Exercise Answer

Exercise Answer1.1 1.

1.1 (1) $2\boldsymbol{A} = 2(-\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\boldsymbol{k}) = -2\:\boldsymbol{i} + 4\:\boldsymbol{j} + 6\:\boldsymbol{k}$. (2) $3\boldsymbol{B} - 2\boldsymbol{A} = 3(-3\:\boldsymbol{j} + 2\:\boldsymbol{k}) -... ...+ 2\:\boldsymbol{j} + 3\:\boldsymbol{k} = 2\:\boldsymbol{i} -13\:\boldsymbol{j}$. (3) $\vert\boldsymbol{A}\vert = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{14}.$ (4)

$$\vert\boldsymbol{A} + \boldsymbol{B}\vert = \vert-\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\boldsymbol{k} -3\:\boldsymbol{j} + 2\:\boldsymbol{k}\vert$$

$$= \vert-\boldsymbol{i} - \:\boldsymbol{j} + 5\:\boldsymbol{k}\vert = \sqrt{1+1+25} = \sqrt{27} = 3\sqrt{3}$$

1.2 (1) $\boldsymbol{A} + \boldsymbol{B} = -\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\bol... ...bol{j} + \:\boldsymbol{k} = -2\boldsymbol{i} +4\boldsymbol{j} + 4\boldsymbol{k}$ (2)

$$\boldsymbol{A} \cdot \boldsymbol{B} = \vert\boldsymbol{A}\vert\vert\boldsymbol{B}\vert\cos{\theta} = \sqrt{1+4+9}\sqrt{4+16+16}\cos{\theta}$$

$$= 6\sqrt{14}\cos{\theta}$$

Note that $\boldsymbol{A} \cdot \boldsymbol{B} = -2+8+12 = 18$. Thus,

$$\cos{\theta} = \frac{18}{6\sqrt{14}} = \frac{3}{\sqrt{14}}$$

Therefore, $\theta = \cos^{-1}{\frac{3}{\sqrt{14}}}$.

Exercise Answer1.2

(1)

$$\boldsymbol{A} \cdot \boldsymbol{B} = \vert\boldsymbol{A}\vert\vert\boldsymbol{B}\vert\cos{\theta} = \sqrt{1+9+1}\sqrt{4+16+4}\cos{\theta}$$

$$= \sqrt{11}\sqrt{24}\cos{\theta}$$

Note that $\boldsymbol{A} \cdot \boldsymbol{B} = -2+12-2 = 8$. Thus,

$$\cos{\theta} = \frac{8}{\sqrt{11}\sqrt{24}} = \frac{4}{\sqrt{66}}$$

Therefore, $\theta = \cos^{-1}{\frac{4}{\sqrt{66}}}$.

(2) The unit vector ${\bf u}$ with the direction of $\boldsymbol{A}$ is

$${\bf u} = \frac{-\:\boldsymbol{i} + 3\:\boldsymbol{j} + \:\boldsy... ...}} = \frac{-\:\boldsymbol{i} + 3\:\boldsymbol{j} + \:\boldsymbol{k}}{\sqrt{11}}$$

Exercise Answer1.3

1.3 1.

$${}^t[1\ 2\ 1] \times {}^t[2\ -1\ -2] = \Vert\begin{array}{lll}{\bf i} & {\bf j} &{\bf k}\\ 1 & 2 & 1\\ 2 & -1 & -2 \end{array}\Vert$$

$$= -3{\bf i}+4{\bf j} -5{\bf k}$$

2. (1)

$$\left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 2 & -3 & -1\\ 1 & 4& -2 \end{array}\right\vert = 10{\bf i} + 3{\bf j} + 11{\bf k}$$

(2)

$$(2{\bf A} - 3{\bf B}) \times ({\bf A} + 2{\vert bf B}) = ({\bf i} - 18{\bf j} + 4{\bf k}) \times (4{\bf i} + 5{\bf j} - 5{\bf k})$$

$$= \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 1 & -18 & 4\\ 4 & 5 & -5 \end{array}\right\vert = 70{\bf i} + 21{\bf j} + 77{\bf k}$$

(1) The area of parallelogram is

$$\vert{\bf A} \times {\bf B}\vert = \left\vert\begin{array}{rrr} \... ...ht\vert = \vert-3{\bf i} +4 {\bf j} -5{\bf k}\vert = \sqrt{9+16+25} = \sqrt{50}$$

(2) $\left\vert\begin{array}{rrr} \:\boldsymbol{i} & \:\boldsymbol{j} & \:\boldsymbo... ...= -3\:\boldsymbol{i} + 4\:\boldsymbol{j} + -5\:\boldsymbol{k} = {}^t[-3\ 4\ -5]$

Exercise Answer1.4 1.

$$\vert{\bf r}\vert\cos{\frac{\pi}{4}}{\bf i} + \vert{\vert bf r}\v... ...\bf r}\vert\cos{\frac{\pi}{6}} = 3\sqrt{2}{\bf i} + 3{\bf j} + 3\sqrt{3}{\bf k}$$

Exercise Answer1.5 1. $\boldsymbol{A}\cdot(\boldsymbol{B}\times\boldsymbol{C}) = \left\vert\begin{arra... ...}1 & 2 & -1\\ 2 & -1 & -1\\ -1 & 3 & 4 \end{array}\right\vert = -20 \neq 0$．Thus they are not coplanar．

2.

$${\bf B} \times {\bf C} = \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 2 & -... ... -1 & 1 & 2 \end{array}\right\vert = (-2-1){\bf i} -(4+1){\bf j} + (2-1){\bf k}$$

$$= -3{\bf i} - 5{\bf j} + {\bf k}$$

Then

$${\bf A} \cdot ({\bf B} \times {\bf C}) = ({\bf i} + 2{\bf j} + {\bf k})\cdot (-3{\bf i} - 5{\bf j} + {\bf k}) = -12$$

3. Note that $\vert{\bf A} \times {\bf B}\vert^2 = ({\bf A} \times {\bf B}) \cdot ({\bf A} \times {\bf B})$. Let ${\bf C} = {\bf A} \times {\bf B}$. Then

$${\bf C} = (A_2 B_3 - A_3B_2){\bf i} - (A_1 B_3 - A_3 B_1){\bf j} + (A_1B_2 0 A_2 B_1){\bf k}$$

Thus,

$$\vert{\bf A} \times {\bf B}\vert^2 = (A_2 B_3 - A_3B_2)^2 + (A_1 B_3 - A_3B_1)^2 + (A_1 B_2 - A_2 B_1)^2$$

$$= A_2^2 B_3^2 + A_3^2 B_2^2 + A_1^2B_3^2 + A_3^2B_1^2+A_1^2 B_2^2 + A_2^2 B_1^2 - (2A_2A_3B_2B_3 + 2A_1 A_3B_1B_3 + 2A_1 A_2 B_1B_2)$$

$$= (A_1^2 + A_2^2 + A_3^2)(B_1^2 + B_2^2 + B_3^2) - (A_1^2B_1^2 + A_2^2B_2^2 + A_3^2 B_3^2 + 2A_2A_3B_2B_3 + 2A_1 A_3B_1B_3 + 2A_1 A_2 B_1B_2)$$

$$= ({\bf A} \cdot {\bf A})({\bf B} \cdot {\bf B}) - ({\bf A} \cdot {\bf B})^2$$

Exercise Answer2.1 1. The components of $\boldsymbol{F}(t)$ are $x = t\cos{t}, y = t\sin{t}, z = t^2$. Then $z = x^2 + y^2$ and the trace of $\boldsymbol{F}(t)$, $(x(t),y(t),z(t))$ is on the paraboloid $z = x^2 + y^2$.

2. Differentiate eachcomponents, we have

$$\boldsymbol{F}^{\prime}(t) = 2t\:\boldsymbol{i} + \:\boldsymbol{j} + 3t^{2}\:\boldsymbol{k}$$

3. $F' = (5t^2)'\:\boldsymbol{i} + (t)'\:\boldsymbol{j} + - (t^2)' \:\boldsymbol{k}... ... (\cos{t})'\:\boldsymbol{j} = \cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j}$.

(1)

$$(\boldsymbol{F}\cdot \mbox{\boldmath$G$} )' = \boldsymbol{F}'\cdot \mbox{\boldmath$G$} + \boldsymbol{F}\cdot \mbox{\boldmath$G$} '$$

$$= (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\c... ... - t^2\:\boldsymbol{k})\cdot(\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$$

$$= (5t^2 - 1)\cos{t} + 11\sin{t}$$

(2)

$$(\boldsymbol{F}\times \mbox{\boldmath$G$} )' = \boldsymbol{F}'\times \mbox{\boldmath$G$} + \boldsymbol{F}\times \mbox{\boldmath$G$} '$$

$$= (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\t... ... t^2\:\boldsymbol{k})\times (\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$$

$$= \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...oldsymbol{k}\\ 5t^2 & t & -t^2\\ \cos{t} & \sin{t} & 0 \end{array}\right\vert$$

$$= (t^2 \sin{t} - 2t \cos{t})\:\boldsymbol{i} - (t^2 \cos{t} + 2t \sin{t})\:\boldsymbol{j} + (5t^2 \sin{t} - \sin{t} - 11t\cos{t})\:\boldsymbol{k}$$

4. Let $\vert\boldsymbol{F}(t)\vert = c$. Then $\vert\boldsymbol{F}(t)\vert^2 = \boldsymbol{F}(t) \cdot\boldsymbol{F}(t) = c^{2}$. Thus, by the derivative of the vector function, we have

$$(\boldsymbol{F}(t) \cdot\boldsymbol{F}(t))^{\prime} = 2\boldsymbol{F}^{\prime}(t) \cdot\boldsymbol{F}(t) = 0$$

Thus the inner product is 0 and $\boldsymbol{F}(t)$ and $\boldsymbol{F}^{\prime}(t)$ are orthogonal.

5. $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \boldsymbol{F}\cdot\boldsymbol{F} - \int\boldsymbol{F}'\cdot\boldsymbol{F}\;dt$．Therefore,， $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \frac{1}{2}\boldsymbol{F}\cdot\boldsymbol{F}$.

6. $(\boldsymbol{F}\cdot \boldsymbol{F})' = 2\boldsymbol{F}' \cdot \boldsymbol{F}$. Then

$$\int \boldsymbol{F}\cdot \boldsymbol{F'} dt = \frac{1}{2}\int \boldsymbol{R} \cdot \boldsymbol{R} dt = \frac{1}{2}\vert\boldsymbol{F}\vert^2$$

Thus

$$\int_{2}^{3}\boldsymbol{F} \cdot \boldsymbol{F'}dt = \frac{1}{2}\... ...boldsymbol{F} \mid_{2}^{3} = \frac{1}{2}(16+4+9 - (4+1 +4)) = \frac{20}{2} = 10$$

Exercise Answer2.2 1. The straight line to be found has a starting point of $(-1,0,2)$ and a direction of ${}^t [1,4,3]-{}^t [-1,0,2] = {}^t [2,4,1]$. Let the arbitrary point on the line as $(x,y,z)$. Then，

$$(x,y,z) = (-1,0,2) + (2,4,1)t, t \in R$$

Solve this for $t$, we have

$$\frac{x+1}{2} = \frac{y}{4} = \frac{z-2}{1}$$

3. The straight line to be found has a starting point of $(-1,0,2)$ and a direction of ${}^t [1,4,3]-{}^t [-1,0,2] = {}^t [2,4,1]$. Let the arbitrary point on the line as $(x,y,z)$. Then，

$$(x,y,z) = (-1,0,2) + (2,4,1)t, t \in R$$

Solve this for $t$, we have

$$\frac{x+1}{2} = \frac{y}{4} = \frac{z-2}{1}$$

4. Since $x = \cos{t}, y = \sin{t}, z = t$，we have $x^2 + y^2 = 1$. Thus, $\boldsymbol{r}(t)$ is a smooth curve that spirally rotates around a cylinder with a radius of 1.

$$\frac{d \boldsymbol{r}}{dt} = -\sin{t}\:\boldsymbol{i} + \cos{t}\:\boldsymbol{j} + \:\boldsymbol{k}$$

implies

$$\vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{(-\sin{t})^{2} + (\cos{t})^{2} + 1} = \sqrt{1 + 1} = \sqrt{2}$$

Therefore,

$$s = \int_{0}^{2\pi}\sqrt{2}dt = 2\sqrt{2}\pi$$

5.

$$\boldsymbol{r}(t) = (\cos\pi t, \sin\pi t, t)$$

$${\bf v}(t) = \boldsymbol{r}'(t) = (-\pi\sin\pi t, \pi\cos\pi t, 1)$$

$$\boldsymbol{A}(t) = \boldsymbol{r}''(t) = (-\pi^{2}\cos\pi t, -\pi^{2}\sin\pi t, 0)$$

Then for $t = 1$,

$${\bf v}(1) = (0, \pi, 1)$$

$$\boldsymbol{A}(1) = (\pi^{2}, 0, 0)$$

Then

$$v = \vert{\bf v}(1)\vert = \sqrt{\pi^{2} + 1}$$

$${\bf t} = \frac{{\bf v}(1)}{\vert{\bf v}(1)\vert} = \frac{(0, \pi, 1)}{\sqrt{\pi^{2} + 1}}$$

Now we have many ways to find $\boldsymbol{n}$. Here we will consider a method that is easy to calculate．

$$\boldsymbol{A} = a_{{\bf t}} + a_{\boldsymbol{n}},a_{{\bf t}} = (\boldsymbol{A}\cdot{\bf t}){\bf t} = 0$$

implies

$$a_{\boldsymbol{n}} = \boldsymbol{A} - \boldsymbol{A}_{{\bf t}} = (\pi^{2}, 0, 0)$$

Thus，

$$\boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1,0, )$$

Other way is

$$v = \vert{\bf v}(t)\vert = \sqrt{\pi^{2}\sin^{2}\pi t + \pi^{2}\cos^{2}\pi t + 1} = \sqrt{\pi^{2} + 1}$$

$$a_{{\bf t}} = \frac{dv}{dt} = 0$$

Thus, $${\boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1, 0, 0)}$$

(a) $${\frac{d \boldsymbol{r}}{dt} = 4a \sqrt{1 - t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}$$ implies

$$\vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{16a^2 (1- t^2) + 16a^2 t^2 + 16a^2} = \sqrt{32 a^2} = 4\sqrt{2} a$$

