Lembretes Matemática

34
Further Mathematical Methods Summer Term 2008 1 Useful Mathematical Results 1.1 Trigonometric Identities We recall the following useful trigonometric identities: Figure 1: tan θ = sin θ cos θ , cosecθ = 1 sin θ , sec θ = 1 cos θ , cot θ = 1 tan θ = cos θ sin θ . From the well known identity cos 2 θ + sin 2 θ =1 (1.1) we have 1 + tan 2 θ = sec 2 θ and cot 2 θ + 1 = cosec 2 θ. Addition and subtraction formulae: sin (α ± β) = sin α cos β ± cos α sin β and cos (α ± β) = cos α cos β sin α sin β If α = β then from the above addition and subtraction formulae we have sin 2α = 2 sin α cos α and cos 2α = cos 2 α - sin 2 α. (1.2) 1

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Transcript of Lembretes Matemática

Page 1: Lembretes Matemática

Further Mathematical Methods

Summer Term 2008

1 Useful Mathematical Results

1.1 Trigonometric Identities

We recall the following useful trigonometric identities:

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Figure 1:

tan θ =sin θ

cos θ, cosecθ =

1sin θ

, sec θ =1

cos θ, cot θ =

1tan θ

=cos θ

sin θ.

From the well known identitycos2 θ + sin2 θ = 1 (1.1)

we have1 + tan2 θ = sec2 θ

andcot2 θ + 1 = cosec2θ.

Addition and subtraction formulae:

sin (α± β) = sin α cos β ± cos α sinβ

andcos (α± β) = cos α cos β ∓ sinα sinβ

If α = β then from the above addition and subtraction formulae we have

sin 2α = 2 sin α cos α

andcos 2α = cos2 α− sin2 α. (1.2)

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Using (1.1) in (1.2) we havecos 2α = 1− 2 sin2 α

⇒ sin2 α = 12 (1− cos 2α) . (1.3)

Alternatively we have

cos 2α = 2 cos2 α− 1

⇒ cos2 α = 12 (1 + cos 2α) . (1.4)

The formulas (1.3) and (1.4) are useful when trying to integrate cos2 θ or sin2 θ.

1.2 Trigonometric functions

In Figure 2 we display graphs of the trigonometric functions sin θ, cos θ, tan θ, csc θ, sec θ and cot θ.

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Figure 2:

1.3 Determinants of Matrices

For a 2× 2 matrix we have

A =[

α βγ δ

]detA =

∣∣∣∣ α βγ δ

∣∣∣∣ = αδ − γβ.

For a 3× 3 matrix we have

A =

a b cd e fg h i

detA = |A| = a

∣∣∣∣e fh i

∣∣∣∣− b

∣∣∣∣d fg i

∣∣∣∣+ c

∣∣∣∣d eg h

∣∣∣∣= a (ei− hf)− b (di− gf) + c (dh− ge) .

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1.4 Vectors

Vectors have magnitude and direction.The vector i has magnitude 1 and points in the increasing x direction. The vector j has magnitude 1and points in the increasing y direction. The vector k has magnitude 1 and points in the increasing zdirection.The vectors i, j and k are unit vectors. All unit vectors have magnitude 1 and usually they are writtenwith hats, e.g. u or v.

A standard vector

F = Fxi + Fyj + Fzk

has 3 components, Fx, Fy and Fz that are all scalars. Fx is the x component of F, Fy is the y componentof F and Fz is the z component of F.

The magnitude of the vector F is denoted by |F| and is defined by

|F| =√

F 2x + F 2

y + F 2z .

A unit vector in the direction of F has magnitude 1 and points in the same direction as F, it is denotedby

F =1|F|

F =1|F|

(Fxi + Fyj + Fzk) .

Example 1.1 For the vectorG = 3i + 2j

find the three components Gx, Gy and Gz, then determine the magnitude of G and the unit vector thatpoints in the same direction as G.

Solution:The three components Gx, Gy and Gz are

Gx = 3, Gy = 2 and Gz = 0.

Hence the magnitude of G is

|G| =√

G2x + G2

y + G2z =

√32 + 22 + 02 =

√13

and the unit vector that points in the same direction as G is

G =1|G|

G =1√13

(3i + 2j) .

Example 1.2 For the vectorF = 2yi− 3zj + 9z2k

find the three components Fx, Fy and Fz and then determine the magnitude of F.

Solution:The three components Fx, Fy and Fz are

Fx = 2y, Fy = −3z, and Fz = 9z2

and hence the magnitude of F is

|F| =√

(2y)2 + (−3z)2 + (9z2)2

=√

4y2 + 9z2 + 81z4.

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1.4.1 The scalar (or dot) product

The scalar product of 2 vectors A and B is given by

A ·B = |A| |B| cos θ

where θ is the angle between the two vectors.

Since cos 0 = 1 and cos π2 = 0 we have

i · i = |i| |i| cos 0 = 1× 1× 1 = 1, j · j = 1 and k · k = 1.

Furthermorei · j = |i| |j| cos

π

2= 1 · 1 · 0 = 0, i · k = 0 and j · k = 0.

IfA = Axi + Ayj + Azk and B = Bxi + Byj + Bzk

A ·B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk)= Axi·Bxi+Axi·Byi+Axi·Bzi

+Ayi·Bxi+Ayi·Byi+Ayi·Bzi

+Azi·Bxi+Azi·Byi+Azi·Bzi

= AxBx (i · i) + AxBy (i · j) + AxBz (i · k)+AyBx (j · i) + AyBy (j · j) + AyBz (j · k)+AzBx (k · i) + AzBy (k · j) + AzBz (k · k)

= AxBx + AyBy + AzBz.

Example 1.3 ForF = 3yi− 2xk and G = i + 2zj− 3k

calculate F ·G.

Solution: We have

F ·G = 3y · 1 + 0 · 2z + (−2x) (−3)= 3y + 6x.

