벡터 미적분 전체 문제

(a) by direct evaluation: The curl of $\mathbf{G}$ is

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{G} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ x & -z & y \end{matrix} \right] =2\hat{\boldsymbol{\imath}} \end{aligned}$$

The part of $S$ in the first octant is sketched in the figure on the left below. $S$ consists of two parts — the cylindrical surface

$$S_1 = \big\{ \ (x,y,z)\ \big|\ y^2+z^2=9,\ 0\le x\le 5\ \big\}$$

and the disc

$$S_2 = \big\{ \ (x,y,z)\ \big|\ x=0,\ y^2+z^2\le 9\ \big\}$$

The normal $\hat{\mathbf{n}}$ to $S_1$ always points radially outward from the cylinder and so always has $\hat{\boldsymbol{\imath}}$ component zero. The normal of $S_2$ is $-\hat{\boldsymbol{\imath}}$. So the flux is

$$\begin{aligned} \iint _S { \boldsymbol{\nabla} }\times \mathbf{G} \cdot \hat{\mathbf{n}}\,\mathrm{d}S &= \iint _{S_1} 2\hat{\boldsymbol{\imath}} \cdot \hat{\mathbf{n}}\,\mathrm{d}S + \iint _{S_2} 2\hat{\boldsymbol{\imath}} \cdot (-\hat{\boldsymbol{\imath}})\,\mathrm{d}S \\ &= -2 \iint _{S_2} \mathrm{d}S \\ &= -2 \big(\pi 3^2\big) = -18\pi \end{aligned}$$

(a) using Stokes' theorem: Let's use Stokes' theorem. The boundary $\partial S$ of $S$ is the cirlce $y^2+z^2=9$, $x=5$, oriented clockwise when viewed from far down the $x$-axis. We'll parametrize it by $\mathbf{r}(\theta)=\big(5, 3\cos\theta, -3\sin\theta\big)$. Then Stokes' theorem gives

$$\begin{aligned} \iint _S { \boldsymbol{\nabla} }\times\mathbf{G}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\oint_{\partial S}\mathbf{G}\cdot\mathrm{d}\mathbf{r} \\ &=\int_0^{2\pi} \big(5\,,\,3\sin\theta,3\cos\theta\big)\cdot \big(0\,,\,-3\sin\theta\,,\,-3\cos\theta\big)\,\mathrm{d}\theta \\ &=\int_0^{2\pi} \big(-9\sin^2\theta-9\cos^2\theta\big)\,\mathrm{d}\theta \\ &=-18\pi \end{aligned}$$

(b) This time we'll use the divergence theorem. The surface $S$ is not closed. So we'll use the auxilary surface formed by “topping $S$ off” with the cap $T=\big\{ \ (5,y,z)\ \big|\ y^2+z^2\le 9\ \big\}$. If we give $T$ the normal vector $\hat{\boldsymbol{\imath}}$, this auxiliary surface, the union of $S$ and $T$, is the boundary of $V=\big\{ \ (x,y,z)\ \big|\ y^2+z^2\le 9,\ 0\le x\le 5\ \big\}$. So the divergence theorem gives

$$\begin{aligned} \iint _S \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S +\iint _T \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &= \iint _{\partial V} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \\ &=\iiint _V { \boldsymbol{\nabla} }\cdot \mathbf{F}\ \mathrm{d}V \\ &=0 \end{aligned}$$

since ${ \boldsymbol{\nabla} }\cdot \mathbf{F} = 0$. Thus the flux of interest is

$$\begin{aligned} \iint _S \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\iint _T \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =-\iint _T \mathbf{F}\cdot\hat{\boldsymbol{\imath}}\,\mathrm{d}S =-\iint _T (2+z)\,\mathrm{d}S \\ &=-2\iint _T \,\mathrm{d}S \qquad\text{since $\iint _T z\,\mathrm{d}S=0$, because $z$ is odd}\\ &= -18\pi \qquad\text{since $T$ has area $9\pi$} \end{aligned}$$

(a) Since

• $\frac{y}{x}$ is defined when $x\ne 0$ and
• $x^{1+x^2}=e^{(1+x^2)\ln x}$ is defined when $\ln x$ is defined, which is when $x>0$ (assuming that we are not allowed to use complex numbers) and
• $y^{1+y^2}=e^{(1+y^2)\ln y}$ is defined when $\ln y$ is defined, which is when $y>0$ and
• $\cos^5(\ln z)$ is defined when $\ln z$ is defined, which when $z>0$

the domain of $\mathbf{F}$ is

$$D=\big\{ \ (x,y,z)\ \big|\ x>0,\ y>0,\ z>0\ \big\}$$

(b) The domain $D$ is both connected (any two points in $D$ can be joined by a curve that lies completely in $D$) and simply connected (any simple closed curve in $D$ can be shrunk to a point continuously in $D$).