Therefore，

$$s = \int_{t_{1}}^{t_{2}} \vert \frac{d \boldsymbol{r}}{dt}\vert dt = 4\sqrt{2} a (t_{2} - t_{1})$$

(b)

$${\bf t} = \frac{d \boldsymbol{r}}{ds} = \frac{d \boldsymbol{r}/dt... ...-t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}{4\sqrt{2} a}$$

(c) $${\frac{d {\bf t}}{dt} = \frac{1}{\sqrt{2}}(\frac{-t}{\sqrt{1 - t^2}}\:\boldsymbol{i} + \boldsymbol{j}), \frac{ds}{dt} = 4\sqrt{2} a}$$ implies

$$\kappa = \vert\frac{d {\bf t}}{ds}\vert = \vert\frac{d {\bf t}/dt... ...vert = \frac{1}{8a}\sqrt{\frac{t^2}{1 - t^2} + 1} = \frac{1}{8a \sqrt{1 - t^2}}$$

Also， $${\boldsymbol{n} = \frac{1}{\kappa}\frac{d {\bf t}}{ds}}$$ implies

$$\boldsymbol{n} = -t\:\boldsymbol{i} + \sqrt{1 - t^2}\:\boldsymbol{j}$$

(d)

$$\boldsymbol{B} = {\bf t} \times \boldsymbol{n} = \frac{1}{\sqrt{2... ...{\sqrt{2}}(-\sqrt{1 - t^2}\:\boldsymbol{i} -t\:\boldsymbol{j} + \boldsymbol{k}$$

Also，

$$\frac{d \boldsymbol{B}}{ds} = \frac{d \boldsymbol{B}/dt}{ds/dt} =... ...frac{t}{\sqrt{1- t^2}}\:\boldsymbol{i} -\boldsymbol{j}) = - \tau \boldsymbol{n}$$

implies $${\tau = \frac{1}{8a \sqrt{1 - t^2}}}$$

Exercise Answer3.1

1. Let the level surface through the point $(x_{0},y_{0},z_{0})$ be

$$f(x,y,z) = c$$

Let any curve that passes through the point $(x_{0},y_{0},z_{0})$ on this level surface be

$$\boldsymbol{r} = \boldsymbol{r}(t) = x(t)\:\boldsymbol{i} + y(t)\:\boldsymbol{j} + z(t)\:\boldsymbol{k}$$

Then $f(\boldsymbol{r}(t)) = f(x(t),y(t),z(t)) = c$. Thus differentiate both sides by $t$.

$$f_{x}x_{t} + f_{y}y_{t} + f_{z}z_{t} = (f_{x},f_{y},f_{z})\cdot(x_{t},y_{t},z_{t}) = \nabla f \cdot\frac{d \boldsymbol{r}}{dt} = 0$$

Therefore, the gradient is orthogonal to the level surface because it is orthogonal to the tangents of the curves on all level surfaces.

2. Let

$$f(x,y,z)=x^{2}+2y^{2}-z^{2}$$

Then the level surface is $f(x,y,z)=0$. Also, ， the normal vector is given by $\nabla f$. Thus

$$\nabla f = (2x, 4y, -2z), \nabla f \mid_{(1,2,3)} = (2,8,-6)$$

Therefore,

$$\boldsymbol{n} = \frac{\nabla f}{\vert\nabla f\vert} = \frac{(2,8,-6)}{\sqrt{4+64 +36}} = \frac{(2,8,-6)}{\sqrt{104}}$$

Next, find the directional unit vector to find the directional derivative in the $(1,3,-1)$ direction at the point $(1,2,3)$. $${{\bf u} = \frac{(1,3,-1)}{\sqrt{11}}}$$．Then the directional derivative is

$$\frac{\partial f}{\partial u} = \nabla f \cdot{\bf u} = (2,8,-6) \cdot\frac{(1,3,-1)}{\sqrt{11}} = \frac{32}{\sqrt{11}}$$

Also, the equation of the tangent plane is

$$2(x-1) + 8(y-2) - 6(z - 3) = 0$$   that is$$2x + 8y - 6z = 0$$

3. Let $f(x,y) = c$ be a streamline equation. Then $\nabla f(x,y) = f_{x}\:\boldsymbol{i} + f_{y}\:\boldsymbol{j}$ expresses the normal vector of $f(x,y) = c$.

$$\nabla f(x,y) \cdot\boldsymbol{F}(x,y) = 0 \ Thus, \ (f_{x}\:\bol... ...{i} + f_{y}\:\boldsymbol{j}) \cdot(-y\:\boldsymbol{i} + x\:\boldsymbol{j}) = 0$$

This implies that， $-y f_{x} + xf_{y} = 0$．Note that the slope of the tangent line of $f(x,y) = c$ is

$$f_{x}dx + f_{y} dy = 0$$   implies$$\frac{dy}{dx} = - \frac{f_{x}}{f_{y}}$$

Then

$$\frac{dy}{dx} = - \frac{f_{x}}{f_{y}} = - \frac{x}{y}$$

implies $x dx + y dy = 0$．Thus $f_{x} = x, f_{y} = y$ and

$$f(x,y) = \int f_{x}dx = \frac{x^2}{2} + c(y) \$$ (5.1)

Next differentiate the equation 5.1 with respect to $y$. Then

$$f_{y} = c'(y) = y$$   implies$$c(y) = \frac{y^2}{2}$$

Therefore, the equation of the streamline is

$$f(x,y) = \frac{x^2}{2} + \frac{y^2}{2} = c$$

4.

$$f_{x} = \frac{\partial}{\partial x}(x^{2}+y^{2}+x^{2})^{\frac{1}{2}} = \f... ...}{2}(x^{2}+y^{2}+z^{2})^{-\frac{1}{2}}(2x) = \frac{x}{\vert\boldsymbol{r}\vert}$$

$$f_{y} = \frac{y}{\vert\boldsymbol{r}\vert}$$

$$f_{z} = \frac{z}{\vert\boldsymbol{r}\vert}$$

implies $-\nabla f=(f_{x},f_{y},f_{z}) = \boldsymbol{F}$． Thus， $\boldsymbol{F}$ is a conservative field.

$$f(x,y,z) = {\vert\boldsymbol{r}\vert} = (x^{2}+y^{2}+z^{2})^{1/2}$$

is the scalar potential of $\boldsymbol{F}$.

(1)

$$\nabla r = (\frac{\partial}{\partial x}\:\boldsymbol{i} + \frac{\... ...ol{j} + 2z\:\boldsymbol{k}}{2\sqrt{x^2 + y^2 + z^2}} = \frac{\boldsymbol{r}}{r}$$

(2)

$$\nabla r^{n} = \frac{\partial}{\partial x}(r^{n})\: \boldsymbol{i} + \frac{\part... ...y}(r^{n})\: \boldsymbol{j} + \frac{\partial}{\partial z}(r^{n})\:\boldsymbol{k}$$

$$= \frac{\partial}{\partial r}(r^{n})\frac{\partial r}{\partial x}\:... ...frac{\partial }{\partial r}(r^{n})\frac{\partial r}{\partial x}\:\boldsymbol{k}$$

$$= nr^{n-1} \left(\frac{\partial r}{\partial x}\boldsymbol{i} + \fra... ...= nr^{n-1}\nabla r = nr^{n-1} \frac{\boldsymbol{r}}{r} = nr^{n-2}\boldsymbol{r}$$

(1)

$${\bf A} = \nabla \left(2r^2 - 4\sqrt{r} + \frac{6}{3\sqrt{r}}\rig... ...oldsymbol{r}) = (4 - \frac{2}{r\sqrt{r}} - \frac{1}{r^2\sqrt{r}})\boldsymbol{r}$$

(2)

$${\bf B} = \nabla (r^2 e^{-r}) = (\nabla r^{2})e^{-r} + r^{2} (\na... ...boldsymbol{r} -r^2 e^{-r}\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$$

Alternate Answer Using $\nabla f(\phi) = \frac{d f}{d\phi} \nabla \phi$, we have

$${\bf B} = \nabla (r^2 e^{-r}) = \frac{d (r^2 e^{-r})}{dr} \nabla r = (2re^{-r} -r^2 e^{-r})\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$$

7. Note that $\nabla \phi$ of $\phi(x,y,z) = x^2 y + 2xz$ is orthogonal to $\phi(x,y,z) = 4$．Therefore the unit normal vector $\boldsymbol{n}$ is

$$\boldsymbol{n} = \frac{(\nabla \phi)_{P}}{\vert\nabla \phi\vert _{P}}$$

Here， $(\nabla \phi)_{P} = (2xy +2z)\boldsymbol{i} + x^2 \boldsymbol{j} + 2x\boldsymbol{k})\mid_{(2,-2,3)} = -2\boldsymbol{i} + 4\boldsymbol{j} + 4\boldsymbol{k}$ implies

$$\boldsymbol{n} = \frac{-2\boldsymbol{i} + 4\boldsymbol{j} + 4\bol... ... + 16 + 16}} = \frac{1}{3}(-\boldsymbol{i} + 2\boldsymbol{j} + 2\boldsymbol{k})$$

8. $\nabla = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial }{\partial z}$ implies

$$\boldsymbol{i}\frac{\partial}{\partial x} (\frac{\phi}{\psi}) = \boldsymbol{i}\frac{\phi_{x}\psi - \phi \psi_{x}}{\psi^2} = \frac{\psi \phi_{x}\boldsymbol{i} - \phi \psi_{x}\boldsymbol{i}}{\psi^2}$$

$$\boldsymbol{j}\frac{\partial}{\partial y} (\frac{\phi}{\psi}) = \boldsymbol{j}\frac{\phi_{y}\psi - \phi \psi_{y}}{\psi^2} = \frac{\psi \phi_{y}\boldsymbol{j} - \phi \psi_{y}\boldsymbol{j}}{\psi^2}$$

$$\boldsymbol{k}\frac{\partial}{\partial z} (\frac{\phi}{\psi}) = \boldsymbol{k}\frac{\phi_{z}\psi - \phi \psi_{z}}{\psi^2} = \frac{\psi \phi_{z}\boldsymbol{k} - \phi \psi_{z}\boldsymbol{k}}{\psi^2}$$

Here， $\phi_{x}\boldsymbol{i} + \phi_{y}\boldsymbol{j} + \phi_{z}\boldsymbol{k} = \nabla \phi$, $\psi_{x}\boldsymbol{i} + \psi_{y}\boldsymbol{j} + \psi_{z}\boldsymbol{k} = \nabla \psi$. Thus，

$$\nabla \left(\frac{\phi}{\psi}\right) = \frac{\phi \nabla \psi - \psi \nabla \phi}{\phi^2}$$

Exercise Answer3.2

Exercise3.2

1. (1) $\phi = x^2 z + e^{y/x}$implies

$$\nabla \phi = \frac{\partial \phi}{\partial x}\boldsymbol{i} + \f... ...x^2}))\boldsymbol{i} + (e^{y/x}(\frac{1}{x}))\boldsymbol{j} + x^2\boldsymbol{k}$$

$\psi = 2z^2 y - xy^2$ implies，

$$\nabla \psi = \frac{\partial \psi}{\partial x}\boldsymbol{i} + \f... ...symbol{k} = -y^2\boldsymbol{i} + (2z^2 - 2xy)\boldsymbol{j} + 4zy\boldsymbol{k}$$

(2) $\nabla (\phi \psi) = (\nabla \phi)\psi + \phi \nabla \psi$. (1) implies

$$\nabla (\phi \psi) = ((2xz + e^{y/x}(-\frac{y}{x^2}))\boldsymbol{i} + (e^{y/x}(\frac{1}{x}))\boldsymbol{j} + x^2\boldsymbol{k})(2z^2 y - xy^2)$$

$$+ (x^2 z + e^{y/x})(-y^2\boldsymbol{i} + (2z^2 - 2xy)\boldsymbol{j} + 4zy\boldsymbol{k})$$

implies the value at the point P$(1,0,-2)$ is

$$\nabla (\phi \psi)_{P} = (-4\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k})(0) + (-2+1)(8)\boldsymbol{j} = -8\boldsymbol{j}$$

3. $\nabla \frac{1}{r} = -\frac{\boldsymbol{r}}{r^3}$ implies $\boldsymbol{F} = -\nabla \phi$. or，

$$\nabla \frac{1}{r} = \frac{\partial (r^{-1})}{\partial x}\:\boldsymbol{i} + \frac{\par... ...} + \frac{\partial (r^{-1})}{\partial r}\frac{d(r)}{\partial z}\:\boldsymbol{k}$$

$$= -r^{-2}(\frac{x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k}}{r}) = - \frac{\boldsymbol{r}}{r^3}$$

Thus， $\boldsymbol{F} = - \nabla \frac{1}{r}$ and the vector field $\boldsymbol{F}$ has a potential $\phi = \frac{1}{r}$. Therefore, the potential energy at P is $\frac{1}{r}$．

3. Use $\nabla r^{n} = nr^{n-1}\nabla r = nr^{n-1}\frac{\boldsymbol{r}}{r} = nr^{n-2}\boldsymbol{r}$

(1)

$${\bf A} = \nabla \left(2r^2 - 4\sqrt{r} + \frac{6}{3\sqrt{r}}\rig... ...oldsymbol{r}) = (4 - \frac{2}{r\sqrt{r}} - \frac{1}{r^2\sqrt{r}})\boldsymbol{r}$$

(2)

$${\bf B} = \nabla (r^2 e^{-r}) = (\nabla r^{2})e^{-r} + r^{2} (\na... ...boldsymbol{r} -r^2 e^{-r}\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$$

Alternate Answer Using $\nabla f(\phi) = \frac{d f}{d\phi} \nabla \phi$, we have

$${\bf B} = \nabla (r^2 e^{-r}) = \frac{d (r^2 e^{-r})}{dr} \nabla r = (2re^{-r} -r^2 e^{-r})\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$$