1.4.2 The vector (or cross) product

The vector product of 2 vectors A and B is given by

A×B = |A| |B| sin θn

where n is a unit vector perpendicular to A and B, pointing in the direction given by the right handscrew rule (i.e. the direction in which a screw would advance if it were turned from A through the angleθ to B. The cross product can be evaluated by calculating the determinant of the 3× 3 matrix

A×B =

∣∣∣∣∣∣i j k

Ax Ay Az

Bx By Bz

∣∣∣∣∣∣= i

∣∣∣∣Ay Az

By Bz

∣∣∣∣− j∣∣∣∣Ax Az

Bx Bz

∣∣∣∣+ k∣∣∣∣Ax Ay

Bx By

∣∣∣∣= i (AyBz −ByAz)− j (AxBz −BxAz) + k (AxBy −BxAy) .

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Example 1.4 ForA = 3xi− 4k and B = 2i− 9xj + k

evaluate A×B.

Solution: We have

A×B =

∣∣∣∣∣∣i j k

3x 0 −42 −9x 1

∣∣∣∣∣∣ = i∣∣∣∣ 0 −4−9x 1

∣∣∣∣− j∣∣∣∣3x −42 1

∣∣∣∣+ k∣∣∣∣3x 02 −9x

∣∣∣∣= i ((0) (1)− (−9x) (−4))− j ((3x) (1)− (2) (−4)) + k ((3x) (−9x)− (2) (0))= i (−36x)− j (3x + 8) + k

(−27x2

)= −36xi− (8 + 3x) j− 27x2k.

Remark: A×B 6= B×A

1.5 Differentiation

Given a function f(x), (or f(t)) of a single variable x (or t), we denote the derivative of f(x) by f ′ (x)or df

dx (or the derivative of f(t) by f(t) or dfdt ) The following table gives the derivatives of some common

functions.

f(x) f ′ (x)xn nxn−1

sinx cos xcos x − sinxtanx sec2 xln |x| 1

xex ex

1.5.1 The product rule

If u = u(x) and v = v(x) then

d (uv)dx

=du

dxv + u

dv

dx= u′v + uv′.

1.5.2 The quotient rule

If u = u(x) and v = v(x) thend

dx

(u

v

)=

u′v − uv′

v2.

1.5.3 The chain rule

If f = f (u(x)) then

df

dx=

df

du

du

dx.

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Example 1.5 Forf(x) = x3 cos x

calculate f ′(x).

Solution: Sincef(x) = x3 cos x we set

u = x3 and v = cos x

and hencedu

dx= 3x2 and

dv

dx= − sinx

so from the product rule we have

df

dx=

d (uv)dx

=du

dxv + u

dv

dx

= 3x2 cos x + x3(− sinx)= 3x2 cos x− x3 sinx.

Example 1.6 For

f(x) =ex

sinx

calculate f ′(x).

Solution: Settingu = ex and v = sinx

we havedu

dx= ex and

dv

dx= cos x.

Hence from the quotient rule we have

df

dx=

d(

uv

)dx

=dudxv − u dv

dx

v2

=ex sinx− ex cos x

sin2 x.

Example 1.7 Forf(x) = cos 3x2

calculate f ′(x).

Solution: Setting u = 3x2 we have f(x) = cos 3x2 = cos u and hence

du

dx= 6x and

df

du=

d cos u

du= − sinu.

Thus from the chain rule we have

df

dx=

df

du

du

dx= − sinu · 6x = −6x sinu = −6x sin 3x2.

RemarksWe note the following useful identities

ea+b = eaeb, eab = (ea)b

and for a, b > 0ln(ab) = ln a + ln b, ln

(a

b

)= ln a− ln b and ln ba = a ln b.

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1.6 Integration

The following table gives the integrals of some common functions.

f(x)∫

f(x)dx

kxn kn+1xn+1 + c, n 6= −1

sin kx - 1k cos kx + c

cos kx 1k sin kx + c

ekx 1kekx + c

kx k ln |x|+ c

1.7 Integration by parts

Recalling the product rule we have

d

dx(fg) = f ′g + fg′

and hence by integration we obtain∫ b

a

d

dx(fg) dx =

∫ b

a

f ′gdx +∫ b

a

fg′dx

⇒ [fg]ba =∫ b

a

f ′gdx +∫ b

a

fg′dx.

Manipulating the above equation gives the standard form of integration by parts formula∫ b

a

f ′gdx = [fg]ba −∫ b

a

fg′dx.

Example 1.8 Use integration by parts to integrate∫ 2

1

xexdx.

Solution: We have∫ 2

1

xexdx = [xex]21 −∫ 2

1

exdx, g = x ⇒ g′ = 1, f ′ = ex ⇒ f = ex

= [xex]21 − [ex]21= 2e2 − e−

(e2 − 1

)= e2.

1.7.1 Integration using substitution

∫ b

a

f(x)dx =∫ u(b)

u(a)

f(u)dudx

du=∫ u(b)

u(a)

(f(u)

dx

du

)du.

Example 1.9 Use integration using substitution to evaluate the following integral∫ 1

0

x(1 + x2

)3dx.

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Solution: Setting u = 1 + x2 we have

du

dx= 2x ⇒ dx

du=

12x

, u(0) = 1 + 02 = 1 and u (1) = 1 + 12 = 2

and hence ∫ 1

0

x(1 + x2

)3dx =

∫ u(1)

u(0)

xu3 dx

dudu,

=∫ 2

1

xu3 12x

du

=12

∫ 2

1

u3du

=12

[u4

4

]21

=18[24 − 14

]=

158

.

1.8 Polar coordinates

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Figure 3:

There are two unit vectors associated with polar coordinates: eρ and eφ. They have magnitude 1 andpoint in the directions of increasing ρ and φ respectively, also they are perpendicular so that

eρ ⊥ eφ.

Remark 1 We note that eρ and eφ are position dependent unlike i, j and k (i.e. the direction in whichthey point depends on where they are in the (x, y) plane).

