Name: Karol Koziol

Solution to Practice Midterm

1. Compute the following double integral by switching the order of integration ˆ 1ˆ 1 x3 p dydx. x4 + y 2 0 x2 (Hint: sketch the domain of integration first.) Solution. The given boundaries of integration are 0 6 x 6 1, x2 6 y 6 1, which is a region of type I in the plane. If we draw a picture and express the domain as a type II region, we get 0 6 y 6 1, √ 0 6 x 6 y. The integral then becomes ˆ

1ˆ

0

0

√

y

ˆ

x3

p dxdy = x4 + y 2 = = = =

2. Compute the integral

! p x 4 + y 2 √y |x=0 dy 2 0 ˆ 1√ 2 1 y − y dy 2 2 0 ˆ 1√ 2−1 y dy 2 0 √ 2−1 2 1 y |0 √ 4 2−1 . 4 1

˚ x dV, E

where E is the tetrahedron with vertices (0, 0, 0), (2, 0, 0), (0, 1, 0), (0, 0, 1). Solution. We write the tetrahedron E as a region of type 1 in 3D space. Its projection D is the triangle in the xy-plane described by 0 6 x 6 2, x 06y 61− . 2 The tetrahedron E lies above D between the graph of the function u1 (x, y) = 0 and the plane passing through the points (2, 0, 0), (0, 1, 0), (0, 0, 1). To write down the equation of this plane, we 1

choose two vectors on this plane h−2, 0, 1i, h−2, 1, 0i and compute their cross product, which is h−1, −2, −2i. It follows that the equation of the plane is x + 2y + 2z = 2. (Alternatively, we could just start with the equation ax + by + cz + d = 0, and find values of a, b, c, d such that the three points (2, 0, 0), (0, 1, 0), (0, 0, 1) lie on the plane.) Solving for z, we see that the plane is the graph of the function x u2 (x, y) = 1 − − y. 2 The triple integral then becomes ˆ 0

2 ˆ 1−x/2 ˆ 1−x/2−y 0

0

ˆ

2 ˆ 1−x/2

x2 x− x dzdydx = − xy 2 0 0 2 x=2 x x3 x4 1 = − + = . 4 6 32 x=0 6

3. Compute the integral

ˆ

2

−2

ˆ √4−y2

e−(x

2 +y 2 )

ˆ

2

dydx = 0

x x3 − x2 + 2 8

dx

dxdy,

0

by switching to polar coordinates. Solution. The domain of integration is the right half of a disc of radius 2, and therefore is described in polar coordinates by 0 6 r 6 2, −π/2 6 θ 6 π/2. Switching the integral into polar coordinates gives ˆ 2 ˆ √4−y2 ˆ 2ˆ 2 2 e−(x +y ) dxdy = −2

ˆ =

2

re−r dθdr

−π/2

0

0

π/2

2

−r2

re

ˆ dr

dθ −π/2

0 2

e−r 2 = π − | 2 0 π = (1 − e−4 ). 2 4. Evaluate the line integral

!

π/2

!

ˆ F~ · d~r, C

where F~ (x, y, z) = hy 3 , 3xy 2 + ez , yez i, and the curve C is parametrized by ~r(t) = het , t5 , ln(1 + ln(t10 − t5 + 1))i,

2

with 0 6 t 6 1. (Hint: there is an easy way and a hard way to do this problem. Don’t do it the hard way!) Solution. It is very hard to compute this line integral directly. If the vector field F~ was a gradient field, then we could use the Fundamental Theorem for Line Integrals and compute this answer quickly. Let’s see if we can solve the equations ∂f = y3, ∂x

∂f = 3xy 2 + ez , ∂y

∂f = yez . ∂z

Integrating partially the first equation with respect to x gives f = xy 3 + g(y, z). Taking partial derivatives of this equation with respect to y and z we get ∂f ∂g = 3xy 2 + = 3xy 2 + ez , ∂y ∂y

∂f ∂g = = yez , ∂z ∂z

therefore g satisfies ∂g = ez , ∂y

∂g = yez . ∂z

Integrating partially the first equation with respect to y gives g = yez + h(z). Taking a partial derivative with respect to z gives ∂g = yez + h0 (z) = yez , ∂z therefore h0 (z) = 0 so h(z) = C any constant and g(y, z) = yez + C and f (x, y, z) = xy 3 + yez + C. Therefore, we see that F~ = ∇f , and we get ˆ F~ · d~r =

ˆ ∇f · d~r

C

C FTLI

=

f (e, 1, 0) − f (1, 0, 0)

=

e + 1.

