: k = 1, k = 2, k = 3, k = 4. . .

3. Compute the theoretical time of navigation between Lisbon and New York if we follow a trajectory ofconstant direction and we sail at speed of 30 knots.

Resolution: In this exercise we need to compute a great-circle distance, which is the shortest distancebetween two points on the surface of a sphere, measured along the surface of the sphere. We are goingto establish that Lisbon is our Point 1 and New York is the Point 2.

Let θ1, ϕ1 and θ2, ϕ2, be the geographical longitude and latitude in radians of the two points, and∆θ,∆ϕ be their absolute differences.

2

Illustration of the central angle α, between two points P1 and P2.

The actual arc length d on a sphere of radius r can be trivially computed as d = Rα, where α is theangle between its two endpoints P1 and P2, measured from the center O of the sphere. If −π ≤ θj ≤ πand −π/2 ≤ ϕj ≤ π/2, then:

cosα =〈 ~OP1, ~OP2〉∥∥∥ ~OP1

∥∥∥∥∥∥ ~OP2

∥∥∥ = cosϕ1 cosϕ2 (cos θ1 cos θ2 + sin θ1 sin θ2) + sinϕ1 sinϕ2

= sinϕ1 sinϕ2 + cosϕ1 cosϕ2 cos(∆θ).

[Obs.: This formula can be directly applying the spherical law of cosines to the spherical trianglewhose vertices are P1, P2 and one of the poles an auxiliary third point; if the North Pole is chosen,the angles of this triangle are π/2− ϕ1, π/2− ϕ2 and α. ]

First let’s collect some data that we need to solve the problem, first of all we need the mean radiusof the Earth which is approximately R ≈ 6371 km, then we need the coordinates of Lisbon whichare (θ1, ϕ1) = (9o8′W, 38o43′N) ≈ (0.1594067, 0.6757333) rad, and also the coordinates of New Yorkwhich are (θ2, ϕ2) = (74o0′W, 40o43′N) ≈ (1.2915436, 0.7106399) rad.

Using such data and the formula obtained above, we can calculate the distance, which is d = Rα =0.8510653 · 6371 = 5422.14 km. Now that we know the distance, we only need to finish the problem bycalculating the time we need to go from Lisbon to New York at a constant speed of 30 knots, knowingthat a knot is 1.852 km/h the theoretical time of navigation t will be:

t =5422.14

30 · 1.852= 97.59 h,

which is a little bit more than 4 days.

4. A cassette tape with thickness of 16µm (micrometers) is rolled in a spool with internal radius r0 =1.11cm and external radius r1 = 2.46cm. How much does the tape of the spool measure? [Hint:Approximate the form of the spool by the curve described in polar coordinates through the equationr = g(θ) = cθ,θ0 ≤ θ ≤ θ1, being c > 0 a constant to be determined, so that rj = cθj .]

Resolution: To solve this problem we shall look at it as if we were to find the length of the curve(C) as an Archimedean spiral. See more information about Archimedean spiral in wikipedia.

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Archimedean spiral

Our data are r0 = 1.11 · 10−2m, r1 = 2.46 · 10−2m and the increment per revolution is 16 · 10−6m. Weare given C in polar coordinates as r = r(θ) for θ0 ≤ θ ≤ θ1, so in cartesian coordinates C can beparameterized as C = σ([θ0, θ1]):

(x, y) = σ(θ) = (r(θ) cos θ, r(θ) sin θ), θ0 ≤ θ ≤ θ1.

To find θ0 and θ1 first we have to determine c. We will do this imposing that the radius increases16 · 10−6m for each rotation. So, with any rotation

16 · 10−6 = g(θ + 2π)− g(θ) = 2πc,

so that

c =8 · 10−6

π.

Now we can calculate:

θ0 =r0c

=0.0111 · π8 · 10−6

,

θ1 =r1c

=0.0246 · π8 · 10−6

.

We can define both r(θ) and r′(θ) because it will be useful for us to work with them during thedefinition of the parameterization:

r (θ) = c · θ and r′ (θ) = c

Finally, we know that the arc length in polar coordinates is:

d` =

√r2 +

(dr

dθ

)2

dθ,

and so, the total arc length of the tape is:

ˆC

d` =

ˆ θ1

θ0

√r(θ)2 + r′(θ)2 dθ =

ˆ θ1

θ0

√(c · θ)2 + c2 dθ = c

ˆ θ1

θ0

√1 + θ2 dθ

= c ·[

1

2· θ ·

√1 + θ2 +

1

2ln∣∣∣θ +

√1 + θ2

∣∣∣]θ=θ1θ=θ0

≈ 94.6m,

where we have used the primitive

ˆ √t2 + a2 dt =

t

2

√t2 + a2 +

a2

2ln∣∣∣t+

√t2 + a2

∣∣∣ + ctant, which

can be found either by the change of variables t = a sinh(x) or the change by parts u =√t2 + a2,

dv = dt.

4

5. Mass of a spiral of the helix of radius R and height h parameterized by

σ(t) =

(R cos t, R sin t,

ht

2π

), 0 ≤ t ≤ 2π,

if the density at each point is proportional to the distance from the origin, being k > 0 the constant.

Resolution: Notice that this helix is a 3D-curve with height h. Its mass M is

ˆCρd`, where we

are said that ρ(x, y, z) = k√x2 + y2 + z2. Along C, ‖σ(t)‖ =

√R2 +

h2t2

4π2=

h

2π

√t2 + a2, where

a =2πR

h, so that ρ(σ(t)) =

kh

2π

√t2 + a2. Moreover, σ′(t) =

(−R sin t, R cos t,

h

2π

)so that

∥∥σ′(t)∥∥ =√R2 +

h2

4π2=

h

2π

√1 + a2. Finally

M =kh2

4π2

√1 + a2

ˆ 2π

0

√t2 + a2 dt =

kh2

4π2

√1 + a2

[t

2

√t2 + a2 +

a2

2ln∣∣∣t+

√t2 + a2

∣∣∣]t=2π

t=0

=kh2

4π2

√1 + a2

(π√

4π2 + a2 + 2π2 ln∣∣∣2π +

√4π2 + a2

∣∣∣− a2 ln a), a =

2πR

h,

where we have used that

ˆ √t2 + a2 dt =

t

2

√t2 + a2 +

a2

2ln∣∣∣t+

√t2 + a2

∣∣∣ + ctant, which can be

found either by the change of variables t = a sinhx or the change by parts u =√t2 + a2, dv = dt.

6. Mass of the cardioid defined in polar coordinates by r = R (1 + cos θ) if the density at each point isproportional to the square root of the distance from the origin, being k the constant.

Resolution: First of all we should define what a cardioid is. A cardioid is a plane curve traced bya point on the perimeter of a circle that is rolling around a fixed circle of the same radius (R). Seemore information about the cardiod in wikipedia.

Cardiod with R= 0.5 (polar coordinates)

We define both r(θ) and r′(θ) because it will be useful for us to work with them during the definitionof the parameterization.

r(θ) = R(1 + cos θ) , r′(θ) = −R sin θ.

5

Polar coordinates are defined to be: (x, y) = (r cos θ, r sin θ), but in this exercise we are given r as afunction of θ, so we have to substitute r = r(θ). As we can see in the figure, θ goes from 0 to 2π. Wecan define the curve in the following way:

C = σ([0, 2π]) with σ(θ) = (r(θ) cos θ, r(θ) sin θ) .

We know that the line integral along a piecewise smooth curve is defined as:

ˆCf d` =

ˆ b

af(σ(t)) ·

∥∥σ′(t)∥∥ dt.

So, now we have to compute the derivative and norm of σ(θ):

σ′(θ) =(r′(θ) cos θ − r(θ) sin θ, r′(θ) sin θ + r(θ) cos θ

)=

(cos θ − sin θsin θ cos θ

)(r′(θ)r(θ)

).

If we compute the determinant of the matrix, we can see that it is equal to 1. So, the norm of thederivative of the parameterization is:∥∥σ′(θ)∥∥ =

√r(θ)2 + r′(θ)2 =

√R2(1 + cos θ)2 +R2 sin2 θ = R

√1 + cos2 θ + 2 cos θ + sin2 θ

= R√

2 + 2 cos θ = R√

2√

1 + cos θ.

We define the density of the cardiod as ρ. In the statement of the problem it is said that ρ isproportional to the square root of the distance from the origin, which can be written as:

ρ(x, y) = k

√√x2 + y2.

If we apply the parameterization

ρ(σ(θ)) = k

√√r(θ)2 cos2 θ + r(θ)2 sin2 θ = k

√r(θ) = k

√R(1 + cos θ),

the mass is defined to be

M =

ˆCρd` =

2πˆ

0

ρ(σ(θ)) ·∥∥σ′(θ)∥∥ dθ = 2

π

0

ρ(σ(θ)) ·∥∥σ′(θ)∥∥ dθ

= 23/2kR

π

0

√R√

1 + cos θ ·√

1 + cos θ dθ = k(2R)3/2π

0

(1 + cos θ) dθ

= k(2R)3/2[θ + sin θ

]π0

= πk(2R)3/2.

The mass of the cardioid if the density at each point is proportional to the square root of the distancefrom the origin, being k the constant, is πk(2R)3/2.

7. Mass of the arc of the conic helix parameterized by

σ(t) =(Ret cos t, Ret sin t, Ret

)that joins the points A = (R, 0, R) and B = (0, Reπ/2, Reπ/2) if its linear density is given by ρ(σ(t)) =ket, being k > 0.

Resolution: In this case we see that 0 ≤ t ≤ π/2, as σ(0) = (R, 0, R) = A and σ(π/2) =(0, Reπ/2, Reπ/2) = B.

6

Arc of the conic helix between A and B for R= 1.

The conic helix is a 3D curve and its mass is:

M =

ˆ

C

ρdl =

π/2ˆ

0

ρ(σ(t))∥∥σ′(t)∥∥ dt.

We are already given ρ(σ(t)) so now we must find ‖σ′(t)‖:

σ′(t) = (Ret cos t−Ret sin t, Ret sin t+Ret cos t, Ret),∥∥σ′(t)∥∥ =

√R2e2t(cos2 t−((((((2 cos t sin t+ sin2 t+ sin2 t+((((

((2 sin t cos t+ cos2 t+ 1)

= Ret√

3.

Now we can finally compute its mass:

M =

ˆ

C

ρ dl =

π/2ˆ

0

ρ(σ(t))∥∥σ′(t)∥∥ dt = kR

√3

π/2ˆ

0

e2t dt

=kR√

3

2

π/2ˆ

0

2e2t dt =kR√

3

2

[e2t]t=π/2t=0

=kR√

3

2(eπ − 1) .

8. Mass of the semi-circle of radius R if its linear density is proportional to the cube of the distance fromthe line that divides the circle into two halves, being k > 0 the constant.

Resolution: To compute the mass of the semicircle we compute the following integral

ˆCρdl, where

ρ is the linear density of the semicircle C. The statements says that it is proportional to the cube ofthe distance that divides the semicircle into two halves, so, we have to find that distance d to computethe density.d = R sin θ so that ρ = kR3 sin3 θ.We parameterize our semicircle in polar coordinates, knowing that 0 ≤ θ ≤ π:

σ(θ) = (R cos θ,R sin θ),

σ′(θ) = (−R sin θ,R cos θ),∥∥σ′(θ)∥∥ =√

(−R sin θ)2 + (R cos θ)2 =√R2 = R.

7

We now proceed to the computation of the mass of the semi-circleˆCρ dl =

ˆ π

0kR3 sin3 θ R dθ = kR4

ˆ π

0

3

4sin θ − 1

4sin 3θ dθ

= kR4

(3

4[− cos θ]θ=πθ=0 +

1

4

[cos 3θ

3

]θ=πθ=0

)=

= kR4

(3

2− 1

6

)=

4

3kR4.

Finally, we get that the mass of the semicircle is4

3kR4, being R the radius and k > 0 the constant.

9. Area and mean height of a fence with a base on a quarter of an astroid inside a circle of radius 30parameterized by

σ(t) = (30 cos3 t, 30 sin3 t), 0 ≤ t ≤ π/2,and whose height is given by the function h(x, y) = 1 + y/3.

Resolution: An astroid is a hypocycloid with 4 cusps: the trace of a fixed point on a small circle ofradius r that rolls within a larger circle of radius 4r.

Astroid x2/3 + y2/3 = 302/3

It satisfies the equation x2/3 + y2/3 = a2/3 or (x1/3)2 + (y1/3)2 = (a1/3)2, so, similarly to the case of acircle, it is readily parameterized by x1/3 = a1/3 cos t, y1/3 = a1/3 sin t or x = a cos3 t, y = a sin3 t. Inthis exercise, a = 30, and we deal only with a quarter of an astroid, so we use the parameterizationσ(t) = (30 cos3 t, 30 sin3 t), 0 ≤ t ≤ π/2.

From σ′(t) = (3 · 30 cos2 t (− sin t), 3 · 30 · sin2 t (cos t)) = 90 sin t cos t(− cos t, sin t), we have that‖σ′(t)‖ = 90 |sin t cos t| = 90 sin t cos t, since 0 ≤ t ≤ π/2, and h(σ(t)) = 1 + 10 sin3 t.

The area of the fence is simply the integral of h along C = σ(t), t ∈ [0, π/2], that is

A(C) :=

ˆChd` =

ˆ π/2

0h(σ(t))

∥∥σ′(t)∥∥ dt =

ˆ π/2

0(1 + 10 sin3 t) 90 sin t cos t dt

= 90

(ˆ π/2

0sin t cos tdt+ 10

ˆ π/2

0sin4 t cos tdt

)

= 90

([sin2 t

2

]t=π/2t=0

+ 10

[sin5 t

5

]t=π/2t=0

)= 90

(1

2+ 2

)= 225.

8

The mean height h =

ˆChd`

ˆC

d`

=A(C)

L(C)=

225

45= 5, since

L(C) =

ˆC

d` =

ˆ π/2

090 sin t cos t dt = 45.

10. Compute the circulation

˛Cx dy − y dx, being C the positively oriented (traveled counter-clockwise)

boundary of a square D with side 2L and center (x0, y0). [ Obs.: This problem allows us to motivateGreen’s theorem.]

Resolution: We have a piecewise smooth curve C = C1 ∪ C2 ∪ C3 ∪ C4, so that

˛C

=4⋃i=1

ˆCi

.

On C1, x0 − L ≤ x ≤ x0 + L, y = y0 − L, so

ˆC1

x dy − y dx = −ˆ x0+L

x0−Ly0 − Ldx = −2L(y0 − L),

on C2, x = x0 + L, y0 − L ≤ y ≤ y0 + L, so

ˆC2

x dy − y dx =

ˆ y0+L

y0−Lx0 + Ldy = 2L(x0 + L),

on C3, x0 − L ≤ x ≤ x0 + L, y = y0 + L, so

ˆC3

x dy − y dx = −ˆ x0−L

x0+Ly0 + Ldx = 2L(y0 + L),

on C4, x = x0 + L, y0 − L ≤ y ≤ y0 + L, so

ˆC4

x dy − y dx =

ˆ y0+L

y0−Lx0 + Ldy = −2L(x0 − L).

Domain D with positive oriented boundary

Summing up,

˛Cx dy − y dx = 8L2, which is just the double of the area of the square: 1

2

˛∂D

x dy −

y dx =

¨D

dx dy, which is a consequence of Green’s theorem.

9

11. Circulation of the vector field F (x, y, z) = (z, x, y) along the curve C parameterized by σ(t) = (t, t2, t3),0 ≤ t ≤ 1.

Resolution: In this exercise, we will apply just the definition of the circulation of a vector field alonga curve: ˆ

C〈F , d`〉 =

ˆ 1

0〈F (σ(t)),σ′(t)〉dt.

We need to compute the scalar product of the vector field evaluated on the parameterized curve andthe derivative of the parameterization

F (σ(t)) = (t3, t, t2), σ′(t) = (1, 2t, 3t2), 〈F (σ(t)),σ′(t)〉 = t3 + 2t2 + 3t4.

Let’s do the computations to get the final result.

ˆC〈F , d`〉 =

ˆ 1

0t3 + 2t2 + 3t4 dt =

[t4

4+

2t3

3+

3t5

5

]t=1

t=0

=1

4+

2

3+

3

5=

91

60.

As we are given the parameterization, we are also given the orientation of the curve C, which doesneed to be checked.

The integral theorems I

12. Compute the circulation of the vector field:(a) F (x, y, z) = (2xy + z3, x2, 3xz2) along the helix parameterized by

σ(t) = (cos t2, sin t2, t2), 0 ≤ t ≤√π.

(b) F (x, y, z) = (ey, xey, 2z) along the curve parameterized by

σ(t) =

(1 + t

ˆ t

1eu

2du, t, t2

), 0 ≤ t ≤ 1.

Resolution: (a) In order to compute this circulation, first we will check if F is conservative. Thishappens when ∇∧ F = 0 and F ∈ C1.

∇∧ F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

2xy + z3 x2 3xz2

∣∣∣∣∣∣ =((0− 0), (3z2 − 3z2), (2x− 2x)

)= 0.

Also, F ∈ C1 because all of its components are continuously differentiable (polynomials and exponen-tials). Since both conditions meet, we can find a potential f for F such that F = ∇f . We will useNewton-Leibnitz Theorem to find the result:

ˆσ〈F , dl〉 =

ˆ b

a〈F (σ(t)),σ′(t)〉dt =

ˆ b

a〈∇f(σ(t)),σ′(t)〉dt =

ˆ b

a(f σ)′(t)dt = f(σ(b))− f(σ(a)).

This shows that the circulation of F doesn’t depend on the path, but only on the initial and finalpoints. Now, we will search for a function f that satisfies:

∂f

∂x= 2xy + z3,

∂f

∂y= x2,

∂f

∂z= 3xz2.

10

For this, we will integrate one by one the components to find the common and non-common results:

Ix =

ˆ2xy + z3dx = x2y + z3x+K1,

Iy =

ˆx2dy = x2y +K2,

Iz =

ˆ3xz2dz = z3x+K3.

Out of all these results, we can conclude that f(x, y, z) = x2y + z3x + K. We will now substitute inour original formula:ˆ

σ〈F , dl〉 = f(σ(

√π))− f(σ(0)) = f(−1, 0, π)− f(1, 0, 0) = (−π3)− (0) = −π3.

(b) First, we check if F is conservative, that is if ∇∧ F = 0 and F ∈ C1.

∇ ∧ F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

ey xey 2z

∣∣∣∣∣∣ = (0− 0, 0− 0, ey − ey) = (0, 0, 0) = 0.

And F ∈ C1 because all the components of F are continously diferenciable (since they are polinomialsand exponentials which do not cancel out at any point). As F is conservative, we will look for itspotential function f so that F =∇f , and by using Newton-Leibnitz Theorem we know:

ˆσ〈F , d`〉 =

ˆ b

a〈F (σ(t)),σ′(t)〉dt =

ˆ b

a〈∇f(σ(t)),σ′(t)〉 dt =

ˆ b

a(f σ)′(t) dt = f(σ(b))− f(σ(a)),

this means that the circulation of F does not depend on the path, but only on the initial and finalpoints.