1.9 3-D coordinate systems

In 3-d there are three commonly used coordinate systems:

coordinate system variables unit vectorsCartesian (x, y, z) (i, j,k)Cylindrical (ρ, φ, z) (eρ, eφ, ez)Spherical (r, θ, φ) (er, eθ, eφ)

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Figure 4:

1.9.1 Cylindrical polar coordinates

A point P (x, y, z) can be written in terms of the cylindrical polar coordinates ρ, φ, z such that P (x, y, z) =P (ρ, φ, z). One can think of P (ρ, φ, z) as being a point on the surface of a cylinder with radius ρ, seeFigure 4.

Remark 2 The unit vector ez points in the increasing z-direction (and is in fact k !).

1.9.2 Spherical polar coordinates

A point P (x, y, z) can be written in terms of the spherical polar coordinates r, θ, φ such that P (x, y, z) =P (r, θ, φ). One can think of P (r, θ, φ) as being a point on the surface of a sphere with radius r, see Figure5.

Remark 3 The unit vectors are such that er points in the increasing r-direction, eθ points in the in-creasing θ-direction, eφ points in the increasing φ-direction.

2 2D Surface Integrals

2.1 Cartesian coordinates

The area A under a graph y = f(x) between two points x = a and x = b (see Figure 6) can be calculatedby evaluating the integral

A =∫ b

a

f(x)dx.

The area A can be approximated by the sum of the areas of N thin strips Si centred at xi with width∆x and height f(xi) (see Figure 6)) hence we have

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Figure 5:

A '∑

area of strips Si =N∑

i=1

f(xi)∆x.

In the limit as the ∆x → 0 i.e. as the number of strips tends to infinity we have

N∑i=1

f(xi)∆x →∫ b

a

f(x)dx = A.

Alternatively we could divide each strip into blocks with centre (xi, yj), width ∆x and height ∆y

A 'N∑

i=1

(area of Si)

where

area of Si =Mi∑j=1

(area of Bj).

Hence

A 'N∑

i=1

(area of Si)

=N∑

i=1

Mi∑j=1

(area of Bj)

=N∑

i=1

Mi∑j=1

∆y∆x︸ ︷︷ ︸area of Bj

.

Note that the number of blocks Mi, depends on the height of Si, i.e. the value of f(xi). In the limit as

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y=f(x)

a b

A

f(x i)

xi

ba

y=f(x)

Figure 6:

the number of blocks tends to infinity and ∆x and ∆y tend to zero we have

N∑i=1

Mi∑j=1

∆y∆x︸ ︷︷ ︸area of Bj

→∫ b

a

∫ f(x)

0

dydx = A.

In general the area of a 2D surface S can be approximated by

A =∑

(area of the surface elements)

=∑

dA.

In the limit as the area of the surface element tends to zero we have

A =∫

S

dA.

Here the integral∫

Sis a double integral i.e.

∫∫, and the dA is a double integrand i.e. dxdy

(1,0)

(0,1)

(0,0)

T

Figure 7:

Example 2.1 Calculate the area of the triangle T bounded by the lines x = 0, y = 0 and y = 1− x.

Solution:First we draw the triangle T , see Figure 7. We have that

area of T =∫

T

dA =∫ (∫

dy

)dx

The inner integral relates to the area of a strip with width ∆x. The base of this strip is at y = 0, the topvaries depending on the position that the strip is along the x-axis (i.e. it is a function of x).

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From Figure 7 we see that the top of the strip is 1− x, so the limits on the inner integral are

area of T =∫ (∫ 1−x

0

dy

)dx.

For the limits on the outer integral we recall that this integral is associated with the sum of the strips.The left hand strip is centred at x = 0 and the right hand one is at x = 1, so we have

area of T =∫ 1

0

∫ 1−x

0

dydx.

To evaluate a double integral we first evaluate the inner integral and then we evaluate the resultingintegral

A =∫ 1

0

(∫ 1−x

0

dy

)dx =

∫ 1

0

[y]1−x

0dx

=∫ 1

0

[(1− x)− 0] dx

=∫ 1

0

(1− x)dx =[x− x2

2

]10

=12.

2.1.1 Polar coordinates

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Figure 8:

If we want to find the area of a circular or part circular surface we use surface elements derived frompolar coordinates.

The area of a surface element can be approximated by (see Figure 8)

∆A ' ρ∆ρ∆φ.

Thus the area of the surface

A '∑

(area of the surface elements)

'∑∑

ρ∆ρ∆φ.

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In the limit as the area of each surface element tends to zero we have

A =∫

S

dA =∫ ∫︸ ︷︷ ︸

S

ρdρdφ︸ ︷︷ ︸dA

.

The limits on the integrals depend on the surface that you are integrating over.

(1,0) (2,0)

(0,2)

(0,1) S

Figure 9:

Example 2.2 Calculate the area of the annular region S by 0 ≤ φ ≤ π/2 and 1 ≤ ρ ≤ 2.

Solution: First we draw the annular region S, see Figure 9. We see that

area of S =∫

S

dA =∫ (∫

ρdρ

)dφ

From Figure 9 we see that the limits for the ρ integral are 1 and 2, while the limits for the φ integral are−π/2 and π/2. Thus

area of S =∫ π

2

0

∫ 2

1

ρdρdφ

=∫ π

2

0

(∫ 2

1

ρdρ

)dφ

=∫ π

2

0

[ρ2

2

]21

=∫ π

2

0

[2− 1

2

]dφ

=∫ π

2

0

32dφ

=32

[φ]π20

=32

(π − 0)

=3π

4.

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2.2 Calculating mass and charge density

The techniques that we have used to calculate the area of surfaces can be applied to calculate the massof a 2D object or the charge on a 2D object.