4. Note that $\nabla \phi$ of $\phi(x,y,z) = x^2 y + 2xz$ is orthogonal to $\phi(x,y,z) = 4$．Therefore the unit normal vector $\boldsymbol{n}$ is

$$\boldsymbol{n} = \frac{(\nabla \phi)_{P}}{\vert\nabla \phi\vert _{P}}$$

Here， $(\nabla \phi)_{P} = (2xy +2z)\boldsymbol{i} + x^2 \boldsymbol{j} + 2x\boldsymbol{k})\mid_{(2,-2,3)} = -2\boldsymbol{i} + 4\boldsymbol{j} + 4\boldsymbol{k}$ implies

$$\boldsymbol{n} = \frac{-2\boldsymbol{i} + 4\boldsymbol{j} + 4\bol... ... + 16 + 16}} = \frac{1}{3}(-\boldsymbol{i} + 2\boldsymbol{j} + 2\boldsymbol{k})$$

5. $\nabla = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial }{\partial z}$ implies

$$\boldsymbol{i}\frac{\partial}{\partial x} (\frac{\phi}{\psi}) = \boldsymbol{i}\frac{\phi_{x}\psi - \phi \psi_{x}}{\psi^2} = \frac{\psi \phi_{x}\boldsymbol{i} - \phi \psi_{x}\boldsymbol{i}}{\psi^2}$$

$$\boldsymbol{j}\frac{\partial}{\partial y} (\frac{\phi}{\psi}) = \boldsymbol{j}\frac{\phi_{y}\psi - \phi \psi_{y}}{\psi^2} = \frac{\psi \phi_{y}\boldsymbol{j} - \phi \psi_{y}\boldsymbol{j}}{\psi^2}$$

$$\boldsymbol{k}\frac{\partial}{\partial z} (\frac{\phi}{\psi}) = \boldsymbol{k}\frac{\phi_{z}\psi - \phi \psi_{z}}{\psi^2} = \frac{\psi \phi_{z}\boldsymbol{k} - \phi \psi_{z}\boldsymbol{k}}{\psi^2}$$

Here， $\phi_{x}\boldsymbol{i} + \phi_{y}\boldsymbol{j} + \phi_{z}\boldsymbol{k} = \nabla \phi$, $\psi_{x}\boldsymbol{i} + \psi_{y}\boldsymbol{j} + \psi_{z}\boldsymbol{k} = \nabla \psi$. Thus，

$$\nabla \left(\frac{\phi}{\psi}\right) = \frac{\phi \nabla \psi - \psi \nabla \phi}{\phi^2}$$

6.

(1) $r = \sqrt{(\xi -x)^2 + (\eta - y)^2 + (\zeta - z)^2}$， $\nabla_{Q} = \boldsymbol{i}\frac{\partial}{\partial \xi} + \boldsymbol{j}\frac{\partial}{\partial \eta} + \boldsymbol{k}\frac{\partial}{\partial \zeta}$, $\nabla_{P} = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial}{\partial z}$，Then

$$\nabla_{Q}r = \boldsymbol{i}\frac{\partial r}{\partial \xi} + \boldsymbol{j}\frac{\partial r}{\partial \eta} + \boldsymbol{k}\frac{\partial r}{\partial \zeta}$$

$$= \boldsymbol{i}\frac{2(\xi-x)}{2r} + \boldsymbol{j}\frac{2(\eta -y)}{2r} + \boldsymbol{k}\frac{2(\zeta - z)}{2r}$$

$$\nabla_{P}r = \boldsymbol{i}\frac{\partial r}{\partial x} + \boldsymbol{j}\frac{\partial r}{\partial y} + \boldsymbol{k}\frac{\partial r}{\partial z}$$

$$= \boldsymbol{i}\frac{-2(\xi-x)}{2r} + \boldsymbol{j}\frac{-2(\eta -y)}{2r} + \boldsymbol{k}\frac{-2(\zeta - z)}{2r} = -\nabla_{Q}r$$

(2)

$$\nabla_{Q}(\frac{1}{r}) = \boldsymbol{i}\frac{\partial r^{-1}}{\partial \xi} + \boldsymbol{... ...l r^{-1}}{\partial \eta} + \boldsymbol{k}\frac{\partial r^{-1}}{\partial \zeta}$$

$$= \boldsymbol{i}(-r^{-2}\frac{(\xi-x)}{r}) + \boldsymbol{j}(-r^{-2}\frac{(\eta-y)}{r}) + \boldsymbol{k}(-r^{-2}\frac{(\zeta-z)}{r})$$

$$\nabla_{P}(\frac{1}{r}) = \boldsymbol{i}\frac{\partial r^{-1}}{\partial x} + \boldsymbol{j}... ...\partial r^{-1}}{\partial y} + \boldsymbol{k}\frac{\partial r^{-1}}{\partial z}$$

$$= \boldsymbol{i}(r^{-2}\frac{(\xi-x)}{r}) + \boldsymbol{j}(r^{-2}\f... ...-y)}{r}) + \boldsymbol{k}(r^{-2}\frac{(\zeta-z)}{r}) = -\nabla_{Q}(\frac{1}{r})$$

Exercise Answer3.3

1. $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = (t^2 + 1)\boldsymbol{i} + 2t^2\boldsymbol{j} + t^3 \boldsymbol{k}$implies

$$d\boldsymbol{r} = ( 2t\boldsymbol{i} + 4t\boldsymbol{j} + 3t^2 \boldsymbol{k})dt$$

Also， ${\mathbf F} = 3xy\boldsymbol{i} - 5z\boldsymbol{j} + 10x \boldsymbol{k} = 3(t^2+1)(2t^2)\boldsymbol{i} - 5(t^3)\boldsymbol{j} + 10(t^2 + 1)\boldsymbol{k}$． Therefore，

$$W = \int_{C}{\mathbf F}\cdot d\boldsymbol{r} = \int_{1}^{2}\left(12t^3(t^2+1) - 20t^4 + 30(t^2+1)t^2\right)dt$$

$$= \int_{1}^{2} (12t^5 + 10t^4 + 12t^3 + 30t^2 )dt = \left[2t^6 + 2t^5 + 3t^4 + 10t^3\right]_{1}^{2}$$

$$= 2(64 -1) + 2(32 - 1) + 3(16-1) + 10(8-1) = 303$$

2.

(1) $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = t^2\boldsymbol{i} + 2t\boldsymbol{j} + t^3\boldsymbol{k}\ (0 \leq t \leq 1)$ implies

$$d\boldsymbol{r} = 2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k}$$

Also， $\phi = 2xyz^2 = 2t^2(2t)(t^3)^2 = 4t^9$．Therefore，

$$\int_{C}\phi d\boldsymbol{r} = \int_{0}^{1}4t^{9} (2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k})dt$$

$$= \int_{0}^{1}(8t^{10}\boldsymbol{i} + 8t^{9}\boldsymbol{j} + 12t^{... ...bol{i} + \frac{8}{10}t^{10}\boldsymbol{j} + t^{12}\boldsymbol{k}\right]_{0}^{1}$$

$$= \frac{8}{11}\boldsymbol{i} + \frac{4}{5}\boldsymbol{j} + \boldsymbol{k}$$

(2) $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = t^2\boldsymbol{i} + 2t\boldsymbol{j} + t^3\boldsymbol{k}\ (0 \leq t \leq 1)$ implies

$$d\boldsymbol{r} = 2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k}$$

Also， ${\mathbf F} = xy\boldsymbol{i} - z\boldsymbol{j} + x^2 \boldsymbol{k} = 2t^3\boldsymbol{i} - t^3\boldsymbol{j} + t^4\boldsymbol{k}$．Therefore,

$$\int_{C}{\mathbf F} \times d\boldsymbol{r} = \int_{0}^{1}\left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsy... ...dsymbol{i} - (6t^5 - 2t^5)\boldsymbol{j} + (4t^3 + 2t^4)\boldsymbol{k}\right)dt$$

$$= \left[(-\frac{t^6}{2} - \frac{2t^5}{5})\boldsymbol{i} - \frac{4t^6}{6}\boldsymbol{j} + (t^4 + \frac{2t^5}{5})\boldsymbol{k}\right]_{0}^{1}$$

$$= (-\frac{1}{2} - \frac{2}{5})\boldsymbol{i} - \frac{2}{3}\boldsymb... ...ac{9}{10}\boldsymbol{i} - \frac{2}{3}\boldsymbol{j} + \frac{7}{5}\boldsymbol{k}$$

3. Note that if ${\bf A}$ satisfies ${\bf A} = -\nabla \phi$，then ${\bf A}$ has a scalar potential and $\int_{C}{\bf A}\cdot d\boldsymbol{r} = 0$. Thus we find $\phi$ so that， $\boldsymbol{r} = \nabla \phi$．

$$\nabla (\boldsymbol{r} \cdot\boldsymbol{r}) = \nabla r^2 = 2 \boldsymbol{r}$$

implies $\boldsymbol{r} = \frac{1}{2} \nabla (\boldsymbol{r} \cdot\boldsymbol{r})$．Therefore，

$$\int_{C}\boldsymbol{r}\cdot d\boldsymbol{r} = \frac{1}{2}\int_{C}\nabla (\boldsymbol{r} \cdot\boldsymbol{r})d\boldsymbol{r} = 0$$

4. The force field ${\mathbf F}$ has a potential $U$. Then ${\mathbf F} = -\nabla U$．This suggest that the equation of motion of this mass point is

$$m \frac{d{\bf v}}{dt} = {\mathbf F} = -\nabla U$$

Also， ${\bf v} = \frac{d\boldsymbol{r}}{dt}$．Then

$$m \frac{d^2{\bf v}}{dt^2} \cdot{\bf v} = -\nabla U \cdot\frac{d\boldsymbol{r}}{dt}$$

Thus，

$$m{\bf v}\cdot\frac{d{\bf v}}{dt} + \frac{d\boldsymbol{r}}{dt} \cdot\nabla U = 0$$

Here calculate $\frac{d\boldsymbol{r}}{dt} \cdot\nabla U$.

$$\frac{d\boldsymbol{r}}{dt} \cdot\nabla U = (\frac{dx}{dt}\boldsym... ...al y}\frac{dy}{dt} + \frac{\partial U}{\partial z}\frac{dz}{dt} = \frac{dU}{dt}$$

Therefore，

$$m{\bf v}\cdot\frac{d{\bf v}}{dt} + \frac{dU}{dt} = \frac{d}{dt}(\frac{1}{2}m{\bf v}\cdot{\bf v} + U)= 0$$

This implies that，

$$\frac{1}{2}mv^2 + U = C \ (C$$   constant$$)$$

Therefore,，

$$\frac{1}{2}mv_{A}^2 + U(A) = \frac{1}{2}mv_{B}^2 + U(B)$$

5. If the origin O is centered on this $xy$ plane and the circle with radius $a$ is $C$, the equation of motion of the mass point is parametrized by $C : x = a\cos{t}, y = a\sin{t}, \ 0 \leq t \leq 2\pi$．Thus，

$$\nabla \phi = \frac{-\frac{y}{x^2}}{1 + (\frac{y}{x})^2}\boldsymb... ...dsymbol{j} = \frac{-\sin{t}}{a}\boldsymbol{i} + \frac{\cos{t}}{a}\boldsymbol{j}$$

Also，

$$d\boldsymbol{r} = (-a\sin{t}\boldsymbol{i} + a\cos{t}\boldsymbol{j})dt$$

Therefore，

$$\int_{C}(\nabla \phi) \cdot d\boldsymbol{r} = \int_{0}^{2\pi}(-\s... ...} +\cos{t}\boldsymbol{j})\cdot(-\sin{t} + \cos{t})dt = \int_{0}^{2\pi}dt = 2\pi$$

6. Let $x = 4\cos{t}, y = 3\sin{t}$. Then

$$\boldsymbol{r} = \boldsymbol{r}(t) = 4\cos{t}\:\boldsymbol{i} + 3... ...\ d\boldsymbol{r}(t) = (-4\sin{t}\:\boldsymbol{i} + 3\cos{t}\:\boldsymbol{j})dt$$

Also,

$$\boldsymbol{F} = (12\cos{t} - 12\sin{t})\:\boldsymbol{i} + (16\cos{t} + 6\sin{t})\:\boldsymbol{j} -36\sin^{2}{t}\:\boldsymbol{k}$$

Thus

$$\int_{C} \boldsymbol{F} \cdot d \boldsymbol{r} = \int_{0}^{2\pi} [48 - 30\sin{t}\cos{t}]dt = 96\pi$$

7.

$$\int_{C}\boldsymbol{F}\cdot{\bf t}ds = \int_{C}\boldsymbol{F}\cdot d\boldsymbol{r}$$

$$= \int_{C} (x^2 y\boldsymbol{i} + (x+y)\:\boldsymbol{j})\cdot(dx\:\boldsymbol{i} + dy\:\boldsymbol{j}) = \int_{C} (x^2 ydx + (x+y)dy)$$

$$= \int_{0}^{2} x^2 x^2 dx + \int_{0}^{2}(x + x^2)(2xdx) = \left[\fr... ...} + \left[\frac{2}{3}x^{3} + \frac{2}{4}x^{4} \right ]_{0}^{2} = \frac{296}{15}$$

8. The directional derivative of $\phi$ at P$(2,-1,2)$ in the direction of ${\bf u}$ is，

$$\frac{\partial \phi(2,-1,2)}{\partial u} = \nabla \phi(1,0,-2) \cdot{\bf u}$$

HHere，

$$\nabla \phi(2,-1,2) = (4z^3 - 6xyz)\boldsymbol{i} + (-3x^2 z)\boldsymbol{j} + (12xz^2 - 3x^2y)\boldsymbol{k}\mid_{(2,-1,2)}$$

$$= (32 + 24)\boldsymbol{i} + (-24)\boldsymbol{j} +(96 + 12))\boldsymbol{k}$$

implies

$$\frac{\partial \phi(2,-1,2)}{\partial u} = (56\boldsymbol{i} - 24... ...ldsymbol{j} + 6\boldsymbol{k}}{7} = \frac{1}{7}(112 + 72 + 648) = \frac{832}{7}$$

Exercise Answer3.4

1.