Then, we look for a function f(x, y, z) which satisfies:

∂f

∂x= ey,

∂f

∂y= xey,

∂f

∂z= 2z.

We begin by integrating each of the components by the different variables and getting the commonand non-common results from each of them but just once (non repeated):

Ix =

ˆey dx = xey +K1

Iy =

ˆxey dy = xey +K2

Iz =

ˆ2z dz = z2 +K3

So we have that F = ∇f with f(x, y, z) = xey + z2 +K, where the K’s are constants. Therefore,ˆσ〈F , d`〉 = f (σ (1)))− f (σ(0)) = f(1, 0, 1)− f(1, 0, 0) = (e + 1 + k)− (1 + k) = e,

as

σ(1) =

(1 +

ˆ 1

1eu

2du, 1, 1

)= (1, 1, 1) ,

σ(0) =

(1 + 0

ˆ 0

1eu

2du, 0, 1

)= (1, 0, 0) .

11

13. Prove, using Green’s theorem, the area of the cycloid of radius R is 3 times the area of the circle ofalso radius R.

Resolution: The cycloid Ccyc is parameterized by σ(t) = (R(t− sin t), R(1− cos t)), 0 ≤ t ≤ 2π, but

with an opposite orientation to the curve C1 of the figure. To compute A(D) =

¨D

dx dy, we look

first for functions P , Q, such that Qx − Py = 1, for instance P (x, y) = −y, Q(x, y) = 0, and we applyGreen’s theorem to the domain D and its boundary C = ∂D = C0 ∪ C1, counter-clockwise orientedwith respect to D:

A(D) =

‰C=∂D

−y dx = −ˆC1

y dx−ˆC0⊂y=0

y dx =

ˆCcyc

y dx =

ˆ 2π

0R(1− cos t)︸ ︷︷ ︸

y(t)

R(1− cos t) dt︸ ︷︷ ︸x′(t) dt

= R2

ˆ 2π

0(1− cos t)2 dt = R2

ˆ 2π

01− 2 cos t+ cos2 t︸ ︷︷ ︸

12+ cos 2t

2

dt = R2

ˆ 2π

0

3

2− 2 cos t+

cos 2t

2dt

= R2

[3

2t−2 sin t+

sin 2t

4

]t=2π

t=0

= 3πR2.

14. Prove, using Green’s theorem, that the area of the rose given in polar coordinates by the equationr = R cos kθ is the half (respectively, the quarter) of the area of the circle of radius R when k is aneven number (respectively, when k is odd).

[Obs.: There are two interpretations of the meaning of the equation r = g(θ). In the first one,the distance from the origin cannot be negative, so the curve is formed by the points that in polarcoordinates are those that meet r = g(θ) > 0. In the second one, we can parameterize the curvethrough the application σ(θ) = (g(θ) cos θ, g(θ) sin θ) and consider it well defined even when g(θ) < 0.]

Resolution: We use the Green Theorem to show that we can compute the area computing F throughthe length of a curve. Using F (x, y) = (−y, x):

A(D) =1

2

‰∂D

x dy − y dx.

We will parameterize as follows:

σ(θ) = (x(θ), y(θ)) = (R cos kθ cos θ,R cos kθ sin θ) ,

which is in principle 2π-periodic: σ(θ+ 2π) = σ(θ) for any θ, and therefore in principle one considersintegrals from 0 to 2π to compute lengths, etc. So now we put our parameterization in the vector fieldand derive the parameterization:

F (σ(θ)) = (−R cos(kθ) sin(θ), R cos(kθ) cos(θ))

σ′(θ) = (−Rk sin(kθ) cos(kθ)−R cos(kθ) sin(θ),−Rk sin(kθ) sin(θ) +R cos(θ)R cos(θ))

From rose in Wikipedia:

How the parameter k affects shapes:

In the form k = n, for integer n, the shape will appear similar to a flower. If n is odd, halfof these will overlap, forming a flower with n petals. However, if n is even, the petals willnot overlap, forming a flower with 2n petals.

12

This simply happens because, although the parameterization in cartesian coordinates is in principle2π-periodic, in some cases it is actually π-periodic. Indeed, using that for any θ

cos k(θ + π) = cos(kθ + kπ)) = cos kθ cos kπ − sin kθsin kπ = (−1)k cos kθ

as well ascos(θ + π) = − cos θ, sin θ + π = sin θ cosπ + cos θsinπ = − sin θ,

we get

x(θ + π) = cos k(θ + π) cos θ + π = (−1)k+1 cos kθ cos θ,

y(θ + π) = cos k(θ + π) sin θ + π = (−1)k+1 cos kθ sin θ,

so that σ(θ+π) = (−1)k+1σ(θ). Therefore for odd k σ is already π-periodic, and we have to considerit only defined for 0 ≤ θ ≤ π, because it overlaps itself from π to 2π.

Therefore, when k is even and k 6= 0 (for k = 0, we have a circle, and the formula is different),

D = σ(θ) : 0 ≤ θ ≤ 2π,

so that

A(D) :=1

2

‰C〈F , d`〉 =

1

2

ˆ 2π

0−R cos(kθ) sin(θ) (−Rk sin(kθ) cos(kθ)−R cos(kθ) sin(θ))

+R cos(kθ) cos(θ) (−Rk sin(kθ) sin(θ) +R cos(θ)R cos(θ)) dθ

=1

2

ˆ 2π

0R2 cos2(kθ) sin2(θ) +R2 cos2(kθ) cos2(θ) dθ.

Taking common factor and the constants out and knowing that cos2(θ) + sin2(θ) = 1 we obtain thisintegral:

A(D) =R2

2

ˆ 2π

0cos2(kθ) dθ.

Using that cos2(θ) =1

2+

cos(2θ)

2,

A(D) :=R2

2

ˆ 2π

0

1

2+

cos(2kθ)

2dθ =

R2

2

[θ

2+

sin(2kθ)

4k

]θ=2π

θ=0

=πR2

2

Polar plot of one rose with 8 petals.

13

When k is odd and k 6= 0:D = σ(θ) : 0 ≤ θ ≤ 2π

The procedure it’s the same, however, the limits of integration have being changed:

A(D) :=R2

2

ˆ π

0

1

2+

cos(2kθ)

2dθ =

R2

2

[θ

2+

sin(2kθ)

4k

]θ=πθ=0

=πR2

4

Polar plot of one rose with 7 petals.

We have proved that the area of the rose given in polar coordinates by the equation r = R cos(kθ)is the half (respectively, the quarter) of the area of the circle of radius R when k is an even number(respectively, when k is odd).

15. We consider the vector field F = (y, 0) and the circle C of radius 1 centered in the origin, orientedcounter-clockwise.

(a) Deduce, by graphical reasoning, what is the sign of circulation

˛C〈F , dl〉.

(b) Calculate the circulation using Green’s Theorem.

Resolution: (a) First, let us write the circle C equation, centered at (0, 0) and of radius 1 andoriented counter-clockwise: C = x2 + y2 = 1.

The circulation of a vector field around acurve is equal to the line integral of the vec-tor field around the curve. We can see in thegraph that the vector field points in the oppo-site direction of the orientation of the curveC. Consequently, the sign of the circulationis negative.

Graphic representation of the circulation

˛C

〈F , dl〉 along

C: the blue arrows indicate the vector field and the circu-lation is in black.

14

(b) Now, let us compute the circulation. On the one hand, we can compute the circulation usingGreen’s theorem since the original integral was a line integral

‰C〈F , d`〉 =

‰CP dx+Qdy =

¨D

∂Q

∂x− ∂P

∂ydx dy,

In other words, we will change the integral into a double integral considering that the partial derivativesof (P (x, y), Q(x, y)) = F (x, y) = (y, 0) are:

∂Q

∂x= 0,

∂P

∂y= 1 =⇒ ∂Q

∂x− ∂P

∂y= −1.

Therefore, ‰C〈F , d`〉 =

¨D

(−1) dx dy.

We notice that

¨D

dx dy is the area of the disk D of radius r = 1, which is A(D) = πr2 = π, therefore

‰C〈F , d`〉 = −

¨D

dx dy = −π.

On the other hand, we can compute the circulation by definition:

‰C〈F , d`〉 =

ˆ b

a〈F (σ(t)),σ′(t)〉 dt.

We parameterize the circle C with the standard counter-clockwise parameterization

(x, y) = σ(t) = (cos t, sin t), 0 ≤ θ ≤ 2π,

Its derivative isσ′(t) = (− sin t, cos t)

If we evaluate the vector field in this concrete parameterization,

F (σ(t)) = (sin t, 0).

So,

‰C〈F , d`〉 =

ˆ 2π

0〈(sin t, 0), (− sin t, cos t)〉dt =

ˆ 2π

0− sin2 tdt

= −ˆ 2π

0cos2 t− cos 2t dt = −

ˆ 2π

0

1

2− cos 2t

2− cos 2tdt

= −ˆ 2π

0

1

2− cos 2t

2=

[sin 2t− 2t

4

]2π0

= −π.

16. Verify Greens theorem with the vector field F (x, y) = (2x3 − y3, x3 + y3) and the annulus

D =

(x, y) ∈ R2 : a2 ≤ x2 + y2 ≤ b2.

15

Domain D of Exercise 16, with the boundary oriented for Green’s theorem.

Resolution: Green’s theorem gives the relationship between a double integral over a planar region Dand the line integral around its boundary ∂D. If Q and P are functions of (x, y) defined on an openregion containing D and having continuous partial derivatives there, then‰

∂DP dx+Qdy =

¨DQx − Py dx dy,

where the path of integration along the boundary ∂D is positive, that is, counter-clockwise with respectto D (leaves the region D on the left side).

Left hand of the formula: Consider first a circle CR =

(x, y) : x2 + y2 = R2

parameterized by

(x, y) = σ(θ) = (R cos θ,R sin θ), (x′, y′) = σ′(θ) = (−R sin θ,R cos θ), 0 ≤ θ ≤ 2π,

which is a counter-clockwise parameterization. Then for P (x, y) = 2x3 − y3, Q(x, y) = x3 + y3‰CR

P dx+Qdy =

ˆ 2π

0P (x(θ), y(θ))x′(θ) +Q(x(θ), y(θ))y′(θ) dθ

=

ˆ 2π

0

(2R3 cos3 θ −R3 sin3 θ

)(−R sin θ) +

(R3 cos3 θ +R3 sin3 θ

)R cos θ dθ

=

ˆ 2π

0

((((

((((((

−2R4 cos3 θ sin θ +R4 sin4 θ +R4 cos4 θ +(((((((

R4 sin3 θ cos θ)

dθ

= R4

(ˆ 2π

0

(1− 2 sin2 θ cos2 θ

)dθ

)=

3π

2R4.

To compute now the circulation of F = (P,Q) along the boundary of D, we notice that ∂D = Cb∪Ca,where Cb must have a counter-clockwise orientation whereas Ca must have a clockwise orientation.Therefore ‰

∂DP dx+Qdy =

‰Cb

P dx+Qdy −‰Ca

P dx+Qdy =3π

2(b4 − a4).

Right hand of the formula: If we introduce polar coordinates (x, y) = T (r, θ) = (r cos θ, r sin θ whichsatisfy |detDT (r, θ)| = r, then D = T (R), where R = (r, θ) : 0 ≤ θ ≤ 2π, a ≤ r ≤ b. Therefore¨

DQx − Py dx dy =

¨D

3x2 + 3y2 dx dy = 3

ˆ 2π

0dθ

ˆ b

ar · r2 dr = 6π

b4 − a4

4=

3π

2(b4 − a4).

16

Result: We have checked that both sides of Green’s theorem give the same result3π

2(b4− a4). Hence

Green’s theorem has been verified.

17. Let C be the boundary of the domain with holes D ⊂ R2 obtained removing three disks of radii 1with centers at (−3, 0), (0, 0) and (3, 0) from the disk of radius 10 centered at (1, 0). Compute thecirculation ˛

C

(y + x3 cosx2

)dx+

(ey

2+ 2x

)dy

orienting clockwise the largest circle of C and counter-clockwise the three smallest ones.

Resolution: For the domain D with the boundary C = ∂D with positive orientation with respect toD, that is, leaving region D on the left side, we can apply Green’s theorem‰

∂Dy + x3 cosx2 dx+ ey

2+ 2x dy =

¨D

2− 1 dx dy = A(D) = 100π − 3π = 97π.

But we are given a circulation of the four circles of the boundary just in the opposite direction;

therefore

˛Cy + x3 cosx2 dx+ ey

2+ 2x dy = −97π.

18. The polar planimeter is a mechanical instrument that allows us to measure two areas of planar domains(see Wikipedia and a Java app). It has the shape of a ruler with two arms (see the figure). One ofthem connects the fixed extreme O = (0, 0) with a mobile point E = (a, b), and the other one connectsthe point E with another mobile point M = (x, y). In the extreme M there is a small wheel whichis perpendicular to the arm EM . We suppose that the arms OE and EM have lengths l and L,respectively. So the point M determines the point E as the intersection between the circle of radiusl centered in O and the one with radius L centered in M . This intersection is unique if the anglebetween both arms is lower to π rad. Therefore, we can consider that a = a(x, y) and b = b(x, y).

Sketch of a polar planimeter.

The method to determine the area of the planar simply connected domain S ⊂ R2 consists in movingthe point M over the boundary C = ∂S in a counter-clockwise direction. The wheel in M measuresthe displacement in the orthogonal direction with respect to the arm. In this problem, we will see thatthe searched area is L times the total displacement of the wheel.

(a) Prove that ax + by ≡ 1 computing the derivatives of the equations a2 + b2 ≡ l2 and (x−a)2 + (y−b)2 ≡ L2.

(b) F (x, y) = (P (x, y), Q(x, y)) is the vector of norm L which is orthogonal to the arm EM in theextreme M = (x, y), given by P (x, y) = b(x, y)− y and Q(x, y) = x− a(x, y). Prove that˛

C+

〈F , d`〉 = Area(D).

17

(c) d is the covered distance by the wheel at the moment it has completed the itinerary along the curveC = ∂D counter-clockwise. Argue that

˛C+

〈F , d`〉 = Ld.

Resolution: (a) We start the proof by differentiating the equations with respect to x and also withrespect to y. From a2 + b2 ≡ l2 we get

∂

∂x(a2 + b2) =

∂

∂x(l2) =⇒ aax + bbx = 0,

∂

∂y(a2 + b2) =

∂

∂y(l2) =⇒ aay + bby = 0.

From (x− a)2 + (y − b)2 ≡ L2 we get

∂

∂x

((x− a)2 + (y − b)2

)=

∂

∂x(L2) =⇒ (x− a)(1− ax) + (y − b)(−bx) = 0,

∂

∂y

((x− a)2 + (y − b)2

)=

∂

∂y(L2) =⇒ (x− a)(−ay) + (y − b)(1− by) = 0.

Isolating the term (x− a) in the last two equations, we can equalize them and we get

−(y − b)(−bx)

(1− ax)=−(y − b)(1− by)

(−ay)=⇒ (−ay)

(1− ax)=−(y − b)(1− by)−(y − b)(−bx)

=⇒

=⇒ aybx = (1− ax)(1− by) =⇒ aybx = 1− ax − by + axby =⇒ ax + by = 1 + axby − aybx.

Now that we have isolated the term ax + by, we can substitute ay and bx with the expressions we find

from isolating those terms in the first two equations we found previously: bx =−aaxb

and ay =−bbya

.

So, we get

ax + by = 1 + axby −bby

aaax

b= 1 +

axby −axby = 1.

Indeed, we have proven that ax + by = 1.

(b) According to Green’s theorem,

˛C+

〈F , d`〉 =

˛C+

P dx+Qdy =

¨D

(Qx − Py) dx dy.

Since Area(D) =

¨D

dx dy, that means that necessarily (Qx − Py) = 1. Let us check this:

Qx =∂Q

∂x= 1− ax and Py =

∂P

∂y= by − 1,

so(Qx − Py) = 1− ax − by + 1 = 2− (ax + by) = 2− 1 = 1,

where we have used that ax + by = 1. Therefore, since (Qx − Py) = 1, we have proven that

˛C+

〈F , d`〉 =

¨D

dx dy = Area(D).

18

(c) We note that the vector field F is parallel to the tangent vector to the curve and we are told thatit has constant norm L. Then,

˛C+

〈F , d`〉 =

˛C+

FT d` = L

˛C+

d` = Ld,

where we have used that

˛C+

d` = d, since we are told that d is the distance traveled by the wheel

along the curve C = ∂D (counter-clockwise).

Surface integrals

19. Area of the helicoid parameterized by

ϕ(r, θ) = (r cos θ, r sin θ, hθ), (r, θ) ∈ D = [0, R] ∧ [0, 2π].

Resolution: We start by computing the associated normal vector to this parameterization of thehelicoid:

ϕr =

cos θsin θ

0

, ϕθ =

−r sin θr cos θh

, ϕr ∧ϕθ =

∣∣∣∣∣∣i j k

cos θ sin θ 0−r sin θ r cos θ h

∣∣∣∣∣∣ =

h sin θ−h cos θ

r

,

from which it follows that ‖ϕr ∧ϕθ‖ =√h2 + r2. Once we have the norm of this normal vector, we

can now substitute it in the integral in order to find the area:

A =

¨D

√h2 + r2 dr dθ =

ˆ 2π

0dθ

ˆ R

0h2 + r2 dr

= 2π

ˆ R

0

√h2 + r2 dr = 2π

r√h2 + r2 + h2 ln(√

h2 + r2 + r)

2

r=Rr=0

= π(R√h2 +R2 + h2 ln

(√h2 +R2 +R

)−(

0 ·√h2 + 02 + h2 ln

(√h2 + 02 + 0

)))= π

(R√h2 +R2 + h2 ln

(√h2 +R2 +R

)− h2 ln(

√h2))

= π

(R√h2 +R2 + h2 ln

(√h2 +R2 +R

h

)).

20. Area of the surface S formed by the piece of the plane x + y + z = 1 that is over the ellipseD = (x, y) ∈ R2 : x2 + 2y2 ≤ 1.

Resolution: We can isolate locally one variable from x+ y+ z = 1, for instance z, to get z = f(x, y),and use the cartesian parameterization ϕ(x, y) = (x, y, z(x, y)), with z = f(x, y) = 1−x−y and (x, y)in a domain D ∈ R2 such that S = ϕ(D). Computing,

ϕx =

10−1

, ϕy =

01−1

, ϕx ∧ϕy =

∣∣∣∣∣∣i j k1 0 −10 1 −1

∣∣∣∣∣∣ =

−(−1)−(−1)

1

,

∥∥ϕx ∧ϕy∥∥ =√

1 + (−1)2 + (−1)2 =√

3.

