Chapter 15

When $C$ is a curve in the plane and $f$ is a surface defined over $C$, then ${\int }_{C}f\left(s\right)𝑑s$ describes the area under the spatial curve that lies on $f$, over $C$.

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

$12\sqrt{2}$

$40\pi$

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 $17/6$.

$1/2$

$23$

$14/5$

$\left(17\sqrt{17}-5\sqrt{5}\right)/3$

${\int }_0^1\left(5t^2{+}_{2}t+2\right)\sqrt{{\left(4t+1\right)}^2+1}𝑑t\approx 17.071$

${\oint }_0^{2\pi }\left(10-4{\cos }^{2}t-{\sin }^{2}t\right)\sqrt{{\cos }^{2}t+4{\sin }^{2}t}𝑑t\approx 74.986$

$7\sqrt{26}/3$

$8{\pi }^3$

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

$24\pi$

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

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

Correct answers should look similar to

Correct answers should look similar to

$\mathrm{div}\overrightarrow{F}=1+2y$; $\mathrm{curl}\overrightarrow{F}=0$

$\mathrm{div}\overrightarrow{F}=x\cos \left(xy\right)-y\sin \left(xy\right)$; $\mathrm{curl}\overrightarrow{F}=y\cos \left(xy\right)+x\sin \left(xy\right)$

$\mathrm{div}\overrightarrow{F}=3$; $\mathrm{curl}\overrightarrow{F}=⟨-1,-1,-1⟩$

$\mathrm{div}\overrightarrow{F}=1+2y$; $\mathrm{curl}\overrightarrow{F}=0$

$\mathrm{div}\overrightarrow{F}=2y-\sin z$; $\mathrm{curl}\overrightarrow{F}=\overrightarrow{0}$

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

True.

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

$11/6$. (One parametrization for $C$ is $\overrightarrow{r}\left(t\right)=⟨3t,t⟩$ on $0\le t\le 1$.)

$0$. (One parametrization for $C$ is $\overrightarrow{r}\left(t\right)=⟨\cos t,\sin t⟩$ on $0\le t\le \pi$.)

$12$. (One parametrization for $C$ is $\overrightarrow{r}\left(t\right)=⟨1,2,3⟩+t⟨3,1,-1⟩$ on $0\le t\le 1$.)

$1$

$0$

$-1/30$

$1/3$

2

$5/6$ joules. (One parametrization for $C$ is $\overrightarrow{r}\left(t\right)=⟨t,t⟩$ on $0\le t\le 1$.)

$24$ ft-lbs.

$2/3$ joules

No.

No.

No.

No.

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

Since $\overrightarrow{F}$ is conservative, it is the gradient of some potential function. That is, $\nabla f=⟨f_x,f_y,f_z⟩=\overrightarrow{F}=⟨M,N,P⟩$. In particular, $M=f_x$, $N=f_y$ and $P=f_z$.

Note that $\mathrm{curl}\overrightarrow{F}=⟨P_y-N_z,M_z-P_x,N_x-M_y⟩=⟨f_{zy}-f_{yz},f_{xz}-f_{zx},f_{yx}-f_{xy}⟩$, which, by Theorem 13.3.1, is $⟨0,0,0⟩$.

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

$0$; no

along, across

the curl of $\overrightarrow{F}$, or $\mathrm{curl}\overrightarrow{F}$

$\mathrm{curl}\overrightarrow{F}$

12

$-2/3$

$1/2$

The line integral ${\oint }_C\overrightarrow{F}\cdot 𝑑\overrightarrow{r}$, 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 $\mathrm{curl}\overrightarrow{F}=2$ over $R$ also has value $8/3$.

Three line integrals need to be computed to compute ${\oint }_C\overrightarrow{F}\cdot 𝑑\overrightarrow{r}$. 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 $\mathrm{curl}\overrightarrow{F}$ over $R$ also has value 2.

$16/15$

$-5\pi$

$-1/2$

$-\pi /2$

$4/3$

Any choice of $\overrightarrow{F}$ is appropriate as long as $\mathrm{curl}\overrightarrow{F}=1$. When $\overrightarrow{F}=⟨-y/2,x/2⟩$, the integrand of the line integral is simply 6. The area of $R$ is $12\pi$.

Any choice of $\overrightarrow{F}$ is appropriate as long as $\mathrm{curl}\overrightarrow{F}=1$. The choices of $\overrightarrow{F}=⟨-y,0⟩$, $⟨0,x⟩$ and $⟨-y/2,x/2⟩$ each lead to reasonable integrands. The area of $R$ is $16/15$.

The line integral ${\oint }_C\overrightarrow{F}\cdot \overrightarrow{n}𝑑s$, 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 $\mathrm{div}\overrightarrow{F}=2$ over $R$ also has value $8/3$.

Three line integrals need to be computed to compute ${\oint }_C\overrightarrow{F}\cdot \overrightarrow{n}𝑑s$. 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 $\mathrm{div}\overrightarrow{F}$ over $R$ also has value $2/3$.