(1) When the curved surface $S$ is projected onto the $xy$ plane，$S$ maps to $\Omega = \{(x,y): 2x + 2y \leq 2, x \geq 0, y \geq 0\}$. Also，from the surface $S : 2x + 2y + z = 2$, if the corresponding $\boldsymbol{r}$ is the position vector,

$$\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + (2-2x-2y)\boldsymbol{k}$$

Thus we find the normal vector of $S$ which is $\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}$，

$$\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = \left\vert\begin{a... ... -2 \end{array}\right\vert = 2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}$$

Note that $dS = \vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert dx dy = \sqrt{4 + 4 + 1}dxdy = 3dxdy$. Then

$$\int_{S}fdS = 3\int_{\Omega}(x^2 + 2y + z - 1)dxdy = 3\int_{0}^{1}\int_{y=0}^{1-x}(x^2 + 2y + (2-2x-2y) -1)dydx$$

$$= 3\int_{0}^{1}\int_{0}^{1-x}(x^2 - 2x + 1)dydx = 3\int_{0}^{1}\int_{0}^{1-x}(x-1)^2\;dydx$$

$$= 3\int_{0}^{1}\left[(x-1)^2y\right]_{0}^{1-x}dx = 3\int_{0}^{1}(1-x)^3\;dx = -3\left[\frac{1}{4}(1-x)^{4}\right]_{0}^{1} = \frac{3}{4}$$

(2) Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + (2-2x-2y)\boldsymbol{k}$. Then $\boldsymbol{n} = \frac{\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}}{\vert\bold... ...bol{r}_{y}\vert} = \frac{2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}}{3}$. Thus，

$${\bf A} \cdot\boldsymbol{n} = (x^2 \boldsymbol{i} + z\boldsymbol{... ...rac{2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}}{3} = \frac{2x^2 + z}{3}$$

Therefore，

$$\int_{S}{\bf A} \cdot\boldsymbol{n}dS = \int_{\Omega}\frac{2x^2 + z}{3}3dxdy = \int_{\Omega}(2x^2 + (2-2x-2y))dxdy$$

$$= \int_{0}^{1}\int_{0}^{1-x}(2x^2 - 2x - 2y + 2)dy dx = \int_{0}^{1}\left[2x^2 y - 2xy -y^2 + 2y\right]_{0}^{1-x}dx$$

$$= \int_{0}^{1}(2x^2(1-x) - 2x(1-x) - (1-x)^2 + 2(1-x))dx$$

$$= \int_{0}^{1}(1-x)(2x^2 - 2x - (1-x) + 2)dx = \int_{0}^{1}(1-x)(2x^2 - x + 1)dx$$

$$= \int_{0}^{1}(-2x^3 +3x^2 - 2x + 1)dx = \left[-\frac{x^4}{2} + x^3 - x^2 + x\right]_{0}^{1}$$

$$= -\frac{1}{2} + 1 - 1 + 1 = \frac{1}{2}$$

2. Since the surface $S$ is region on the $xy$ plane，the unit normal vector is $\boldsymbol{n} = \boldsymbol{k}$．

$${\bf A} \times \boldsymbol{n} = (x\boldsymbol{i} + (x-y)\boldsymb... ...{xy}\\ 0 & 0& 1 \end{array}\right\vert = (x-y)\boldsymbol{i} - x\boldsymbol{j}$$

Also，$S$ is on the $xy$ plane. Then $S = \Omega$，$dS = dxdy$．Here，$\Omega$ is a disk, so we use the polar coordinates, then $x = r\cos{\theta}, y = r\sin{\theta}, 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq a$ and

$$\int_{S}{\bf A} \times \boldsymbol{n}dS = \int_{0}^{\pi/2}\int_{0}^{a}((r\cos{\theta} - r\sin{\theta})\boldsymbol{i} - r\cos{\theta}\boldsymbol{j})rdr d\theta$$

$$= \int_{0}^{\pi/2}\left[\frac{r^3}{3}\right]_{0}^{a}\left((\cos{\theta} - \sin{\theta})\boldsymbol{i} - \cos{\theta}\boldsymbol{j}\right)d\theta$$

$$= \frac{a^3}{3}\left[(\sin{\theta} + \cos{\theta}\boldsymbol{i} - \... ...3}{3}((1-1)\boldsymbol{i} - \boldsymbol{j}) = -\frac{a^3}{3}\boldsymbol{j} F3-4$$

3. By projecting $S$ onto $xy$ plane, then $S$ aps to $\Omega = \{(x,y):y^2 = 4, 0 \leq x \leq 3, y \geq 0\}$. Next by letting ${\bf r}$ be a position vector ${\bf r} =x{\bf i} + y{\bf j} + \sqrt{4-y^2}{\bf k}$. Thus，the normal vector of the surface $S$ is ${\bf r}_{x} \times {\bf r}_{y} = \frac{y}{\sqrt{4-y^2}}{\bf j} +{\bf k}$． Therefore,，

$$\iint_{S}(3x{\bf i} + 4z{\bf j} +2y{\bf k}) \cdot {\bf n} dS = \i... ...bf j} +2y{\bf k})\cdot ({\bf r}_{x} \times {\bf r}_{y})dxdy = \iint_{S}(6y)dxdy$$

Here，express $\Omega$ using vertical simple, we have，

$$\iint_{S}(6y)dxdy = \int_{0}^{3}\int_{0}^{2}6y dy dx = \int_{0}^{3}3y^2\mid_{0}^{2} = \int_{0}^{3}12dx = 36$$

4. By projecting the surface $S$ onto $yz$ plane, $S$ maps to $\Omega = \{(y,z):y^2 = a^2, 0 \leq z \leq 1\}$. Next，let ${\bf r}$ be a postion vector ${\bf r} =x{\bf i} + \sqrt{a^2 -x^2}{\bf j} + z{\bf k}$. Thus，the normal vector of $S$ is ${\bf r}_{z} \times {\bf r}_{x} = \frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}$.

$${\bf n} = \frac{{\bf r}_{z} \times {\bf r}_{x}}{\vert{\bf r}_{z} \times {\b... ...} = \frac{\frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}}{\sqrt{\frac{y^2}{4-y^2} + 1}}$$

$$= \frac{\frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}}{\sqrt{\frac{4}{4-y^2}}}$$

Therefore,，

$$\iint_{S}(3x{\bf i} + 4z{\bf j} +2y{\bf k}) \cdot {\bf n} dS = \iint_{S}{\bf r}_{x} \times {\bf r}_{y}dxdy = \iint_{S}(6y)dxdy$$

Now express，$\Omega$ using the vertical simple,

$$\iint_{S}(6y)dxdy = \int_{0}^{3}\int_{0}^{2}6y dy dx = \int_{0}^{3}3y^2\mid_{0}^{2} = \int_{0}^{3}12dx = 36$$

5. The position vector is $\boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + (4-x^2 - y^2)\:\boldsymbol{k}$. Then

$$\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS = \iint_{\Omega}\lef... ... 1 & -2y \end{array} \right\vert dx dy = \iint_{\Omega}(4 - x^ 2 + y^2) dx dy$$

Here we use the polar coordinate transformation, $\Omega : x^2 + y^2 \leq 4$. Then $x = r\cos{\theta}, \ y = r\sin{\theta}$ implies

$$\iint_{\Omega}(4 - x^ 2 + y^2) dx dy = \int_{0}^{2\pi} \int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta})r dr... ...eft[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2}\theta \right ]_{0}^{2} d\theta$$

$$= \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$$

6. Let $\boldsymbol{r}$ be the position vector corresponds to $S : x^2 + y^2 + z = 4$. Then

$$\boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + (4 - x^2 - y^2)\:\boldsymbol{k}$$

Next we find the normal vecor $\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}$ of $S$.

$$\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = (\boldsymbol{i} -2x\:\boldsymbol{k}) \times (\boldsymbol{j} -2y\:\boldsymbol{k})$$

$$= \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...nd{array}\right\vert = 2x\:\boldsymbol{i} + 2y\:\boldsymbol{j} + \boldsymbol{k}$$

Thus

$$\vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert = \sqrt{4x^2 + 4y^2 + 1}$$

Here，we use the polar coordintate transformation $${\Omega : 4 - (x^2 + y^2) \geq 0}$$. Then

$$\iint_{S} f(x,y,z) dS = \iint_{\Omega}\frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}\vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert dx dy$$

$$= \iint_{\Omega}(2y^2 + z)dx dy = \int_{0}^{2\pi}\int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta}) r dr d\theta$$

$$= \int_{0}^{2 \pi}\left[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2... ...i} d\theta = \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$$

8. Since $x^2 + y^2 + z^2 = a^2$，for $z \geq 0$，the position vector is $\boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + \sqrt{a^2-x^2 - y^2}\:\boldsymbol{k}$. Thus，

$$\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS = \iint_{\Omega}\lef... ...\frac{z}{r^3} \\ 1 & 0 & -2x \\ 0 & 1 & -2y \end{array} \right\vert dx dy$$

10. For the plane DEFG: since $x = 1$, the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = \boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}$. Here the positive direction is from the bck of the plane to the front of DEFG. Thus, the unit normal vector is

$$\boldsymbol{n} = \frac{\boldsymbol{r}_{y} \times \boldsymbol{r}_{... ...boldsymbol{k}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right\vert = \boldsymbol{i}$$

Also，the orthogonal projection on to $yz$ plane is $\Omega_{yz} = \{(y,z) : 0 \leq y \leq 1, 0 \leq z \leq 1\}$. Then，

$$\iint_{DEFG}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\in... ...;dz = \int_{0}^{1}(2-z)dx = \left[2z - \frac{z^2}{2}\right]_{0}^1 = \frac{3}{2}$$

For the plane ABCO: since $x = 0$, the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = y\boldsymbol{j} + z\boldsymbol{k}$. Here，the positive direction is the direction of the back of plane ABCO to the front．Thus，the unit normal vector is

$$\boldsymbol{n} = \frac{\boldsymbol{r}_{z} \times \boldsymbol{r}_{... ...oldsymbol{k}\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{array}\right\vert = -\boldsymbol{i}$$

Also，the orthogonal projection onto $yz$ plane is $\Omega_{yz} = \{(y,z) : 0 \leq y \leq 1, 0 \leq z \leq 1\}$. Then

$$\iint_{ABCO}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\in... ...y=0}^{1}\;dz = \int_{0}^{1}zdx = \left[\frac{z^2}{2}\right]_{0}^1 = \frac{1}{2}$$

For the plane ABEF: since $y = 1$, the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = x\boldsymbol{i} + \boldsymbol{j} + z\boldsymbol{k}$. Here the positive direction is the direction of the back of the plane ABEF to the front. Thus, the unit normla vector is

$$\boldsymbol{n} = \frac{\boldsymbol{r}_{z} \times \boldsymbol{r}_{... ...boldsymbol{k}\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{array}\right\vert = \boldsymbol{j}$$

Also，the orthogonal projection onto $xz$ plane is $\Omega_{xz} = \{(x,z) : 0 \leq x \leq 1, 0 \leq z \leq 1\}$. Then

$$\iint_{ABEF}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\int_{0}^{1}x^2dxdz = \frac{1}{3}$$

Similarly，if you do the surface integral of the remaining surface，we have

$$\int \boldsymbol{F}\cdot\boldsymbol{n}dS = \frac{11}{6}$$

(1) $w = f(r), r = r(x,y,z)$implies

$$\nabla f(r) = \frac{d f}{dr}(\frac{\partial r}{\partial x}\boldsy... ...ol{j} + \frac{\partial r}{\partial z}\boldsymbol{k}) = \frac{d f}{dr} \nabla r.$$

Thus，

$$\nabla^2 f(r) = \nabla \cdot(\frac{d f}{dr} \nabla r) = \nabla (\frac{d f}{dr}) \cdot\nabla r + \frac{d f}{dr} \nabla \cdot\nabla r.$$

Note that $\nabla r \cdot\nabla r = \frac{\boldsymbol{r}}{r} \cdot\frac{\boldsymbol{r}}{r} = 1$. Also， $\nabla \cdot\nabla r = \nabla \cdot\frac{\boldsymbol{r}}{r} = \nabla \cdot(r^{-... ...symbol{r} = -r^{-2}\boldsymbol{r} \cdot\boldsymbol{r} + r^{-1}(3) = \frac{2}{r}$. Therefore，

$$\nabla^2 f(r) = \frac{d^{2}f}{dr^2} + \frac{2}{r}\frac{df}{dr}.$$

(2) Since $\nabla^2 f(r) = 0$，we have

$$\frac{d^{2}f}{dr^2} + \frac{2}{r}\frac{df}{dr} = 0.$$

Note that if we let $w = \frac{d f}{dr}$，then $\frac{d w}{dr} = - \frac{2}{r}w$ and it is separable．Thus， $\frac{d w}{w} = -\frac{2}{r}dr$ and integrate both sides, we have

$\log\vert w\vert = -2\log\vert r\vert + c$. Thus， $w = cr^{-2}$. $w = \frac{d f}{dr}$ and $\frac{d f}{dr } = cr^{-2}$．Therefore， $f = c_{0}r^{-1} + c_{1}$. Exercise Answer3.5 Basic formula Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \sqrt{x^2 + y^2 + z^2}$. Then (1) $\nabla r = \frac{\boldsymbol{r}}{r}\ (2)\ \nabla r^{n} = nr^{n-1}\nabla r = nr^... ...dsymbol{A}= (\nabla \phi) \cdot\boldsymbol{A} + \phi \nabla \cdot\boldsymbol{A}$

1.

(1)

$$\nabla \cdot(2x^2 z\boldsymbol{i} - xy^2 z \boldsymbol{j} + 3yz^2... ...ial (-xy^2)}{\partial y} + \frac{\partial (3yz^2)}{\partial z} = 4xz -2xy + 6yz$$

(2)

$$\nabla^2(3x^2 z - y^2 z^3 + 4x^2 y) = \frac{\partial^2 (3x^2 z - y^2 z^3 + 4x^2 y)}{\partial x^2} + \fr... ...2 y)}{\partial y^2} + \frac{\partial (3x^2 z - y^2 z^3 + 4x^2 y)}{\partial z^2}$$

$$= 6z + 8y - 2z^3 -6y^2 z$$

(3)

$$\nabla(\nabla \cdot\boldsymbol{F}) = \nabla (6xy + 4y^3 -4x^2z) = (6y-8xz)\boldsymbol{i} + (6x+12y^2)\boldsymbol{j} - 4x^2\boldsymbol{k}$$

2.