(c) The curl of $\mathbf{F}$ is

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ \frac{y}{x} +x^{1+x^2} & x^2-y^{1+y^2} & \cos^5(\ln z) \end{matrix} \right] =\big(2x-{1}/{x}\big)\hat{\mathbf{k}} \end{aligned}$$

(d) The integrand for direct evaluation looks very complicated. On the other hand ${ \boldsymbol{\nabla} }\times\mathbf{F}$ is quite simple. So let's try Stokes' thoerem. Denote

$$S=\big\{ \ (x,y,z)\ \big|\ 2\le x\le 4,\ 2\le y\le 4,\ z=2\ \big\}$$

The boundary of $S$ is $C$. Because of the clockwise orientation of $C$, we assign the normal vector $-\hat{\mathbf{k}}$ to $S$. See the sketch below

Then, by Stokes' theorem,

$$\begin{aligned} \oint_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} &=\iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S = \iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot(-\hat{\mathbf{k}})\,\mathrm{d}S = -\iint _S \big(2x-\tfrac{1}{x}\big)\,\mathrm{d}S \\ &= -\int_2^4\mathrm{d}x\int_2^4\mathrm{d}y\ \big(2x-\tfrac{1}{x}\big) =-\int_2^4\mathrm{d}x\ 2\big(2x-\tfrac{1}{x}\big) = - 2\Big[x^2-\ln x\Big]_2^4 \\ &=-2\big[12-\ln 2\big] =2\ln 2-24 \end{aligned}$$

(e) Since ${ \boldsymbol{\nabla} }\times\mathbf{F}$ is not $\mathbf{0}$, $\mathbf{F}$ cannot be conservative.

Considering that there are ten line segments in $C$, it is probably not very efficient to use direct evaluation. Two other possible methods come to mind. If $\mathbf{F}$ is conservative, we can use $\mathbf{F}$'s potential. Even if $\mathbf{F}$ is not conservative, it may be possible to efficiently use Stokes' (or Green's) theorem. So let's compute

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ y & 2x-10 & 0 \end{matrix} \right] =\hat{\mathbf{k}} \end{aligned}$$

As ${ \boldsymbol{\nabla} }\times\mathbf{F}\ne\mathbf{0}$, the vector field $\mathbf{F}$ is not conservative. As ${ \boldsymbol{\nabla} }\times\mathbf{F}\ne\mathbf{0}$ is very simple, it looks like Stokes' theorem could provide an efficient way to compute the integral. The left figure below contains a sketch of $C$.

The curve $C$ is not closed, and so is not the boundary of a surface, so we cannot apply Stokes' theorem directly. But we can easily come up with a surface whose boundary contains $C$. Let $R$ be the shaded region in the figure on the right above. The boundary $\partial R$ of $R$ consists of two parts — $C$ and the line segment $L$. The normal of $R$ for $-\hat{\mathbf{k}}$ (since $\partial R$ is oriented clockwise). So Stokes' theorem gives

$$\begin{aligned} \int_C \mathbf{F} \cdot\mathrm{d}\mathbf{r} + \int_L \mathbf{F} \cdot\mathrm{d}\mathbf{r} = \iint _R { \boldsymbol{\nabla} }\times\mathbf{F} \cdot(-\hat{\mathbf{k}})\ \mathrm{d}S = \iint _R (\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{k}})\ \mathrm{d}S = -\text{Area}(R) \end{aligned}$$

