Exercise Answer

Exercise Answer3.2

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

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

2. 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

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

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詳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，

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

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,

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

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

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

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.

合成関数の微分法implies

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

Exercise詳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.

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．Usiing 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.

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

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

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

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.