Solutions To Selected Problems Chapter N Index

Chapter O

Exercises O.1

1.

When $C$ is a curve in the plane and $f$ is a surface defined over $C$ , then $\int_{C}f(s)\operatorname{d}\!s$ describes the area under the spatial curve that lies on $f$ , over $C$ .
2.

The evaluation is the same. The $\oint$ notation signifies that the curve $C$ is a closed curve, though the evaluation is the same.
3.

The variable $s$ denotes the arc-length parameter, which is generally difficult to use. The Key Idea allows one to parameterize a curve using another, ideally easier-to-use, parameter.
4.

Answers will vary.
5.

$12\sqrt{2}$
6.

$41\sqrt{10}/2$
7.

$40\pi$
8.

$10\pi^{2}$
9.

Over the first subcurve of $C$ , the line integral has a value of $3/2$ ; over the second subcurve, the line integral has a value of $4/3$ . The total value of the line integral is thus $17/6$ .
10.

Over the first subcurve of $C$ , the line integral has a value of $2\sqrt{2}/3$ ; over the second subcurve, the line integral has a value of $\pi-2$ . The total value of the line integral is thus $\pi+2\sqrt{2}/3-2$ .
11.

$1/2$
12.

$(\ln 2)/2$
13.

$23$
14.

$4\pi$
15.

$14/5$
16.

$2\sqrt{2}\pi^{2}$
17.

$(17\sqrt{17}-5\sqrt{5})/3$
18.

$2/5$
19.

$\int_{0}^{1}(5t^{2}+_{2}t+2)\sqrt{(4t+1)^{2}+1}\operatorname{d}\!t\approx 17.071$
20.

$\int_{0}^{\pi}t\sqrt{1+\cos^{2}t}\operatorname{d}\!t\approx 6.001$
21.

$\oint_{0}^{2\pi}\big(10-4\cos^{2}t-\sin^{2}t\big)\sqrt{\cos^{2}t+4\sin^{2}t}% \operatorname{d}\!t\approx 74.986$
22.

$\int_{-1}^{1}\big(3t^{3}+2t+5\big)\sqrt{9t^{4}+1}\operatorname{d}\!t\approx 15% .479$
23.

$7\sqrt{26}/3$
24.

$2\pi$
25.

$8\pi^{3}$
26.

$5/2$
27.

$M=8\sqrt{2}\pi^{2}$ g; center of mass is $(0,-1/(2\pi),8\pi/3)$ .
28.

$M\approx 0.237$ g; center of mass is approximately $(0.173,0.099,0.065)$ .
29.

$24\pi$
30.
31.

Exercises O.2

1.

Answers will vary. Appropriate answers include velocities of moving particles (air, water, etc.); gravitational or electromagnetic forces.
2.

Specific answers will vary, though should relate to the idea that “more of the vector field is moving into that point than out of that point.”
3.

Specific answers will vary, though should relate to the idea that the vector field is spinning clockwise at that point.
4.

No; to be incompressible, the divergence needs to be 0 everywhere, not just at one point.
5.

Correct answers should look similar to
6.

Correct answers should look similar to
7.

Correct answers should look similar to
8.

Correct answers should look similar to
9.

$\operatorname{div}\vec{F}=1+2y$ ; $\operatorname{curl}\vec{F}=0$
10.

$\operatorname{div}\vec{F}=0$ ; $\operatorname{curl}\vec{F}=1+2y$
11.

$\operatorname{div}\vec{F}=x\cos(xy)-y\sin(xy)$ ; $\operatorname{curl}\vec{F}=y\cos(xy)+x\sin(xy)$
12.

$\operatorname{div}\vec{F}=\frac{4}{(x^{2}+y^{2})^{2}}$ ; $\operatorname{curl}\vec{F}=0$
13.

$\operatorname{div}\vec{F}=3$ ; $\operatorname{curl}\vec{F}=\left\langle-1,-1,-1\right\rangle$
14.

$\operatorname{div}\vec{F}=2x+2y+2z$ ; $\operatorname{curl}\vec{F}=\left\langle 2y,2z,2x\right\rangle$
15.

$\operatorname{div}\vec{F}=1+2y$ ; $\operatorname{curl}\vec{F}=0$
16.

$\operatorname{div}\vec{F}=2y$ ; $\operatorname{curl}\vec{F}=0$
17.