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

$\overrightarrow{r}(u,v)=⟨0,u,v⟩$ with $0\le u\le 2$, $0\le v\le 1$.

$\overrightarrow{r}(u,v)=⟨3\sin u\cos v,2\sin u\sin v,4\cos u⟩$ with $0\le u\le \pi$, $0\le v\le 2\pi$.

Answers may vary.

For $z=\frac{1}{2}\left(3-x\right)$: $\overrightarrow{r}(u,v)=⟨u,v,\frac{1}{2}\left(3-u\right)⟩$, with $1\le u\le 3$ and $0\le v\le 2$.

For $x=1$: $\overrightarrow{r}(u,v)=⟨0,u,v⟩$, with $0\le u\le 2$, $0\le v\le 1$

For $y=0$: $\overrightarrow{r}(u,v)=⟨u,0,v/2\left(3-u\right)⟩$, with $1\le u\le 3$, $0\le v\le 1$

For $y=2$: $\overrightarrow{r}(u,v)=⟨u,2,v/2\left(3-u\right)⟩$, with $1\le u\le 3$, $0\le v\le 1$

For $z=0$: $\overrightarrow{r}(u,v)=⟨u,v,0⟩$, with $1\le u\le 3$, $0\le v\le 2$

Answers may vary.

For $z=2y:\overrightarrow{r}(u,v)=⟨u,v\left(4-u^2\right),2v\left(4-u^2\right)⟩$ with $-2\le u\le 2$ and $0\le v\le 1$.

For $y=4-x^2:\overrightarrow{r}(u,v)=⟨u,4-u^2,2v\left(4-u^2\right)⟩$ with $-2\le u\le 2$ and $0\le v\le 1$.

For $z=0$: $\overrightarrow{r}(u,v)=⟨u,v\left(4-u^2\right),0⟩$ with $-2\le u\le 2$ and $0\le v\le 1$.

Answers may vary.

For $x^2+y^2/9=1$: $\overrightarrow{r}(u,v)=⟨\cos u,3\sin u,v⟩$ with $0\le u\le 2\pi$ and $1\le v\le 3$.

For $z=1$: $\overrightarrow{r}(u,v)=⟨v\cos u,3v\sin u,1⟩$ with $0\le u\le 2\pi$ and $0\le v\le 1$.

For $z=3$: $\overrightarrow{r}(u,v)=⟨v\cos u,3v\sin u,3⟩$ with $0\le u\le 2\pi$ and $0\le v\le 1$.

Answers may vary.

For $z=1-x^2$: $\overrightarrow{r}(u,v)=⟨u,v,1-u^2⟩$ with $-1\le u\le 1$ and $-1\le v\le 2$.

For $y=-1$: $\overrightarrow{r}(u,v)=⟨u,-1,v\left(1-u^2\right)⟩$ with $-1\le u\le 1$ and $0\le v\le 1$.

For $y=2$: $\overrightarrow{r}(u,v)=⟨u,2,v\left(1-u^2\right)⟩$ with $-1\le u\le 1$ and $0\le v\le 1$.

For $z=0$: $\overrightarrow{r}(u,v)=⟨u,v,0⟩$ with $-1\le u\le 1$ and $-1\le v\le 2$.

$S=2\sqrt{14}$.

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

$4\pi r^2$

$S={\int }_0^3{\int }_0^{2\pi }\sqrt{v^2+4v^4}𝑑u𝑑v=\left(37\sqrt{37}-1\right)\pi /6\approx 117.319$.

$S={\int }_0^1{\int }_{-1}^1\sqrt{{\left(5u^2-2uv-5\right)}^2+u^4+{\left(1-u^2\right)}^2}𝑑u𝑑v\approx 7.084$.

curve; surface

outside

$240\sqrt{3}$

$24$

$0$

$-1/2$

$0$; the flux over ${𝒮}_1$ is $-45\pi$ and the flux over ${𝒮}_2$ is $45\pi$.

$3$

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.

Curl.

Circulation on $C$: ${\oint }_C\overrightarrow{F}\cdot 𝑑\overrightarrow{r}=\pi$

${\iint }_{𝒮}\left(\mathrm{curl}\overrightarrow{F}\right)\cdot \overrightarrow{n}𝑑S=\pi$.

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 $0+\left(-6\right)+6=0$.

${\iint }_{𝒮}\left(\mathrm{curl}\overrightarrow{F}\right)\cdot \overrightarrow{n}𝑑S={\iint }_{𝒮}0𝑑S=0$.

$216\pi$

$12\pi /5$

$128/225$

$8192/105\approx 78.019$

$5/3$

$23\pi$

Each field has a divergence of 1; by the Divergence Theorem, the total outward flux across $𝒮$ is ${\iint }_{D}1𝑑S$ for each field.

Answers will vary. Often the closed surface $𝒮$ is composed of several smooth surfaces. To measure total outward flux, this may require evaluating multiple double integrals. Each double integral requires the parametrization 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.