Solutions to Practice Problems for Test 4

Solutions to Practice Problems for Test 4 1. Let C be the line segment Zfrom the point (0, 1, 1) to the point (2, 2, 3). Evaluate the line integral y ds. C

Answer: First, we parametrize the line segment from (0, 1, 1) to (2, 2, 3): x(t) = (0, 1, 1) + t(2, 1, 2) = (2t, 1 + t, 1 + 2t) for 0 ≤ t ≤ 1. Thus, x0 (t) = (2, 1, 2) Thus, ||x0 (t)|| =

√

4+1+4=3

We can now evaluate the line integral: · ¸1 Z Z 1 t2 9 = y ds = (1 + t)3 dt = 3 t + 2 0 2 C 0 2. Let C be the curve in R3 defined by the equations r = 1, z = θ with endpoints (x, y, z) = (1, 0, 0) and (x, y, z) = (1, 0, 2π). Evaluate the following line integral: Z xz dx + yz dy + z 2 dz C

Answer: We can parameterize this curve using the parameter t = θ. Since r = 1, we have the parametric equations: x = cos t y = sin t z = t

0 ≤ t ≤ 2π

The bounds for t come from the endpoints (x, y, z) = (1, 0, 0) and (x, y, z) = (1, 0, 2π). Thus: dx = − sin t dt dy = cos t dt dz = dt 1

We can now compute the vector line integral: Z Z 2π ¡ ¢ 2 xz dx + yz dy + z dz = −t cos t sin t + t cos t sin t + t2 dt C Z0 2π = t2 dt · =

0

1 3 t 3

¸2π = 0

8π 3 3

3. Consider the vector field F(x, y, z) = (sin(x2 ) + y) i + (x + z) j + y k. (a) Is F conservative? (b) Let C be the curve parameterized by x = sin t y = cos t −π ≤t≤π z = t2 I Use your answer to part (a) to evaluate F · ds. C

Answer: (a) To see if F is conservative, we compute ∇ × F: ¯ ¯ i j ¯ ¯ ¯ ∂ ∂ ∇×F = ¯ ¯ ∂x ∂y ¯ ¯ sin(x2 ) + y x + z

¯ k ¯¯ ¯ ∂ ¯ ¯ ∂z ¯¯ y ¯

= (1 − 1) i + 0 j + (1 − 1) k = 0 Yes, F is conservative. (b) The trick to this part is to realize that C is a closed loop. At t = −π, the parametric equations give x = 0, y = 1, and z = π 2 , and at t = π, the parametric equations give the same values. Since F is conservative, there exists some function f such that ∇f = F. Thus: I F · ds = f (0, 1, π 2 ) − f (0, 1, π 2 ) = 0 C

2

You can also use Stokes’s Theorem to obtain the same answer (we have to use Stokes’s Theorem instead of Green’s Theorem because the curve is in R3 ). Let S be a surface with C as the boundary (this is where it is important that C is a closed curve — if C was not closed, there could not be a surface with C as the boundary). Then, by Stokes’s Theorem: I ZZ ZZ F · ds = (∇ × F) · dA = 0 · dA = 0 C

S

S

4. Consider the vector field F(x, y, z) = y i+(x + z cos y) j+(ez + sin y) k. (a) Find a function f such that ∇f = F. (b) Let C be the x y z

curve parameterized by = t = t sin t 0≤t≤π = t cos t Z Use your answer to part (a) to evaluate F · ds. C

Answer: (a) We want a function f such that ∂f =y ∂x

and

∂f = x + z cos y ∂y

and

∂f = ez + sin y ∂z

From the first equation, we see that f (x, y, z) = xy + g(y, z) where g(y, z) is some function of y and z. If we differentiate this equation for f with respect to y, we obtain ∂g ∂f =x+ ∂y ∂y Comparing with the previous expression for Thus:

∂f ∂g , we see that = z cos y. ∂y ∂y

g(y, z) = z sin y + h(z) 3

where h(z) is some function of z. Combining with the previous expression for f , we get f (x, y, z) = xy + z sin y + h(z) If we differentiate this equation for f with respect to z, we obtain ∂f dh = sin y + ∂z dz ∂f dh Comparing with the previous expression for , we see that = ez . ∂z dz Thus: h(z) = ez + C Thus: f (x, y, z) = xy + z sin y + ez + C (b) At t = 0, the curve C is at the point (0, 0, 0). At t = π, the curve C is at the point (π, 0, −π). Thus: Z h i(π,0,−π) z F · ds = xy + z sin y + e (0,0,0)