If the density of the rectangular plate S bounded by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1 is given by f(x, y) =(1 + x2)(1 + y) kgm−2 we can approximate the mass of S by dividing it up into small surface elementsand summing the masses of each of these elements.Since for an object with constant density, mass = area x density, we can approximate the mass of asurface element by its area multiplied by the density at its mid-point

i.e. (mass of S.E.)' ∆x∆yf(xi, yj)Thus we have

S =∑

(mass of S.E.) =∑i,j

∆x∆yf(xi, yj) →∫ ∫

f(x, y)dydx.

Adding the required limits for x and y gives

Mass of S =∫ 2

0

∫ 1

0

(1 + x2)(1 + y)dydx

=∫ 2

0

[∫ 1

0

(1 + x2)(1 + y)dy

]dx

=∫ 2

0

[(1 + x2)

∫ 1

0

(1 + y)dy

]dx

=∫ 2

0

(1 + x2)[y +

y2

2

]10

dx

=∫ 2

0

(1 + x2)((

1 +12

)− 0)

dx

=∫ 2

0

32(1 + x2)dx

=32

[x +

x3

3

]20

= 7 kg.

Example 2.3 If the charge density on the semi circular disc S in Figure 10 is f(ρ, φ) = 3ρ Cm−2,calculate the total electric charge on S.

Solution: We havetotal charge onS =

∑i

charge on S.E.

charge on S.E. ' area of S.E.× charge density at the mid-point of S.E.(ρi, φj)= f(ρi, φj)ρ∆ρ∆φ.

So

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(0,−1)

(0,1)

(1,0)

S

Figure 10:

total charge onS =∫ π

2

−π2

∫ 1

0

f(ρ, φ)ρdρdφ

=∫ π

2

−π2

∫ 1

0

3ρρdρdφ

=∫ π

2

−π2

(∫ 1

0

3ρ2dρ

)dφ

=∫ π

2

−π2

[ρ3]10dφ

=∫ π

2

−π2

= [φ]π2−π2

= π C.

2.3 General Formulas

We recall the following useful formulas.

Area of S =∫

S

dA,

Mass of S =∫

S

fdA where f = density,

charge on S =∫

S

fdA where f = charge density .

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2.4 Double integrals from a mathematical point of view

So far we have been looking at things from a physical point of view. We could have asked the previousquestions in the following mathematical ways:

Example 2.4 Evaluate the integral of the function f(x, y) = (1 + x2)(1 + y) over the two dimensionalrectangular region with 0 6 x 6 2, 0 6 y 6 1.

Solution: ∫ 2

0

∫ 1

0

f(x, y)dydx =∫ 2

0

∫ 1

0

(1 + x2)(1 + y)dydx = 7.

Example 2.5 Find the integral of f(ρ, φ) = 3ρ over the semi-circular region 0 6 ρ 6 1, −π2 6 φ 6 π

2 .

Solution: ∫ π2

−π2

∫ 1

0

3ρρdρdφ = π.

3 Surface Areas in 3D

To evaluate the area of a 3D surface we use

A =∫

S

dA.

As in the case of 2D surface integrals∫

Sis a double integral i.e.

∫∫, and dA is a double integrand i.e.

dxdy.

3.1 Cartesian coordinates

If we wish to evaluate the surface area (or the mass, or the charge) of the hollow block 0 ≤ x ≤ 7,0 ≤ y ≤ 5 and 1 ≤ z ≤ 3, we need to break the block up into six 2D surfaces, i.e. its six faces.

On the top and bottom faces of the block x and y vary, and z is constant, so we have

area of top =∫ 5

0

∫ 7

0

dxdyand area of bottom =∫ 5

0

∫ 7

0

dxdy.

On the left and right hand faces x and z vary and y is constant, so we have

area of left hand face =∫ 3

1

∫ 7

0

dxdzand area of right hand face =∫ 3

1

∫ 7

0

dxdz.

On the remaining two faces y and z vary, and x is constant, so

area of front =∫ 3

1

∫ 5

0

dydzand area of back =∫ 3

1

∫ 5

0

dydz.

Example 3.1 Find the integral of f(x, y, z) = 1 + x + y + z over the surface of the hollow block S.

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Solution: We have∫S

fdA =∫

top

fdA +∫

bottom

fdA +∫

left side

fdA +∫

right side

fdA +∫

front

fdA +∫

back

fdA

=∫ 5

0

∫ 7

0

(1 + x + y + z

(=1)

)dxdy +

∫ 5

0

∫ 7

0

(1 + x + y + z

(=3)

)dxdy

+∫ 3

1

∫ 7

0

(1 + x + y

(=0)

+ z

)dxdz +

∫ 3

1

∫ 7

0

(1 + x + y

(=5)

+ z

)dxdz

+∫ 3

1

∫ 5

0

(1 + x

(=0)+ y + z

)dydz +

∫ 3

1

∫ 5

0

(1 + x

(=7)+ y + z

)dydz

=∫ 5

0

∫ 7

0

(1 + x + y + 1) dxdy +∫ 5

0

∫ 7

0

(1 + x + y + 3) dxdy

+∫ 3

1

∫ 7

0

(1 + x + z) dxdz +∫ 3

1

∫ 7

0

(1 + x + 5 + z) dxdz

+∫ 3

1

∫ 5

0

(1 + y + z) dydz +∫ 3

1

∫ 5

0

(1 + 7 + y + z) dydz

=∫ 5

0

∫ 7

0

(6 + 2x + 2y) dxdy +∫ 3

1

∫ 7

0

(7 + 2x + 2z) dxdz +∫ 3

1

∫ 5

0

(9 + 2y + 2z) dydz

=∫ 5

0

[6x + x2 + 2yx

]70dy +

∫ 3

1

[7x + x2 + 2zx

]70dz +

∫ 3

1

[9y + y2 + 2zy

]50dz

=∫ 5

0

(91 + 14y) dy +∫ 3

1

(98 + 14z) dz +∫ 3

1

(70 + 10z) dz

=[91y + 7y2

]50

+[91z + 7z2

]31

+[70z + 5z2

]31

= 630 + 336− 98 + 255− 75 = 1048.