5. Calculate the work done by the vector field F~ (x, y) = hxy, 3y 2 i on a particle moving along the path ~r(t) = h11t4 , t3 i from t = 0 to t = 1. Solution. To calculate the work, we must evaluate the integral ˆ ˆ 1 ~ F · d~r = F~ (~r(t)) · ~r 0 (t) dt. C

0

3

We have F~ (~r(t)) = h11t7 , 3t6 i ~r 0 (t) = h44t3 , 3t2 i F~ (~r(t)) · ~r 0 (t) = 484t10 + 9t8 Therefore

ˆ F~ · d~r

Work =

C ˆ 1

=

F~ (~r(t)) · ~r 0 (t) dt

0

ˆ

1

484t10 + 9t8 dt

= 0

= 44t11 + t9 |10 = 45. ˚

6. Evaluate

xyz dV, E

where E is the region in the first octant with x2 + y 2 + z 2 6 4 (recall that in the first octant we have x > 0, y > 0, z > 0). Solution. The region E can be easily described in spherical coordinates: 06ρ62 0 6 θ 6 π/2 0 6 φ 6 π/2 We change variables into spherical coordinates (and remember the factor of ρ2 sin(φ)), and we get ˚

ˆ xyz dV

E

2 ˆ π/2 ˆ π/2

(ρ sin(φ) cos(θ))(ρ sin(φ) sin(θ))(ρ cos(φ))ρ2 sin(φ) dφdθdρ 0 0 0 ! ˆ ! ˆ 2 ˆ π/2 π/2 5 3 = ρ dρ sin(φ) cos(φ) dφ sin(θ) cos(θ) dθ

=

0

= = = 7. Evaluate

ρ6

0

|20

6 64 1 1 · · 6 4 2 4 . 3

0

sin(φ)4 4

π/2

|0

sin(θ)2 2

π/2

|0

ffi y 4 dx + 2xy 3 dy, C

4

where C is the ellipse x2 + 2y 2 = 2. (Hint: use Green’s theorem.) Solution We have P (x, y) = y 4 and Q(x, y) = 2xy 3 . Both of these functions are defined on the entire xy-plane, so we may apply Green’s Theorem. We have ∂Q ∂P − = 2y 3 − 4y 3 = −2y 3 . ∂x ∂y Therefore, we have

¨

ffi

−2y 3 dA,

y 4 dx + 2xy 3 dy = D

C

x2 + 2y 2

where D is the region inside the ellipse = 2. Notice that this region is symmetric about the x-axis, and that the function −2y 3 is odd with respect to y. Therefore, by symmetry, we conclude that ¨ −2y 3 dA = 0. D

5

Solution to Practice Midterm

1. Compute the following double integral by switching the order of integration ˆ 1ˆ 1 x3 p dydx. x4 + y 2 0 x2 (Hint: sketch the domain of integration first.) Solution. The given boundaries of integration are 0 6 x 6 1, x2 6 y 6 1, which is a region of type I in the plane. If we draw a picture and express the domain as a type II region, we get 0 6 y 6 1, √ 0 6 x 6 y. The integral then becomes ˆ

1ˆ

0

0

√

y

ˆ

x3

p dxdy = x4 + y 2 = = = =

2. Compute the integral

! p x 4 + y 2 √y |x=0 dy 2 0 ˆ 1√ 2 1 y − y dy 2 2 0 ˆ 1√ 2−1 y dy 2 0 √ 2−1 2 1 y |0 √ 4 2−1 . 4 1

˚ x dV, E

where E is the tetrahedron with vertices (0, 0, 0), (2, 0, 0), (0, 1, 0), (0, 0, 1). Solution. We write the tetrahedron E as a region of type 1 in 3D space. Its projection D is the triangle in the xy-plane described by 0 6 x 6 2, x 06y 61− . 2 The tetrahedron E lies above D between the graph of the function u1 (x, y) = 0 and the plane passing through the points (2, 0, 0), (0, 1, 0), (0, 0, 1). To write down the equation of this plane, we 1

choose two vectors on this plane h−2, 0, 1i, h−2, 1, 0i and compute their cross product, which is h−1, −2, −2i. It follows that the equation of the plane is x + 2y + 2z = 2. (Alternatively, we could just start with the equation ax + by + cz + d = 0, and find values of a, b, c, d such that the three points (2, 0, 0), (0, 1, 0), (0, 0, 1) lie on the plane.) Solving for z, we see that the plane is the graph of the function x u2 (x, y) = 1 − − y. 2 The triple integral then becomes ˆ 0

2 ˆ 1−x/2 ˆ 1−x/2−y 0

0

ˆ

2 ˆ 1−x/2

x2 x− x dzdydx = − xy 2 0 0 2 x=2 x x3 x4 1 = − + = . 4 6 32 x=0 6

3. Compute the integral

ˆ

2

−2

ˆ √4−y2

e−(x

2 +y 2 )

ˆ

2

dydx = 0

x x3 − x2 + 2 8

dx

dxdy,

0

by switching to polar coordinates. Solution. The domain of integration is the right half of a disc of radius 2, and therefore is described in polar coordinates by 0 6 r 6 2, −π/2 6 θ 6 π/2. Switching the integral into polar coordinates gives ˆ 2 ˆ √4−y2 ˆ 2ˆ 2 2 e−(x +y ) dxdy = −2

ˆ =

2

re−r dθdr

−π/2

0

0

π/2

2

−r2

re

ˆ dr

dθ −π/2

0 2

e−r 2 = π − | 2 0 π = (1 − e−4 ). 2 4. Evaluate the line integral

!