$\operatorname{div}\vec{F}=2y-\sin z$ ; $\operatorname{curl}\vec{F}=\vec{0}$
18.

$\operatorname{div}\vec{F}=\frac{2}{(x^{2}+y^{2}+z^{2})^{2}}$ ; $\operatorname{curl}\vec{F}=\vec{0}$
19.
20.
21.
22.
23.
24.
25.
26.

Exercises O.3

1.

False. It is true for line integrals over scalar fields, though.
2.

The input of $\vec{F}$ should be a point in the plane, not a two dimensional vector.
3.

True.
4.

False.
5.

We can conclude that $\vec{F}$ is conservative.
6.

By the Fundamental Theorem of Line Integrals, since $\vec{F}$ is conservative, $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}=f(B)-f(A)$ , where $f$ is a potential function for $\vec{F}$ and $A$ and $B$ are the initial and terminal points of $C$ , respectively. Since $C$ is a closed curve, $A=B$ , and hence $f(B)-f(A)=0$ .
7.

$11/6$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle 3t,t\right\rangle$ on $0\leq t\leq 1$ .)
8.

$5/3$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle t,t^{2}\right\rangle$ on $0\leq t\leq 1$ .)
9.

$0$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle\cos t,\sin t\right\rangle$ on $0\leq t\leq\pi$ .)
10.

$2/5$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle t,t^{3}\right\rangle$ on $-1\leq t\leq 1$ .)
11.

$12$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle 1,2,3\right\rangle+t\left\langle 3,1,-1\right\rangle$ on $0\leq t\leq 1$ .)
12.

$1$ .
13.

$1$
14.

$-2\pi$
15.

$0$
16.

$2\pi$
17.

$-1/30$
18.

$0$
19.

$1/3$
20.

$-1/2$ ; $1/2$
21.

2
22.

$2\pi(\pi-1)$
23.

0
24.

$67/15$
25.

0
26.

$6$
27.

$5/6$ joules. (One parameterization for $C$ is $\vec{r}(t)=\left\langle t,t\right\rangle$ on $0\leq t\leq 1$ .)
28.

$13/15$ joules. (One parameterization for $C$ is $\vec{r}(t)=\left\langle t,\sqrt{t}\right\rangle$ on $0\leq t\leq 1$ .)
29.

$24$ ft-lbs.
30.

$24$ ft-lbs.
31.

$2/3$ joules
32.

$8/15$ joules
33.

No.
34.

Yes. $f(x,y)=x^{2}/2-y^{2}/2$ .
35.

No.
36.

Yes. $f(x,y)=xy^{2}+x^{3}$
37.

No.
38.

Yes. $f(x,y)=4x^{2}y+2y^{2}+3x$
39.

No.
40.

Yes. $f(x,y,z)=ax+by+cz$ .
41.

Yes. $f(x,y,z)=x^{2}/x+xy+z^{3}/3$ .
42.

No
43.
- $f(x,y)=xy+x$
  
  $\operatorname{curl}\vec{F}=0$ .
  
  $1$ . (One parameterization for $C$ is $\vec{r}(t)=\left\langle t,t-1\right\rangle$ on $0\leq t\leq 1$ .)
  
  $1$ ( $f(0,1)=0$ and $f(1,0)=1$ )
44.
- $f(x,y)=x^{2}+xy+y^{2}$
  
  $\operatorname{curl}\vec{F}=0$ .
  
  $0$ .
  
  $0$ ( $f(0,0)=0$ )
45.
- $f(x,y)=x^{2}yz$
  
  $\operatorname{curl}\vec{F}=\vec{0}$ .
  
  $250$ .
  
  $250$ ( $f(1,-1,0)=0$ $f(5,5,2)=250$ )
46.
- $f(x,y)=x^{2}+y^{2}+z^{2}$
  
  $\operatorname{curl}\vec{F}=\vec{0}$ .
  
  $0$ .
  
  $0$ ( $f(1,0,0)=1$ )
47.
- $f(x,y)=xy^{2}+x^{3}/3$
  
  $\operatorname{curl}\vec{F}=\vec{0}$
  
  $0$ .
  
  $0$ ( $f(1,0)=1/3$ )
48.
- $f(x,y)=xy^{2}+x^{3}/3$
  
  $\operatorname{curl}\vec{F}=\vec{0}$
  
  $2/3$ .
  