C

=

e

−π

−1

5. Consider the region R bounded by the parabolas y = x2 and x = y 2 . Let C be the boundary of R oriented counterclockwise. Use Green’s Theorem to evaluate I

√

(y + e

x

¡ ¢ ) dx + 2x + cos(y 2 ) dy

C √ ¡ ¢ Answer: The vector field is F = y + e x , 2x + cos(y 2 ) . Thus,

rot(F) = 2 − 1 = 1 Thus by Green’s Theorem, 4

Z

Z

1

Z

√

1

1 dy dx =

F · ds = C

0

· =

Z

x

x2

2 3/2 1 3 x − x 3 3

0

¸1 = 0

Z h i√x y dx = x2

1

¡√

¢ x − x2 dx

0

2 1 1 − = 3 3 3

6. Let R be the region bounded by the curve x = 3t2 − t3 , y = 6t − 2t2 : y

x

Use Green’s Theorem to find the area of the region R. Answer: First, we need to find the bounds for t. In particular, we want to know when the curve passes through the origin. Setting 3t2 − t3 = 0 and 6t − 2t2 = 0 and solving, we get t = 0 and t = 3. Thus, our bounds are 0 ≤ t ≤ 3. If you plot some points for the parametric equation, you will notice that the parametrization goes clockwise around the region instead of the counterclockwise that is required for Green’s Theorem. This means, we will need to negate the integral. It is okay to not notice this at this point in the problem — if we didn’t notice, then we would get a negative answer, and since the area must be positive, we would realize that the orientation was incorrect and make the answer positive. We need to find a vector field F such that rot(F) = 1. The vector field F = −y i will work (as would x j or lots of other possibilities). 5

Then, by Green’s Theorem: ZZ

I dA =

R

−y dx ∂D

We have: x = 3t2 − t3 y = 6t − 2t2

dx = (6t − 3t2 ) dt dy = (6 − 4t) dt

with 0 ≤ t ≤ 3. Thus: ZZ I dA = −y dx R ∂D Z 3 = (6t − 2t2 )(6t − 3t2 ) dt 0

=

81 10

7. Evaluate the surface integral ZZ z dA S

where S is the surface z =

p

x2 + y 2 , 0 ≤ z ≤ 2.

Answer: First we parameterize the surface. The surface has equation z = r; we will use the parameters t = r and u = θ. Then, the parametric equations for the surface are: x = t cos u y = t sin u z = t

0≤t≤2 0 ≤ u ≤ 2π

The tangent vectors to the surface are: Tt = (cos u, sin u, 1) Tu = (−t sin u, t cos u, 0) 6

And the normal vector to the surface is ¯ ¯ ¯ i j k ¯¯ ¯ sin u 1 ¯¯ Tt × Tu = ¯¯ cos u ¯ −t sin u t cos u 0 ¯ = −t cos u i − t sin u j + t k Thus:

p

t2 cos2 u + t2 sin2 u + t2 dt du √ √ √ = t2 + t2 dt du = 2t2 dt du = t 2 dt du

dA = kTt × Tu k dt du =

Thus: ZZ

Z

2π

Z

z dA = S

Z

2

2π

Z

2

t kTt × Tu k dt du = Z

0

0

2π

= 0

√ √ 8 2 16π 2 du = 3 3

0

√ t2 2 dt du

0

8. Use geometric reasoning to evaluate the following surface integrals: (a) The integral

ZZ p x2 + y 2 + z 2 dA S 2

where S is the surface x + y 2 + z 2 = 4, z ≥ 0. (b) The integral

ZZ (x i + y j) · dA S

where S is the surface x2 + y 2 = 4, 1 ≤ z ≤ 3, oriented with unit normal vectors pointing outwards. (c) The integral

ZZ (z k) · dA S

where S is the disc x2 + y 2 ≤ 9 on the plane z = 4, oriented with unit normal vectors pointing upwards. 7

Answer: (a) The surface is a the top half of a sphere p of radius 2 centered at the origin. On the surface, the value of x2 + y 2 + z 2 is 2. Thus: ZZ p ZZ x2 + y 2 + z 2 dA = 2 dA = 2 (surface area of S) S

S

4πr2 The area of the top half of the sphere is = 2π(22 ) = 8π. 2 Thus: ZZ p x2 + y 2 + z 2 dA = 2(8π) = 16π S

ZZ F · N dA

(b) We want to write the integral as a scalar surface integral S

where N is the unit normal vector to the surface. Since S is a cylinder about the z-axis, the vector x i + y j is normal to S and points in the outward direction. So: xi + yj N= p x2 + y 2 On the surface S, x2 + y 2 = 4, so N=

xi + yj 2

Thus,

x2 + y 2 (x, y, 0) = =2 2 2 Thus, we want to evaluate the integral ZZ 2 dA = 2(surface area of cylinder) F · N = (x, y, 0) ·