(1) $\nabla r^{-3} = \frac{d r^{-3}}{dr}\nabla r = -3r^{-4}\frac{\boldsymbol{r}}{r} = -3r^{-5}\boldsymbol{r}$implies $r\nabla r^{-3} = -3r^{-4}\boldsymbol{r}$. Therefore，

$$\nabla \cdot(r \nabla r^{-3}) = \nabla \cdot(-3r^{-4}\boldsymbol{... ...oldsymbol{r}}{r} \cdot\boldsymbol{r} -3r^{-4}(3) = 12r^{-4} - 9r^{-4} = 3r^{-4}$$

(2) $\nabla\cdot\left(\frac{\boldsymbol{r}}{r^2}\right) = \nabla \cdot r^{-2}\boldsy... ...boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{-2}(3) = -2r^{-2} + 3r^{-2} = r^{-2}$. Therefore,

$$\nabla^{2}\left\{\nabla\cdot\left(\frac{\boldsymbol{r}}{r^2}\right)\right\} = \nabla \cdot(\nabla(r^{-2})) = \nabla \cdot(-2r^{-3}\frac{\boldsymbol{r}}{r} = \nabla \cdot(-2r^{-4}\boldsymbol{r}$$

$$= \nabla(-2r^{-4}) \cdot\boldsymbol{r} + -2r^{-4} \nabla \cdot\bold... ...oldsymbol{r}}{r} \cdot\boldsymbol{r} - 2r^{-4}(3) = 8r^{-4} - 6r^{-4} = 2r^{-4}$$

3. Let ${\bf w} = w_{1}\boldsymbol{i} + w_{2}\boldsymbol{j} + w_{3}\boldsymbol{k}$. Then

$${\bf w} \times \boldsymbol{r} = \left\vert\begin{array}{ccc} \bol... ...oldsymbol{i} + (w_{3}x -w_{1}z)\boldsymbol{j} + (w_{1}y - w_{2}x)\boldsymbol{k}$$

Therefore，

$$\nabla \cdot({\bf w} \times \boldsymbol{r}) = 0$$

4.

(1) By the basic formula，

$$\nabla^2(\phi \psi) = \nabla \cdot\nabla(\phi \psi) = \nabla \cdot((\nabla \phi)\psi + ... ...(\nabla \psi) + (\nabla \phi)\cdot(\nabla\psi) + \phi \nabla \cdot(\nabla \psi)$$

$$= \psi + (\nabla^2 \phi) + 2(\nabla \phi)\cdot(\nabla \psi) + \phi \nabla^2 \psi$$

(1) By the basic formula,

$$\nabla \cdot(\phi \nabla \psi) = (\nabla \phi) \cdot(\nabla \psi)... ...abla \cdot(\nabla \psi) = (\nabla \phi) \cdot(\nabla \psi) + \phi \nabla^2 \psi$$

(3) (2) implies， $\nabla \cdot(\phi \nabla \psi) = (\nabla \phi) \cdot(\nabla \psi) + \phi \nabla^2 \psi$. By symmetry $\nabla \cdot(\psi \nabla \phi) = (\nabla \psi) \cdot(\nabla \phi) + \psi \nabla^2 \phi$. Therefore，

$$\nabla \cdot(\phi \nabla \psi - \psi \nabla \phi) = \phi \nabla^2 \psi - \psi \nabla^2 \phi$$

5. $\nabla \phi = \frac{\partial \phi}{\partial x}\boldsymbol{i} + \frac{\partial \... ...l z} = 2xyz^3\boldsymbol{i} + x^2 z^3 \boldsymbol{j} + 3x^2 y z^2\boldsymbol{k}$ implies

$$\frac{\partial \phi}{\partial x} = 2xyz^3,\ \frac{\partial \phi}{\partial y} = x^2 z^3,\ \frac{\partial \phi}{\partial z} = 3x^2 yz^2$$

Thus， $\phi(x,y,z) = x^2 yz^3 + c(y,z)$．Here using $\frac{\partial \phi}{\partial y} = x^2 z^3$， $\frac{\partial c(y,z)}{\partial y} = 0$. Then $c(y,z) = c(z)$．Finally， $\frac{\partial \phi}{\partial z} = 3x^2 yz^2$ implies $\frac{\partial c(z)}{\partial z} = 0$. Then $c(z) = c$．Thus， $\phi(x,y,z) = x^2 yz^3 + c$. Now by the initial value $\phi(1,-2,2) =4$， we have $-16+c = 4$ and $c = 20$.

6.

$$(\nabla U) \times (\nabla V) = \left\vert\begin{array}{ccc} \bold... ..._{z}V_{x} - U_{x}V_{z})\boldsymbol{j} + (U_{z}V_{y} - U_{y}V_{z})\boldsymbol{k}$$

Therefore，

$$\nabla \cdot((\nabla U) \times (\nabla V)) = \frac{\partial(U_{y}V_{z} - U_{z}V_{y})}{\partial x} + \frac{\par... ...U_{x}V_{z})}{\partial y} + \frac{\partial(U_{x}V_{y} - U_{y}V_{x})}{\partial z}$$

$$= U_{yx}V_{z} + U_{y}V_{zx} - U_{zx}V_{y} - U_{z}V_{yx} + U_{zy}V_{x} + U_{z}V_{xy} - U_{xy}V_{z} - U_{x}V_{zy}$$

$$+ U_{xz}V_{y} + U_{x}V_{yz} - U_{yz}V_{x} - U_{y}V_{xz} = 0$$

7.

$$\frac{d\boldsymbol{r}}{dt}\cdot\nabla \phi = (\frac{dx}{dt}\bolds... ...artial \phi}{\partial z}\frac{\partial z}{\partial t} = \frac{d}{dt}\phi(x,y,z)$$

8. Differentiation of composite function implies

$$\frac{d \phi}{dt} = \frac{\partial \phi}{\partial x}\frac{dx}{dt}... ...\nabla \phi \cdot\frac{d \boldsymbol{r}}{dt} + \frac{\partial \phi}{\partial t}$$

9.

$${\rm div} \boldsymbol{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}$$

$$= 3y + x^{2} + y^{2}$$

．

(1)

$$\nabla \cdot\boldsymbol{F} = \frac{\partial}{\partial x}(3xyz^2) ... ...\partial y}(2xy^3) + \frac{\partial}{\partial z}(-x^2yz) = 3yz^2 + 6xy^2 - x^2y$$

(2) $\nabla \phi = 6x\:\boldsymbol{i} - z\:\boldsymbol{j} - y\:\boldsymbol{k}$ implies

$$\boldsymbol{F} \cdot\nabla \phi = (3xyz^2)(6x) + (2xy^3)(-z) + (-x^2yz)(-y) = 18x^2yz^2 - 2xy^3z + x^2y^2z$$

(1)

$$\nabla \cdot\boldsymbol{r} = (\frac{\partial}{\partial x}\boldsym... ...{k})\cdot(x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}) = 1 + 1 + 1 = 3.$$

(2) By the differentiation of composite functions,，

$$\nabla \cdot(r^{n}\boldsymbol{r}) = (\nabla r^{n}) \cdot\boldsymb... ...boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{n}(3) = nr^{n} + 3r^{n} = (n+3)r^{n}$$

(3)

$$\nabla \cdot(\frac{\boldsymbol{r}}{r^3}) = \nabla \cdot r^{-3}\bo... ...mbol{r} = -3r^{-4}\frac{\boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{-3}(3) = 0.$$

12

$$\nabla^2 \phi = \frac{\partial^2}{\partial x^2}(3x^2 y - y^3z^2) ... ...y^3z^2) + \frac{\partial^2}{\partial z^2}(3x^2 y - y^3z^2) = 6y - 6yz^2 - 2y^3.$$

Exercise Answer3.6

1.

(1)

$$\nabla \times \boldsymbol{A} = \left\vert\begin{array}{ccc} \bold... ...j}(4xz-3z^3) + \boldsymbol{k}(0) = y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j}$$

(2)

$$\nabla \times (\phi \boldsymbol{A}) = (\nabla \phi) \times \boldsymbol{A} + \phi \nabla \times \boldsymbol{A}$$

$$= \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ... y} & \frac{\partial}{\partial z}\\ 2xz^2 & -yz & 3xz^3 \end{array}\right\vert$$

$$= (3x^3z^4 + x^2 y^2z)\boldsymbol{i} + (2x^3yz^2 -6x^2yz^4)\boldsym... ...^2-2x^3z^3)\boldsymbol{k} + x^2yz(y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j})$$

$$= (3x^3z^4 + 2x^2y^2)\boldsymbol{i} + (6x^3 yz^2 -9x^2 yz^4)\boldsymbol{i} - (2xy^2z^2 + 2x^3z^3)\boldsymbol{k}$$

(3) (1)implies $\nabla \times \boldsymbol{A} = y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j}$.

$$\nabla \times (\nabla \times \boldsymbol{A}) = \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...^3 & 0 \end{array}\right\vert = (-4x+9z^2)\boldsymbol{i} + (4z-1)\boldsymbol{k}$$

2.

$$\nabla \times {\bf v} = \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...vert = (c+1)\boldsymbol{i} + (a-4)\boldsymbol{j} + (b-2)\boldsymbol{k} = {\bf0}$$

implies $a = 4, b = 2, c = -1$

3.

$$\nabla \cdot(\boldsymbol{A} \times \boldsymbol{B}) = (\frac{\partial}{\partial x}\boldsymbol{i} + \frac{\partial}{\par... ...\partial}{\partial z}\boldsymbol{k})\cdot(\boldsymbol{A} \times \boldsymbol{B})$$

$$= \boldsymbol{i}\cdot\frac{\partial}{\partial x}(\boldsymbol{A}\tim... ...dsymbol{k}\cdot\frac{\partial}{\partial z}(\boldsymbol{A}\times \boldsymbol{B})$$

$$= \boldsymbol{i}\cdot(\frac{\partial \boldsymbol{A}}{\partial x} \t... ...ldsymbol{B} + \boldsymbol{A} \times \frac{\partial \boldsymbol{B}}{\partial z})$$

$$= \boldsymbol{B}\cdot(\boldsymbol{i} \times \frac{\partial \boldsym... ...\frac{\partial \boldsymbol{A}}{\partial z})\ \hskip 1cm ([A\ B\ C] = [C\ A\ B])$$

$$- \boldsymbol{A}\cdot(\boldsymbol{i} \times \frac{\partial \boldsym... ...ymbol{B}\cdot(\boldsymbol{k} \times \frac{\partial \boldsymbol{B}}{\partial z})$$

$$= \boldsymbol{B}\cdot(\nabla \times \boldsymbol{A}) - \boldsymbol{A}\cdot(\nabla \times \boldsymbol{B}) = 0$$

(1)

$$\nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$$

を用いると，

$$\nabla(\boldsymbol{C} \cdot\boldsymbol{A}) = (\boldsymbol{A}\cdot... ...la \times \boldsymbol{C}) + \boldsymbol{C}\times (\nabla \times \boldsymbol{A})$$

ここで， $(\boldsymbol{A}\cdot\nabla)\boldsymbol{C} = 0, \boldsymbol{A} \times(\nabla \times \boldsymbol{C})= {\bf0}$implies

$$\nabla(\boldsymbol{C}\cdot\boldsymbol{A}) = (\boldsymbol{C}\cdot\nabla)\boldsymbol{A} + \boldsymbol{C} \times (\nabla \times \boldsymbol{A})$$

(2)

$$\nabla \cdot(\boldsymbol{C} \times \boldsymbol{A}) = (\frac{\partial }{\partial x}\boldsymbol{i} + \frac{\partial}{\pa... ...partial}{\partial z}\boldsymbol{k}) \cdot(\boldsymbol{C} \times \boldsymbol{A})$$

$$= \boldsymbol{i}\cdot\frac{\partial}{\partial x}(\boldsymbol{C} \ti... ...symbol{k}\cdot\frac{\partial }{\partial z}\boldsymbol{C} \times \boldsymbol{A})$$

$$= \boldsymbol{i}\cdot(\frac{\partial C}{\partial x} \times \boldsym... ...ldsymbol{A} + \boldsymbol{C} \times \frac{\partial \boldsymbol{A}}{\partial z})$$

$$= -\boldsymbol{C} \cdot(\boldsymbol{i} \times \frac{\partial \bolds... ...mbol{C} \cdot(\boldsymbol{k} \times \frac{\partial \boldsymbol{A}}{\partial z})$$

$$= -\boldsymbol{C} \cdot(\nabla \times \boldsymbol{A})$$

(3)

$$\nabla \times (\boldsymbol{C} \times \boldsymbol{A}) = (\boldsymbol{i}\frac{\partial }{\partial x} + \boldsymbol{j}\frac... ...l{k}\frac{\partial }{\partial z}) \times (\boldsymbol{C} \times \boldsymbol{A})$$

$$= \boldsymbol{i} \times \frac{\partial (\boldsymbol{C}\times\boldsy... ... (\boldsymbol{C}\times\boldsymbol{A})}{\partial z} \hskip 0.5cm (\boldsymbol{C} )$$

$$= \boldsymbol{i} \times (\boldsymbol{C} \times \frac{\partial \bold... ...ldsymbol{C})\boldsymbol{B} - (\boldsymbol{A}\cdot\boldsymbol{B})\boldsymbol{C})$$

$$= (\boldsymbol{i} \cdot\frac{\partial \boldsymbol{A}}{\partial x})\... ... (\boldsymbol{k} \cdot\boldsymbol{C})\frac{\partial \boldsymbol{A}}{\partial z}$$

$$= \boldsymbol{C}(\nabla \cdot\boldsymbol{A}) - (\boldsymbol{C} \cdot\nabla)\boldsymbol{A}$$

5. Using the formula

$$\nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$$

$$\nabla(\boldsymbol{A} \cdot\boldsymbol{A}) = (\boldsymbol{A}\cdot\nabla)\boldsymbol{A} + (\boldsymbol{A}\cdot\... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{A})$$

$$= 2(\boldsymbol{A}\cdot\nabla)\boldsymbol{A} + 2\boldsymbol{A} \times(\nabla \times \boldsymbol{A})$$

Thus，

$$(\boldsymbol{A}\cdot\nabla)\boldsymbol{A} = \frac{1}{2}\nabla \vert\boldsymbol{A}\vert^2 - \boldsymbol{A} \times (\nabla \times \boldsymbol{A})$$