19

In this case we have that the interior of an ellipse with semi-axes a = 1 and b = 1/√

2 and and thereforeA(D) = πab = π/

√2, so that

A(S) =

¨D

∥∥ϕx ∧ϕy∥∥ dx dy =√

3A(D) =√

3π√2

= π

√3

2.

21. The solid angle of a portion S of the sphere of radius R is Ω(S) := Area(S)/R2, measured in sr( steradians). Particularly, the complete sphere has 4πsr.(a) SR,α is the portion of the sphere with radius R which is inside the cone with vertex in the centreof the sphere and semi-angle α. For what values of the semi-angle α is Ω(SR,α) = 1?(b) Prove that the area of the spherical cap of radius R and height h < R defined by

CR,h = (x, y, z) ∈ R3 : x2 + y2 + z2 = R2, z ≤ R− h

is the same that the area of the lateral of a radius R and height h cylinder.

Resolution: (a) We know that the solid angle of a portion a portion S of the sphere is

Ω(S) :=Area(S)

R2,

and we want to find α such thatΩ(SR,α) = 1,

so we haveArea(SR,α)

R2= 1.

First of all, we have to parameterize the sphere with the restrictions of the cone. Therefore, we willuse the parameterization

ϕ(u, v) = (R cosu cos v,R sinu cos v,R sin v).

We will calculate the area of the portion SR,α = ϕ(DR,α) of the sphere as

Area(SR,α) =

¨DR,α

‖ϕu ∧ϕv‖ dudv.

Let us compute the partial derivatives of the parameterization.

ϕu =∂ϕ

∂u(u, v) = (−R sinu cos v,R cosu cos v, 0) = R cos v(− sinu, cosu, 0),

ϕv =∂ϕ

∂v(u, v) = (−R cosu sin v,−R sinu sin v,R cos v) = R(− cosu sin v,− sinu sin v, cos v).

Now we calculate the cross product of the partial derivatives and its norm

ϕu ∧ϕv = R2 cos v(cosu cos v, sinu cos v, sin v),

‖ϕu ∧ϕv‖ = R2 cos v

√cos2 v(sin2 u+ cos2 u) + sin2 v = R2 cos v

√cos2 v + sin2 v = R2 cos v.

Once we have found the norm of the normal vector, we have to find the domain DR,α to have SR,α =ϕ(DR,α). Following the spherical change, we know that 0 ≤ u ≤ 2π, and regarding at the restrictions

of the cone, we can conclude thatπ

2− α ≤ v ≤ π

2so that

DR,α =

(u, v) : 0 ≤ u ≤ 2π,π

2− α ≤ v ≤ π

2

.

20

Finally, we can calculate our integral

Area(SR,α) =

¨S‖ϕu ∧ϕv‖ dv du =

ˆ 2π

0du

ˆ π2

π2−α

R2 cos v dv = 2πR2 [sin v]v=π/2v=π/2−α

= 2πR2(

1− sinπ

2− α

)= 2π(1− cosα)R2.

Once we have found the area of our surface in function of the angle α, let us compare with the equation

we had for the solid angle. FromArea(SR,α)

R2= 1 we get

2π(1− cosα)R2

R2= 1 ⇐⇒ 1− cosα =

1

2π⇐⇒ cosα = 1− 1

2π⇐⇒ α = arccos

(1− 1

2π

).

In conclusion, when α = arccos(1− 1/2π), the solid angle of the portion of the sphere SR,α is 1.(b)

It is enough to observe that thespherical cap CR,h equals to aportion of SR,h being h = (1 −cosα)R.It is easier to understand ittrhough an image of a sphericalcap as a portion of a sphere, asthe one on the right. Spherical cap of radius R and height h.

22. Area and volume of the intersection of two cylinders with radius R whose axis are cut perpendicularly(plot done taking R=1). (This is the Steinmetz solid: see Wikipedia and MathWorld.)

Resolution:

Intersection of two cylinders with radius R whose axis arecut perpendicularly, from exercise 22.

Steinmetz solid (source Wikipedia), from exer-cise 22.

To do this exercise, we are going to split it in two parts: part 1 to solve the area and part 2 to solvethe volume. But first, we have to evaluate the data given. We have two cylinders represented whichcreate a solid (the Steinmetz solid) represented in the figure on the right side. These cylinders are:

C1 : x2 + y2 = R2

C2 : x2 + z2 = R2

21

Area: To calculate the surface area of a curve, we know that A(W ) =

ˆW‖ϕx ∧ ϕy‖ dx dy, and to do

this, we need to parameterize one of the two equations, using cartesian parameterization, and calculatethe vector product of its derivatives, and then find the boundaries of the other equation. First, wecompute the parameterization and the vector product:

x2 + z2 = R2 =⇒ z = ±√R2 − x2,

ϕ(x, y) =(x, y,±

√R2 − x2

),

ϕx =

(1, 0,± −2x

2√R2 − x2

)and ϕy = (0, 1, 0),

ϕx ∧ϕy =

(∓ x√

R2 − x2, 0, 1

).

Now, we calculate the integration limits, by looking at the boundaries. Taking C2 as the surface areacalculated, C1 has to be the frontier in which C2 starts and end. From the figures below, fixing thevariable x, the limits for C2 are:

−R ≤ x ≤ R

−√R2 − x2 ≤ y ≤

√R2 − x2

Now, we can compute the surface area of the intersection of the two cylinders. Below, we take intoaccount only z > 0.

A(C2 > 0) =

ˆC2>0

‖ϕx ∧ ϕy‖ dx dy =

ˆ R

−R

ˆ √R∗2−x2−√R∗2−x2

√(− x√

R2 − x2

)2

+ 02 + 12 dy dx =

=

ˆ R

−R

ˆ √R∗2−x2−√R∗2−x2

√x2

R2 − x2+ 1 dy dx =

ˆ R

−R

ˆ √R∗2−x2−√R∗2−x2

√x2 +R2 − x2R2 − x2

dy dx =

=

ˆ R

−R

ˆ √R∗2−x2−√R∗2−x2

R√R2 − x2

dy dx =

ˆ R

−R

R√R2 − x2

2√R2 − x2 dx =

ˆ R

−R2R dx =

= 2R · 2R = 4R2.

4R2 is the surface area of the cylinder C2 for z > 0. So the total surface area for the cylinder C2 istwice this area, for positive and negative z: 2 · 4R2 = 8R2.

However, this is not the total surface area of the Steinmetz solid. Because of the symmetry of thesolid, the total surface area would be twice the area of one cylinder, that is:

A(Steinmetz solid) = 2 · 8R2 = 16R2

Volume: For the volume, we can find the limits of integration by splitting the plot in two planes:Fixing the x variable, and isolating the y and the z variables in each circumference equation fromfigures above, we can find the integration limits:

R ≤ x ≤ R

−√R2 − x2 ≤ y ≤

√R2 − x2

−√R2 − x2 ≤ z ≤

√R2 − x2

So, the volume of the intersection of two cylinders with radius R whose axis are cut perpendicularly

22

Plane xy of the intersection of the Steinmetz solid (Tak-ing R2 = 1), from exercise 22.

Plane xz of the intersection of the Steinmetz solid (Tak-ing R2 = 1), from exercise 22.

is:

V (Steinmetz solid) =

ˆ R

−R

ˆ √R2−x2

−√R2−x2

ˆ √R2−x2

−√R2−x2

1 dz dy dx =

ˆ R

−R2√R2 − x2 · 2

√R2 − x2 dx =

=

ˆ R

−R4(R2 − x2) dx = 4 ·

[R2x− x3

3

]R−R

= 4 ·[R3 − R3

3−(−R3 − −R

3

3

)]=

= 4 ·[2R3 − 2

3R3

]= 4 · 4

3R3 =

=16

3R3

Another easier method to compute the volume is considering that the cross-section Wx for −R ≤ x ≤ Ris a square of area A(x) = 4(R2 − x2) so by Cavalieri’s principle,

V (Steinmetz solid) =

ˆ R

−RA(x) dx =

ˆ R

−R4(R2 − x2) dx =

16

3R3.

From this exercise, we can say that the surface area and the volume of the Steinmetz solid are 16R2

and 163 R

3, respectively.

23. Applying the Pappus’ centroid theorem, compute the centroids of the semi-circle and the semi-disk ofradius R.

Resolution:

The first Pappus’ centroid theorem states that the surface area A of a surface of revolution generatedby rotating a plane curve C about an axis external e to C and on the same plane is equal to theproduct of the length L(C) of the curve C and the distance d = 2πdist(c, e) traveled by the geometriccentroid c of C:

A = L(C)d = 2πdist(c, e)L(C).

23

Therefore for the semi-circle of radius R and center at the origin, its centroid takes the form c = (r, 0).When we rotate around the z-axis, by the first Pappus’ centroid theorem, 4πR2 = dπR, where d = 2πr,

so that the centroid of the semi-circle is c =

(2R

π, 0

).

The second Pappus’ centroid theorem states that the volume V of a solid of revolution generated byrotating a plane domain D about an external axis e is equal to the product of the area A(D) and thedistance d = 2πdist(c, e) traveled by the geometric centroid c of D. (The centroid of D is usuallydifferent from the centroid of its boundary curve C.) That is

V = A(D)d = 2πdist(c, e)A(D).

For the semi-disk of radius R and center at the origin, its centroid takes the form c = (r, 0). When

we rotate around the z-axis, by the second Pappus’ centroid theorem,4πR3

3= d

πR2

2, where d = 2πr,

so that the centroid of the semi-disk is c =

(4R

3π, 0

).

24. Compute the volume V and the area A of the band which is obtained after drilling a ball of radiusR with a drill bit (a cylinder) of radius r < R that goes through its center. Check that if h is theheight of the band, then V only depends on h (V = πh3/6), whereas A is the same as the sum ofthe areas of the sides of two cylinders with radius r and R, both with height h (A = 2πh(R + r))(“ napkin ring problem”).

Resolution:

Side view of the napkin ring for napkin ring problem Cross-section top view for napkin ring problem

First we compute the volume of the napkin ring by means of Cavalieri’s principle. To find the areaA(y) of the horizontal cross-section we subtract the area of the smaller circle with radius r from thearea of the bigger circle with radius rb

A(y) = πr2b − πr2 = π(√

R2 − y2)2− π

(√(R2 −

(h2

)2)2

= π(R2 − y2 −R2 +

(h2

)2)= π

((h2

)2− y2

)It can be clearly seen that this area A(y) does not depend on the radius of the sphere R. If we put

24

the axis at the center of the sphere at (0, 0, 0) and we apply Cavalieri’s principle, we can say that

V =

ˆ h/2

−h/2A(y) dy =

ˆ h/2

−h/2π((h

2

)2− y2

)dy

= π

[(h2

)2y − y3

2

]h/2−h/2

=πh3

4

(1− 1

3

)=πh3

6.

To find the area of the napkin ring, we will add the area of the sphere and of the cylinder and subtractthe area of the spherical caps at the top and bottom.

Different areas of the napkin ring for napkin ring problem

The area for the sphere and cylinder are

Asphere = 4πR2

Acylinder = 2πrh

For the area of the spherical cap we will use the formula related to Exercise V34

A(S) = 2π

ˆ b

ar

√1 +

(dr

dz

)2

dz

which is valid for any surface of revolution around the z-axis. Indeed, given a surface of revolutionS = (x, y, z) : r = g(z) ≥ 0, a ≤ z ≤ b, where r =

√x2 + y2, we can parameterize it as S = ϕ(D),

by means of the parameterization ϕ(θ, z) = (g(z) cos θ, g(z) sin θ, z), D = (θ, z) ∈ R2 : a ≤ z ≤ b, 0 ≤θ ≤ 2π introduced in Exercise 34. Writing the formula for the area of this surface, we obtain theformula above (writing r = g(z) and dr

dz = g′(z)) as

A(S) =

¨S

dS =

¨D‖ϕθ ∧ϕz‖ dθ dz = 2π

ˆ b

ag(z)

√1 + g′(z)2 dz

where

ϕθ ∧ϕz = g(z))

cos θsin θ−g′(z)

25

and‖ϕθ ∧ϕz‖ = g(z)

√1 + g′(z)2

For a sphere we have that r = g(z) =√R2 − z2 and so

dr

dz= − z√

R2 − z2= −z

r.

Also √1 +

( dr

dz

)2=

√1 +

(− z

r

)2=

√r2 + z2

r=R

r

and so then

A(S) = 2π

ˆ b

ar

√1 +

(dr

dz

)2

dz = 2π

ˆ R

h/2rR

rdz

= 2πR[z]Rh/2

= πR(2R− h)

Adding the area of the cylinder and sphere, and subtracting the area of the spherical cap twice (asthere are two caps) we get

A = 4πR2 + 2πrh− 2(π2R2 − πRh)) = 2πh(R+ r)

25. We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a point and goesthrough the center of the sphere. In coordinates, we can describe the sphere as E = x2 + y2 + z2 = 4R2

and the cylinder as Q = (x−R)2 + y2 = R2.(a) The intersection E∩Q results in the Viviani’s curve, that has length cR for a constant c. Providean integral for c (no need to compute it).(b) We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a point andgoes through the center of the sphere. In coordinates, we can describe the sphere as E = x2 +y2 +z2 =4R2 and the cylinder as Q = (x−R)2 + y2 = R2. Compute the area of the portions S+

1 and S−1 of thesphere which are contained in the interior of the cylinder (Viviani’s vault).(c) We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a pointand goes through the center of the sphere. In coordinates, we can describe the sphere as E =x2 + y2 + z2 = 4R2

and the cylinder as Q =

(x−R)2 + y2 = R2

. Compute the area of the por-

tion S2 of the cylinder contained in the interior of the sphere.

Resolution: (a) We know that the formula to compute the length of a curve C = σ ([a, b]) is

ˆC

d` =

ˆ b

a

∥∥σ′(t)∥∥ dt

We have to parameterize the Viviani’s curve E ∩Q, and there are at least two ways to do it.

26

Viviani’s curve. A cylinder going through the sphere.

The first way of parameterization, based on a cylindrical parameterization for Q, is σ(θ) =(R+R cos θ,R sin θ, z), where

z = ±√

4R2 − x2 − y2 = ±√

4R2 −R2 − 2R2 cos θ −R2 cos2 θ −R2 sin2 θ

= ±√R2(4− 1− 2 cos θ − (cos2 θ + sin2 θ) = ±

√R2(2− 2 cos θ) = ±R

√2(1− cos θ),

so that E ∩Q is a concatenation of two paths

σ(θ) =(R+R cos θ,R sin θ,±R

√2(1− cos θ)

), 0 ≤ θ ≤ 2π,

with

σ′(θ) =

(−R sin θ,R cos θ,± R sin θ√

2(1− cos θ)

),∥∥σ′(θ)∥∥ = R

√1 +

sin2 θ

2(1− cos θ).

Since ‖σ′(θ)‖ is the same for both paths, we can consider only one and multiply by 2 to compute itslength: ˆ

E∩Qd` = 2

ˆ 2π

0

∥∥σ′(θ)∥∥ dθ = 2R

ˆ 2π

0

√1 +

sin2 θ

2(1− cos θ)dθ = cR,

where

c = 2

ˆ 2π

0

√1 +

sin2 θ

2(1− cos θ)dθ.

The second way of parameterization is based on first parameterizing the sphere E by (x, y, z) =(r cosϕ cos θ, r cosϕ sin θ, r sinϕ), for 0 ≤ θ ≤ 2π, −π/2 ≤ φ ≤ π/2, and then substituting in theequation x2 + y2 − 2Rx = 0 of Q, to get cosφ(cosφ − cos θ) = 0 which implies cosφ = cos θ whichimplies θ = ±φ, because of the trigonometric identity

cosφ− cos θ = 2 sin

(θ + φ

2

)sin

(θ − φ

2

).

Again, E ∩Q a concatenation of two paths θ = ±φ. Choosing for instance θ = φ, we get a parameter-ization of half Viviani’s curve:

(x, y, z) = σ(φ) = (r cosφ cosφ, r cosφ sinφ, r sinφ), −π/2 ≤ φ ≤ π/2,

27

withσ′(φ) =

(−4R sinφ cosφ, 2R cos2 φ sin2 φ, 2R cosφ

),∥∥σ′(φ)

∥∥ = 2R√

1 + cos2 φ.

Since ‖σ′(φ)‖ is the same for θ = −φ, we can consider only the path θ = φ and multiply by 2 tocompute the length of Viviani’s curve E ∩Q:

ˆE∩Q

d` = 2

ˆ 2π

0

∥∥σ′(φ)∥∥ dφ = 2 · 2R

ˆ π2

−π2

√1 + cos2 φ dφ = cR,

where

c = 4

ˆ π2

−π2

√1 + cos2 φ dφ.

Computing both integrals with MATLAB, we get a value of c ≈ 15, 28 . Using both parameterizatons isa good way to check whether our result is right or not:

c = 2

ˆ 2π

0

√1 +

sin2 θ

2(1− cos θ)dθ = 4

ˆ π2

−π2

√1 + cos2 φ dφ.

(b) To compute both surfaces we will use the following formula for a parameterized surface S = ϕ(D):

A(S) =

¨D‖ϕu ∧ϕv‖ dudv

First, we need to parameterize the sphere E. To do so, we will use spherical coordinates and the datagiven in the statement (r = 2R), such that

(x, y, z) = ϕ(θ, φ) = (2R cos θ cosφ, 2R sin θ cosφ, 2R sinφ), where − π

2≤ φ ≤ π

2,−π ≤ θ ≤ π.

Then, we have to differentiate the parameterization with respect to θ and φ and calculate the normof their cross products:

ϕθ = (−2R cosφ sin θ, 2R cos θ cosφ, 0),

ϕφ = (−2R cos θ sinφ,−2R sinφ sin θ, 2R cosφ),∥∥ϕθ ∧ϕφ∥∥ =

√16R4 cos4 φ+ 16R4 cos2 φ sin2 φ =

√16R4 cos2 φ = 4R2 cosφ.

Before calculating the area of the sphere surface we need to know that, when checking if the sphericalcurve fulfills the cylinder, we get that cos θ = cosφ and therefore, φ = ±θ. therefore, we are going totake the domain of the parameterization ϕ to compute the area as

D =

(θ, φ) : − π

2≤ φ ≤ π

2, 0 ≤ θ ≤ φ

,

which will parameterize only half of the upper part S+1 of the sphere’s surface contained in the cylinder,

and therefore the area computed with this parameterization will be one half of the upper area of thesphere’s surface contained in the cylinder, what is the same, one quarter of the total surface area insidethe cylinder. To understand it better, one can look at the front view in the figure below.

Viviani’s window seen from different perspectives

28

A(S+1 ) = 2

(¨D

4R2 cosφ dφ dθ

)= 2

(4R2

ˆ π2

0

ˆ φ

0cosφ dθ dφ

)= 8R2

ˆ π2

0[cosφθ]θ=φθ=0 dφ

= 8R2

ˆ π2

0φ cosφ dφ = 8R2

[φ sinφ−

ˆsinφ dφ

]φ=π2

φ=0

= 8R2 [φ sinφ+ cosφ]φ=π

2φ=0

= 8R2[π

2− 0 + 0− 1

]= 8R2

(π

2− 1

)= 4πR2 − 8R2.