S

The surface area of the cylinder is 2πrh = 2π(2)(3 − 1) = 8π. Thus, ZZ (x i + y j) · dA = 2(8π) = 16π S

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ZZ (c) We want to write the integral as a scalar surface integral

F · N dA S

where N is the unit normal vector to the surface. Since S is disc on the plane z = 4, the vector N = k is normal to S and points in the upward direction. So: F · N = (0, 0, z) · (0, 0, 1) = z On the plane z = 4, we have F · N = z = 4. Thus, we want to evaluate the integral ZZ 4 dA = 4(area of the disc) S

The area of the disc is πr2 = π(3)2 = 9π. Thus, ZZ (z k) · dA = 4(9π) = 36π S

9. Let S be the surface defined by z = 5 − x2 − y 2 with z ≥ 1, oriented with downward pointing normal. Consider the following vector field: F = (ez + 1) i + y j + y cos(x2 ) k ZZ Use Stokes’s Theorem to evaluate (∇ × F) · dA. S

Answer: By Stokes’s Theorem, ZZ Z (∇ × F) · dA = S

F · ds ∂S

The boundary of S is defined by the equations 1 = 5−x2 −y 2 , z = 1 (this is the intersection of the surface with the plane z = 1). Thus, boundary is the circle x2 + y 2 = 4 in the plane z = 1. We can parameterize this curve as follows: x = 2 cos t y = 2 sin t z = 1 9

0 ≤ t ≤ 2π

Thus: dx = −2 sin t dt dy = 2 cos t dt dz = 0 Since the surface is oriented with unit normals pointing downwards, the curve needs to be oriented clockwise. We have currently oriented it counterclockwise, so we will need to negate the integral. Now, we can compute the integral: Z Z F · ds = (ez + 1) dx + y dy + y cos(x2 ) dz ∂S ∂S Z 2π ¡ ¢ = − (e + 1)(−2 sin t) + 2 sin t(2 cos t) + 2 sin t cos(4 cos2 t)(0) dt Z0 2π = − (−2(e + 1) sin t + 4 sin t cos t) dt 0 £ ¤2π = − 2(e + 1) cos t + 2 sin2 t 0 =

0

Note: Here is a slightly quicker way to see that the integral is 0. By Stokes’s Theorem, if two surfaces have the same boundary, then the two surface integrals of ∇ × F must be the equal (since both surface integrals equal the line integral on the boundary). In this problem, we could consider the disc inside the circle x2 + y 2 = 4 in the plane z = 1. The surface integral on this disc must be equal to the surface integral that we want to compute. On this disc, z = 1, so F = (e + 1) i + y j + y cos(x2 ) k Also, the unit normal vector to the disc is N = k. And, if we compute (∇ × F) · N we get 0. So the surface integral must be 0.

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10. (a) Find a vector field F such that ∇ × F = x2 k. (b) Use Stokes’s Theorem to evaluate ZZ ¡ 2 ¢ x k · dA S

where S be the surface z = sin(r2 π), 0 ≤ r ≤ 1, oriented with unit normal vectors pointing upwards. Answer: (a) One such vector field is F = −yx2 i. (Another such vector field is 1 3 x j, and there are many more.) 3 (b) By Stokes’s Theorem and part (a): ZZ Z ¡ 2 ¢ (−yx2 i) · ds x k · dA = ∂S

S

The boundary of S is the curve defined by the equations z = sin(π) = 0 and r = 1. This is the circle x2 + y 2 = 1 in the xy-plane. We can parameterize the boundary of S as follows: x = cos t y = sin t z = 0

0 ≤ t ≤ 2π

Thus: dx = − sin t dt dy = cos t dt dz = 0 Since the surface is oriented with unit normals pointing upwards, the curve needs to be oriented counterclockwise, which is the way it is oriented by this parametrization. Thus: Z Z 2π 2 (−yx i) · ds = −yx2 dx ∂S Z0 2π = − sin t cos2 t(− sin t) dt Z0 2π = sin2 t cos2 t dt 0

11

1 (1 − cos(2t)), we have: 2 Z Z 2π Z 2π 1 2 2 2 2 (−yx i) · ds = sin t cos t dt = sin (2t) dt 4 ∂S 0 0 · µ ¶¸2π Z 2π 1 1 1 = (1 − cos(4t)) dt = t − sin(4t) 8 8 4 0 0 π = 4 Since sin(2t) = 2 sin t cos t and sin2 t =

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