6. $\nabla \times (\rho \boldsymbol{F}) = (\nabla \rho)\times \boldsymbol{F} + \rho(\nabla \times \boldsymbol{F})$ また， $\nabla \times (\rho \boldsymbol{F}) = \nabla \times (\nabla p) = 0$implies $\nabla \times \boldsymbol{F} = -\frac{\nabla \rho}{\rho} \times \boldsymbol{F}$．Therefore，

$$\boldsymbol{F} \cdot(\nabla \times \boldsymbol{F}) = \boldsymbol{... ...l{F}) = \frac{\nabla \rho}{\rho}\cdot(\boldsymbol{F} \times \boldsymbol{F}) = 0$$

7.

$$\nabla \times \boldsymbol{A} = \left\vert\begin{array}{ccc} \bold... ...1)\boldsymbol{i} - (3z^2 - 3z^2)\boldsymbol{j} + (6x-6x)\boldsymbol{k} = {\bf0}$$

Note that there exits $\phi$ so that $\nabla \times \boldsymbol{A} = {\bf0}$implies $\boldsymbol{A} = -\nabla \phi$.

$$\boldsymbol{A} = (6xy + z^3)\boldsymbol{i} + (3x^2 - z)\boldsymbo... ...hi}{\partial y}\boldsymbol{j} + \frac{\partial \phi}{\partial z}\boldsymbol{k})$$

implies

$$-\frac{\partial \phi}{\partial x} = 6xy + z^3,\ -\frac{\partial \phi}{\partial y} = 3x^2 -z,\ -\frac{\partial \phi}{\partial z} = 3xz^2 - y$$

First，- $\frac{\partial \phi}{\partial x} = 6xy + z^3$implies $-\phi(x,y,z) = 3x^2y + xz^3 + f(y,z)$．Now partially differentiate $\phi$ by $y$，

$$-\frac{\partial \phi}{\partial y} = 3x^2 + f_{y}(y,z)$$

On the one hand，

$$-\frac{\partial \phi}{\partial y} = 3x^2 -z$$

implies

$$f_{y}(y,z) = -z$$

which also implies $f(y,z) = -yz + g(z)$．Thus， $-\phi(x,y,z) = 3x^2 y + xz^3 - yz + g(z)$. Here partially differentiate $\phi(x,y,z)$ by $z$，

$$-\frac{\partial \phi}{\partial z} = 3xz^2 - y + g'(z)$$

On the other hand，

$$-\frac{\partial \phi}{\partial z} = 3x^2 -y$$

implies

$$g'(z) = 0$$

Thus，$g(z) = c$ and $\phi(x,y,z) = -(3x^2 y + xz^3 - yz + c)$

8. ${\rm curl} \boldsymbol{F} = \left\vert\begin{array}{ccc} \boldsymbol{i} & \bol... ... \end{array} \right\vert = 2\boldsymbol{i} + \boldsymbol{j} + 2x\boldsymbol{k}$

8. Since $\boldsymbol{F}$ is conservative， there exists $f$ so that $\boldsymbol{F} = \nabla f$．Now we find ${\rm curl} \boldsymbol{F}$. Then

$$\nabla \times \boldsymbol{F} = \nabla \times \nabla f = \left\ver... ...} + (f_{xz} - f_{zx})\boldsymbol{j} + (f_{yx} - f_{xy})\boldsymbol{k} = {\bf0}$$

(2) Let ${\bf w} = w_{1}\boldsymbol{i} + w_{2}\boldsymbol{j} + w_{3}\boldsymbol{k}$. Then

$${\bf w} \times \boldsymbol{r} = \left\vert\begin{array}{ccc} \bol... ...oldsymbol{i} + (w_{3}x -w_{1}z)\boldsymbol{j} + (w_{1}y - w_{2}x)\boldsymbol{k}$$

Thus，

$$\nabla \times ({\bf w} \times \boldsymbol{r}) = \left\vert\begin{... ...= 2w_{1}\boldsymbol{i} + 2w_{2}\boldsymbol{j} + 2w_{3}\boldsymbol{k} = 2{\bf w}$$

(1)

$$\nabla \times (2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:... ...y} & \frac{\partial}{\partial z}\\ 2x+yz & zx & xy \end{array}\right\vert = 0$$

Then

$$\oint_{C}((2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}) \cdot d\boldsymbol{r} = 0$$

(2) Since $2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}$ has a scalar potential, we find the scalar potential.

$$(2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}... ...\partial y}\:\boldsymbol{j} + \frac{\partial \phi}{\partial z}\:\boldsymbol{k})$$

implies $\phi = -(x^2 + xyz) + C$．Note that

$$d\boldsymbol{r} = \frac{d\boldsymbol{r}}{dt}dt = (\frac{dx}{dt}\:\boldsymbol{i} + \frac{dy}{dt}\:\boldsymbol{j} + \frac{dz}{dt}\:\boldsymbol{k})dt$$

Then，

$$\int_{C}((2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}) \cdot d\boldsymbol{r} = \int_{C}(-\nabla \phi \cdot(\frac{dx}{dt}\:\boldsymbol{i} + \frac{dy}{dt}\:\boldsymbol{j} + \frac{dz}{dt}\:\boldsymbol{k}))dt$$

$$= \int_{{\rm P}}^{{\rm Q}}(-d \phi) = -\phi]_{P}^{Q} = (x^2 + xyz)]_{P}^{Q} = 4-6 - 1 = -3$$

Exercise Answer4.1

1.

(1) Gauss's divergence theorem implies

$$\int_{S} \frac{\boldsymbol{r}}{r^2}\cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot\frac{\boldsymbol{r}}{r^2}\;dV$$

Here，

$$\nabla \cdot\frac{\boldsymbol{r}}{r^2} = \nabla \cdot(r^{-2}\bold... ...c{\boldsymbol{r}}{r}\cdot\boldsymbol{r} + 3r^{-2} = -2r^{-2} + 3r^{-2} = r^{-2}$$

Therefore，

$$\int_{S} \frac{\boldsymbol{r}}{r^2}\cdot\boldsymbol{n}\;dS = \int_{V}\frac{1}{r^2}\;dV$$

(2) Tranform $\int_{S}\boldsymbol{r} \times \boldsymbol{n}\;dS$ into the surface integral．Using any constant vetor $\boldsymbol{C}$ and the triple scalar product,

$$\int_{S}\boldsymbol{C} \cdot(\boldsymbol{r} \times \boldsymbol{n})\;dS = \int_{S}\boldsymbol{n} \cdot(\boldsymbol{C} \times \boldsymbol{r})$$

$$= \int_{V}\nabla \cdot(\boldsymbol{C} \times \boldsymbol{r})\;dV = ... ...rtial y} & \frac{\partial }{\partial z}\\ x & y & z \end{array}\right\vert = 0$$

Here， $\boldsymbol{C}$ is a constant vector implies

$$\int_{S}\boldsymbol{r} \times \boldsymbol{n}\;dS = 0$$

(3) Gauss's dievergence theorem implies

$$\int_{S} r^{n} \boldsymbol{r}\cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot r^{n}\boldsymbol{r}\;dV = \int_{V} \left((\nabla r^{n})\cdot\boldsymbol{r} + r^{n}\nabla \cdot\boldsymbol{r}\right)\;dV$$

$$= \int_{V} (nr^{n-1}\nabla r \cdot\boldsymbol{r} + 3r^{n})\;dV = \int_{V}(nr^{n} + 3r^{n})\;dV$$

$$= (3+n)\int_{V}r^{n}\;dV$$

(4) Transfrom $\int_{S}r^{n}\boldsymbol{n}\;dS$ into surface integral．Using a constant vector $\boldsymbol{C}$,

$$\int_{S}\boldsymbol{C} \cdot(r^{n}\boldsymbol{n})\;dS = \int_{S}r^{n}\boldsymbol{C}\cdot\boldsymbol{n}\;dS = \int_{V}\nab... ...\int_{V}(\nabla r^{n}\cdot\boldsymbol{C} + r^{n}\nabla \cdot\boldsymbol{C})\;dV$$

$$= \int_{V}nr^{n-1}\frac{\boldsymbol{r}}{r}\cdot\boldsymbol{C}\;dV = \int_{V}\boldsymbol{C} \cdot nr^{n-2}\boldsymbol{r}\;dV$$

これimplies

$$\int_{S}r^{n}\boldsymbol{n} = \int_{V}nr^{n-2}\boldsymbol{r}\;dV$$

(5) Transform $\int_{S}r^{n}\boldsymbol{r} \times \boldsymbol{n}\;dS$ into surface integral．Using a constant vector $\boldsymbol{C}$ and the triple scalar product,

$$\int_{S}\boldsymbol{C}\cdot(r^{n}\boldsymbol{r} \times \boldsymbol{n})\;dS = \int_{S}\boldsymbol{n} \cdot\boldsymbol{C} \times (r^{n}\boldsymb... ...l{r})\;dS = \int_{V}\nabla \cdot(\boldsymbol{C} \times r^{n}\boldsymbol{r})\;dV$$

$$= \int_{V}(-\boldsymbol{C} \cdot(\nabla \times r^{n}\boldsymbol{r})... ...ot(\nabla r^{n} \times \boldsymbol{r} + r^{n} \nabla \times \boldsymbol{r})\;dV$$

$$= -\int_{V}\boldsymbol{C} \cdot(nr^{n-1}\frac{\boldsymbol{r}}{r} \times \boldsymbol{r} + 0)\;dV = 0$$

(6) Transform $\int_{S}r^{2}\boldsymbol{n}\;dS$ into surface integral．Using a constant vector $\boldsymbol{C}$, we have

$$\int_{S}\boldsymbol{C}\cdot(r^{2}\boldsymbol{n})\;dS = \int_{S}r^{2}\boldsymbol{C} \cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot(r^{2}\boldsymbol{C})\;dV$$

$$= \int_{V} (\nabla r^{2} \cdot\boldsymbol{C} + r^{2}\nabla \cdot\boldsymbol{C})\;dV = \int_{V} (2r\frac{\boldsymbol{r}}{r} \cdot\boldsymbol{C})\;dV$$

$$= \boldsymbol{C}\cdot\int_{V}2\boldsymbol{r}\;dV$$

Therefore，

$$\int_{S}r^{2}\boldsymbol{n}\;dS = 2\int_{V}\boldsymbol{r}\;dV$$

(7) $\int_{S}F(r)\boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$を用いると，

$$\int_{S}\boldsymbol{C}\cdot(F(r)\boldsymbol{n})\;dS = \int_{S}F(r)\boldsymbol{C} \cdot\boldsymbol{n}\;dS = \int_{V}\nab... ...\int_{V} (\nabla F(r) \cdot\boldsymbol{C} + F(r)\nabla \cdot\boldsymbol{C})\;dV$$

$$= \int_{V} \frac{dF}{dr}\frac{\boldsymbol{r}}{r} \cdot\boldsymbol{C})\;dV = \boldsymbol{C}\cdot\int_{V}\frac{dF}{dr}\frac{\boldsymbol{r}}{r}\;dV$$

Therefore，

$$\int_{S}F(r)\boldsymbol{n}\;dS = \int_{V}\frac{dF}{dr}\frac{\boldsymbol{r}}{r}\;dV$$

2. Using $\boldsymbol{C}$，we write into surfacee integral.

$$\int_{S}\boldsymbol{C} \cdot(\boldsymbol{n}\times (\nabla \phi))\;dS = \int_{S} \boldsymbol{n} \cdot(\nabla \phi \times \boldsymbol{C})\;dS = \int_{V} \nabla \cdot(\nabla \phi \times \boldsymbol{C})\;dV$$

$$= \int_{V}\left(\boldsymbol{C} \cdot(\nabla \times \nabla \phi) + \nabla \phi \cdot(\nabla \times \boldsymbol{C})\right)\;dV$$

Note that $(\nabla \times \nabla \phi = 0, \nabla \times \boldsymbol{C} = {\bf0})$.

$$\int_{S}\boldsymbol{C} \cdot(\boldsymbol{n}\times (\nabla \phi))\;dS = 0$$

3. Let $C$ be the boundary of the curved surface $S$, and let $S_{1}$ and $S_{2}$ be the curved surfaces separated by the boundary. Also，let the unit normal vector of $S_{1}$ be $\boldsymbol{n}_{1}$，the unt normal vector of $S_{2}$ be $\boldsymbol{n}_{2}$．Then let the unit normal vector of the surface $S$ be $\boldsymbol{n}$. We have $\boldsymbol{n}_{1} = \boldsymbol{n}, \boldsymbol{n}_{2} = -\boldsymbol{n}$ or $\boldsymbol{n}_{1} = -\boldsymbol{n}, \boldsymbol{n}_{2} = \boldsymbol{n}$．Here, ，

$$\int_{S_{1}}\boldsymbol{A}\cdot\boldsymbol{n}_{1}\;dS + \int_{S_{... ...mbol{A}\cdot\boldsymbol{n}_{2}\;dS = \int_{V}\nabla \cdot\boldsymbol{A}\;dV = 0$$

implies

$$\int_{S_{1}}\boldsymbol{A}\cdot\boldsymbol{n}_{1}\;dS = -\int_{S_{2}}\boldsymbol{A}\cdot\boldsymbol{n}_{2}\;dS$$

4.