By symmetry, we can see that A(S+1 ) = A(S−1 ) = 4πR2 − 8R2.

Hence, the total area of the sphere surface contained in the interior of the cylinder is:

A(S1) = 2(4πR2 − 8R2) = 8πR2 − 16R2.

(c) First of all we must visualize how the sphere and the cylinder intersect with each other. It’s betteran image than a thousand words.

Viviani’s curve of Exercise 25c.

Secondly, we shall use the formula to compute the area of a surface S = ϕ(D), that is:

A(S) =

¨D‖ϕu ∧ ϕv‖ du dv.

We must first parameterize the cylinder Q with cylindrical coordinates that will depend on θ and z

(x, y, z) = ϕ(θ, z) = (R+R cos θ,R sin θ, z).

Then, as we made a parameterization that depends on z and θ, we’ll find the limits of integration ofthose. We substitute the parameterization in the equation of the sphere E:

(R+R cos θ)2 + (R sin θ)2 + z2 = 4R2 ⇐⇒ R2 + 2R2 cos θ +R2 cos2 θ +R2 sin2 θ + z2 = 4R2

⇐⇒ z2 = 2R2 − 2R2 cos θ ⇐⇒ z = ±√

2R2 − 2R2 cos θ.

In the end we have the following domainD of definition of the parameterization ϕ such that S2 = ϕ(D):

D =

(θ, z) : 0 ≤ θ ≤ 2π, −√

2R2 − 2R2 cos θ ≤ z ≤√

2R2 − 2R2 cos θ.

The next step is to differentiate the parameterization with respect to θ and z, make their respectivecross product and compute its norm.

ϕθ = R(− sin θ, cos θ, 0), ϕz = (0, 0, 1), ϕθ ∧ϕz = R(cos θ,− sin θ, 0), ‖ϕθ ∧ϕz‖ = R.

29

Now we are ready to compute the area of the portion of the cylinder in the interior of the sphere.

A(S2) =

2πˆ

0

√2R2−2R2 cos θˆ

−√2R2−2R2 cos θ

R dz dθ = R

2πˆ

0

2√

2R2 − 2R2 cos θ dθ = R

2πˆ

0

2√

2R√

1− cos θ dθ

= R

2πˆ

0

2√

2R√

2 sinθ

2dθ = 4R2

2πˆ

0

sinθ

2dθ = −8R2

[cos

θ

2

]θ=2π

θ=0

dθ = 16R2 ,

where we used the trigonometric relation sin2 α

2=

1− cosα

2, which can also be expressed as

√1− cosα =

√2 sin

α

2.

26. Compute the centroid of the hemisphere (surface) and the semi-ball (solid) of radius R.

Resolution: The hemisphere is defined as

S =

(x, y, z) ∈ R3 : x2 + y2 + z2 = R2, z ≥ 0.

In order to compute the centroid of this surface we have to use the formula and the procedure fromvector calculus. By looking at the figure, we can say that x and y will be zero by symmetry, so weonly have to calculate the z coordinate of the centroid:

z =

ˆSzρdS

ˆSρdS

=

¨Dz (Ψ(θ, ϕ)) ρ (Ψ(θ, ϕ)) ‖Ψθ ∧Ψϕ‖ dθ dϕ¨Dρ (Ψ(θ, ϕ)) ‖Ψθ ∧Ψϕ‖ dθ dϕ

,

where Ψ(θ, ϕ) is the parameterization of S, that is, S = Ψ(D).

First, let us write the parameterization of the hemisphere in spherical coordinates and find its domain:

(x, y, z) = Ψ(θ, ϕ) = (R cos θ sinϕ,R sin θ cosϕ,R sinϕ), (θ, ϕ) ∈ D,

where we need to find the domain D to have S = Ψ(D). As the figure gives a total turn in the xyaxis, θ goes from 0 to 2π, while it only gives a quarter of a turn in the zy axis, so ϕ goes from 0 to π

2 :

D =

0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π

2

.

Then, let us compute ‖ Ψθ ∧Ψϕ ‖:

Ψθ = R cosϕ(− sin θ, cos θ, 0), Ψϕ = R(− cos θ sinϕ,− sin θ sinϕ, cosϕ)

Ψθ ∧Ψϕ = R2 cosϕ(cos θ cosϕ, sin θ cosϕ, sinϕ), ‖Ψθ ∧Ψϕ‖ = R2 cosϕ.

Since there is no density given in the statement, we will assume ρ = 1, so we can apply the formula tofind z:

z =

¨Dz(Ψ(θ, ϕ))ρ(Ψ(θ, ϕ)) ‖ Ψθ ∧Ψϕ ‖ dθ dϕ¨Dρ(Ψ(θ, ϕ)) ‖ Ψθ ∧Ψϕ ‖ dθ dϕ

=

ˆ 2π

0

ˆ π2

0R sinϕR2 cosϕdϕdθ

ˆ 2π

0

ˆ π2

0R2 cosϕdϕdθ

=

R3

ˆ 2π

0dθ

ˆ π2

0sinϕ cosϕdϕ

R22π[sinϕ]ϕ=1ϕ=0

=R32π

[sin2 ϕ

2

]ϕ=π2

ϕ=0

2πR2=

R3π

2πR2=R

2.

30

Therefore the centroid of the hemisphere is

(0, 0,

R

2

).

The semi-ball of R3 is defined as

W =

(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ R2, z ≥ 0

In order to compute the centroid of this solid we have to use the formula and the procedure from tripleintegrals. Like in the previous case, we can see by symmetry that x and y are 0, so we only have tocalculate z:

z =

˚Wρz dx dy dz

V (W ).

Let us use the change T to spherical coordinates and specify the domain B where W = T (B):

(x, y, z) = T (r, θ, ϕ) = (r cos θ sinϕ, r sin θ cosϕ, r sinϕ), (r, θ, ϕ) ∈ B, |detDT (r, θ, ϕ)| = r2 cosϕ,

whereB =

0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π

2, 0 ≤ r ≤ R

.

We also need the volume of the semi-ball W . Since it is a half of a ball of radius R, we will use theformula for the volume of a ball of radius R divided by 2: V (W ) = 2

3πR3. Since there is no density

given in the statement, we will assume that ρ = 1, so we can now apply the formula to find z:

z =

˚Wρz dx dy dz

V (W )=

ˆ 2π

0

ˆ π2

0

ˆ R

0r sinϕ r2 cosϕdr dϕdθ

23πR

3=

2π[sin2 ϕ

2

]ϕ=π2

ϕ=0

[r4

4

]r=Rr=0

23πR

3=

2π 12R4

423πR

3=

=6R4π

16=

3R

8.

Therefore the centroid of the semi-ball is

(0, 0,

3R

8

).

27. Total charge of the lateral side of the cone with height h and radius R if the charge density is propor-tional to the distance from the base, being k > 0 the constant.

Resolution: To calculate the total charge, we have to integrate the charge density and so, we mustcompute the following integral:

Q =

ˆSρdS =

ˆSkz dS =

ˆDkz ‖ϕθ ∧ϕz‖ dz dθ,

where ρ is the charge density on each point and S = ϕ(D).

Firstly, we obtain the cone by turning, respect to the z axis,the red line in the figure which is parameterized by

σ(z) =

(Rz

h, 0, z

), for 0 ≤ z ≤ h.

31

The parameterization of the cone S is

(x, y, z) = ϕ(θ, z) =

(Rz

hcos θ,

Rz

hsin θ, z

), (θ, z) ∈ D,

where D is the rectangleD = (θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h .

We proceed now to compute ‖ϕθ ∧ϕz‖:

ϕθ =Rz

h(− sin θ, cos θ, 0),

ϕz =

(R

hcos θ,

R

hsin θ, 1

),

ϕθ ∧ϕz =Rz

h2(h cos θ, h sin θ,−R),

‖ϕθ ∧ϕz‖ =Rz

h2

√h2 +R2.

We finally have all the ingredients to compute the total charge:

Q =

ˆDkz‖ϕθ ∧ϕz‖ dz dθ =

ˆ 2π

0dθ

ˆ h

0kzRz

h2

√R2 + h2 dz = 2π

kR

h2

√R2 + h2

ˆ h

0z2 dz

=2πkRh

√R2 + h2

3.

28. Total charge of the surface

S =

(x, y, z) ∈ R3 : z2 = x2 + y2 + a2, a ≤ z ≤ a√

2

(piece of the hyperboloid of two sheets) if the charge density is e(x, y, z) = kz, with k > 0.

Resolution: Notice that a hyperboloid of two sheets is a surface of revolution. Therefore, wewill parameterize it with cylindrical coordinates. That is, r2 = x2 + y2. Then, S is given byr = g(z) :=

√z2 − a2 for a ≤ z ≤ a

√2. So, S can be parameterized as S = ϕ(D) for ϕ(θ, z) =

(g(z) cos θ, g(z) sin θ, z), D = (θ, z) ∈ R2 : 0 ≤ θ ≤ 2π, a ≤ z ≤ a√

2.We now compute ϕθ and ϕz in order to obtain the normal vector associated to this parameterization.

ϕθ = (−g(z) sin θ, g(z) cos θ, 0)>

ϕz = (g′(z) cos θ, g′(z) sin θ, 1)>

Now we compute ϕθ ∧ϕz:

ϕθ ∧ϕz =

∣∣∣∣∣∣i j k

−g(z) sin θ g(z) cos θ 0g′(z) cos θ g′(z) sin θ 1

∣∣∣∣∣∣ =

g(z) cos θg(z) sin θ

−g(z)g′(z) sin2 θ − g(z)g′(z) cos2 θ

=

g(z) cos θg(z) sin θ−g(z)g′(z)

= g(z)

cos θsin θ−g′(z)

.

32

Therefore, the normal vector is

‖ϕθ ∧ϕz‖ = g(z)

√cos2 θ + sin2 θ + (−g′(z))2 = g(z)

√1 + (g′(z))2

=√z2 − a2 ·

√1 +

(z√

z2 − a2

)2

=√z2 − a2 ·

√1 +

z2

z2 − a2

=

√

(z2 − a2) · (z2 − a2 + z2)

z2 − a2

=√

2z2 − a2.

We now have everything we need to compute the total charge surface, which is:

Q(S) =

¨Se dS =

¨De (ϕ(θ, z)) · ‖ϕθ ∧ϕz‖ dθ dz

=

ˆ 2π

0dθ

ˆ a√2

akz ·

√2z2 − a2 dz = 2πk

ˆ a√2

az ·√

2z2 − a2 dz.

We will separately solve the last integral for practical reasons.We will do so by doing a change ofvariables.

2πk

ˆz ·√

2z2 − a2 dz =

2z2 − a2 = u2 → u =

√2z2 − a2

4z dz = 2udu→ z dz = udu2

= 2πk

ˆ √u2 · u du

2= πk

ˆu2 du = πk

[u3

3

]+ C.

Now we undo the change of variable writing the final expressions in terms of z and thus obtining thetotal charge of S.

Q(S) = 2πk

ˆ a√2

az ·√

2z2 − a2 dz = πk

(√

2z2 − a2)3

3

z=a√2

z=a

= πk

(√

2 · (a√

2)2 − a2)3

3−

(√2 · a2 − a2

)33

= πk

(√

2 · 2a2 − a2)3

3−

(√a2)3

3

= πk

(√

3a2)3

3− a3

3

= πk

[a3√

33

3− a3

3

]= πka3

[3√

3

3− 1

3

]= πka3

(√3− 1

3

).

So, the total charge of the surface S is:

Q(S) = πka3(√

3− 1/3).

29. Let us consider the ellipsoid

S = (x, y, z) ∈ R3 : x2/a2 + y2/b2 + z2/c2 = 1

oriented by its unit and outward-pointing normal vector N to the solid W = (x, y, z) ∈ R3 : x2/a2 +y2/b2 + z2/c2 ≤ 1 enclosed by S = ∂W . Consider the function f : S → R+ given by f(p) =dist(O, TpS), where O = (0, 0, 0) is the origin and TpS is the tangent plane to the ellipsoid in the point

33

p.

(a) Compute

¨Sf dS.

(b) Prove that the normal component of the vector field F (x, y, z) =

(x2

a2,y2

b2,z2

c2

)is FN := 〈F,N〉 =

1/f . Compute the flux

‹S〈F , dS〉. Notice the symbol typically for the flux along the boundary S of a

closed solid W .

(c) Compute the flux

‹S〈G, dS〉, where G(x, y, z) = zF (x, y, z), and F (x, y, z) =

( xa2,y

b2,z

c2

).

Resolution: (a) First, we will parameterize S = ψ(D) with an ellipsoidal parameterization

p = (x, y, z) = ψ(θ, ϕ) = (a cos θ cosϕ, b sin θ cosϕ, c sinϕ)

where, as we are considering the whole ellipsoid, D = (θ, ϕ) : 0 ≤ θ ≤ 2π,−π/2 ≤ ϕ ≤ π/2. For anypoint p = ψ(θ, ϕ) ∈ S, we have the two independent tangent vectors:

ψθ = (−a sin θ cosϕ, b cos θ cosϕ, 0)> ψϕ = (−a cos θ sinϕ,−b sin θ sinϕ, c cosϕ)>,

and the associated normal vector, which turns out to be outward-pointing:

ψθ ∧ψϕ =

∣∣∣∣∣∣i j k

−a sin θ cosϕ b cos θ cosϕ 0−a cos θ sinϕ −b sin θ sinϕ c cosϕ

∣∣∣∣∣∣ =

bc cos2 ϕ cos θac cos2 ϕ sin θab cosϕ sinϕ

= abc cosϕ

cos θ cosϕ

asin θ cosϕ

bsinϕ

c

= abc cosϕ

a cos θ cosϕ

a2b sin θ cosϕ

b2c sinϕ

c2

.

Therefore, the tangent plane TpS for p = (x, y, z) = ψ(θ, ϕ) has as unit outward-pointing normal

vector N(p) :=ψθ ∧ψϕ∥∥ψθ ∧ψϕ∥∥ .

Ellipsoid S and the tangent plane TpS to S = ∂W in a point p ∈ S. The red line is the segment between theorigin and the closest point of TpS. This segment has the direction of N(p) and length f(p) = |〈N(p),p〉|.

The distance between the origin and a point p of the surface can be expressed by |〈N(p),p〉|, whereN(p) is the unit normal vector to S at p, so that f(p) = |〈N(p),p〉|.

34

For p = (x, y, z) = ψ(θ, ϕ) we have f(ψ(θ, ϕ))∥∥ψθ ∧ψϕ∥∥ =

∣∣⟨ψθ ∧ψϕ,ψ⟩∣∣, where

∣∣⟨ψθ ∧ψϕ,ψ⟩∣∣ =

∣∣∣∣∣⟨abc cosϕ

(a cos θ cosϕ

a2,b sin θ cosϕ

b2,c sinϕ

c2

), (a cos θ cosϕ, b sin θ cosϕ, c sinϕ)

⟩∣∣∣∣∣= abc cosϕ

(a2 cos2 θ cos2 ϕ

a2+b2 sin2 θ cos2 ϕ

b2+c2 sin2 ϕ

c2

)

= abc cosϕ

(x2

a2+y2

b2+z2

c2

)= abc cosϕ,

so thatf(ψ(θ, ϕ))

∥∥ψθ ∧ψϕ∥∥ = abc cosϕ.

We are now ready to compute the integral of f on S = ψ(D):

¨Sf dS =

¨Df(ψ(θ, ϕ))

∥∥ψθ ∧ψϕ∥∥ dϕdθ =

ˆ 2π

0

ˆ π/2

−π/2abc cosϕdϕdθ

= abc

ˆ 2π

0dθ

ˆ π/2

−π/2cosϕdϕ = 2πabc [sinϕ]

ϕ=π/2ϕ=−π/2 = 2πabc · 2 = 4πabc.

(b)

The ellipsoid of Exercise 29b with p located at the North pole.

For any point of the surface of the ellipsoid p = (x, y, z) ∈ S ⇐⇒ g(x, y, z) = 0 where g(x, y, z) =x2/a2 + y2/b2 + z2/c2− 1. According to the resolution of Exercise V27, ∇g(p) = 2(x/a2, y/b2, z/c2) =2F (p) is a normal vector to S in p, moreover pointing outward. Therefore the outward-pointingunitary normal vector is

N(p) =∇g(p)

‖∇g(p)‖=

F (p)

‖F (p)‖.

It is easy to check that 〈F (p),p〉 = 1 for any p = (x, y, z) ∈ S. The distance from a point q to a planeΠ with unitary normal vector N passing though a point p is |〈N ,p− q〉|. Therefore the distance fromthe origin (q = 0) to the tangent plane TpS to S in p is

f(p) = |〈N(p),p〉| = |〈F (p),p〉|‖F (p)‖

=1

‖F (p)‖.

We can now compute FN:

FN = 〈F (p),N(p)〉 =〈F (p),F (p)〉‖F (p)‖

=‖F (p)‖2

‖F (p)‖= ‖F (p)‖ =

1

f(p).

35

To compute the flux we can use Gauss theorem‹S〈F , dS〉 =

˚W

divF dx dy dz =

˚W〈∇,F 〉dV

where the ellipsoid S is oriented by its unit and outward-pointing normal vector N and W is the solid

enclosed by the ellipsoid S = ∂W . As 〈∇,F 〉 =

(1

a2+

1

b2+

1

c2

)V (W ) =

4π

3abc, we get

‹S〈F , dS〉 =

˚W

divF dx dy dz =

(1

a2+

1

b2+

1

c2

)˚W

dx dy dz

=

(1

a2+

1

b2+

1

c2

)V (W ) =

(1

a2+

1

b2+

1

c2

)· 4π

3abc

=4π

3

(bc

a+ac

b+ab

c

).

(c) First of all we compute G:

G(x, y, z) =

(xz

a2,yz

b2,z2

c2

)= (P (x, y, z), Q(x, y, z), R(x, y, z)) ,

with

P =xz

a2, Q =

yz

b2, R =

z2

c2.

To apply Gauss theorem ˆS+

〈G, dS〉 =

˚W

divGdx dy dz

we compute the divergence

divG =∂P

∂x+∂Q

∂y+∂R

∂z=

z

a2+z

b2+

2z

c2= z

(1

a2+

1

b2+

2

c2

)and make a change W = T (B) to adapted spherical coordinates

x = ar cosφ cos θy = br cosφ sin θz = cr sinφ

, B :

−π

2 ≤ φ ≤π2

0 ≤ θ ≤ 2π0 ≤ r ≤ 1

, |detDT (r, θ, φ)| = abcr2 cosφ.