  $2/3$ ( $f(1,0)=1/3$ , $f(-1,0)=-1/3$ )
49.
- $f(x,y)=x^{2}y^{2}/2$
  
  $\operatorname{curl}\vec{F}=\vec{0}$
  
  $625\pi^{4}/2048$
  
  $625\pi^{4}/2048$ ( $f(0,0)=0$ ,
  $f(-5\pi/4\sqrt{2},-5\pi/4\sqrt{2})=625\pi^{4}/2048$
50.
- $f(x,y)=y^{2}+x^{2}e^{y}$
  
  $\operatorname{curl}\vec{F}=\vec{0}$
  
  $-1$
  
  $-1$ ( $f(1,0)=1$ , $f(0,0)=0$ )
51.

Since $\vec{F}$ is conservative, it is the gradient of some potential function. That is, $\nabla f=\left\langle f_{x},f_{y},f_{z}\right\rangle=\vec{F}=\left\langle M,N,% P\right\rangle$ . In particular, $M=f_{x}$ , $N=f_{y}$ and $P=f_{z}$ .

Note that $\operatorname{curl}\vec{F}=\left\langle P_{y}-N_{z},M_{z}-P_{x},N_{x}-M_{y}% \right\rangle=\left\langle f_{zy}-f_{yz},f_{xz}-f_{zx},f_{yx}-f_{xy}\right\rangle$ , which, by Theorem 13.3.1, is $\left\langle 0,0,0\right\rangle$ .
52.
53.
54.
55.
56.
57.

(b) No. Hint: Think of how $f$ is defined.
58.

Yes. $f(x,y)=axy+bx+cy+d$ .
59.

$0$ ; no
60.

$7/10$ ; $7/10$ ; no

Exercises O.4

1.

along, across
2.

It is the measure of flow around the entirety of a closed curve $C$ .
3.

the curl of $\vec{F}$ , or $\operatorname{curl}\vec{F}$
4.

the divergence of $\vec{F}$ , or $\operatorname{div}\vec{F}$
5.

$\operatorname{curl}\vec{F}$
6.

$\operatorname{div}\vec{F}$
7.

12
8.

$12$
9.

$-2/3$
10.

$10/3$
11.

$1/2$
12.

$1/2$
13.

The line integral $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}$ , over the parabola, is $38/3$ ; over the line, it is $-10$ . The total line integral is thus $38/3-10=8/3$ . The double integral of $\operatorname{curl}\vec{F}=2$ over $R$ also has value $8/3$ .
14.

Both the line integral and double integral have value of $2\pi$ .
15.

Three line integrals need to be computed to compute $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}$ . It does not matter which corner one starts from first, but be sure to proceed around the triangle in a counterclockwise fashion.

From $(0,0)$ to $(2,0)$ , the line integral has a value of 0. From $(2,0)$ to $(1,1)$ the integral has a value of $7/3$ . From $(1,1)$ to $(0,0)$ the line integral has a value of $-1/3$ . Total value is 2.

The double integral of $\operatorname{curl}\vec{F}$ over $R$ also has value 2.
16.

Two line integrals need to be computed to compute $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}$ . Along the parabola, the line integral has value $25.5$ . Along the line, the line integral has value $-21$ . Together, the total value is $4.5$

The double integral of $\operatorname{curl}\vec{F}$ over $R$ also has value $4.5$ .
17.

$16/15$
18.
19.

$-5\pi$
20.
21.

$-1/2$
22.

$1/3$
23.

$-\pi/2$
24.

$-45\pi/32$
25.

0
26.

$0$
27.

$4/3$
28.

$-4/3$
29.

Any choice of $\vec{F}$ is appropriate as long as $\operatorname{curl}\vec{F}=1$ . When $\vec{F}=\left\langle-y/2,x/2\right\rangle$ , the integrand of the line integral is simply 6. The area of $R$ is $12\pi$ .
30.

Any choice of $\vec{F}$ is appropriate as long as $\operatorname{curl}\vec{F}=1$ . The choices of $\vec{F}=\left\langle-y,0\right\rangle$ and $\left\langle 0,x\right\rangle$ each lead to reasonable integrands. The area of $R$ is $4/3$ .
31.