3.2 Cylindrical polar coordinates

a

x

y

z

4

0

Figure 11:

What about if we wish to find the mass of a hollow cylinder S in Figure 11 whose surface density isf(ρ, φ, z) = 1 + z2?

We use Mass =∫S

fdA, but what is dA? First we split the surface of the cylinder into three parts; the

top, the base and the curved surface (C.S.), so

mass =∫

top

fdA +∫

base

fdA +∫

C.S.

fdA.

17

Page 18: Lembretes Matemática

The top and the base are 2D surfaces, so as in Section 2 we have (see Figure 12)∫top

fdA =∫ 2π

0

∫ a

0

fρdρdφ and∫

base

fdA =∫ 2π

0

∫ a

0

fρdρdφ

with z constant on both surfaces.

On the curved surface we have ρ is constant, and in fact ρ = a where a is the radius of the cylinder.

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Figure 12:

We know that dA is related to the area of a surface element on the curved surface of the cylinder seefigure 12.

So area = a∆φ∆z and hence dA = adφdz.

Thus ∫C.S.

fdA =∫ 4

0

∫ 2π

0

fadφdz.

So

mass =∫ 2π

0

∫ a

0

fρdρdφ +∫ 2π

0

∫ a

0

fρdρdφ +∫ 4

0

∫ 2π

0

fadφdz.

Since f = 1 + z2 we have

mass =∫ 2π

0

∫ a

0

(1 + z2

(=16)

)ρdρdφ +

∫ 2π

0

∫ a

0

(1 + z2

(=0)

)ρdρdφ +

∫ 4

0

∫ 2π

0

(1 + z2

)adφdz

=∫ 2π

0

∫ a

0

17ρdρdφ +∫ 2π

0

∫ a

0

ρdρdφ +∫ 4

0

∫ 2π

0

(1 + z2

)adφdz

=∫ 2π

0

172 a2dφ +

∫ 2π

0

a2

2dφ +

∫ 4

0

2πa(1 + z2

)dz

= 17πa2 + πa2 + 2πa

[z +

z3

3

]40

= 18πa2 + 50 23πa.

18

Page 19: Lembretes Matemática

3.3 Spherical polar coordinates

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Figure 13:

If we want to evaluate the integral of f(r, θ, φ) over the surface of a sphere with radius R we note thatsince r is constant (= R) and θ and φ vary on the surface of the sphere we have (see Figure 13)

area of S.E. = R∆θ ·R sin θ∆φ

= R2 sin θ∆φ∆θ

and hencedA = R2 sin θdθdφ.

Remark 4 The angle θ only varies from 0 to π, while the angle φ varies from 0 to 2π (for the wholesurface of a sphere).

Example 3.2 Given a hollow sphere with centre at the origin, radius 2, with charge density f(R, θ, φ) =sin2 θ + cos2 θ, calculate the total charge on the surface of the sphere.

Solution: We have

total charge =∫S

fdA (f is charge density)

=∫ π

0

∫ 2π

0

(sin2 θ + cos2 θ

)R2 sin θdφdθ

=∫ π

0

∫ 2π

0

1 � 22 sin θdφdθ

=∫ π

0

∫ 2π

0

4 sin θ [φ]2π0 dθ

= 8π

∫ π

0

sin θdθ

= 16π.

19

Page 20: Lembretes Matemática

3.4 Scalar and vector fields

A scalar (vector) field is a distribution of scalar values (vector values) on/in a specified surface/region inspace, such that there is a unique scalar (vector) associated with each point on/in the surface/region.

Examples of scalar fields are temperature in a room or pressure in a room.Examples of vector fields are magnetic flux around a bar magnet or water velocity on the surface of ariver.

3.5 Flux of a vector field

The flux of a vector field F = Fxi+Fyj+Fzk (F = Fρeρ + Fφeφ + Fzez, F = Frer + Fθeθ + Fφeφ),through a surface S is equal to the surface integral of the component of F normal to the surface.

Mathematically we have that

Flux = Φ =∫

S

FndA =∫

S

F · ndA

where Fn denotes the component of F that points in the direction of n.If n points out of the surface we have outward flux, if n point in to the surface we have the inwardflux.

Example 3.3 Evaluate the outward flux of F = 3eρ − 2ez through the base of a cylinder centred at theorigin, with height H and radius R.

Solution:On the base of the cylinder n =− ez and hence

F · n = (3eρ − 2ez) · (−ez)= (3)(0) + (0)(0) + (−2)(−1) = 2

SinceFlux = Φ =

∫S

F · ndA

we haveΦ =

∫S

2dA.

But what is dA? On the base of a cylinder ρ and φ vary and we have dA = ρdρdφ so,

Φ =∫ 2π

0

∫ R

0

2ρdρdφ =∫ 2π

0

[2ρ2

2

]R

0

dφ = 2πR2.

Alternatively, since we know that the area of S =∫

SdA, we have Φ = 2× area of S = 2πR2.

If we want to calculate the outward flux of G = ρzeρ + cos(φ)eφ − ez through the curved surface of thecylinder in Example 3.3, we have n = eρ and so

G · n = (2ρzeρ + cos(φ)eφ − ez) · eρ

= (2ρz)(1) + cos(φ)(0) + (−1)(0)= 2ρz

20

Page 21: Lembretes Matemática

and hence the outward flux

Φ =∫

S

2ρzdA

=∫ H

2

−H2

∫ 2π

0

2ρzRdφdz

=∫ H

2

−H2

∫ 2π

0

2RzRdφdz

=∫ H

2

−H2

∫ 2π

0

2R2zdφdz

=∫ H

2

−H2

[2R2zφ

]2π

0dz

=∫ H

2

−H2

4πR2zdz

= 4πR2

[z2

2

]H2

−H2

= 0.

4 Volume Integrals

4.1 Cartesian, cylindrical polar and spherical polar coordinates

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Figure 14:

The integral of a function g(x, y, z), or g(ρ, φ, z) or g(r, θ, φ), over a 3D object is given by

21

Page 22: Lembretes Matemática

∫object

gdV

where dV is the limit of the volume of a small volume element in the object.