(1) Gauss's divergence theorem implies

$$\int_{S}\phi \boldsymbol{A}\cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\phi \boldsymbol{A}\;dV$$

Note that， $\nabla \cdot(\phi\boldsymbol{A})= (\nabla \phi) \cdot\boldsymbol{A} + \phi \nabla \cdot\boldsymbol{A}$.

$$\int_{S}\phi \boldsymbol{A}\cdot\boldsymbol{n}dS = \int_{V}(\nabla \phi) \cdot\boldsymbol{A}\;dV + \int_{V} \phi \nabla \cdot\boldsymbol{A}\;dV$$

(2) Gauss's divergence theorem implies

$$\int_{S}(\boldsymbol{B} \times \boldsymbol{A})\cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\boldsymbol{B} \times \boldsymbol{A})\;dV$$

ここで，Exercise#i#>mplies $\nabla \cdot(\boldsymbol{B} \times \boldsymbol{A}) = \boldsymbol{A}\cdot(\nabla \times \boldsymbol{B}) - \boldsymbol{B}\cdot(\nabla \times \boldsymbol{A})$．Thus，

$$\int_{S}(\boldsymbol{B} \times \boldsymbol{A})\cdot\boldsymbol{n}... ...symbol{B})\;dV - \int_{V}\boldsymbol{B} \cdot(\nabla \times \boldsymbol{A})\;dV$$

(3) Gauss's divergence theorem implies

$$\int_{S}((\nabla \phi) \times \boldsymbol{A})\cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot((\nabla \phi) \times \boldsymbol{A})\;dV$$

Here，

$$\nabla \cdot((\nabla \phi) \times \boldsymbol{A}) = \boldsymbol{A... ...bla \times \boldsymbol{A}) = -(\nabla \phi) \cdot(\nabla \times \boldsymbol{A})$$

Therefore，

$$\int_{S}((\nabla \phi) \times \boldsymbol{A})\cdot\boldsymbol{n}dS = -\int_{V}(\nabla \phi) \cdot(\nabla \times \boldsymbol{A}\;dV$$

(4) Gauss's divergence theorem implies

$$\int_{S}\phi \boldsymbol{A} \cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\phi \boldsymbol{A})\;dV$$

Here， $\boldsymbol{A} = \nabla \phi, \nabla^2 \phi = 0$implies

$$\nabla \cdot(\phi \boldsymbol{A}) = (\nabla \phi)\cdot\boldsymbol... ... = \vert\boldsymbol{A}\vert^2 + \phi \nabla^2 \phi = \vert\boldsymbol{A}\vert^2$$

Therefore，

$$\int_{S}\phi \boldsymbol{A} \cdot\boldsymbol{n}dS = \int_{V}\vert\boldsymbol{A}\vert^2\;dV$$

5.

(1)
labelenshu:4-1-5-1 $\frac{\partial \phi}{\partial n}$ is the directional derivative of the direction of unit normal vector $\boldsymbol{n}$ implies $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$．Thus，

$$\int_{S}\psi\frac{\partial \phi}{\partial n}dS = \int_{S}\psi \nabla \phi \cdot\boldsymbol{n}dS$$

Here using Gauss's divergence theorem,

$$\int_{S}\psi \nabla \phi \cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\psi \nabla \phi)\;dV$$

Moreover， $\nabla \cdot(\psi \nabla \phi) = (\nabla \psi)\cdot(\nabla \phi) + \psi \nabla \cdot(\nabla \phi) = (\nabla \psi)\cdot(\nabla \phi) + \psi \nabla^2 \phi$implies

$$\int_{S}\psi \nabla \phi \cdot\boldsymbol{n}dS = \int_{V}\{\psi \nabla^2 \phi + (\nabla \psi) \cdot(\nabla \phi)\}\;dV$$

(2) $\frac{\partial \phi}{\partial n}$ is the directional derivative of the direction of unit normal vector implies $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$．Thus，

$$\int_{S}\phi\frac{\partial \phi}{\partial n}dS = \int_{S}\phi \nabla \phi \cdot\boldsymbol{n}dS$$

Here using Gauss's divergence theorem,

$$\int_{S}\phi \nabla \phi \cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\phi \nabla \phi)\;dV$$

なお， $\nabla \cdot(\phi \nabla \phi) = (\nabla \phi)\cdot(\nabla \phi) + \phi \nabla \cdot(\nabla \phi) = \vert\nabla \phi\vert^2 + \phi \nabla^2 \phi$implies

$$\int_{S}\phi \nabla \phi \cdot\boldsymbol{n}dS = \int_{V}\{\phi \nabla^2 \phi + \vert\nabla \psi\vert^2\}\;dV$$

(3) The result of (1) subtracts the result of (2). Then，

$$\int_{S}(\psi\frac{\partial \phi}{\partial n} - \phi\frac{\partial \phi}{\partial n})dS = \int_{V}\{\psi\nabla^2\phi - \phi \nabla^2 \phi\}\;dV$$

(4) $\phi$ is harmonic function means that $\nabla^2 \phi = 0$．Therefore，using (2),

$$\int_{S}\phi\frac{\partial \phi}{\partial n}dS = \int_{V}\vert\nabla \phi\vert^2\;dV$$

(5) $\phi, \psi$ are harmonic functions means that $\nabla^2 \phi = \nabla^2 \psi = 0$．Therefore，using (3),

$$\int_{S}(\psi\frac{\partial \phi}{\partial n} - \phi\frac{\partial \phi}{\partial n})dS = \int_{V}\{\psi\nabla^2\phi - \phi \nabla^2 \phi\}\;dV = 0$$

(6) $\phi$ is harmonic function. Then (4)implies

$$\int_{S}\phi\frac{\partial \phi}{\partial n}dS = \int_{V}\vert\nabla \phi\vert^2\;dV$$

Let $\phi = 0$ on $S$. Then $\int_{V}\vert\nabla \phi\vert^2\;dV = 0$ and $\vert\nabla \phi\vert^2 = 0$．Thus， $\vert\nabla \phi\vert = 0$. Hence, $\nabla \phi = 0$．Therefore，$\phi$は定数．

6. If $\int_{S}{\bf A}\cdot\boldsymbol{n}dS = 0$，then Gauss's divergence theorem implies

$$\int_{S}{\bf A}\cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot{\bf A}\;dV = 0$$

Therefore， $\nabla \cdot{\bf A} = 0$. The theorem 3.4 implies ${\bf A}$ has a vector potential.

7. $\int_{S}{\bf A} \times \boldsymbol{n}dS = {\bf0}$．Here, using a constant vector ${\bf C}$, rewrite into surface integral.

$$\int_{S}{\bf C} \cdot({\bf A} \times \boldsymbol{n})\;dS = \int_{S} \boldsymbol{n} \cdot({\bf C} \times {\bf A})\;dS$$

Gauss's divergence theorem implies

$$\int_{S} \boldsymbol{n} \cdot({\bf C} \times {\bf A})\;dS = \int_{V}\nabla \cdot({\bf C} \times {\bf A})dV = 0$$

Here，

$$0 = \nabla \cdot({\bf C} \times {\bf A}) = {\bf A} \cdot\nabla \t... ...C} - {\bf C} \cdot\nabla \times {\bf A} = -{\bf C} \cdot(\nabla \times {\bf A})$$

Therefore， $\nabla \times {\bf A} = 0$ and ${\bf A}$ has a scalar potential.

(1) $\boldsymbol{F} = xz^2\boldsymbol{i} + (x^2 y - z^3)\boldsymbol{j} + (2xy + y^2 z)\boldsymbol{k}$. Then we find $\boldsymbol{F}$

$${\rm div}\boldsymbol{F} = \nabla \cdot(xz^2\boldsymbol{i} + (x^2 y - z^3)\boldsymbol{j} + (2xy + y^2 z)\boldsymbol{k}) = z^2 + x^2 + y^2$$

Thus

$$\iint_{S} xz^2 dydz + (x^2 y - z^3)dzdx + (2xy + y^2 z)dxdy = \iiint_{V}(x^2 + y^2 + z^2)dV$$

Now $V$ is an upper sphere with the radius $a$. Then by the spherical coordinate transformation $(\rho,\phi,\theta)$, we have

$$\int_{\theta = 0}^{2\pi}\int_{\phi = 0}^{\frac{\pi}{2}}\int_{\rho = 0}^{a}\rho^2 \rho^2 \sin{\phi}d\rho d\phi d\theta = \int_{\theta = 0}^{2\pi} d\theta \int_{\phi = 0}^{\frac{\pi}{2}} ... ...} \int_{\rho = 0}^{a}\rho^4 d\rho = 2\pi(1)(\frac{a^5}{5}) = \frac{2\pi a^5}{5}$$

(1)

$$\iint_{S_{1}}xz^2 dydz + (x^2 y - z^3)dzdx + (2xy + y^2 z)dxdy = \iint_{S_{1}}xz^2 dydz + \iint_{S_{1}}(x^2 y -z^3)dxdy$$

$$+ \iint_{S_{1}}(2xy + y^2 z)dxdy$$

Since $z = \sqrt{a^2 - x^2 - y^2}$，projection onto $yz$ plane，$x = 0$ implies $\Omega_{1} = \{(y,z) : y^2 + z^2 \leq a^2, z \geq 0\}$. Here，we consier $x > 0$ and $x < 0$.

$$\iint_{S_{1}}xz^2 dydz = \int_{y=-a}^{a}\int_{z =0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 -... ... - \int_{y=-a}^{a}\int_{z =0}^{\sqrt{a^2 -y^2}}- z^2 \sqrt{a^2 - y^2 - z^2}dzdy$$

$$= 4\int_{y=0}^{a}\int_{z = 0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 - z^2}dzdy$$

Next projection onto $xz$ plane. Then $y = 0$ implies $\Omega_{1} = \{(x,z) : x^2 + z^2 \leq a^2, z \geq 0\}$．Thus，for $y > 0$ and $y < 0$,

$$\iint_{S_{1}}(x^2y -z^3)dzdx = 2\int_{x=-a}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}(x^2\sqrt{a^2 - x^2 ... ...zdx = 4\int_{x=0}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}x^2\sqrt{a^2 - x^2 - z^2}dzdx$$

Lastly，projection onto $xy$ plane. Then $x = 0$ implies $\Omega_{1} = \{(x,y) : x^2 + y^2 \leq a^2\}$．

$$\iint_{S_{1}}(2xy + y^2 z)dzdx = \int_{x=-a}^{a}\int_{y=-\sqrt{z^2 - x^2}}^{\sqrt{a^2 - x^2}}(2xy ... ...{a}\int_{y=-\sqrt{z^2 - x^2}}^{\sqrt{a^2 - x^2}} y^2\sqrt{a^2 - x^2 - y^2} dydx$$

$$= 4\int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - y^2}dydx$$

Next consider the surface integral for $S_{2} = \{(x,y) : x^2 + y^2 \leq a^2\}$．First projection onto，$yz$ plane. $S_{2}$ maps to $\Omega_{2} = \{(y,z) : -a \leq y \leq a, z= 0\}$.

$$\iint_{S_{2}}xz^2 dydz = 0$$

By Projecting onto $xz$ plane，$S_{2}$ maps to $\Omega_{2} = \{(x,z) : -a \leq x \leq a, z= 0\}$.，

$$\iint_{S_{2}}(x^2 y - z^3) dzdx = 0$$

By projecting onto $xy$ plane, $S_{2}$ maps to $\Omega_{2} = \{(x,y) : x^2 + y^2 \leq a^2\}$.

$$\iint_{S_{2}}(2xy + y^2z)dzdy = \iint_{S_{2}}2xy dzdx = \int_{-a}^{a}\int_{-\sqrt{a^2 -x^2}}^{\sqrt{a^2 -x^2}}2xydydx = 0$$

Adding these，

$$\int_{S_{1}}\boldsymbol{F}\cdot\boldsymbol{n}\;dS = 4\int_{y=0}^{a}\int_{z = 0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 ... ...zdy + 4\int_{x=0}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}x^2\sqrt{a^2 - x^2 - z^2}dzdx$$

$$+ 4\int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - ... ...x = 12 \int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - y^2}dydx$$

Here using the polar coorinates，we have $x = r\cos{\theta}, \ y = r\sin{\theta}$. Thus

$$\int_{S_{1}}\boldsymbol{F}\cdot\boldsymbol{n}\;dS = 12\int_{\theta = 0}^{\pi/2}\int_{r = 0}^{a}r^2 \sin^{2}{\theta}\s... ...2(\int_{0}^{a}r^3 \sqrt{a^2 - r^2}dr) (\int_{0}^{\pi/2}\sin^{2}{\theta}d\theta)$$

$$= 12(\int_{t=a^2}^{0} \sqrt{t}(a^2-t)(\frac{dt}{-2}) ) (\frac{\pi}{4}) = (3\pi)(\frac{1}{2}\int_{0}^{a^2}(a^2 t^{1/2} - t^{3/2})dt)$$

$$= (\frac{3\pi}{2})\left[\frac{2}{3}a^2 t^{3/2} - \frac{2}{5}t^{5/2}... ...^{a^2} = (\frac{3\pi}{2})(\frac{2}{3}a^5 - \frac{2}{5}a^5) = \frac{2\pi a^2}{5}$$

(1) Using the arbitrary constant, we express by the surface integral and apply Gauss's divergence theorem. Then

$$\int_{S}\boldsymbol{C} \cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot\boldsymbol{C}\;dV = 0$$

Therefore， $\int_{S}\boldsymbol{C} \cdot\boldsymbol{n}\;dS = 0$

(2) By Gauss's divergence theorem,

$$\int_{S}\boldsymbol{r}\cdot\boldsymbol{n}\;dS = \int_{V}\nabla \c... ...al x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z})\;dV = 3V$$

(3) Using a constant vector，we express by the surface integral and apply the triple scalar product and Gauss's divergance theorem,

$$\int_{S}\boldsymbol{C} \cdot(\boldsymbol{n} \times \boldsymbol{r})\;dS = \int_{S}\boldsymbol{n} \cdot(\boldsymbol{r} \times \boldsymbol{C})\;dS = \int_{V}\nabla \cdot(\boldsymbol{r} \times \boldsymbol{C})\;dV$$

$$= \int_{V}(\boldsymbol{C} \cdot\nabla \times \boldsymbol{r} + \boldsymbol{r} \cdot\nabla \times \boldsymbol{C})\;dV = 0$$

(1) The problem of surface integral is always rewrite into $\int_{S}\boldsymbol{A}\cdot\boldsymbol{n}\;dS$．In this case， $\frac{\partial \phi}{\partial n}$ isa directional derivative in the direction of the unit normal vector $\boldsymbol{n}$. Then $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$．Therefore,，

$$\int_{S}\frac{\partial \phi}{\partial n} \;dS = \int_{S}\nabla \phi \cdot\boldsymbol{n}\;dS$$

Now using Gauss's divergence theorem,

$$\int_{S}\frac{\partial \phi}{\partial n}dS = \int_{S}\nabla \phi \cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot(\nabla \phi)\;dV = \int_{V}\nabla^2 \phi\;dV$$

(2) $\phi$ is harmonic function. Then $\nabla^2 \phi = 0$. Thus，(1) implies

$$\int_{S}\frac{\partial \phi}{\partial n}\;dS = \int_{V}\nabla^2 \phi\;dV = 0$$

Exercise Answer4.2

1.