Any choice of $\vec{F}$ is appropriate as long as $\operatorname{curl}\vec{F}=1$ . The choices of $\vec{F}=\left\langle-y,0\right\rangle$ , $\left\langle 0,x\right\rangle$ and $\left\langle-y/2,x/2\right\rangle$ each lead to reasonable integrands. The area of $R$ is $16/15$ .
32.

Any choice of $\vec{F}$ is appropriate as long as $\operatorname{curl}\vec{F}=1$ . The choice of $\vec{F}=\left\langle-y/2,x/2\right\rangle$ leads to a reasonable integrand after simplification. The area of $R$ is $41\pi/10$ .
33.

The line integral $\oint_{C}\vec{F}\cdot\vec{n}\operatorname{d}\!s$ , over the parabola, is $-22/3$ ; over the line, it is $10$ . The total line integral is thus $-22/3+10=8/3$ . The double integral of $\operatorname{div}\vec{F}=2$ over $R$ also has value $8/3$ .
34.

Both the line integral and double integral have value of $0$ .
35.

Three line integrals need to be computed to compute $\oint_{C}\vec{F}\cdot\vec{n}\operatorname{d}\!s$ . It does not matter which corner one starts from first, but be sure to proceed around the triangle in a counterclockwise fashion.

From $(0,0)$ to $(2,0)$ , the line integral has a value of 0. From $(2,0)$ to $(1,1)$ the integral has a value of $1/3$ . From $(1,1)$ to $(0,0)$ the line integral has a value of $1/3$ . Total value is $2/3$ .

The double integral of $\operatorname{div}\vec{F}$ over $R$ also has value $2/3$ .
36.

Two line integrals need to be computed to compute $\oint_{C}\vec{F}\cdot\vec{n}\operatorname{d}\!s$ . Along the parabola, the line integral has value $159/20$ . Along the line, the line integral has value $6$ . Together, the total value is $279/20$ .

The double integral of $\operatorname{div}\vec{F}$ over $R$ also has value $279/20$ .
37.

Exercises O.5

1.

Answers will vary, though generally should meaningfully include terms like “two sided”.
2.

Many possible answers exist; the one given by the book is the Möbius band.
3.
- $\vec{r}(u,v)=\left\langle u,v,3u^{2}v\right\rangle$ on $-1\leq u\leq 1$ , $0\leq v\leq 2$ .
  
  $\vec{r}(u,v)=\\ \langle 3v\cos u+1,3v\sin u+2,$
  $3(3v\cos u+1)^{2}(3v\sin u+2)\rangle$ , on $0\leq u\leq 2\pi$ , $0\leq v\leq 1$ .
  
  $\vec{r}(u,v)=\left\langle u,v(2-2u),3u^{2}v(2-2u)\right\rangle$ on $0\leq u,v\leq 1$ .
  
  $\vec{r}(u,v)=\left\langle u,v(1-u^{2}),3u^{2}v(1-u^{2})\right\rangle$ on $-1\leq u\leq 1$ , $0\leq v\leq 1$ .
4.
- $\vec{r}(u,v)=\left\langle u,v,4u+2u^{2}\right\rangle$ on $1\leq u\leq 4$ , $5\leq v\leq 7$ .
  
  $\vec{r}(u,v)=\\ \left\langle 4v\cos u,3v\sin u,16v\cos u+2(3v\sin u)^{2}\right\rangle$ , on $0\leq u\leq 2\pi$ , $0\leq v\leq 1$ .
  
  $\vec{r}(u,v)=\\ \left\langle u,u+v(4-2u),4u+2\big(u+v(4-2u)\big)^{2}\right\rangle$ on
  $0\leq u\leq 2$ , $0\leq v\leq 1$ .
  
  $\vec{r}(u,v)=\left\langle v\cos u,v\sin u,4v\cos u+2(v\sin u)^{2}\right\rangle$ on $0\leq u\leq 2\pi$ , $2\leq v\leq 5$ .
5.

$\vec{r}(u,v)=\left\langle 0,u,v\right\rangle$ with $0\leq u\leq 2$ , $0\leq v\leq 1$ .
6.

$\vec{r}(u,v)=\left\langle u,0,1-u+vu\right\rangle$ with $0\leq u\leq 1$ , $0\leq v\leq 1$ .
7.

$\vec{r}(u,v)=\left\langle 3\sin u\cos v,2\sin u\sin v,4\cos u\right\rangle$ with $0\leq u\leq\pi$ , $0\leq v\leq 2\pi$ .
8.