In Cartesian coordinates dV = dxdydz.

In cylindrical polar coordinates dV = ρdρdφdz (see Figure 14).

In spherical polar coordinates dV = r2 sin(θ)dθdφdr (see Figure 15).

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Figure 15:

Volume integrals can be used to calculate the volume of 3D objects, or the mass or charge of/in a 3Dobject. Also they can be used to calculate the moment of inertia of a 3D object. For the objects in Figure16 we have:

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Figure 16:

In cartesian coordinates

22

Page 23: Lembretes Matemática

∫V

gdV =∫ c

0

∫ b

0

∫ a

0

gdxdydz.

In cylindrical polar coordinates ∫V

gdV =∫ H

0

∫ 2π

0

∫ a

0

gρdρdφdz.

In spherical polar coordinates ∫V

gdV =∫ 2π

0

∫ π

0

∫ R

0

gr2 sin(θ)drdθdφ.

Example 4.1 Evaluate the volume integral of g(r, θ, φ) = 4r cos(φ/8) over a sphere centred at the originwith radius 2.

Solution: We have ∫sphere

f(ρ, φ, z)dV =∫ 2

0

∫ π

0

∫ 2π

0

g(r, θ, φ)r2 sin(θ)drdθdφ

with g(r, θ, φ) = 4r cos(φ/8). Thus∫sphere

f(ρ, φ, z)dV =∫ 2π

0

∫ π

0

∫ 2

0

4r3 cos(φ/8) sin(θ)drdθdφ

=(∫ 2π

0

cos(φ/8)dφ

)(∫ π

0

sin(θ)dθ

)(∫ 2

0

4r3dr

)= [8 sin(φ/8)]2π

0 [− cos(θ)]π0[r4]20

= 8(

1√2− 0)· (−(−1− 1)) · (16− 0)

=256√

2.

4.2 Moment of Inertia

The moment of inertia of a mass m about the z-axis is given by I = mρ2, where ρ =√

x2 + y2 is thedistance from the z-axis.The moment of inertia of a solid object V about the z-axis is equal to the sum of the moments of inertiaof the individual volume elements in the object.

Since the volume elements are small we can approximate their mass by their volume multiplied by thedensity at their mid points pi. Also we can approximate their distance from the z-axis by the distance oftheir mid points pi from the z-axis.

Hence the moment of inertia of a volume element is approximated by

Ii = ρ2Mass ' f(pi)∆Viρ2i .

Thus we have that the moment of inertia I is such that

I 'N∑

i=1

Ii =N∑

i=1

f(pi)∆Viρ2i →

∫object

fρ2dV.

23

Page 24: Lembretes Matemática

Example 4.2 Calculate the moment of inertia of a cylinder with radius 1, height H, centre the originand density f = ρz2.

Solution: We have

I =∫ H

2

−H2

∫ 2π

0

∫ 1

0

ρ2ρz2ρdρdφdz

=∫ H

2

−H2

∫ 2π

0

∫ 1

0

ρ4z2dρdφdz

=∫ H

2

−H2

∫ 2π

0

∫ 1

0

[ρ5

5

]10

z2dφdz

=∫ H

2

−H2

∫ 2π

0

15z2dφdz

=∫ H

2

−H2

5z2dz

=[2π

15z3

]H2

−H2

=πH3

120.

5 Line Integrals

5.1 Work done by a force

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Figure 17:

Line integrals are used to calculate the work done by a force F (or a vector field) F = Fxi+Fyj+Fzk, ona particle in moving it along a curve C =

−−→AB see Figure 17.

The work done by a constant force F on a body undergoing linear displacement is given by the distance|AB| multiplied by the component of F in the direction AB, i,e, in the direction S, see Figure 18.Thus we have

W = F · S = |F| |S| cos(θ)

and sinceF = FsS + Fs⊥ S⊥

withFs = |F| cos(θ)

24

Page 25: Lembretes Matemática

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Figure 18:

we haveW = Fs |S| .

For a non-constant force F acting on a particle moving along a non-linear path C we can approximatethe work done by approximating the path C by small linear path segments, and then on each of thesesegments we can approximate the force F by the value of F at the start of each linear segment (i.e. by aconstant value).

Figure 19:

On the ith segment along the curve that has start point at ri = xii + yij + zik and at end pointri+1 = xi+1i + yi+1j + zi+1k (see figure 19), we have that the work done by the constant force F(ri) inmoving a particle along the segment is

Wi = F(ri) ·∆ri.

SoW ≈

∑i

Wi =∑

i

F(ri) ·∆ri.

In the limit as the size of the line segments tends to zero, we have

W =∫

C

F(r) · dr.

25

Page 26: Lembretes Matemática

Since F = Fxi+Fyj+Fzk and r = xi + yj + zk we have

F · dr = (Fxi+Fyj+Fzk) · (dxi + dyj + dzk)= Fxdx + Fydy + Fzdz.

Hence

W =∫

C

Fxdx + Fydy + Fzdz. (5.5)

We need to write (5.5) as a definite integral involving a single variable, say x, y, z, t or φ.

5.2 Line integrals in the x-y plane

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Figure 20:

Example 5.1 Calculate the work done by F = 3yi − 5xj+100xyk on a particle moving along the pathAB shown in Figure 20.

Solution: On the path AB we have y = 12x and z = 3 and hence dy

dx = 12 and dz

dx = 0.We can write (5.5) as a definite integral in x:

W =∫

C

Fxdx + Fydy + Fzdz =∫ XB

XA

(Fx(x)dx

dx+ Fy(x)

dy

dx+ Fz(x)

dz

dx)dx

where XA is the x coordinate at the start point A and XB is the x coordinate at the end B.Thus XA = 0 and XB = 4. Furthermore since

Fx(x) = 3y =32x, Fy(x) = −5x, Fz(x) = 100xy = 50x

anddx

dx= 1,

dy

dx=

12

anddz

dx= 0

we have

W =∫ 4

0

[(32x

)(1)− (5x)

(12

)+ (50x)(0)

]dx

=∫ 4

0

−xdx =[−x2

2

]40

= −8.