We can now solve the integral:

˚W

divG dx dy dz =

ˆ 2π

0dθ

ˆ π2

−π2

dφ

ˆ 1

0abcr2 cosφ · z

(1

a2+

1

b2+

2

c2

)dr

= abc

(1

a2+

1

b2+

2

c2

)ˆ 2π

0dθ

ˆ π2

−π2

cosφ sinφ dφ

ˆ 1

0r2 dr

= abc

(1

a2+

1

b2+

2

c2

)2π

ˆ π2

−π2

1

3cosφ sinφ

= abc

(1

a2+

1

b2+

2

c2

)2π

3

[sin2 φ

2

]φ=π2

φ=−π2

= 0.

We could expect this result. As divG is odd in z and the solid W is symmetric with respect to z, itis clear that the integral I of divG on W is zero. Indeed, under the change of variables (x, y, z) =(u, v,−w) one can see that I = −I.

36

30. When a water tap is opened slowly, water flowing out from the opening becomes thinner as it floweddown, and a water jet is formed, with a radius that decreases with the distance from the tap. Computethe form of the water jet as a function of the initial radius r0, the initial speed v0 and the gravity g,assuming that the vertical component of the speed is constant in each horizontal section.

Resolution: To analyze this problem, we first have to take four considerations about the fluid:

(a) No viscosity: there is no friction inside it.

(b) Stationary: the velocity at a given point is constant with respect to time.

(c) Incompressible: the density remains constant.

(d) Irrotational: there cannot be whirlpools, there are not angular movement with respect any point.

Under these assumptions, we can continue our analysis. The first step is to know the Continuityequation which is synthesized in the following formula, where v1 is the fluid velocity in the point Awith a section S1 and v2 is the velocity at the point B with section S2:

v1 · S1 = v2 · S2

Then, let us calculate the velocity of a section of the jet using physics concepts:

∆Em = 0 thus Em0 = Emf .

Then, knowing the two components of the mechanical energy, we can reach the following formula. Wewill take the final height equal to zero (h = 0):

1

2mv20 +mgh0 =

1

2mv2 +mg · 0.

Now, the mass vanishes, and reorganizing all the terms, we obtain the formula just below, where v isthe section velocity, v0 the initial section velocity, g is the gravity that we will consider constant andfinally h0 is the initial height:

v2 = v20 + 2gh0.

From the two previous equations, we can obtain the radius r in function of the initial radius r0, theinitial section velocity v0 and the height h.

r = r04

√v20

v20 + 2gh0.

31. Flux of the vector field F (x, y, z) = (x2,−y2, z2) through the following surfaces:(a)The boundary of:

W =

(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ 3R2, 0 ≤ z ≤√x2 + y2 −R2

,

oriented by an outward-pointing normal.

37

W on the (r, z) plane and on the (x, y, z) space with outward-pointing normal vectors for R = 1.

(b) The octant of sphere:

S =

(x, y, z) ∈ R3 : x2 + y2 + z2 = R2, x, y, z ≥ 0,

oriented by an outward-pointing normal.

First octant of the sphere W for R = 1 with outward-pointing normal N .

Resolution: (a) W =

(x, y, z) ∈ R3 : r2 + z2 ≤ 3R2, 0 ≤ z ≤√r2 −R2

, where r2 = x2 + y2, so in

cylindrical coordinates (x, y, z) = T (r, θ, z) = (r cos θ, r sin θ, z), W = T (B) with

B = (r, θ, z) : r2 + z2 ≤ 3R2, 0 ≤ z ≤√r2 −R2, 0 ≤ θ ≤ 2π

= (r, θ, z) : r2 + z2 ≤ 3R2, z2 ≤ r2 −R2, z ≥ 0, 0 ≤ θ ≤ 2π,

thus W is a solid of revolution around the z-axis, and its boundary consists of 3 parts: ∂W =S1 ∪ S2 ∪ S3, where these surfaces are defined by the equations

S0 = z = 0, R ≤ r ≤√

3R,

S1 = z =√r2 −R2, R ≤ r ≤

√2R, S2 = z =

√3R2 − r2,

√2R ≤ r ≤

√3R,

38

or more precisely, they can be parameterized as

S0 = ϕ0(D0), ϕ0(r, θ) = (r cos θ, r sin θ, 0), D0 = (r, θ) : R ≤ r ≤√

3R, 0 ≤ θ ≤ 2π,

S1 = ϕ1(D1), ϕ1(r, θ) = (r cos θ, r sin θ,√r2 −R2), D1 = (r, θ) : R ≤ r ≤

√2R, 0 ≤ θ ≤ 2π,

S2 = ϕ2(D2), ϕ2(r, θ) = (r cos θ, r sin θ,√

3R2 − r2), D2 = (r, θ) :√

2R ≤ r ≤√

3R, 0 ≤ θ ≤ 2π.

For a parameterization S = ϕ(D) with ϕ(r, θ) = (r cos θ, r sin θ, f(r)), D = (r, θ) : rm ≤ r ≤ rM,0 ≤ θ ≤ 2π of a curve C given by z = f(r), the tangent vectors and the normal vector are given by

ϕr =

cos θsin θf ′(r)

, ϕθ =

−r sin θr cos θ

0

, ϕr ∧ ϕθ =

∣∣∣∣∣∣i j k

cos θ sin θ f ′(r)−r sin θ r cos θ 0

∣∣∣∣∣∣ = r

−f ′(r) cos θ−f ′(r) sin θ

1

,

where ϕr ∧ ϕθ is always an upward-pointing vector, since its z-component is positive: r > 0.

As F (x, y, z) = (x2,−y2, z2), (F ϕ) (r, θ) = (r2 cos2 θ,−r2 sin2 θ, f(r)2) and 〈F ϕ, ϕr ∧ ϕθ〉 =−r3f ′(r)

(cos3 θ − sin3 θ

)+ rf(r)2, so that the flux of F through S pointing outward S is given by

¨S〈F , dS〉 =

¨S〈F ,N ext dS〉 =

¨D〈F ϕ,ϕr ∧ϕθ〉 dr dθ

= −

ˆ 2π

0dθ(cos3 θ − sin3 θ

) ˆ rM

rm

r3f ′(r) dr + 2π

ˆ rM

rm

rf(r)2 dr

= 2π

ˆ rM

rm

rf(r)2 dr,

since

ˆ 2π

0cos3 θ dθ =

ˆ 2π

0sin3 θ dθ = 0

We now apply the formula above to the 3 surfaces S0, S1, S2 with outward-pointing normal vectors toW : on S0 with downward-pointing normal vector (negative induced orientation), and on S1 andS2 with upward-pointing normal vectors:

On S0,

¨S0

〈F , dS〉 = −2π

ˆ √3RR

r 02 dr = 0,

on S1,

¨S1

〈F , dS〉 = 2π

ˆ √2RR

r(r2 −R2

)dr = 2π

[r4

4− R2r2

2

]r=√2Rr=R

=π

2R4,

on S2,

¨S2

〈F , dS〉 = 2π

ˆ √3R√2R

r(3R2 − r2

)dr = 2π

[3R2r2

2− r4

4

]r=√3Rr=√2R

=π

2R4.

Summing up,

‹∂W〈F , dS〉 = πR4.

Alternatively, we could use the parameterization S = ψ(D) with ψ(θ, z) = (g(z) cos θ, g(z) sin θ, z) forS = S1 or S = S2 with D = (θ, z) : 0 ≤ z ≤ R, 0 ≤ θ ≤ 2π and r = g(z) =

√R2 + z2 on S1, with

r = g(z) =√

3R2 − z2 on S2. Then the tangent vectors and the normal vector are given by

ψθ=

−g(z) sin θg(z) cos θ

0

,ψz=g′(z) cos θg′(z) sin θ

1

,ψθ ∧ψz=∣∣∣∣∣∣

i j k−g(z) sin θ g(z) cos θ 0g′(z) cos θ g′(z) sin θ 1

∣∣∣∣∣∣=g(z)

cos θsin θ−g′(z)

, (1)

where now ψθ ∧ ψz is an outward-pointing vector if g(z)g′(z) < 0, and an inward-pointing vector ifg(z)g′(z) > 0.

39

On S1, g′(z) =

z√R2 + z2

=z

g(z), so g′(z)g(z) = z on S1,

on S2, g′(z) =

z√3R2 − z2

= − z

g(z), so g′(z)g(z) = −z on S2.

Therefore ψθ ∧ψz points inward in S1 and outward in S2,

(F ψ) (θ, z) = (g(z)2 cos2 θ,−g(z)2 sin2 θ, z2),

〈F ψ,ψθ ∧ψz〉 = g(z)3(cos3 θ − sin3 θ

)− z2g(z)g′(z)

and ¨S1

〈F , dS〉 =

¨S1

〈F ,N ext dS〉 = −¨D〈F ψ,ψr ∧ψθ〉 dr dθ

= −

(

ˆ 2π

0dθ(cos3 θ − sin3 θ

) ˆ R

0g(z)3 dz − 2π

ˆ R

0z2g(z)g′(z) dz

)

= 2π

ˆ R

0z2g(z)g′(z) dz = 2π

ˆ R

0z3 dz = 2π

R4

4=πR4

2,

whereas ¨S2

〈F , dS〉 =

¨S2

〈F ,N ext dS〉 =

¨D〈F ψ,ψr ∧ψθ〉 dr dθ

=

ˆ 2π

0dθ(cos3 θ − sin3 θ

) ˆ R

0g(z)3 dz − 2π

ˆ R

0z2g(z)g′(z) dz

= −2π

ˆ R

0z2g(z)g′(z) dz = 2π

ˆ R

0z3 dz =

πR4

2.

On S0 we cannot apply this parameterization r = g(z) since z ≡ 0, but directly F (x, y, 0) =

(x2,−y2, 0),N ext = (0, 0,−1)>, so that FN = 〈F ,N ext〉 = 0 and

¨S0

〈F , dS〉 =

¨S0

〈F ,N ext〉 dS = 0.

Summing up again,

‹∂W〈F , dS〉 = πR4.

(b)We use a parameterization S = ψ(D) based on spherical coordinates taking ρ = R with (x, y, z) =ψ(θ, φ) = (R cosφ cos θ,R cosφ sin θ,R sinφ) and D = (θ, φ) : 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2 for theoctant of the sphere. The tangent vectors and the normal vector are given by

ψθ =

−R cosφ sin θR cosφ cos θ

0

, ψφ =

−R sinφ cos θ−R sinφ sin θ

R cosφ

,∥∥ψθ ∧ψφ∥∥ = R2 |cosφ| = R2 cosφ,

with ψθ ∧ψφ pointing outward the sphere since

ψθ ∧ψφ =

∣∣∣∣∣∣i j k

−R cosφ sin θ R cosφ cos θ 0−R sinφ cos θ −R sinφ sin θ R cosφ

∣∣∣∣∣∣ = R2 cosφ

cosφ cos θcosφ sin θ

sinφ

.

As F (x, y, z) = (x2,−y2, z2), (F ψ)(θ, φ) = R2(cos2 φ cos2 θ,− cos2 φ sin2 θ, sin2 φ

). We have all the

ingredients to compute the flux of the vector field F through the surfaces S oriented by an outward-

40

pointing normal:

¨S〈F , dS〉 =

¨D〈F ψ,ψθ ∧ψφ〉 dθ dφ

= R4

¨D

(cos4 φ cos3 θ − cos4 φ sin3 θ + cosφ sin3 φ

)dθ dφ

= R4

ˆ π/2

0

(cos3 θ − sin3 θ

)dθ

ˆ π/2

0cos4 φ dφ+

ˆ π/2

0dθ

ˆ π/2

0sin3 φ cosφ dφ

= R4π

2

[sin4 φ

4

]φ=π/2φ=0

=πR4

8,

where we have used that

ˆ π/2

0cos3 θ dθ =

ˆ π/2

0(1− sin2 θ) cos θ dθ =

[sin θ − sin3 θ

3

]θ=π/2θ=0

= 2/3.

Analogously,

ˆ π/2

0sin3 θ dθ = 2/3 so that

ˆ π/2

0cos3 θ − sin3 θ dθ = 0.

Alternatively, we could use the cartesian parameterization ϕ(x, y) = (x, y, f(x, y)), where z = f(x, y) =√R2 − x2 − y2, and D = (x, y) : x2 + y2 ≤ R2, x ≥ 0, y ≥ 0 chosen such that S = ϕ(D). The

associated normal vector is ϕx ∧ ϕy = (−fx,−fy, 1)>, where fx =−x√

R2 − x2 − y2= −x

f, and fy =

−yf

, which is a vertical vector and therefore points outward the sphere. Recalling that F (x, y, z) =

(x2,−y2, z2), we can compute again the flux of F through the piece S of the sphere oriented by anoutward-pointing normal vector as

¨S〈F , dS〉 =

¨D〈F ϕ,ϕx ∧ϕy〉 dx dy

=

¨D

(−x2fx + y2fy + f2

)dx dy

=

¨D

x3 − y3

fdx dy +

¨Df2 dx dy =

πR4

8,

where

¨D

x3 − y3

fdx dy =

x = r cos θy = r sin θ

=

ˆ π/2

0

(cos3 θ − sin3 θ

)dθ

ˆ R

0

r4 dr√R2 − r2

= 0,

and

¨Df2 dx dy =

ˆ π/2

0dθ

ˆ R

0

(R2 − r2

)r dr =

π

2

[R2r2

2− r4

4

]r=Rr=0

=π

2

(R4

2− R4

4

)=πR4

8.

32. Flux of the vector field F (x, y, z) = (2x,−y, 0) through the surface

S =

(x, y, z) ∈ R3 : x2 + y2 = R2, x, y ≥ 0, 0 ≤ z ≤ h

(a quarter of cylinder) oriented by its outward-pointing normal.

41

Surface and corresponding vectors.

Resolution: We use a parameterization S = ϕ(D) based on cylindrical coordinates, in which r = R,and: ϕ(θ) = (R cos θ,R sin θ, z). The domain is: D = (θ, z) : 0 ≤ θ ≤ π/2, 0 ≤ z ≤ h, since x andy are positive, and the maximum height is h for the quarter of the cylinder. The tangent vectors aregiven by the respective partial derivatives as follows:

ϕθ =

−R sin θR cos θ

0

, ϕz =

001

,

We have to take into account that the normal vector points outward the sphere since the cross productindicates a positive direction in the first quadrant:

ϕθ ∧ϕz = (R cos θ,R sin θ, 0)

Now that we have all the necessary information, we can proceed to integrate the field using ourparameterization. If we develop the calculations:

¨S〈F , dS〉 =

¨D〈F ϕ,ϕθ ∧ϕz〉 dz dθ =

ˆ π/2

0

ˆ h

0〈(2R cos θ,−R sin θ, 0), (R cos θ,R sin θ, 0)〉dz dθ

=

ˆ π/2

0

ˆ h

02R2 cos θ −R2 sin θ dz dθ =

ˆ π/2

0

[2R2 cos θ −R2 sin θ · z

]h0

dθ

= h

ˆ π/2

02R2 cos θ −R2 sin θ dθ = h

ˆ π/2

02R2

(1

2+

1

2cos 2θ

)−R2

(1

2− 1

2cos 2θ

)dθ

= h

ˆ π/2

0

R2

2+

3R2

2cos 2θ dθ =

hR2

2

ˆ π/2

01 + 3 cos 2θ dθ =

hR2

2

[θ +

3

2sin 2θ dθ

]π/20

=R2hπ

4.

33. Flux of the vector field F (x, y, z) = (x2, y2, z2) through the surface

S =

(x, y, z) ∈ R3 : z = h(x2 + y2)/R2, z ≤ h,

(piece of the elliptic paraboloid) oriented by its interior normal.

42

Resolution: We use a parameterization S = ϕ(D) based on cylindrical parameters with ϕ(r, θ) =(r cos θ, r sin θ, h

r2

R2

)and D = (r, θ) : 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π for the elliptic paraboloid.

ϕr =

cos θsin θhR2 2r

, ϕθ =

−r sin θr cos θ

0

,

ϕr ∧ϕθ =

∣∣∣∣∣∣i j k

cos θ sin θ hR2 2r

−r sin θ r cos θ 0

∣∣∣∣∣∣ =

−2r2 hR2 cos θ

−2r2 hR2 sin θr

.

Also

F ϕ(r, θ) =

(r2 cos2 θ, r2 sin2 θ, h2

r4

R4

).

Therefore: ¨S〈F , dS〉 =

¨D〈F ϕ,ϕr ∧ϕθ〉dr dθ

=

¨D

2r4 cos3 θh

R2+ 2r4 sin3 θ

h

R2− r5 h

2

R4dr dθ

=

¨D

2r4 cos3 θh

R2+

¨D

2r4 sin3 θh

R2−¨Dr5h2

R4dr dθ

= 2π

R

0

r5h2

R4dr dθ = 2π

h2

R4

[r6

6

]R0

=πR2h2

3.

34. Flux of the vector field r(x, y, z) = (x, y, z) through the portion S of the hyperbolic paraboloid z =h(x2 − y2)/L2, cut by the plains x = 0, x = L and z = 0, oriented in the point O = (0, 0, 0) by thevector k = (0, 0, 1).

Resolution: First, S is the piece of the hyperbolic paraboloid z = h(x2 − y2)/L2, cut by the plainsx = 0, x = L and z = 0. So, if we plot the hyperboloid (which will be plotted lately), this means that

S = (x, y, z) ∈ R3 : z = h(x2 − y2)/L2, 0 ≤ x ≤ L, z ≥ 0.

Imposing z ≥ 0 is equivalent to impose x2 − y2 ≥ 0 (we assume that h and L are positive) or,equivalently y2 ≤ x2 or −x ≤ y ≤ x, so we can write

S = (x, y, z) ∈ R3 : z = h(x2 − y2)/L2, 0 ≤ x ≤ L,−x ≤ y ≤ x = ϕ(D),

where ϕ(x, y) = (x, y, f(x, y) := h(x2 − y2)/L2 is a Cartesian parameterization of S, defined on

D =

(x, y) ∈ R2 : 0 ≤ x ≤ L,−x ≤ y ≤ x.

We do a plot of the hyperbolic paraboloid in order to be easier to understand the problem.

43

Hyperbolic paraboloid and surface S.

By equation

ϕx =

10fx

, ϕy =

01fy

, ϕx ∧ϕy =

∣∣∣∣∣∣i j k1 0 fx0 1 fy

∣∣∣∣∣∣ =

−fx−fy1

,∥∥ϕx ∧ϕy∥∥ =

√1 + f2x + f2y ,

we know that the associated normal vector to this parameterization is ϕx ∧ ϕy = (−fx,−fy, 1)> =(−2hx

L2,2hy

L2, 1

)>, which is an upward-pointing normal vector, and for (x, y, z) = (0, 0, 0) coincides

with the normal vector provided in the statement of the problem for the orientation of S, which is(0, 0, 1).

We are given the vector field r(x, y, z) = (x, y, z), so that writing z = f(x, y) = h(x2 − y2)/L2, wehave that 〈r ϕ,ϕx ∧ ϕy〉 = −f(x, y), and finally after some computations we get the result for theflux: ¨

S〈r, dS〉 =

¨D〈r ϕ,ϕx ∧ϕy〉 dS = −

¨Df(x, y) dx dy = − h

L2

ˆ L

0dx

ˆ x

−xx2 − y2 dy

= − h

L2

ˆ L

0

4x3

3dx = − h

L2

[x4

3

]L0

= −hL4

3L2= −hL

2

3.

