MULTIVARIABLE CALCULUS

Sample Midterm Problems

November 27, 2009

INSTRUCTOR: Anar Akhmedov

1. Convert the point P = (4, π/3, 4) from cylindrical to spherical coordinates. Plot the pointP using the spherical coordinate system.

Solution: Notice that the angle θ will not change, so we have θ = π/3. Since ρ2 = r2+z2 =42 + 42 = 32, we get ρ = 4

√2. tanφ = r

z= 1. Since 0 ≤ φ ≤ π, we have φ = π/4. Thus

the spherical coordinates of the point P is given by (4√

2, π3, π

4).

2. Use spherical coordinates to evaluate∫ ∫ ∫

E3e(x2+y2+z2)3

2 dV where E is the region insidethe sphere x2 + y2 + z2 = 2 in the first octant.

Solution: The region E in spherical coordinates is given by 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ π/2and 0 ≤ ρ ≤

√2. Thus we need to compute the following integral

∫ π/20

∫ π/20

∫ √

20 3eρ3

ρ2 sin(φ) dρ dθ dφ =∫ π/20

∫ π/20 eρ3

sin(φ)|√

20 dθ dφ =

∫ π/20

∫ π/20 (e2

√

2 − 1) sin(φ) dθ dφ = π(e2√

2−1)

2

3. Let R be the unit square. Use Green’s theorem to evaluate the line integral∮∂R y2dx+x2dy

Solution: By Green’s theorem, we have∮∂R y2dx + x2dy =

∫ ∫R

∂∂x

(x2) − ∂∂y

(y2) dA =∫ ∫

R(2x − 2y) dA =∫ 10

∫ 10 (2x − 2y)dy dx = 0

4. Find the volume of the solid that lies inside the sphere x2 + y2 + z2 = 2 and outside of thecone z2 = x2 + y2.

Solution: We will use the spherical coordinates to compute the volume. The equation ofthe sphere becomes ρ =

√2. To convert the equation of the cone, add z2 to both sides of

the equation z2 = x2 + y2. We get 2z2 = x2 + y2 + z2 = ρ2. Since z = ρ cos(φ), we get2ρ2 cos2(φ) = ρ2. Solving for φ, we obtain φ = π/4 or φ = 3π/4 for the equation of thecone.

To find the volume, we evaluate V =∫ 2π0

∫ 3π/4π/4

∫ √

20 ρ2 sin(φ) dρ dθ dφ =

2π(ρ3/3)|√

20 (−cos(φ))|3π/4

π/4 = 8π/3.

5. Consider the double integral∫ ∫

Dx−y

(x+2y)2dA where D is the region bounded by the following

four lines: x − y = 0, x − y = 3, x + 2y = 2, and x + 2y = 6. Evaulate the integral usingan appropriate change of variables.

Solution: Use the change of variables u = x + 2y and v = y − x. Solving for x andy, we obtain the inverse transformation given by x = u−2v

3and y = u+v

3. First, note

that the image of the region D under the transformation u = x + 2y, v = y − x isa rectangle with vertices at (2,−3), (6,−3), (6, 0), and (2, 0). Next, we compute the

1

Jacobian ∂(x,y)∂(u,v)

= 1/3. Using the change of variable formula for double integrals, we have∫ ∫

Dx−y

(x+2y)2dA =

∫ 0−3

∫ 62

−vu2 |∂(x,y)

∂(u,v)|du dv = 1/2.

6. Find the area enclosed by the curve r(t) = < cos2(t), cos(t) sin(t) >, where t is in [0, π],using Green’s theorem.

Solution: The area is given by

Area = 12

∮C xdy − ydx = 1

2

∫ π0 (xdy

dt− y dx

dt)dt = 1

2

∫ π0 (cos2(t)(− sin(t) sin(t) + cos(t) cos(t)) −

cos(t) sin(t)(−2 cos(t) sin(t)))dt = 12

∫ π0 cos2(t) dt = π/4.

7. Explain why the line integral∫Cxdx + ydy + zdz is independent of path. Calculate the

integral along two different paths from (0, 0, 0) to (1, 1, 1).

Solution: Note that the vector field F = < x, y, z > is the gradient field of the fuctionf(x, y, z) = 1

2(x2 + y2 + z2). Using the Fundamental Theorem for Line Integrals, we show

that the given line integral is independent of path.

Choose the first path from (0, 0, 0) to (1, 1, 1) by r(t) = < t, t, t >. Denote this path by C1.The line integral over C1 is∫C1

xdx + ydy + zdz =∫ 10 (tdt + tdt + tdt) = 3

∫ 10 t dt = 3/2.

Choose the second path from (0, 0, 0) to (1, 1, 1) by r(t) = < t, t2, t3 >. Denote this pathby C2. The line integral over C2 is∫C2

xdx + ydy + zdz =∫ 10 (tdt + t2dt2 + t3dt3) =

∫ 10 (t + 2t3 + 3t5) dt = 3/2.

2