26

Page 27: Lembretes Matemática

Remark 5

1. The line integral of a vector field F along a path P = P1 + P2 can be written as∫P

F · dr =∫

P1

F · dr+∫

P2

F · dr.

2. ∫ B

A

F · dr =−∫ A

B

F · dr.

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Figure 21:

Example 5.2 Calculate the work done by F = 3xi + 5xj − 2k on a particle moving along the pathAc = AB + BC shown in Figure 21.

Solution: We have∫ C

A

F · dr =∫ B

A

F · dr+∫ C

B

F · dr

=∫ B

A

Fxdx + Fydy + Fzdz +∫ C

B

Fxdx + Fydy + Fzdz.

On path P1 we have

x = −1, z = 0 ⇒ dx

dy= 0 and

dz

dy= 0

and on path P2 we have

x = −1, y = 2 ⇒ dx

dz= 0 and

dy

dz= 0.

Thus we have

∫ B

A

F · dr =∫ YB

YA

(Fx(y)dx

dy+ Fy(y)

dy

dy+ Fz(y)

dz

dy)dy∫ C

B

F · dr =∫ ZC

ZB

(Fx(z)dx

dz+ Fy(z)

dy

dz+ Fz(z)

dz

dz)dz

⇒∫ B

A

F · dr =∫ 2

0

((3x)(0) + (5x)(1)− (2)(0))dy =∫ 2

0

−5dy = −10

⇒∫ C

B

F · dr =∫ 1

0

((3x)(0) + (5x)(0)− (2)(1))dz =∫ 1

0

−2dz = −2

27

Page 28: Lembretes Matemática

and hence ∫ C

A

F · dr =− 10− 2 = −12.

Example 5.3 Calculate the work done by F = −yi + xj on moving a particle along the path AB shownin Figure 22.

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Figure 22:

Solution: The work done is given by

W =∫ B

A

F · dr =∫ B

A

Fxdx + Fydy =∫ XB

XA

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx

and sincey = (4− x2)

12 ⇒ dy

dx=

12(4− x2)(−2x) = − x

(4− x2)12

we have

W =∫ XB

XA

[−(4− x2)

12 (1) + (x)

(−x

(4− x2)12

)]dx

=∫ 2

0

(−(4− x2)

12 − x2(4− x2)−

12

)dx.

This is not an easy integral to solve, so hopefully there is a better way of calculating W .

5.3 Parametric Equations

Any path in 3D may be expressed in terms of three parametric equations x(t), y(t) and z(t) that involvea parameter t that varies from t1 to t2 such that

x(t1) = x1, y(t1) = y1, z(t1) = z1, x(t2) = x2, y(t2) = y2 and z(t2) = z2.

We have (see Figure 23) ∫ P2

P1

F · dr =∫ t2

t1

(Fx(t)dx

dt+ Fy(t)

dy

dt+ Fz(t)

dz

dt)dt.

Example 5.3 (continued)Using polar coordinates we can write the curve C = AB in terms of three parametric equations

28

Page 29: Lembretes Matemática

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Figure 23:

x = 2 cos φ, y = 2 sin φ, z = constant

⇒ dx

dφ= −2 sinφ,

dy

dφ= 2 cos φ,

dz

dφ= 0.

When φ = 0 x = 2, y = 0 and z = constant and hence we are at point A. While when φ = π2 x = 0,

y = 2 and z = constant hence we are at point B.Thus we have

∫ B

A

F · dr =∫ φB

φA

(−ydx

dφ+ x

dy

dφ)dφ

=∫ π

2

0

[(−2 sinφ)(−2 sinφ) + (2 cos φ)(2 cos φ)] dφ

=∫ π

2

0

(4 sin2 φ + 4 cos2 φ)dφ

=∫ π

2

0

4dφ = 2π.

Example 5.4 Evaluate the integral of F = 2xi + 3zj− 5k along the curve given parametrically by

x(t) = t, y(t) = 2t, z(t) = −t2

where t varies from 0 to 1.

Solution: Since

x(t) = t, y(t) = 2t, z(t) = −t2

⇒ dx

dt= 1,

dy

dt= 2,

dz

dt= −2t

29

Page 30: Lembretes Matemática

we have ∫C

F · dr =∫ t2

t1

(Fx(t)dx

dt+ Fy(t)

dy

dt+ Fz(t)

dz

dt)dt

=∫ 1

0

[(2x)(1) + (3z)(2) + (−5)(−2t)] dt

=∫ 1

0

(2t− 6t2 + 10t)dt

=∫ 1

0

(12t− 6t2)dt

=[6t2 − 2t3

]10

= 4.

5.4 Path dependence/independence

A line integral of a field/force F from a point A to a point B is said to be path dependent if it dependson the path taken to get from A to B, otherwise it is said to be path independent. If the line integral ofF is path dependent F is non conservative.

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Figure 24:

Example 5.5 By considering two paths that start from A = (2, 0) and end at B = (0, 2) (see Figure 24),show that the line integral of the vector field F = −yi + xj is path dependent.

Solution: From Example 5.3 we have that ∫P2

F · dr = 2π.

While on P2 we have∫P2

F · dr =∫ B

A

Fxdx + Fydy =∫ XB

XA

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx.

Since the path P2 is defined by y = 2− x we have dydx = −1 and hence

∫P2

F · dr =∫ 0

2

[(−y)(1) + (x)(−1)] dx

=∫ 0

2

[−(2− x)− x] dx

=∫ 0

2

−2dx = − [2x]02 = 4.

30

Page 31: Lembretes Matemática

So∫

P1F · dr 6=

∫P2

F · dr and hence the integral of F is path dependent which tells us that F is non-conservative.