Answers may vary; one solution is

$\vec{r}(u,v)=\left\langle v\cos u,v,4v\sin u\right\rangle$ with $0\leq u\leq 2\pi$ , $-1\leq v\leq 5$ .
9.

Answers may vary.

For $z=\frac{1}{2}(3-x)$ : $\vec{r}(u,v)=\left\langle u,v,\frac{1}{2}(3-u)\right\rangle$ , with $1\leq u\leq 3$ and $0\leq v\leq 2$ .

For $x=1$ : $\vec{r}(u,v)=\left\langle 1,u,v\right\rangle$ , with $0\leq u\leq 2$ , $0\leq v\leq 1$

For $y=0$ : $\vec{r}(u,v)=\left\langle u,0,v/2(3-u)\right\rangle$ , with $1\leq u\leq 3$ , $0\leq v\leq 1$

For $y=2$ : $\vec{r}(u,v)=\left\langle u,2,v/2(3-u)\right\rangle$ , with $1\leq u\leq 3$ , $0\leq v\leq 1$

For $z=0$ : $\vec{r}(u,v)=\left\langle u,v,0\right\rangle$ , with $1\leq u\leq 3$ , $0\leq v\leq 2$
10.

Answers may vary.

For $z=2x+4y-4$ : $\vec{r}(u,v)=\left\langle u,1-u/2+uv/2,2u+4(1-u/2+uv/2)-4\right\rangle$ , with $0\leq u\leq 2$ , $0\leq v\leq 1$ .

For $x=2$ : $\vec{r}(u,v)=\left\langle 2,u,4uv\right\rangle$ , with $0\leq u\leq 1$ , $0\leq v\leq 1$

For $y=1$ : $\vec{r}(u,v)=\left\langle u,1,2uv\right\rangle$ , with $0\leq u\leq 2$ , $0\leq v\leq 1$

For $z=0$ : $\vec{r}(u,v)=\left\langle u,1-u/2+uv/2,0\right\rangle$ , with $0\leq u\leq 2$ , $0\leq v\leq 1$
11.

Answers may vary.

For $z=2y:\vec{r}(u,v)=\left\langle u,v(4-u^{2}),2v(4-u^{2})\right\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1$ .

For $y=4-x^{2}:\vec{r}(u,v)=\left\langle u,4-u^{2},2v(4-u^{2})\right\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1$ .

For $z=0$ : $\vec{r}(u,v)=\left\langle u,v(4-u^{2}),0\right\rangle$ with $-2\leq u\leq 2$ and $0\leq v\leq 1$ .
12.

Answers may vary.

For $y=1-z^{2}$ : $\vec{r}(u,v)=\left\langle u,v(1-u^{2}),\sqrt{1-v(1-u^{2})}\right\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1$ .

For $y=1-x^{2}$ : $\vec{r}(u,v)=\left\langle u,1-u^{2},uv\right\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1$ .

For $x=0$ : $\vec{r}(u,v)=\left\langle 0,v(1-u^{2}),u\right\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1$ .

For $y=0$ : $\vec{r}(u,v)=\left\langle u,0,v\right\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1$ .

For $z=0$ : $\vec{r}(u,v)=\left\langle u,v(1-u^{2}),0\right\rangle$ with $0\leq u\leq 1$ and $0\leq v\leq 1$ .
13.

Answers may vary.

For $x^{2}+y^{2}/9=1$ : $\vec{r}(u,v)=\left\langle\cos u,3\sin u,v\right\rangle$ with $0\leq u\leq 2\pi$ and $1\leq v\leq 3$ .

For $z=1$ : $\vec{r}(u,v)=\left\langle v\cos u,3v\sin u,1\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .

For $z=3$ : $\vec{r}(u,v)=\left\langle v\cos u,3v\sin u,3\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .
14.

Answers may vary.

For $x^{2}+y^{2}=(z-1)^{2}$ : $\vec{r}(u,v)=\left\langle v\cos u,v\sin u,1-v\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .

For $z=0$ : $\vec{r}(u,v)=\left\langle v\cos u,v\sin u,0\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .
15.

Answers may vary.

For $z=1-x^{2}$ : $\vec{r}(u,v)=\left\langle u,v,1-u^{2}\right\rangle$ with $-1\leq u\leq 1$ and $-1\leq v\leq 2$ .