To check this result, we could also use Gauss’ theorem, adding two flat covers S0 = (x, y, z) : 0 ≤ x ≤L,−x ≤ y ≤ x, z = 0, with outward-pointing normal vector N0 = (0, 0, 1)> and SL = (x, y, z) : x =L,−L ≤ y ≤ L, z = f(x, y), with outward-pointing normal vector (1, 0, 0)>, and area A(SL) = 4hL/3,which enclose a solid

W = (x, y, z) ∈ R3 : 0 ≤ x ≤ L,−x ≤ y ≤ x, 0 ≤ z ≤ h(x2 − y2)/L2,

with volume V (W ) = hL2/3, such that ∂W = S ∪ S0 ∪ SL. Gauss’s theorem gives˚W

div r dx dy dz =

‹∂W+

〈r, dS〉 =

¨S+

〈r, dS〉+

¨S+0

〈r, dS〉+

¨S+L

〈r, dS〉.

On the one hand, as div r = 3,

˚W

div r dx dy dz = 3V (W ) = hL2. On the other hand, rN =

〈r,N0〉 = 0 on S0 so that

¨S+0

〈r, dS〉 = 0, and rN = 〈r,NL〉 = L on SL so that

¨S+L

〈r, dS〉 =

¨S+L

〈r,NL dS〉 =

¨S+L

〈r,NL〉 dS = LA(SL) = L · 4hL/3 = 4hL2/3. Putting all this together,

¨S+

〈r, dS〉 =

˚W

div r dx dy dz −¨S+0

〈r, dS〉 −¨S+L

〈r, dS〉 = hL2 − 0− 4hL2

3= −hL

2

3.

44

35. Flux of the vector field F (x, y, z) = (2x2, 3y2, z2) through the boundary of the following solid of revo-lution:

W =

(x, y, z) ∈ R3 :√x2 + y2 ≤ z ≤

√2R2 − x2 − y2

,

oriented by its outward-pointing normal.

Resolution: W in cylindrical coordinates takes the form

B = (r, θ, z) : r2 ≤ z2, 0 ≤ z ≤√

2R2 − r2 = B1 ∪B2

with B1 = (r, θ, z) : r ≤ z/, z ≤ R, cone of base a circle of radius R and height R, and B2 =(r, θ, z) : z2 + r2 ≤ 2R2, z ≥ R, hemisphere of radius

√2R.

Let S1 = (x, y, z) : z =√x2 + y2, x2 +y2 ≤ R2, S2 = (x, y, z) : z =

√2R2 − x2 − y2, x2 +y2 ≤ R2.

On S1 we use a cylindrical parameterization S1 = ϕ(D1), with ϕ(θ, z) = (g(z) cos θ, g(z) sin θ, z) andg(z) = z, g′(z) = 1 on the domain D1 = (θ, z) : 0 ≤ z ≤ R, 0 ≤ θ ≤ 2π. By equation (1), theassociated normal vector is ψθ ∧ ψz = g(z)(cos θ, sin θ,−g′(z))> = z(cos θ, sin θ,−1)>, which pointsoutward W along S1, since has a negative z-component. Therefore the flux of F through S1 orientedby an outward-pointing normal vector is

¨S−1〈F , dS〉 =

¨D1

〈F ϕ,ϕθ ∧ϕz〉dθ dz

=

¨D−1

(2g(z)3 cos3 θ + 3g(z)3 sin3 θ − zg(z)g′(z)

)dθ dz

=

ˆ 2π

0

(2 cos3 θ + 3 sin3 θ

)dθ

ˆ R

0z3 dz −

ˆ 2π

0dθ

ˆ R

0z3 dz

= −πR4

2,

since

ˆ 2π

0cos3 θ dθ =

ˆ 2π

0sin3 θ dθ = 0

On S2 we also use a cylindrical parameterization S2 = ϕ(D2), with g(z) =√

2R2 − z2, g′(z) =−z√

2R2 − z2= − z

g(z)on the domain D2 = (θ, z) : R ≤ z ≤

√2R, 0 ≤ θ ≤ 2π. The associated normal

vector is ψθ ∧ ψz = (g(z) cos θ, g(z) sin θ, z)>, which also points outward W along S2, since has apositive z-component. Therefore the flux of F through S2 oriented by an outward-pointing normalvector is ¨

S2

〈F , dS〉 =

¨D2

〈F ϕ,ϕθ ∧ϕz〉 dθ dz

=

¨D2

(2g(z)3 cos3 θ + 3g(z)3 sin3 θ + z3

)dθ dz

=

ˆ 2π

0

(2 cos3 θ + 3 sin3 θ

)dθ

ˆ √2RR

g(z)3 dz +

ˆ √2RR

z3 dz

=

ˆ √2RR

z3 dz =2π

4

(4R4 −R4

)=

3πR4

2.

Summing up,

‹∂W〈F , dS〉 =

3πR4

2− πR4

2= πR4.

45

The integral theorems II

36. Given the vector field F (x, y, z) = (4xz,−y2, yz), compute its flux through the boundary of the unitcube W = [0, 1]3 oriented by its outward-pointing normal.

Resolution: The cube W has 6 different faces, so the direct computation of the flux of F through∂W involves the computation of 6 integral surfaces. Instead, we are going to apply Gauss’ theorem tocompute one single triple integral using that divF = Px +Qy +Rz = 4z − 2y + y = 4z − y

‹S+

〈F , dS〉 =

˚W

divF dx dy dz =

˚W

(4z − y) dx dy dz

=

ˆ 1

0dx

ˆ 1

0dy

ˆ 1

04z dz −

ˆ 1

0dx

ˆ 1

0y dy

ˆ 1

0dz =

ˆ 1

04z dz −

ˆ 1

0y dy

=[2z2]z=1

z=0−[y2

2

]y=1

y=0

= 2− 1

2=

3

2.

37. Let F be a central force without singularities, so F = h(r)r, where r(x, y, z) = (x, y, z), r2(x, y, z) =(x2 + y2 + z2) and h(r) is a function of class C1 in the interval [0,∞). SR is the sphere with ra-dius R centered in the origen and oriented by its outward-pointing normal. Compute the exit flux¨SR

〈F , dS〉: (a) Integrating the normal component of the vector field over the sphere.

(b) Applying Gauss theorem. Compare the result with Gauss Law.

Resolution: (a) First of all, notice that a sphere of radius R centered in the origin has the followingequation: x2 + y2 + z2 = R2 and that in the statement of the problem we are given that r2(x, y, z) =(x2 + y2 + z2). Hence, on the sphere we have that r2 = R2.Then, we use a parameterization SR = ψ(D) based on spherical coordinates with (x, y, z) = ψ(θ, φ) =(R cosφ cos θ,R cosφ sin θ,R sinφ) and D = (θ, φ) : 0 ≤ θ ≤ 2π,−π/2 ≤ φ ≤ π/2. The tangentvectors and the normal vector are given by

ψθ =

−R cosφ sin θR cosφ cos θ

0

, ψφ =

−R sinφ cos θ−R sinφ sin θ

R cosφ

,∥∥ψθ ∧ψφ∥∥ = R2 |cosφ| = R2 cosφ,

with ψθ ∧ψφ pointing outward the sphere since

ψθ ∧ψφ =

∣∣∣∣∣∣i j k

−R cosφ sin θ R cosφ cos θ 0−R sinφ cos θ −R sinφ sin θ R cosφ

∣∣∣∣∣∣ =

R2 cos2 φ cos θR2 cosφ sin θR2 cosφ sinφ

= R2 cosφ

cosφ cos θcosφ sin θ

sinφ

.

As F (x, y, z) = h(r)r(x, y, z), (F ψ)(θ, φ) = h(R)(R cosφ cos θ,R cosφ sin θ,R sinφ). We now haveall the necessary data to compute the flux of the central force F through the surface SR oriented by

46

its outward-pointing normal:

¨SR

〈F , dS〉 =

¨D〈F ψ,ψθ ∧ψφ〉 dθ dφ

= h(R)

¨D

(R3 cos3 φ cos2 θ +R3 cos3 φ sin2 θ +R3 sin2 φ cosφ

)dθ dφ

= h(R)R3

ˆ π/2

−π/2

ˆ 2π

0

(cos3 φ(cos2 θ + sin2 θ) + sin2 φ cosφ

)dθ dφ

= h(R)R3

ˆ π/2

−π/2

ˆ 2π

0

(cos3 φ+ sin2 φ cosφ

)dθ dφ

= h(R)R3

ˆ π/2

−π/2

ˆ 2π

0

(cosφ(cos2 φ+ sin2 φ)

)dθ dφ

= h(R)R3

ˆ π/2

−π/2

ˆ 2π

0(cosφ) dθ dφ

= h(R)R3

ˆ π/2

−π/2(cosφ) [θ]θ=2π

θ=0 dφ = 2πh(R)R3

ˆ π/2

−π/2(cosφ) dφ

= 2πh(R)R3 [sinφ]π/2−π/2 = 2πh(R)R3(1− (−1) = 4πh(R)R3.

Therefore the exit flux

¨SR

〈F , dS〉 = 4πh(R)R3.

(b) We must apply Gauss’ theorem on the solid W = (x, y, z) : x2 + y2 + z2 = R2. Gauss’ theoremtells us that: ˚

WdivF dx dy dz =

¨SR

〈F , dS〉

We must first find the divergence of the vector field F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)),

divF =∂P

∂x+∂Q

∂y+∂R

∂z

∂P

∂x=d(h(r)x)

dx=

dh

dr

dr

dx· x+ h(r) · 1 =

dh

dr

1

2√x2 + y2 + z2

· 2x · x+ h(r) =dh

dr

1

r· x2 + h(r).

Similarly,∂Q

∂y=

dh

dr

1

r· y2 + h(r),

∂R

∂z=

dh

dr

1

r· z2 + h(r).

Therefore, the divergence is

divF = 3h(r) +dh

dr

1

r(x2 + y2 + z2) = 3h(r) +

dh

dr

1

rr2 = 3h(r) + h′(r)r

Knowing that the sphere is oriented by its outward-pointing normal we can compute

¨SR

〈F , dS〉 =

˚W

divF dx dy dz =

˚W

3h(r) dx dy dz +

˚Wh′(r)r dx dy dz.

To solve the triple integral we can use spherical coordinates. Let’s take B such that T (B) = W :

B =

(r, θ, ρ) : 0 ≤ θ ≤ 2π,

−π2≤ ρ ≤ π

2, 0 ≤ r ≤ R

.

47

Then ¨SR

〈F , dS〉 = 3

˚

B

h(r)r2 cos ρdr dρdθ +

˚

B

h′(r)r3 cos ρdr dρdθ

=

ˆ 2π

0dθ

ˆ π2

−π2

cos ρdρ

ˆ R

0h(r)r2 dr +

ˆ 2π

0dθ

ˆ π2

−π2

cos ρdρ

ˆ R

0h′(r)r3 dr

= 12π

ˆ R

0h(r)r2 dr + 4π

ˆ R

0h′(r)r3 dr,

and finally integrating by parts using

u = h(r) du = h′(r) dr,

dv = r2 v = r3

3 ,

¨SR

〈F , dS〉 = 12π

[[h(r)

r3

3

]R0

−ˆ R

0

r3

3h′(r) dr

]+ 4π

ˆ R

0h′(r)r3 dr

= 4πR3h(R)−

4π

ˆ R

0h′(r)r3 dr +

4π

ˆ R

0h′(r)r3 dr

= 4πR3h(R).

Therefore the solution to this exercise is that

¨SR

〈F , dS〉 = 4πR3h(R). One can also check the

solution calculating the flow directly through sphere, to obtain the same result.

Additionally, if we compare the result with Gauss’ law we can see that the flux (using, for instance,

the expression for the electric field E =r

r3can be computed with an integral over the surfaces of the

sphere which gives

ˆSR

⟨r

r3, dS

⟩=

1

R2

ˆdS =

1

R24πR2 = 4π. It is important to note that in this

example there is a singularity at r = 0, and to treat it properly you have to take out a small ball ofradius ε, apply Gauss’ theorem with two boundaries, and take a limit when ε tends to zero.

The important information to know is that this result is a generalization of the one we have foundusing Gauss’ theorem. Furthermore, Gauss’ theorem can be applied to spheres centered at the originwhereas Gauss’ law is valid for any sphere.

38. Compute the flux of the vector field F (x, y, z) = (4x,−2y2, z2) through the cylinder

S =

(x, y, z) ∈ R3 : x2 + y2 = 4, 0 ≤ z ≤ 3

oriented by its outward-pointing normal. [ Hint: In this problem and from now on, you can cover thesurface in order to be able to apply Gauss’ theorem.]

Resolution: Notice that S is a cylinder of equation x2 + y2 = 4 bounded by 0 ≤ z ≤ 3. So, apartfrom the cylinder we are going to add 2 horizontal flat covers S0 and S3, where the subindex refer tothe value of z at the covers. This solid will be called W :

W =

(x, y, z) ∈ R3 : x2 + y2 = 4, 0 ≤ z ≤ 3

and its boundary ∂W will consist on 3 pieces, ∂W = S ∪ S0 ∪ S3.

48

Solid W of Exercise V48.

As we have to apply Gauss’ theorem, first of all we are going to recall it:

‹S+

〈F , dS〉 =

˚W

divF dx dy dz,

where S+ means that we must use a normal vector pointing outward W . First, we calculate thetriple integral. Since the vector field is F (x, y, z) = (4x,−2y2, z2), divF = 4 − 4y + 2z. Since Wis the interior of a cylinder, we are going to parameterize W with cylindrical coordinates. Noticethat W =

(x, y, z) ∈ R3 : r2 = 4, 0 ≤ z ≤ 3

, where r2 = x2 + y2, so in cylindrical coordinates

(x, y, z) = T (r, θ, z) = (r cos θ, r sin θ, z), we have W = T (B) with

B = (r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ z ≤ 3, 0 ≤ θ ≤ 2π.

We now calculate the divergence integral:

˚W

divF dx dy dz =

2πˆ

0

dθ

3ˆ

0

dz

2ˆ

0

r (4− 4r sin θ + 2z) dr = 2

2πˆ

0

dθ

3ˆ

0

dz

2ˆ

0

2r − 2r2 sin θ + rz dr

= 2

2πˆ

0

dθ

2ˆ

0

[r2 − 2r3 sin θ

3+r2z

2

]r=2

r=0

dz = 2

2πˆ

0

dθ

3ˆ

0

4− 16

3sin θ + 2z dz

= 2

2πˆ

0

[4z − 16

3sin θ z + z2

]z=3

z=0

dθ = 2

2πˆ

0

21 dθ −

2πˆ

0

16 sin θ dθ = 84π.

Now, it is time to calculate the flux through the covers.

49

Notice the unitary outward-pointing normal vector is N0 = (0, 0,−1) in S0 and N3 = (0, 0, 1) in S3.So, this makes very easy and short the computation of the flux through these covers, because, in S3,FN = 〈F ,N〉 = z2 = 9 and

¨S3

〈F , dS〉 =

¨S3

〈F ,N〉 dS =

¨S3

FN dS = 9A(S3) = 9 · π22 = 36π.

And in S0, FN = 〈F ,N〉 = 0 so

¨S0

〈F , dS〉 =

¨S0

〈F ,N〉 dS =

¨S0

FN dS = 0.

Finally,‹S+

〈F , dS〉 =

˚W

divF dx dy dz −¨S0

〈F , dS〉 −¨S3

〈F , dS〉 = 84π − 0− 36π = 48π.

We are asked about the flux through the lateral surface S of the cylinder oriented by its outward-pointing normal, and to apply the Gauss’ theorem, we also use S oriented by its outward-pointingnormal, so we don’t have to change sign.

Therefore, the flux of F through the whole boundary ∂W = S ∪ S0 ∪ S3 is 48π.

39. Compute the flux of the vector field F (x, y, z) = (−x, 0, x + z) through the piece of the sphere givenby S =

(x, y, z) ∈ R3 : x2 + (y − 5)2 + (z − 5)2 = 25, y ≤ 9, z ≥ 1

oriented by the normal vector

N(x, y, z) = (x, y − 5, z − 5).

Sphere S without covers of Exercise 39.

Resolution: We first notice that S is the sphere with center (0, 5, 5) and radius 5 without two caps.We will apply Gauss’ theorem on a solid W , which states that the integral of a vector field over asolid W is equal to the flux of the vector field through its closed boundary surface ∂W . So we have tocompute one single triple integral using divF = Px +Qy +Rz = −1 + 0 + 1 = 0, so that

‹∂W+

〈F , dS〉 =

˚W

divF dx dy dz = 0.

50

We take W as the interior of S:

W =

(x, y, z) ∈ R3 : x2 + (y − 5)2 + (z − 5)2 ≤ 25, y ≤ 9, z ≥ 1,

and its boundary ∂W consists of S plus two flat disks S1 and S9 of radius 3 and thus area 9π:

S1 =

(x, y, z) ∈ R3 : x2 + (y − 5)2 + (z − 5)2 ≤ 25, y ≤ 9, z = 1

=

(x, y, 1) : x2 + (y − 5)2 ≤ 9,

S9 =

(x, y, z) ∈ R3 : x2 + (y − 5)2 + (z − 5)2 ≤ 25, y = 9, z ≤ 1

=

(x, 9, z) : x2 + (z − 5)2 ≤ 9.

So we get

0 =

‹∂W+

〈F , dS〉 =

‹S+

〈F , dS〉+

¨S+1

〈F , dS〉+

¨S+9

〈F , dS〉.

As what we want is the flux of the field on S:‹S+

〈F , dS〉 = −¨S+1

〈F , dS〉 −¨S+9

〈F , dS〉.

In order to compute the flux through the flat covers Sj , j = 1, 9,

¨S+j

〈F , dS〉 =

¨S+1

〈F ,N+j dS〉 =

¨S1

〈F ,N+j 〉dS,

we may specify the unit outward-pointing normal vectors to W . They are N+9 = (0, 1, 0) and N+

1 =(0, 0,−1) (notice that S1 is in the lower part of W ). Recalling that F (x, y, z) = (−x, 0, x+ z), we havethus 〈F ,N+

9 〉 = 0 on S9 and 〈F ,N+1 〉 = −x− z = −x− 1 on S1. Therefore

¨S+9

〈F , dS〉 =

¨S1

〈F ,N+9 〉dS = 0,

¨S+1

〈F , dS〉 =

¨S1

〈F ,N+1 〉dS =

¨S1

−x− 1 dS = −

¨

S1

x dx dy −¨S1

dx dy = −A(S1) = −9π,

where the first integral is zero because x is an odd function integrated on the symmetric disk S1.Finally ‹

S+

〈F , dS〉 = −(−9π)− 0 = 9π,

and we only have to check that the sign is correct, that is, that the normal vector N(x, y, z) =(x, y−5, z−5) of the statement is an outward-pointing vector to W . As the equation of S is g(x, y, z) =0, with g(x, y, z) := x2 + (y − 5)2 + (z − 5)2 − 25, we first recall that any vector proportional to∇g(x, y, z) = 2(x, y − 5, z − 5) is a normal vector vector to S (see Exercise V27), as is the case ofN(x, y, z) = (x, y − 5, z − 5), which clearly points out of the origin and threfore outward W . Forinstance for p = (5, 5, 5) ∈ S, N(p) = (5, 0, 0) points in the increasing direction of the x axis.