5.5 Closed curves (loops)

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Figure 25:

For a closed curve C that comprises of the path C1 from rA to rB (see Figure 25) and the path C2 fromrB to rA, we have that the line integral of F around the closed curve (loop) C is given by∮

C

F · dr=∫

C1

F · dr+∫

C2

F · dr.

The circle around the integral of C indicates that C is a closed curve.

5.6 Conservative fields (or forces)

Remark 6 1. The line integral of a conservative field from a point A to a point B is path independent,i.e. it only depends on the initial point A and the end point B.

2. The line integral of a conservative field around any closed loop is always equal to zero, i.e.∮C

F · dr =0 (5.6)

for a conservative field F and for any closed curve C.

3. If∮

CG · dr 6=0 then G is non conservative.

4. If∮

CG · dr =0 then G is might be conservative.

Example 5.6 Show that a conservative force F satisfies (5.6).

Solution:Since

∮C

F · dr=∫

C1F · dr+

∫C2

F · dr and∫ B

AF · dr = −

∫ A

BF · dr, and for a conservative field

∫C1

F ·dr =−

∫C2

F · dr, we have the following for any conservative field.

∮C

F · dr =∫

C1

F · dr+∫

C2

F · dr

= −∫

C2

F · dr+∫

C2

F · dr = 0.

Example 5.7 Calculate the line integral of F = x2i + yj along the two paths C1 = AP + PB and C2

shown in Figure 26. Say if the field F could be a conservative field.

31

Page 32: Lembretes Matemática

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Figure 26:

Solution:The path C1 consists of the two straight line paths AP and PB. On the first path AP, y = 0 and xvaries from 0 to 1 ⇒ dy

dx = 0.On the second path PB, x = 1 and y varies from 0 to 1 ⇒ dx

dy = 0.Hence we have

∫C1

F · dr =∫ P

A

F · dr+∫ B

P

F · dr

=∫ XP

XA

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx +

∫ YB

YP

(Fx(y)dx

dy+ Fy(y)

dy

dy)dy

=∫ 1

0

[(x2)(1) + (y)(0)

]dx +

∫ 1

0

[(x2)(0) + (y)(1)

]dy

=∫ 1

0

x2dx +∫ 1

0

ydy

=[x3

3

]10

+[y2

2

]10

=13

+12

=56.

On C2, y = x ⇒ dydx = 1

∫C2

F · dr =∫ A

B

Fxdx + Fydy

=∫ XA

XB

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx

=∫ 0

1

[(x2)(1) + (y)(1)

]dx

=∫ 0

1

(x2 + x)dx

=[x3

3+

x2

2

]01

= −56.

Since the path C1 + C2 is a closed curve with∮C1+C2

F · dr =∫

C1

F · dr+∫

C2

F · dr

=56− 5

6= 0

32

Page 33: Lembretes Matemática

it follows that F may be conservative.

6 Partial derivatives

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Figure 27:

The partial derivative of a function F (x, y, z, t, ...) with respect to x is defined by ∂F∂x and is evaluated by

treating the other variables as constant and differentiating with respect to x. Consider a function of twovariables F (x, t), the partial derivatives ∂F

∂x and ∂F∂t at a given point (x, t) of a function F (x, t), represent

the slopes of the 3D graph y = F (x, t) in the direction of the x-axis and the t-axis at the given point(x, t) (see figure 27). We define

∂F

∂x= lim

h→0

F (x + h, t)− F (x, t)h

(6.7)

and∂F

∂t= lim

h→0

F (x, t + h)− F (x, t)h

. (6.8)

Example 6.1 Given that f(x, y, t) = 3x2tey find ∂f∂x , ∂f

∂t and ∂f∂y .

Solution: We have

∂f

∂x= 3tey d

dxx2 = (3tey)(2x) = 6teyx

∂f

∂t= 3x2ey d

dtt = 3x2ey

∂f

∂x= 3x2t

d

dyey = 3x2tey.

6.1 The chain rule for partial derivatives

For a function f(u(x, y, t)) we have

∂xf(u(x, y, t)) =

df

du

∂u

∂x,

∂yf(u(x, y, t)) =

df

du

∂u

∂yand

∂tf(u(x, y, t)) =

df

du

∂u

∂t. (6.9)

Example 6.2 For f = cos(3x2y) find ∂f∂x .

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Solution: Setting u = 3x2y we have

cos(3x2y) = cos u withdf

du= − sinu, and

∂u

∂x= 6xy.

Hence from the chain rule (6.9) we have

∂f

∂x= −6xy sin(3x2y).

6.2 Higher order partial derivatives

From the example above we see that partial derivatives of a function F (x, y, t) are also functions of x, yand t, and so can be partially differentiated themselves.

Second order partial derivatives take the following form

∂2F

∂x2=

∂x

∂F

∂x∂2F

∂y∂x=

∂y

∂F

∂x

∂2F

∂y2=

∂y

∂F

∂y

∂2F

∂x∂y=

∂x

∂F

∂y.

Example 6.3 For F (x, y) = 3ex cos 2y calculate

∂F

∂x,

∂F

∂y,

∂2F

∂x2,

∂2F

∂y∂x,

∂2F

∂y2and

∂2F

∂x∂y.

Solution: We have∂F

∂x= 3ex cos 2y,

∂F

∂y= −6ex sin 2y

and

∂2F

∂x2=

∂x

∂F

∂x=

∂x(3ex cos 2y) = 3ex cos 2y

∂2F

∂y∂x=

∂y

∂F

∂x=

∂y(3ex cos 2y) = −6ex sin 2y

∂2F

∂y2=

∂y

∂F

∂y=

∂y(−6ex sin 2y) = −12ex cos 2y

∂2F

∂x∂y=

∂x

∂F

∂y=

∂x(−6ex sin 2y) = −6ex sin 2y.

Remark 7 It is always the case that∂2F

∂x∂y=

∂2F

∂y∂x.

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