For $y=-1$ : $\vec{r}(u,v)=\left\langle u,-1,v(1-u^{2})\right\rangle$ with $-1\leq u\leq 1$ and $0\leq v\leq 1$ .

For $y=2$ : $\vec{r}(u,v)=\left\langle u,2,v(1-u^{2})\right\rangle$ with $-1\leq u\leq 1$ and $0\leq v\leq 1$ .

For $z=0$ : $\vec{r}(u,v)=\left\langle u,v,0\right\rangle$ with $-1\leq u\leq 1$ and $-1\leq v\leq 2$ .
16.

Answers may vary.

For $z=4-x^{2}-4y^{2}$ : $\vec{r}(u,v)=\left\langle 2v\cos u,v\sin u,4-(2v\cos u)^{2}-4(v\sin u)^{2}\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .

For $z=0$ : $\vec{r}(u,v)=\left\langle 2v\cos u,v\sin u,0\right\rangle$ with $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ .
17.

$S=2\sqrt{14}$ .
18.

$S=\sqrt{6}/2$ .
19.

$S=4\sqrt{3}\pi$ .
20.

$S=3\sqrt{3}\pi$ .
21.

$4\pi r^{2}$
22.

$\pi R\sqrt{h^{2}+R^{2}}$
23.
24.
25.

$S=\int_{0}^{3}\int_{0}^{2\pi}\sqrt{v^{2}+4v^{4}}\operatorname{d}\!u% \operatorname{d}\!v=(37\sqrt{37}-1)\pi/6\approx 117.319$ .
26.

$S=\int_{0}^{1}\int_{0}^{1}\sqrt{v^{2}+4u^{2}v^{2}+4v^{4}}\operatorname{d}\!u% \operatorname{d}\!v\approx 0.931$ .
27.

$S=\int_{-1}^{1}\int_{0}^{1-v^{2}}\sqrt{27}\operatorname{d}\!u\operatorname{d}% \!v=4\sqrt{3}\approx 6.928$ .
28.

$S=\int_{0}^{1}\int_{0}^{2\pi}\sqrt{v^{2}+4v^{4}}\operatorname{d}\!u% \operatorname{d}\!v=(5\sqrt{5}-1)\pi/6\approx 5.330$ .

Exercises O.6

1.

curve; surface
2.

Answers will vary; in general, it means that more of the vector field passes through the surface opposite the direction of the normal vector than in the same direction of the normal vector.
3.

outside
4.

0
5.

$240\sqrt{3}$
6.

$40\pi$
7.

$24$
8.

$15$
9.

$0$
10.

$0$
11.

$-1/2$
12.

$\pi$
13.

$0$ ; the flux over $\mathcal{S}_{1}$ is $-45\pi$ and the flux over $\mathcal{S}_{2}$ is $45\pi$ .
14.

$9\pi/8$ ; the flux over $\mathcal{S}_{1}$ is $3\pi/4$ (use $\vec{r}(u,v)=\left\langle\sin u\cos v,\sin u\sin v,\cos u\right\rangle$ on $\pi/3\leq u\leq\pi$ , $0\leq v\leq 2\pi$ ) and the flux over $\mathcal{S}_{2}$ is $3\pi/8$ (use $\vec{r}(u,v)=\left\langle v\sqrt{3}\cos(u)/2,v\sqrt{3}\sin(u)/2,1/2\right\rangle$ for $0\leq u\leq 2\pi$ , $0\leq v\leq 1$ .
15.

$3$
16.

$7/4$

Exercises O.7

1.

Answers will vary; in Section 15.4, the Divergence Theorem connects outward flux over a closed curve in the plane to the divergence of the vector field, whereas in this section the Divergence Theorem connects outward flux over a closed surface in space to the divergence of the vector field.
2.

Divergence.
3.

Curl.
4.

Green’s Theorem.
5.

Outward flux across the plane $z=2-x/2-2y/3$ is 22; across the plane $z=0$ the outward flux is $-8$ ; across the planes $x=0$ and $y=0$ the outward flux is 0.

Total outward flux: $14$ .

$\iint_{D}\operatorname{div}\vec{F}\operatorname{d}\!V=\int_{0}^{4}\int_{0}^{3-% 3x/4}\int_{0}^{2-x/2-2y/3}(2x+2y)\operatorname{d}\!z\operatorname{d}\!y% \operatorname{d}\!x=14$ .
6.