1. Stokes' theorem implies

$$\int_{C}(\nabla \phi \psi)\cdot d\boldsymbol{r} = \int_{S}(\nabla \times (\nabla \phi \psi))\cdot\boldsymbol{n}dS = 0$$

Here，

$$\nabla \phi \psi = \psi (\nabla \phi) + \phi (\nabla \psi)$$

implies

$$\int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r} = -\int_{C}\psi(\nabla \phi)\cdot d\boldsymbol{r}$$

2.

(1) Rewrite a line integral into $\int_{C}{\bf A}\cdot d\boldsymbol{r}$．To do so, using a constant vector ${\bf C}$ and the triple scalar product，we have

$$\int_{C}{\bf C} \cdot\boldsymbol{r} \times d\boldsymbol{r} = \int_{C}d\boldsymbol{r} \cdot{\bf C} \times \boldsymbol{r}$$

Here, using Stokes'theorem,

$$\int_{C}d\boldsymbol{r} \cdot{\bf C} \times \boldsymbol{r} = \int_{S} (\nabla \times ({\bf C} \times \boldsymbol{r}))\cdot\boldsymbol{n}dS$$

Now

$$\nabla \times ({\bf C} \times \boldsymbol{r}) = (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...oldsymbol{k}\frac{\partial}{\partial z}) \times ({\bf C} \times \boldsymbol{r})$$

$$= \boldsymbol{i} \times \frac{\partial ({\bf C} \times \boldsymbol{... ...oldsymbol{k} \times \frac{\partial ({\bf C} \times \boldsymbol{r})}{\partial z}$$

$$= \boldsymbol{i} \times ({\bf C} \times \boldsymbol{r}_{x}) + \bold... ...\boldsymbol{r}_{y}) + \boldsymbol{k} \times ({\bf C} \times \boldsymbol{r}_{z})$$

$$= (\boldsymbol{i} \cdot\boldsymbol{r}_{x}) {\bf C} - (\boldsymbol{i... ...dot\boldsymbol{r}_{z}){\bf C} - (\boldsymbol{k} \cdot{\bf C})\boldsymbol{r}_{z}$$

$$= (\boldsymbol{i} \cdot\boldsymbol{r}_{x} + \boldsymbol{j} \cdot\bo... ...ot{\bf C})\boldsymbol{r}_{y} + (\boldsymbol{k} \cdot{\bf C})\boldsymbol{r}_{z})$$

$$= (\nabla \cdot\boldsymbol{r}){\bf C} - ({\bf C} \cdot\nabla)\boldsymbol{r} +$$

Here，

$$\{({\bf C} \cdot\nabla)\boldsymbol{r}\} \cdot\boldsymbol{n} = {\bf C} \cdot\nabla (\boldsymbol{r} \cdot\boldsymbol{n})$$

holds，

$$\nabla \times ({\bf C} \times \boldsymbol{r}) \cdot\boldsymbol{n} = (\nabla \cdot\boldsymbol{r}){\bf C} \cdot\boldsymbol{n} - ({\bf C} \cdot\nabla){\bf A} \cdot\boldsymbol{n}$$

$$= {\bf C} \cdot\{(\nabla \cdot\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}$$

This implies

$$\int_{C}{\bf C} \cdot\boldsymbol{r} \times d\boldsymbol{r} = \int... ...t\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}dS$$

Nete that $C$ is a constant vector. Then，

$$\int_{C} \boldsymbol{r} \times d\boldsymbol{r} = \int_{S} \{(\nab... ...t\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}dS$$

Finally,

$$\nabla (\boldsymbol{r} \cdot\boldsymbol{n}) = (\boldsymbol{n} \cd... ...bol{r}) + \boldsymbol{r} \times (\nabla \times \boldsymbol{n}) = \boldsymbol{n}$$

Therefore，

$$\int_{C}\boldsymbol{r} \times d\boldsymbol{r} = \int_{S}(3\boldsymbol{n} - \boldsymbol{n})dS = 2\int_{S}\boldsymbol{n}dS$$

(2) Using Stokes' theorem,

$$\int_{C}r^{k}\boldsymbol{r}\cdot d\boldsymbol{r} = \int_{S} (\nabla \times (r^{k}\boldsymbol{r}))\cdot\boldsymbol{n}dS$$

Note that，

$$(\nabla r^{k})\times \boldsymbol{r} + r^{k} \nabla \times \boldsy... ...bol{r} + r^{k}(0) = kr^{k-2} \frac{\boldsymbol{r}}{r} \times \boldsymbol{r} = 0$$

Then

$$\int_{C}r^{k}\boldsymbol{r}\cdot d\boldsymbol{r} = 0$$

(3) Let $C$ be a constant vector. Then using Stokes' theorem，

$$\int_{C}{\bf C} \cdot\boldsymbol{r}(\nabla \phi) \cdot d\boldsymbol{r} = \int_{S} (\nabla \times ({\bf C} \cdot\boldsymbol{r}(\nabla \phi)))\cdot\boldsymbol{n}dS$$

Here，

$$\nabla \times ({\bf C} \cdot\boldsymbol{r} (\nabla \phi)) = \nabla({\bf C} \cdot\boldsymbol{r}) \times (\nabla \phi) + ({\bf ... ...a \times \nabla \phi = \nabla({\bf C} \cdot\boldsymbol{r}) \times (\nabla \phi)$$

$$\nabla({\bf C} \cdot\boldsymbol{r}) = (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...al y} + \boldsymbol{k}\frac{\partial ({\bf C} \cdot\boldsymbol{r})}{\partial z}$$

$$= \boldsymbol{i} ({\bf C} \cdot\frac{\partial \boldsymbol{r}}{\part... ... y}) + \boldsymbol{k} ({\bf C} \cdot\frac{\partial \boldsymbol{r}}{\partial z})$$

$$= \boldsymbol{i} ({\bf C} \cdot\boldsymbol{i}) + \boldsymbol{j} ({\... ...} \cdot\boldsymbol{j}) + \boldsymbol{k} ({\bf C} \cdot\boldsymbol{k}) = {\bf C}$$

implies

$$\int_{C}{\bf C} \cdot(\nabla (\nabla \phi))\cdot d\boldsymbol{r} = \int_{S} {\bf C} \times ({\bf C} \cdot\boldsymbol{r}(\nabla \phi))\cdot\boldsymbol{n}dS = \int_{S}({\bf C} \times \nabla \phi)\cdot\boldsymbol{n}dS$$

$$= \int_{S}{\bf C} \cdot(\nabla \phi \times \boldsymbol{n})dS$$

Therefore，

$$\int_{C}\boldsymbol{r}(\nabla \phi)\cdot d\boldsymbol{r} = \int_{S}(\nabla \phi \times \boldsymbol{n})dS$$

3. Stokes' theorem implies

$$\int_{C}\phi{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times \phi {\bf A})\cdot\boldsymbol{n}dS$$

Note that

$$\nabla \times (\phi {\bf A}) = (\nabla \phi)\times {\bf A} + \phi \nabla \times {\bf A}$$

Then

$$\int_{C}\phi{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times... ...t\boldsymbol{n}dS + \int_{S}\{\phi \nabla \times {\bf A}\}\cdot\boldsymbol{n}dS$$

4. Rewrite the line integral into $\int_{C}{\bf A}\cdot d\boldsymbol{r}$．Then using a constant vector ${\bf C}$ and the triple scalar product，and Stokes' theorem，we have

$$\int_{C}{\bf C} \cdot\frac{\boldsymbol{r} \times d\boldsymbol{r}}... ...}(\nabla \times \frac{{\bf C} \times \boldsymbol{r}}{r^3})\cdot\boldsymbol{n}dS$$

．

Here using the triple scalar product, we have

$$\nabla \times \frac{{\bf C} \times \boldsymbol{r}}{r^3} = (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...k}\frac{\partial}{\partial z}) \times {\bf C} \times \frac{\boldsymbol{r}}{r^3}$$

$$= \boldsymbol{i} \times {\bf C} \times \frac{\partial}{\partial x}(... ...} \times {\bf C} \times \frac{\partial}{\partial z}(\frac{\boldsymbol{r}}{r^3})$$

$$= \left(\boldsymbol{i}\cdot\frac{\partial}{\partial x}(\frac{\bolds... ...dsymbol{j} \cdot{\bf C})\frac{\partial}{\partial y}(\frac{\boldsymbol{r}}{r^3})$$

$$+ \left(\boldsymbol{k}\cdot\frac{\partial}{\partial z}(\frac{\bolds... ...dsymbol{k} \cdot{\bf C})\frac{\partial}{\partial z}(\frac{\boldsymbol{r}}{r^3})$$

$$= {\bf C}(\nabla \cdot(\frac{\boldsymbol{r}}{r^3})) - {\bf C}\cdot\nabla (\frac{\boldsymbol{r}}{r^3})$$

Furthermore,，

$$\nabla \cdot(\frac{\boldsymbol{r}}{r^3}) = \nabla \cdot(r^{-3}\bo... ...}\frac{\boldsymbol{r}}{r}\cdot\boldsymbol{r} + 3r^{-3} = -3r^{-3} + 3r^{-3} = 0$$

implies

$$\int_{C}{\bf C} \cdot\frac{\boldsymbol{r} \times d\boldsymbol{r}}... ...int_{S}\{{\bf C} \cdot\nabla(\frac{\boldsymbol{r}}{r^3})\}\cdot\boldsymbol{n}dS$$

Here，${\bf C}$ is a constant vector.

$$\int_{C}\frac{\boldsymbol{r} \times d\boldsymbol{r}}{r^3} = -\int_{S}\nabla(\frac{\boldsymbol{r}}{r^3})\cdot\boldsymbol{n}dS$$

Finally，

$$\boldsymbol{n} \cdot\nabla (\frac{\boldsymbol{r}}{r^3}) = (n_{1}\frac{\partial}{\partial x} + n_{2}\frac{\partial}{\partial y} + n_{3}\frac{\partial}{\partial z})(r^{-3}\boldsymbol{r})$$

$$= n_{1}(\frac{\partial}{\partial x}(\frac{\boldsymbol{r}}{r^3}) + \... ...boldsymbol{r}}{r^3}) + \frac{1}{r^3}\frac{\partial \boldsymbol{r}}{\partial z})$$

$$= \boldsymbol{n} \cdot\nabla(\frac{1}{r^3})\boldsymbol{r} + \frac{\... ...oldsymbol{r}}{r}\cdot\boldsymbol{n} \boldsymbol{r} + \frac{\boldsymbol{n}}{r^3}$$

$$= -3\frac{\boldsymbol{n} \cdot\boldsymbol{r}}{t^5}\boldsymbol{r} + \frac{\boldsymbol{n}}{r^3}$$

5. Stokes' theorem implies

$$0 = \int_{C}{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times {\bf A})\cdot\boldsymbol{n}dS$$

Thus， $\nabla \times {\bf A} = 0$．Therefore，${\bf A}$ has a scalar potential.

6 The boundary of $S$, $\partial S$ is a circle $x^2 + y^2 = 4$．Thus the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = 2\cos{t}\boldsymbol{i} + 2\sin{t}\boldsymbol{j}$. Then we find the line integral

$$\oint_{\partial S}\boldsymbol{F}\cdot d\boldsymbol{r} = \oint_{\partial S}(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j} + \boldsymbol{k})\cdot(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j})dt$$

$$= \int_{0}^{2\pi}[4\sin^{2}{t} + 4\cos^{2}{t}]dt = 8\pi$$

Next we find the surface integral.．

$$\nabla \times \boldsymbol{F} = \left\vert\begin{array}{ccc} \bol... ...c{\partial}{\partial z}\\ -y & x & 1 \end{array}\right\vert = 2\boldsymbol{k}$$

We find the unit normal vector. Then

$$\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = \left\vert\begin{a... ...t\vert = \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}$$

Thus,

$$\boldsymbol{n} = \frac{ \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\b... ... \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}\Vert }$$

Therefore,，

$$\iint_{S}(\nabla \times \boldsymbol{F}) \cdot\boldsymbol{n} dS = 2\iint_{S}\boldsymbol{k} \cdot\frac{\boldsymbol{r}_{x} \times \bo... ...bol{r}_{y} \Vert} \Vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} \Vert dx dy$$

$$= 2\iint_{\Omega}\boldsymbol{k} \cdot\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} dx dy$$

$$= \iint_{\Omega}2 dx dy = 2(4\pi) = 8 \pi$$

This shows that Stokes' theorem hols.

7.

$$\nabla \times ((2x+yz)\boldsymbol{i} + zx\boldsymbol{j} + xy\bold... ...\frac{\partial}{\partial z}\\ 2x+yz & zx & xy \end{array}\right\vert = {\bf0}$$

Then，${\bf F}$ has a scalar potential．So, we find $f$ so that ${\bf F} = \nabla f$． $f_{x} = 2x+yz, f_{y} = zx, f_{z} = xy$ implies

$$f = ^int f_{x}dx = x^2 + xyz + g(y,z)$$

Here， $f_{y} = zx$ imlies $f_{y} = xz + g_{y}(y,z)$. Thus，$g_{y} = 0$. in other words， $g(y,z) = h(z)$. Next， $f_{z} = xy$ implies $f_{z} = xy + h'(z)$. Thus $h(z) = 0$．Here，let $h(z) = 0$. Then

$$f = x^2 + xyz$$

Using this, we have

$$\int_{C}((2x+yz)\boldsymbol{i} + zx\boldsymbol{j} + xy\boldsymbol{k}) \cdot d\boldsymbol{r} = f(2,-1,3) - f(1,0,-1) = -2 - 1 = -3$$

8 We first calculate $\nabla \times (\phi \nabla \psi)$.

$$\nabla \times (\phi \nabla \psi) = (\nabla \phi) \times (\nabla \psi) + \phi \nabla \times (\nabla \psi)$$

Here， $\nabla \times (\nabla \psi) = {\bf0}$ implies

$$\nabla \times (\phi \nabla \psi) = (\nabla \phi) \times (\nabla \psi)$$

Therefore, by Stokes' theorem，

$$\iint_{S}\{(\nabla \phi) \times (\nabla \psi)\}\cdot\boldsymbol{n}dS = \int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r}$$