π/2

!

ˆ F~ · d~r, C

where F~ (x, y, z) = hy 3 , 3xy 2 + ez , yez i, and the curve C is parametrized by ~r(t) = het , t5 , ln(1 + ln(t10 − t5 + 1))i,

2

with 0 6 t 6 1. (Hint: there is an easy way and a hard way to do this problem. Don’t do it the hard way!) Solution. It is very hard to compute this line integral directly. If the vector field F~ was a gradient field, then we could use the Fundamental Theorem for Line Integrals and compute this answer quickly. Let’s see if we can solve the equations ∂f = y3, ∂x

∂f = 3xy 2 + ez , ∂y

∂f = yez . ∂z

Integrating partially the first equation with respect to x gives f = xy 3 + g(y, z). Taking partial derivatives of this equation with respect to y and z we get ∂f ∂g = 3xy 2 + = 3xy 2 + ez , ∂y ∂y

∂f ∂g = = yez , ∂z ∂z

therefore g satisfies ∂g = ez , ∂y

∂g = yez . ∂z

Integrating partially the first equation with respect to y gives g = yez + h(z). Taking a partial derivative with respect to z gives ∂g = yez + h0 (z) = yez , ∂z therefore h0 (z) = 0 so h(z) = C any constant and g(y, z) = yez + C and f (x, y, z) = xy 3 + yez + C. Therefore, we see that F~ = ∇f , and we get ˆ F~ · d~r =

ˆ ∇f · d~r

C

C FTLI

=

f (e, 1, 0) − f (1, 0, 0)

=

e + 1.

5. Calculate the work done by the vector field F~ (x, y) = hxy, 3y 2 i on a particle moving along the path ~r(t) = h11t4 , t3 i from t = 0 to t = 1. Solution. To calculate the work, we must evaluate the integral ˆ ˆ 1 ~ F · d~r = F~ (~r(t)) · ~r 0 (t) dt. C

0

3

We have F~ (~r(t)) = h11t7 , 3t6 i ~r 0 (t) = h44t3 , 3t2 i F~ (~r(t)) · ~r 0 (t) = 484t10 + 9t8 Therefore

ˆ F~ · d~r

Work =

C ˆ 1

=

F~ (~r(t)) · ~r 0 (t) dt

0

ˆ

1

484t10 + 9t8 dt

= 0

= 44t11 + t9 |10 = 45. ˚

6. Evaluate

xyz dV, E

where E is the region in the first octant with x2 + y 2 + z 2 6 4 (recall that in the first octant we have x > 0, y > 0, z > 0). Solution. The region E can be easily described in spherical coordinates: 06ρ62 0 6 θ 6 π/2 0 6 φ 6 π/2 We change variables into spherical coordinates (and remember the factor of ρ2 sin(φ)), and we get ˚

ˆ xyz dV

E

2 ˆ π/2 ˆ π/2

(ρ sin(φ) cos(θ))(ρ sin(φ) sin(θ))(ρ cos(φ))ρ2 sin(φ) dφdθdρ 0 0 0 ! ˆ ! ˆ 2 ˆ π/2 π/2 5 3 = ρ dρ sin(φ) cos(φ) dφ sin(θ) cos(θ) dθ

=

0

= = = 7. Evaluate

ρ6

0

|20

6 64 1 1 · · 6 4 2 4 . 3

0

sin(φ)4 4

π/2

|0

sin(θ)2 2

π/2

|0

ffi y 4 dx + 2xy 3 dy, C

4

where C is the ellipse x2 + 2y 2 = 2. (Hint: use Green’s theorem.) Solution We have P (x, y) = y 4 and Q(x, y) = 2xy 3 . Both of these functions are defined on the entire xy-plane, so we may apply Green’s Theorem. We have ∂Q ∂P − = 2y 3 − 4y 3 = −2y 3 . ∂x ∂y Therefore, we have

¨

ffi

−2y 3 dA,

y 4 dx + 2xy 3 dy = D

C

x2 + 2y 2

where D is the region inside the ellipse = 2. Notice that this region is symmetric about the x-axis, and that the function −2y 3 is odd with respect to y. Therefore, by symmetry, we conclude that ¨ −2y 3 dA = 0. D

5