40. Let us consider the piece of the ellipsoid given by

S =

(x, y, z) ∈ R3 : x2 + 2y2/3 + z2 = 1, y ≤ 1, z ≤ 2/3.

(a) Compute the circulation

˛Cz dx+ y dy−x dz for C = S ∩y = 1 oriented by the tangent vector

T (x, 1, z) = (−z, 0, x).

(b) Compute the flux

¨S

dz dx orienting S in a compatible way with (a).

51

Domain W = S ∪ S1 ∪ S2/3 of Exercise 40

Resolution:

(a) The curve C = (x, y, z) : x2 + z2 = 1/3, y = 1 is a circle of radius 1/√

3 centered at the origin

of the plane y = 1 that we parameterize as C = σ(I), with (x, y, z) = σ(t) =(

1√3

cos t, 1, 1√3

sin t)

and I = [0, 2π]. To check its orientation, we compute its tangent vector (x′, y′, z′) = σ′(t) =(− 1√

3sin t, 0, 1√

3cos t

)= (−z, 0, x) = T (x, 1, z), so C with these parameterization is oriented as

we were asked for. We now can proceed directly to compute the circulation of the vector fieldF (x, y, z) = (z, y,−x) along C:

˛Cz dx+ y dy − x dz =

ˆIzx′ + yy

′ − xz′ dt =

ˆ 2π

0− (z2 + x2)︸ ︷︷ ︸

1/3

dt = −2π

3.

Alternatively, we could apply Green’s theorem to the disk S1 = (x, y = 1, z) : x2 + y2 ≤ 1/√

3 ofradius 1

√3 contained in the plane y = 1 with the boundary C = ∂S1 oriented counter-clockwise

with respect to S1, or in other words, leaving the region S1 on the left side, as is the case with theparameterization σ, and the vector field F (x, y, z) = (z, y,−x), taking into account that rotF =∣∣∣∣∣∣i j k∂x ∂y ∂zz y −x

∣∣∣∣∣∣ =

020

:

‰C=∂S1

z dx− x dz =

ˆD

(−1− 1) dx dz = −2A(D) = −2π

3.

Even more, we can apply Stokes’ theorem to the surface S1. For this, we have to orient S1 by theunitary normal vector −N1 = (0,−1, 0), since in this way its boundary curve C+ = ∂S1 is orientedby the vector −N1 leaving S1 on the left side:

˛C+

〈F , d`〉. =

¨S+1

〈rotF , dS〉 =

¨S+1

〈rotF ,−N1〉dS = −2A(S1) = −2π

3.

(b) By definition, if G = (P,Q,R),

¨S〈G, dS〉 =

¨SP dy dz +

¨SQdz dx+

¨SR dx dy, so we are

given the vector field G = (0, 1, 0). (Recall that

¨S

dz dx =

¨D

∣∣∣∣zu zvxu xv

∣∣∣∣ dudv, if we parameterize

S = ϕ(D) and ϕ(u, v) = (x(u, v), y(u, v), z(u, v)).)

Take W =

(x, y, z) ∈ R3 : x2 + 2y2/3 + z2 ≤ 1, y ≤ 1, z ≤ 2/3

, W = S ∪ S1 ∪ S2/3, where S2/3 =(x, y, z = 2/3) : x2 + 2y2/3 ≤ 5/9 is a domain enclosed by an ellipse in the plane z = 2/3, withunitary normal vectors N , N1 = (0, 1, 0), N2/3 = (0, 0, 1) pointing outward W . In particular, the

52

unitary normal vector N on S points also outward W , which is compatible with the orientation of C,which is oriented leaving S on the left side. By Gauss’ theorem,

0 =

˚W

divG dx dy dz =

‹∂W+

〈G, dS〉 =

¨S+

〈G, dS〉+

¨S+1

〈G, dS〉+

¨S+2/3

〈G, dS〉,

and ¨S+1

〈G, dS〉 =1

2

¨S+1

〈rotF , dS〉 =1

2

¨S+1

〈rotF ,N1〉 dS =1

2

¨S+1

2 dS = A(S1) =π

3,

¨S+2/3

〈G, dS〉 =1

2

¨S+2/3

〈rotF , dS〉 =1

2

¨S+2/3

〈rotF ,N2/3〉 dS = 0.

Therefore

¨S

dz dx =

¨S+

〈G, dS〉 = −¨S+1

〈G, dS〉 −¨S+2/3

〈G, dS〉 = −π3

.

41. Flux of the vector field F (x, y, z) =(2e−xy, e−xy2, 1

)through the piece of the sphere given by S =

(x, y, z) ∈ R3 : (x− 2)2 + (y − 2)2 = z4, 1 ≤ z ≤ 2

oriented by the normal vector N(x, y, z) = (x −2, y − 2,−2z3).

Resolution:

Surface S of exercise 41.

We have to apply Gauss’ theorem on a solid W , that states that the integral of a vector field over asolid W is equal to the flux of the vector field through its closed boundary surface. So we have tocompute one triple integral using divF = Px +Qy +Rz = −2e−xy + 2ye−x + 0 = 0:

‹S+

〈F , dS〉 =

˚W

divF dx dy dz = 0.

In our case the solid will be the interior of the surface S:

W =

(x, y, z) ∈ R3 : (x− 2)2 + (y − 2)2 ≤ z4, 1 ≤ z ≤ 2,

and its boundary ∂W will consists on three pieces, the surface S and two flat covers S1 and S2 whichhappen to be disks of radius 1 and 2, respectively:

S1 =

(x, y, z) ∈ R3 : (x− 2)2 + (y − 2)2 ≤ z4, z = 1

=

(x, y, 1) ∈ R3 : (x− 2)2 + (y − 2)2 ≤ 1,

S2 =

(x, y, z) ∈ R3 : (x− 2)2 + (y − 2)2 ≤ z4, z = 2

=

(x, y, 2) ∈ R3 : (x− 2)2 + (y − 2)2 ≤ 4.

53

So we get ‹S+

〈F , dS〉 =

¨S+

〈F , dS〉+

¨S+1

〈F , dS〉+

¨S+2

〈F , dS〉 = 0.

As what we want is the flux of the vector field on S:¨S+

〈F , dS〉 = −¨S+1

〈F , dS〉 −¨S+2

〈F , dS〉.

The unit outward-pointing normal vector of S2 is N2 = (0, 0, 1), and the outward-pointing normalvector of S1 is N1 = (0, 0,−1). The outward-pointing normal vector of S is indeed a downward-pointing vector, by the figure, like the normal vector N given in the statement of the exercise.

Moreover, 〈F , dS〉 = 〈F ,N2〉dS = dS on S2, and 〈F , dS〉 = 〈F ,N1〉dS = −dS on S1. Therefore,the flow through S2 is A(S2) = 16π and the flow through S1 is −A(S2) = −π. So we reach our finalsolution ¨

S+

〈F , dS〉 = π − 16π = −15π.

42. Flux of the vector field F (x, y, z) = (1, 0, 2) through the piece of elliptic paraboloid

S =

(x, y, z) ∈ R3 : z = x2 + 4y2, z ≤ 3y2 + 1

oriented by the normal vector with positive vertical component. [ Hint: You can add a non flat coverto the surface.]

Resolution:

Surface S of Exercise 42

Notice that S is a piece of a parabolic cylinder of equation z = x2 + 4y2 that is cut by a paraboliccylinder of equation z = 3y2 + 1, and satisfies x2 + 4y2 ≤ 3y2 + 1, or x2 + y2 ≤ 1, so we canwrite S = ϕ(D) for the cartesian parameterization ϕ(x, y) = (x, y, f(x, y) = x2 + 4y2) on the circleD = (x, y) : x2+y2 ≤ 1. The associated normal vector is ϕx∧ϕy = (−fx,−fy, 1)> = (−2x,−8y, 1)>,which has positive vertical component, so this parameterization provides the orientation for S that weare asked. Therefore¨

S〈F , dS〉 =

¨D〈F ϕ,ϕx ∧ϕy〉 dx dy

=

¨D〈(1, 0, 2), (−2x,−8y, 1)〉 dx dy =

¨D

2−2x dx dy = 2A(D) = 2π,

54

where we have used that˜D 2x dx dy = 0, because under the change (x, y) = (−u, v) the integrand

2x changes sign but the domain D remains the same.

Domain W closed by S and the cover Sc of Exercise 42

Alternatively, we can try to apply Gauss’ theorem on a solid W with a boundary ∂W containing S.Thus we introduce W = (x, y, z) : x2 + 4y2 ≤ z ≤ 3y2 + 1 whose boundary ∂W = S ∪ Sc contains Sand the “cover” surface Sc = (x, y, z) : z = 3y2 + 1, z ≥ x2 + 4y2 = ψ(D), for the parameterizationψ(x, y) = (x, y, g(x, y) = 3y2 +1) on the same circle D = (x, y) : x2 +y2 ≤ 1. The associated normalvector is ψx∧ψy = (−gx,−gy, 1)> = (0,−6y, 1)>, which has positive vertical component and thereforepoints outward W .

On the other hand, our previous parameterization ϕ of S points now inward W . So if we call ∂W+ =S+ ∪ S+

c the boundary of W oriented by the outward-pointing normal vector to W , the orientation ofS+ is just opposite to the orientation of S, and thus¨

S〈F , dS〉 = −

¨S+

〈F , dS〉.

We now apply Gauss’ theorem on W . Since the vector field F (x, y, z) = (1, 0, 2) is solenoid, that is,divF = 0, the flux of F through the whole boundary ∂W = S ∪ Sc will be zero, so the flux of Fentering W through S will be equal to the flux of F exiting W through Sc:

0 =

˚W

divF dx dy dz =

‹∂W+

〈F , dS〉 =

¨S+

〈F , dS〉+

¨S+c

〈F , dS〉

Then¨S〈F , dS〉 = −

¨S+

〈F , dS〉 =

¨S+c

〈F , dS〉 =

¨D〈(1, 0, 2), (0− 6y, 1)〉 dx dy = 2A(D) = 2π.

43. Compute the circulation

˛C−y2 dx + z dy + x dz where C is the triangle obtained when intersecting

the plane 2x + 2y + z = 6 with the three axis of coordinates, oriented by the unit normal vectorN(x, y, z) = (2/3, 2/3, 1/3).

Resolution: The curve C has 3 different pieces, so the direct computation of the circulation of thevector field F (x, y, z) = (−y2, z, x) along C involves the computation of 3 line integrals. Instead, weare going to apply Stokes’ theorem to compute one single double integral:˛

C+

−y2 dx+ z dy + x dz =

˛∂S+

〈F , d`〉 =

¨S+

〈rotF , dS〉,

55

where rotF =

∣∣∣∣∣∣i j k∂x ∂y ∂z−y2 z x

∣∣∣∣∣∣ =

−1−12y

, and S = (x, y, z) : z = f(x, y) = 6 − 2x − 2y, (x, y) ∈ D,

where D = (x, y) : x + y ≤ 3, x ≥ 0, y ≥ 0. Parameterizing S = ϕ(D) with the cartesian param-eterization ϕ(x, y) = (x, y, f(x, y) on D, the associated normal vector ϕx ∧ ϕy = (−fx,−fy, 1)> =

(2, 2, 1)> =√

9N is oriented like N . Therefore

˛C+

−y2 dx+ z dy + x dz =

¨S+

〈(−1,−1, 2y), (2, 2, 1)〉dx dy =

ˆ 3

0dx

ˆ 3−x

02y − 4 dy

=

ˆ 3

0dx[y2 − 4y

]y=3−xy=0

=

ˆ 3

0(3− x)2 − 4(3− x) dx

=

ˆ 3

0x2 − 2x− 3 dx =

[x3

3− x2 − 3x

]x=3

x=0

= 9− 9− 9 = −9.

Consider the piecewise regular surface S = S1 ∪ S2 ⊂ R3 defined by

S1 = (x, y, z) ∈ R3 : x2 + y2 = 1, 0 ≤ z ≤ 1,

S2 = (x, y, z) ∈ R3 : x2 + y2 + (z − 1)2 = 1, z ≥ 1,

oriented by the outward-pointing normal. Given the vector field F (x, y, z) = (x + xz + yz2, y +

xyz3, x2z4), compute the flux

¨S〈rotF , dS〉.

44. Consider the piecewise regular surface S = S1 ∪ S2 ⊂ R3 defined by

S1 = (x, y, z) ∈ R3 : x2 + y2 = 1, 0 ≤ z ≤ 1,

S2 = (x, y, z) ∈ R3 : x2 + y2 + (z − 1)2 = 1, z ≥ 1,

oriented by the outward-pointing normal. Given the vector field F (x, y, z) = (x + xz + yz2, y +xyz3, x2z4), compute the flux

˜S〈rotF , dS〉.

Resolution: First of all, we plot the surface S. We know that S1 is the surface of a cylinder and thatS2 is the surface of a sphere. Considering their domains, the piecewise regular surface S = S1 ∪ S2looks like this:

Surface S of exercise 44.

56

As we are asked to compute the flux˜S 〈rotF , dS〉 given the vector field F (x, y, z) = (x + xz +

yz2, y + xyz3, x2z4), we will use Stoke’s theorem¨S+

〈rotF , dS〉 =

˛C+

〈F , d`〉,

and so we need to compute the circulation of F along a counter-clockwise oriented curve C+ whichcorresponds to the boundary of S. This curve is defined by C = ∂S = (x, y, z) ∈ R3 : x2 +y2 = 1, z =0. The parameterization of C = ∂S is

(x, y, z) = σ(θ) = (cos θ, sin θ, 0), 0 ≤ θ ≤ 2π,

and thereforeσ′(θ) = (− sin θ, cos θ, 0),

F (σ(θ)) = (cos θ, sin θ, 0).

We then need to check that our parameterization σ is actually oriented counter-clockwise. To do sowe take the point

σ(0) = (cos 0, sin 0, 0) = (1, 0, 0),

and see how the vectorσ′(0) = (− sin 0, cos 0, 0) = (0, 1, 0),

is oriented. As we can see in the picture below, C is indeed oriented counter-clockwise (C+).

Surface S of exercise 44 viewed from below with a tangent vector T = σ′(t) of the boundary C = ∂S.

Finally, by Stoke’s theorem:

¨S+

〈rotF , dS〉 =

˛C+

〈F , d`〉 =

ˆ b

a〈F (σ(θ)),σ′(θ)〉dθ =

ˆ 2π

0〈(cos θ, sin θ, 0), (− sin θ, cos θ, 0)〉 dθ =

=

ˆ 2π

0(− sin θ cos θ + sin θ cos θ) dθ = 0.

Alternatively, we could have also realized that we did not need to compute the parameterization of C,as the vector field on C is F = (x, y, 0). Consequently, the circulation to be computed is

˛C+

〈F , d`〉 =

˛C+

x dx+ y dy = 0,

as x2 + y2 = 1 on C implies x dx+ y dy = 0.

57

45. Verify the Stokes theorem with the vector field F (x, y, z) = (2z, x, y2) and the piece of the circularparaboloid

S = (x, y, z) ∈ R3 : z = 4− x2 − y2, z ≥ 0

oriented so that the boundary C = ∂S is traveled counter-clockwise.

Resolution: Stokes’ theorem states that˛∂S+

〈F , d`〉 =

¨S+

〈rotF , dS〉,

so, in order to verify it we need to compute both sides of the equality and see if they yield the sameresult. Before computing any integrals, let’s represent the exercise graphically.

Graph for exercise 45.

We see that we have a circular paraboloid going downwards on the z axis which is cut by the planez = 0. We are going to study the part of the paraboloid over this plane. Also a normal vector isplotted on it, which should be upward-pointing, that later on we will see why it is useful.

Let’s start with the left part of the equation:

˛∂S+

〈F , d`〉 =

˛C+

〈F , d`〉 =

ˆ b

a〈F (σ(θ)),σ′(θ)〉 dθ,

where C = ∂S = σ([a, b]) is parameterized counter-clockwise by σ, according to the statement of theexercise.

In this case it is easy to use a cylindrical parameterization of the paraboloid S: x = r cos θ, y =r sin θ, z = 4 − x2 − y2 = 4 − r2, or (x, y, z) = ψ(r, θ) = (r cos θ, r sin θ, 4 − r2). If we compute theintersection of the paraboloid and the plane, we get r = 2 for the boundary C = ∂S, which happensto be a circle of radius 2:

C = σ([0, 2π]), σ(θ) = (2 cos θ, 2 sin θ, 0).

Now we proceed to calculate the scalar product between F evaluated at this parameterization and thederivative of σ(θ).

F (σ(θ)) = (0, 2 cos θ, 4 sin2 θ), σ′(θ) = (−2 sin θ, 2 cos θ, 0), 〈F (σ(θ)),σ′(θ)〉 = 4 cos2 θ.

58

Finally we compute the first integral.

˛C+

〈F , d`〉 =

ˆ 2π

04 cos2 θ dθ = 4

ˆ 2π

0

1 + cos (2θ)

2dθ = 4

([θ

2

]θ=2π

θ=0

+

[sin(2θ)

4

]θ=2π

θ=0

)= 4π.

Now we have to compute the other part of the equation of the Stokes theorem and see if it yields thesame result. To do that, first we compute the curl of F

rotF =

∣∣∣∣∣∣i j k∂x ∂y ∂z2z x y2

∣∣∣∣∣∣ = (2y, 2, 1).

Secondly, we use a cartesian parameterization for S = ϕ(D) and we get the associated normal vectorϕx ∧ϕy

ϕ(x, y) = (x, y, 4− x2 − y2), (x, y) ∈ D = (x, y) : (x2 + y2 ≤ 4

ϕx = (1, 0,−2x), ϕy = (0, 1,−2y), ϕx ∧ϕy =

∣∣∣∣∣∣i j k1 0 −2x0 1 −2y

∣∣∣∣∣∣ = (2x, 2y, 1),

and therefore the vector surface element is upward-pointing:

dS = ϕx ∧ϕy dx dy = (2x, 2y, 1) dx dy.

Before doing any more calculations we should make sure that the parameterization ϕ of our surface Sis consistent with the orientation given by C. To do that, we evaluate it at a random point, for examplefor x = 0 and y = 0 we have dS = (0, 0, 1) dx dy, which is an upward-pointing vector and thereforea vector pointing out of the paraboloid as seen in the figure, so we know that our normal vector isconsistent with C, which is traveled counter-clockwise. Let’s keep on going with the calculations then.