Outward flux across the cylinder $x^{2}+y^{2}=1$ is 0; across the plane $z=3$ the outward flux is $3\pi$ ; across the plane $z=-3$ the outward flux is $3\pi$ .

Total outward flux: $6\pi$ .

$\iint_{D}\operatorname{div}\vec{F}\operatorname{d}\!V=\int_{0}^{2\pi}\int_{0}^% {1}\int_{-3}^{3}r\operatorname{d}\!z\operatorname{d}\!r\operatorname{d}\!% \theta=6\pi$ .
7.

Outward flux across the surface $z=xy(3-x)(3-y)$ is 252; across the plane $z=0$ the outward flux is $-9$ .

Total outward flux: $243$ .

$\iint_{D}\operatorname{div}\vec{F}\operatorname{d}\!V=\int_{0}^{3}\int_{0}^{3}% \int_{0}^{xy(3-x)(3-y)}12\operatorname{d}\!z\operatorname{d}\!y\operatorname{d% }\!x=243$ .
8.

Outward flux across the paraboloid is $112\pi/3$ ; across the disk the outward flux is 0.

Total outward flux: $112\pi/3$ .

$\iint_{D}\operatorname{div}\vec{F}\operatorname{d}\!V=\int_{0}^{2\pi}\int_{0}^% {2}\int_{0}^{4-r^{2}}(2z+2)r\operatorname{d}\!z\operatorname{d}\!r% \operatorname{d}\!\theta=112\pi/3$ .
9.

With an upward normal, both integrals are $-3\pi$ .
10.
11.

Circulation on $C$ : $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}=\pi$

$\iint_{\mathcal{S}}\bigl(\operatorname{curl}\vec{F}\bigr)\cdot\vec{n}% \operatorname{d}\!S=\pi$ .
12.

Circulation on $C$ : $\oint_{C}\vec{F}\cdot\operatorname{d}\!\vec{r}=\pi$

$\iint_{\mathcal{S}}\bigl(\operatorname{curl}\vec{F}\bigr)\cdot\vec{n}% \operatorname{d}\!S=\pi$ .
13.

Circulation on $C$ : The flow along the line from $(0,0,2)$ to $(4,0,0)$ is 0; from $(4,0,0)$ to $(0,3,0)$ it is $-6$ , and from $(0,3,0)$ to $(0,0,2)$ it is 6. The total circulation is $0+(-6)+6=0$ .

$\iint_{\mathcal{S}}\bigl(\operatorname{curl}\vec{F}\bigr)\cdot\vec{n}% \operatorname{d}\!S=\iint_{\mathcal{S}}0\operatorname{d}\!S=0$ .
14.

Circulation on $C$ : The flow along the parabola is $-32/15$ ; the flow along the line is $4/3$ . The total circulation is $4/3-32/15=-4/5$ .

$\iint_{\mathcal{S}}\bigl(\operatorname{curl}\vec{F}\bigr)\cdot\vec{n}% \operatorname{d}\!S=-4/5$ .
15.

$216\pi$
16.

3
17.

$12\pi/5$
18.

$0$
19.

$128/225$
20.

$8$
21.

$8192/105\approx 78.019$
22.

$64/3$
23.

$5/3$
24.

$8\pi$
25.

$23\pi$
26.

$0$
27.

Each field has a divergence of 1; by the Divergence Theorem, the total outward flux across $\mathcal{S}$ is $\iint_{D}1\operatorname{d}\!S$ for each field.
28.
- $\operatorname{curl}\vec{F}=1$ .
  
  $\operatorname{curl}\vec{F}\cdot\vec{n}=1$ , where $\vec{n}$ is a unit vector normal to $\mathcal{S}$ .
29.

Answers will vary. Often the closed surface $\mathcal{S}$ is composed of several smooth surfaces. To measure total outward flux, this may require evaluating multiple double integrals. Each double integral requires the parameterization of a surface and the computation of the cross product of partial derivatives. One triple integral may require less work, especially as the divergence of a vector field is generally easy to compute.
30.

Answers will vary. Often the closed curve $C$ is composed of several smooth curves. To measure the total circulation, one may have to evaluate line integrals along each curve. Each line integral requires the parameterization of its curve. It may be less work to evaluate one single double (i.e., surface) integral.
31.
32.
33.
34.