〈rotF , dS〉 = 〈(2y, 2, 1), (2x, 2y, 1)〉 dx dy = 4xy + 4y + 1 dx dy.

To make the last integration on the disk D easier, we will change to polar coordinates.

¨S+

〈rotF , dS〉 =

¨D

4xy + 4y + 1 dx dy =

ˆ 2π

0

ˆ 2

0(4r2 cos θ sin θ + 4r sin θ + 1) r dθ dr.

We can compute this integral as the sum of three different ones. But before doing that, notice thatby symmetry of the cos and the sin integrated between 0 and 2π, the two first integrals will be zero.Therefore, we have a very simple integral to compute

¨S+

〈rotF , dS〉 =

ˆ 2π

0

ˆ 2

0r dr dθ = 4π.

We have verified then Stokes’ theorem.

46. Let us consider the piece of a cylinder given by S = (x, y, z) ∈ R3 : x2 + y2 = 4, 0 ≤ z ≤ 5.(a)Compute the flux of the vector field F (x, y, z) = (2x, y, 3z) through the surface S oriented by theunit normal vector N(x, y, z) = (x/2, y/2, 0).(b)Compute the circulation of the vector field F (x, y, z) = (2x, y, 3z) along the boundary ∂S oriented

by the normal vector N =(x

2,y

2, 0)

.

(c)Compute the circulation of the vector field F (x, y, z) = (2x, y, 3z) along each connected component

59

of the boundary ∂S where S is oriented by the normal vector N =(x

2,y

2, 0)

.

Resolution: (a)We are asked to compute the flux through the surface S so we will use the fluxdefinition formula, however we must first parameterize our surface. As we know our surface is acylinder we will be using cylindrical parameterization with r2 = 4:

(x, y, z) = ϕ(θ, z) = (2 cos θ, 2 sin θ, z), (θ, z) ∈ D = (θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 5 .

With this parameterization S = ϕ(D) and we can now consider the flux as:¨S〈F , dS〉 =

¨D〈F ϕ,ϕθ ∧ϕz〉 dz dθ.

First of all we need to evaluate our field F on the parameterization:

F (ϕ(θ, z)) = (4 cos θ, 2 sin θ, 3z).

We can proceed to find the vector product ϕθ ∧ϕz. Let’s begin by finding the derivatives with respectto θ and z:

ϕθ = (−2 sin θ, 2 cos θ, 0), ϕz = (0, 0, 1).

If we compute the vector product we get the following result:

ϕθ ∧ϕz = (2 cos θ, 2 sin θ, 0).

Substituting in the integral and computing the scalar product 〈F ϕ,ϕθ ∧ϕz〉 we get¨S〈F , dS〉 = 4

ˆ 2π

02 cos2 θ + sin2 θ dθ

ˆ 5

0dz = 20(2π + π) = 60π.

The last step is to check whether this result that we achieved using the vector ϕθ ∧ ϕz is the sameorientation as the vector N given by the statement. Let us consider a point such that θ = π/2 andz = 0. If we substitute in the parameterization we can check that this is the point p = (x = 0, y =2, z = 0). Now we can compute the vector N and the vector ϕθ∧ and check for linear dependence andorientation.

N(p) = (0, 1, 0),

ϕθ ∧ϕz(p) = (0, 2, 0).

When we compute the vectors we can see that they are linearly dependent and they both are outwardnormal vectors. Therefore we must not change the sign of the result. We can now determine that theflux of the vector field F across the surface S oriented by the normal vector N will be equal to 60π.

(b) First of all we must deduce the orientation of the two circles of the boundary ∂S of S with a figureand check the normal vector direction. This is essential as we will need these orientations to strike theproblem afterwards.

Cylinder S with orientations of Exercise 46(b).

60

In the figure we can see the outward-pointing normal vector N , and with the right hand rule we knowit induces a counter-clockwise circle (black circle). We must now look at the tangency of this inducedcircle with the boundaries of S.

Observing the tangency in the top circle C5 at z = 5, we see that it induces on it a clockwise orientation.Observing the tangency in the lower circle C0 at z = 0, we see that it induces on it a counter-clockwiseorientation. This will be very useful when we strike the problem using Stoke’s theorem because wemust know all these orientations to apply:

˛∂S+

〈F , d`〉 =

¨S+

〈rotF , dS〉.

Now that we know the induced orientation of the boundary ∂S we can face the problem. We have threeways of facing this problem that must lead us to the same result, using Stokes’ theorem, computingthe circulation by its definition and finding a potential function.

1. Stokes’ theorem. To be able to apply Stokes’ theorem the following conditions must be met:

We must have a surface S with a boundary ∂S: We do have the surface of a cylinder and thecurves are the upper lower covers perimeter.

∂S must be a curve or union of curves: We have two circles, one located on z = 5 and anotheron z = 0 which are the circular paths on the covers.

The vector F must belong to C1 in S and ∂S: F is C1 everywhere.

∂S must be oriented by N of S leaving S on the left side: The statement of the problem isproviding us with a normal vector.

As the four conditions are met, we can apply Stokes’ theorem. Our plan will be to find rotF and thencompute its scalar product with dS = N dS, as the above Stoke’s theorem states that the integral ofthis scalar product on S will be the same as the circulation of F on the boundary of S. In our casethis theorem will be very useful as the rotational is (0, 0, 0). Indeed

rotF = (∂y(3z)− ∂z(y), −∂x(3z) + ∂z(2x), ∂x(y)− ∂y(2x)) = (0, 0, 0),

so that ˛∂S+

〈F , d`〉 =

¨S+

〈rotF , dS〉 = 0.

2. Computing the circulation. The direct computation of the circulation must be done in thetwo components of ∂S = C0 ∪ C5. If we parameterize both circles with a standard counter-clockwiseparameterization:

C0 = σ0 ([0, 2π]) , with (x, y, z) = σ0(α) = (2 cosα, 2 sinα, 0),

C5 = σ5 ([0, 2π]) , with (x, y, z) = σ5(α) = (2 cosα, 2 sinα, 5),

which only differ on the value of z, then˛∂S+

〈F , d`〉 =

‰C0

〈F , d`〉+

C5

〈F , d`〉 =

˛σ0

〈F , d`〉 −˛σ5

〈F , d`〉,

where we have used a counter-clockwise orientation in C0 and a clockwise orientation in C5. As

σ′j(α) = (−2 sinα, 2 cosα, 0), j = 0, 5,

F (x, y, z) = (2x, y, 3z) and z = j on each circle Cj , j = 0, 5, we have that

F (σj(α)) = (4 cosα, 2 sinα, 3j)), 〈F (σ(α)) ,σ′(α)〉 = −4 cosα sinα,

61

so we have all the ingredients to compute the circulation of F along any circle of ∂S˛σj

〈F , d`〉 =

ˆ 2π

0〈F (σj(α)) ,σ′j〉 dα = −

ˆ 2π

04 cosα sinα dα = −

[2 sin2 α

∣∣α=2π

α=0= 0,

for j = 0, 5 so that

˛∂S+

〈F , d`〉 = 0.

3. Finding a potential function. As rotF = 0, it is easy to the potential function f(x, y, z) =

x2 +y2

2+

3z2

2for F which satisfies ∇~f = F . Therefore the circulation of F along a curve is equal to

the difference of the potential function at the endpoints of C. If C is a closed curve, the endpoints of C

coincide, and therefore the circulation of F along any closed curve is zero. In particular,

˛∂Cj

〈F , d`〉 =

0 along the two circles Cj , j = 0, 5, that form the boundary of S, so that, again,

˛∂S+

〈F , d`〉 = 0.

Summarizing, we can conclude that the circulation of F along the boundary of the cylinder S is zero,having calculated it in three different ways.

(c) We have to consider all the components of the boundary. In our case, we study two curves,represented in the figure below.

Cylinder with both curves indicated from exercise 46(c).

In order to calculate both circulations, we will apply Stokes’ theorem˛∂S+

〈F , d`〉 =

¨S+

〈rotF , dS〉.

So as to apply Stokes’ theorem, we will need an auxiliary invented surface. We will use those thathave as a boundary the previously mentioned curves

C1 = ∂S1 for S1 =

(x, y, z) ∈ R3 : x2 + y2 ≤ 4, z = 5

= ϕ1(D), ϕ1(r, θ) = (r cos θ, r sin θ, 5),

C2 = ∂S2 for S2 =

(x, y, z) ∈ R3 : x2 + y2 ≤ 4, z = 0

= ϕ2(D), ϕ2(r, θ) = (r cos θ, r sin θ, 0),

where the parameterizations ϕ1, ϕ2 of S1, S2 are defined on the common domain

D = (r, θ) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2 .

We compute divF = ∇∧ F for our vectorial field:

∇∧ F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

2x y 3z

∣∣∣∣∣∣ = (0, 0, 0).

We now have everything we need to apply the theorem, so we proceed.

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Circulation through ∂S1 We first check if our parameterization ϕ1 for S1 is well-oriented by cal-culating the associated normal vector ϕ1r ∧ϕ1θ:

ϕ1r = (cos θ, sin θ, 0), ϕ1θ = (−r sin θ, r cos θ, 0), ϕ1r ∧ϕ1θ = (0, 0, r).

Since 0 ≤ r ≤ 2, ϕ1r∧ϕ1θ is upward-pointing, and C1 = ∂S1 should be counter-clockwise orientedfor the application of Stokes’ theorem. Since we want C1 to be clockwise oriented, we will haveto change the sign of the circulation.

˛C1−=∂S1

〈F , d`〉 = −˛C1

+=∂S1

〈F , d`〉 = −¨S+

〈rotF︸ ︷︷ ︸=0

, dS〉 = 0.

Circulation through ∂S2 We repeat the same procedure as before. We first check if our parame-terization ϕ2 for S2 is well-oriented by calculating the associated normal vector ϕ2r ∧ϕ2θ:

ϕ2r = (cos θ, sin θ, 0), ϕ2θ = (−r sin θ, r cos θ, 0), ϕ2r ∧ϕ2θ = (0, 0, r).

Since 0 ≤ r ≤ 2, ϕ2r∧ϕ2θ is upward-pointing, and C2 = ∂S2 should be counter-clockwise orientedfor the application of Stokes’ theorem, as it is.

˛C2

+=∂S2

〈F , d`〉 =

¨S+

〈rotF︸ ︷︷ ︸=0

, dS〉 = 0.

Therefore we can conclude the circulation of the vector field F along each connected component ofthe boundary ∂S is zero.

We notice that the parameterizations of S1, S2 were not really necessary. As long as divF = 0,

Stokes’theorem immediately gives

˛Cj

〈F , d`〉 = 0, j = 1, 2.

47. (a) For the surface S =

(x, y, z) : z = x2 + y2, z ≤ 1, x, y ≥ 0

oriented at the point

(1

2,1

2,1

2

)by

the vector N = (1, 1,−1)/√

3, compute the flux of the vector field F = (1, 0, 0) through S, applyingStokes’ theorem. [ Hint: Use that a constant vector field F = v admits as a potential vector field thelinear vector field G = 1

2v ∧ r.](b) Compute the flux of the vector field F = (1, 0, 0) through S, applying Gauss’ theorem.(c) Compute the flux of the vector field F = (1, 0, 0) through S, just using the definition of flux.

Resolution: (a) Let us state Stokes’ theorem for a vector field G

¨S+

〈rotG, dS〉 =

˛∂S+

〈G, d`〉.

where S ⊂ R3 is a surface oriented by a unit normal vector N , and its boundary ∂S is oriented bythe vector N leaving S on the left side.

63

Surface S and its oriented boundary of Exercise 47(a).

As F = (1, 0, 0) is constant, using the hint for v = (1, 0, 0), the vector field G given by

G =1

2v ∧ r =

1

2(1, 0, 0) ∧ (x, y, z) =

1

2

∣∣∣∣∣∣i j k1 0 0x y z

∣∣∣∣∣∣ =1

2(0,−z, y) ,

satisfies

rotG =∇ ∧G =1

2

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

0 −z y

∣∣∣∣∣∣ = (1, 0, 0) = F .

In accordance to that, now we have by Stokes’ Theorem:¨S+

〈F , dS〉 =

¨S+

〈rotG, dS〉 =

˛∂S+

〈G, d`〉 =1

2

˛∂S+

−z dy + y dz =1

2

˛∂S+

y dz − z dy.

The boundary ∂S has three different curves, so that ∂S = C1 ∪ C2 ∪ C3, with

C1 =

(x, y, z) : z = x2 + y2, z = 1, x, y ≥ 0

=

(x, y, z) = σ1(θ) = (cos θ, sin θ, 1), 0 ≤ θ ≤ π

2

,

C2 =

(x, y, z) : z = x2 + y2, z ≤ 1, x = 0, y ≥ 0

=

(x, y, z) = σ2(y) = (0, y, y2), 0 ≤ y ≤ 1,

C3 =

(x, y, z) : z = x2 + y2, z ≤ 1, x ≥ 0, y = 0

=

(x, y, z) = σ3(x) = (x, 0, x2), 0 ≤ x ≤ 1.

We now compute the circulations

ˆCj

〈G, d`〉 =1

2

ˆσj

y dz − z dy using the parameterizations σj

written above for j = 1, 2, 3, and later on we will study if they follow the correct orientation withrespect to the normal vector N of S.

1

2

ˆσ1

y dz − z dy =1

2

ˆ π/2

0− cos θ dθ = −1

2, (because z = 1, y = cos θ, dy = cos θ dθ, dz = 0),

1

2

ˆσ2

y dz − z dy =1

2

ˆ π/2

02y2 dy − y2 dy =

1

6, (because z = y2, y = y, dy = dy, dz = 2y dy),

1

2

ˆσ3

y dz − z dy = 0, (because z = x2, y = 0, dy = 0, dz = 2x dx).

Now we check the correct orientations for Stokes’ theorem. As the normal vector N = (1, 1,−1)/√

3points down, we should have

64

• the path C3 beginning at (0, 0, 0) and ending at (1, 0, 1) which agrees with σ3,

• the path C1 beginning at (1, 0, 1) and ending at (0, 1, 1) which disagrees with σ1,

• the path C2 beginning at (0, 1, 1) and ending at (0, 0, 0) which disagrees with σ2.

Therefore ¨S+

〈F , dS〉 =1

2

˛∂S+

y dz − z dy

=1

2

ˆC+

1

y dz − z dy +1

2

ˆC+

2

y dz − z dy +1

2

ˆC+

3

y dz − z dy

= −1

2

ˆσ1

y dz − z dy − 1

2

ˆσ2

y dz − z dy +1

2

ˆσ3

y dz − z dy

=1

2− 1

6− 0 =

1

3.

(b) In order to apply Gauss’ theorem a closed surface is needed, it will be achieved by enclosing Swith the intersections of W and the x = 0 plane, the y = 0 plane and the z = 1 plane:

W = (x, y, z) ∈ R3 : x2 + y2 ≤ z ≤ 1, x ≥ 0, y ≥ 0.

The first flat cover S1, as already mentioned, will close S with the intersection of W and the x = 0plane. We introduce x = 0 in W :

S1 = (x, y, z) ∈ R3 : y2 ≤ z ≤ 1, x = 0, y ≥ 0.

Its outward-pointing normal vector N1 and the normal component FN1 of the vector field F are,respectively:

N1 = (−1, 0, 0),

FN1 = 〈F ,N1〉 = 〈(1, 0, 0), (−1, 0, 0)〉 = −1.

The second flat cover S2 will close S with the intersection of W and the y = 0 plane:

S2 = (x, y, z) ∈ R3 : x2 ≤ z ≤ 1, x ≥ 0, y = 0.

Its outward-pointing normal vector N2 and the normal component FN2 of the vector field F are,respectively:

N2 = (0,−1, 0),

FN2 = 〈F ,N2〉 = 〈(1, 0, 0), (0,−1, 0)〉 = 0.

The third, and last, flat cover will close S with the intersection of W and the z = 1 plane:

S3 = (x, y, z) ∈ R3 : x2 + y2 ≤ 1, x ≥ 0, y ≥ 0.

Its outward-pointing normal vector N3 and the normal component FN3 of the vector field F are,respectively:

N3 = (0, 0, 1),

FN3 = 〈F ,N3〉 = 〈(1, 0, 0), (0, 0, 1)〉 = 0.

65

Flat covers S1, S2, S3 with their respective outward-pointing normal vectors.

Applying Gauss’ Theorem:‹∂W+

〈F , dS〉 =

¨S+

〈F , dS〉+¨S+1

〈F , dS1〉+¨S+2

〈F , dS2〉+¨S+3

〈F , dS3〉 =

˚W

divF dx dy dz.

As divF = 0 and 〈F , dS〉 = FN dS:¨S+

〈F , dS〉 = −¨S+1

〈F , dS1〉

−

¨S+2

〈F , dS2〉

−

¨S+3

〈F , dS3〉 = −¨S+1

−1 dS1 = A(S1).

Hence, ¨S+

〈F , dS〉 = A(S1) =

ˆ 1

0dy

ˆ 1

y2dz =

ˆ 1

01− y2 dy =

2

3.

We conclude that the flux going through the surface S oriented by the outward-pointing normal N is2/3.(c) We use the parameterization based on cylindrical coordinates S = ϕ(D) with ϕ(r, θ) =(r cos θ, r sin θ, z = f(r)), f(r) = r2 which satisfies

ϕr =

cos θsin θ2r

, ϕθ =

−r sin θr cos θ

0

, ϕr ∧ϕθ =

∣∣∣∣∣∣i j k

cos θ sin θ 2r−r sin θ r cos θ 0

∣∣∣∣∣∣ =

−2r2 cos θ−2r2 sin θ

r

,

and we see that the normal vector associated to this parameterization ϕr ∧ ϕθ is always an upward-pointing vector, since its z-component is positive: r > 0.

S of Exercise 47(c) on the (r, z) plane and on the (x, y, z) space with outward-pointing normal vectors.

As x, y ≥ 0 and 0 ≤ θ ≤ π2 ,

D =

(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π

2

.

66

The flux is given by ¨D〈F ϕ,ϕr ∧ϕθ〉 dr dθ,

where F ϕ = (1, 0, 0) and 〈F ϕ,ϕr ∧ϕθ〉 = −2r2 cos θ. Then

¨D〈F ϕ,ϕr ∧ϕθ〉 dr dθ =

ˆ π2

0

ˆ 1

0−2r2 cos θ dr dθ =

ˆ π2

0

−2 cos θ

3dθ = −2

3.

Let’s check now the sign of the flux. Substituting (12 ,12 ,

12) = (r cos θ, r sin θ, r2), we get r cos θ =

r sin θ = r2 = 12 , so that on this point the normal vector provided by the parameterization is

ϕr ∧ϕθ = (−2r2 cos θ,−2r2 sin θ, r) = (−r,−r,√z) = − 1√

2(1, 1,−1) = −

√3

2N ,

so we have to change the sign of the flux. The flux is2

3.

Equivalently, we can notice that the given unitary normal vector N is a downward-pointing vector.This already implies the change of sign after the computation of the flux.

67