$R$ is the union of $5$ triangles, each of height $1$ and base $1$. So

$$\text{Area}(R) = 5\times \frac{1}{2}\times 1\times 1 =\frac{5}{2}$$

If we denote by $-L$ the line segment from $(0,0)$ to $(5,5)$, we can parametrize $-L$ by $\mathbf{r}(t) = t(5,5)$, $0\le t\le 1$ and

$$\begin{aligned} \int_{-L} \mathbf{F} \cdot\mathrm{d}\mathbf{r} &=\int_0^1 \overbrace{\big(5t\,\hat{\boldsymbol{\imath}} + (10t-10)\,\hat{\boldsymbol{\jmath}}\big)}^{\mathbf{F}(\mathbf{r}(t))} \cdot\overbrace{(5\,\hat{\boldsymbol{\imath}}+5\,\hat{\boldsymbol{\jmath}})}^{\mathbf{r}'(t)}\ \mathrm{d}t =\int_0^1 5\big(15t-10\big) \mathrm{d}t =25\left(\frac{3}{2}-2\right) \\ &=-\frac{25}{2} \end{aligned}$$

All together

$$\begin{aligned} \int_C \mathbf{F} \cdot\mathrm{d}\mathbf{r} = -\text{Area}(R) - \int_L \mathbf{F} \cdot\mathrm{d}\mathbf{r} = -\text{Area}(R) + \int_{-L} \mathbf{F} \cdot\mathrm{d}\mathbf{r} =-\frac{5}{2}-\frac{25}{2} =-15 \end{aligned}$$

By Stokes' Theorem,

$$\oint_C\mathbf{E}\cdot \mathrm{d}\mathbf{r}=\iint _S({ \boldsymbol{\nabla} }\times\mathbf{E})\cdot \hat{\mathbf{n}}\,\mathrm{d}S$$

so Faraday's law becomes

$$\iint _S\Big({ \boldsymbol{\nabla} }\times\mathbf{E} +\frac{1}{c}\frac{\partial\mathbf{H}}{\partial t}\Big)\cdot\hat{\mathbf{n}}\,\mathrm{d}S=0$$

This is true for all surfaces $S$. So the integrand, assuming that it is continuous, must be zero.

To see this, let $\mathbf{G}=\Big({ \boldsymbol{\nabla} }\times\mathbf{E}+\frac{1}{c}\frac{\partial\mathbf{H}}{\partial t}\Big)$. Suppose that $\mathbf{G}(\mathbf{x}_0)\ne 0$. Pick a unit vector $\hat{\mathbf{n}}$ in the direction of $\mathbf{G}(\mathbf{x}_0)$. Let $S$ be a very small flat disk centered on $\mathbf{x}_0$ with normal $\hat{\mathbf{n}}$ (the vector we picked). Then $\mathbf{G}(\mathbf{x}_0)\cdot\hat{\mathbf{n}}>0$ and, by continuity, $\mathbf{G}(\mathbf{x})\cdot\hat{\mathbf{n}}>0$ for all $\mathbf{x}$ on $S$, if we have picked $S$ small enough. Then $\iint _S\Big({ \boldsymbol{\nabla} }\times\mathbf{E} +\frac{1}{c}\frac{\partial\mathbf{H}}{\partial t}\Big)\cdot \hat{\mathbf{n}}\,\mathrm{d}S>0$, which is a contradiction. So $\mathbf{G}=\mathbf{0}$ everywhere and we conclude that

$${ \boldsymbol{\nabla} }\times\mathbf{E}+\frac{1}{c}\frac{\partial\mathbf{H}}{\partial t}=0$$

The curl of the specified vector field is

$$\begin{aligned} \nabla\times\mathbf{F} &=\nabla\times\big(z\,\hat{\boldsymbol{\imath}}+x\,\hat{\boldsymbol{\jmath}}+y^3z^3\,\hat{\mathbf{k}}\big) \\ &= \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ z & x & y^3z^3 \end{matrix}\right] \\ &=3y^2z^3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}} \end{aligned}$$

For every $t$, we have $x(t)=z(t)$ and $x(t)^2+y(t)^2+z(t)^2=2$. So the specified curve is the intersection of the plane $x=z$ and the sphere $x^2+y^2+z^2=2$. This curve is the boundary of the circular disk

$$D = \big\{ \ (x,y,z)\ \big|\ x=z,\ x^2+y^2+z^2\le 2\ \big\}$$

The curve is oriented so that $\big(x(t),y(t)\big)=\big(\cos t,\sqrt{2}\sin t\big)$ runs in the standard (counterclockwise) direction. So the unit normal to $D$ used in Stokes' theorem has positive $\hat{\mathbf{k}}$ component. Since the plane $x-z=0$ has unit normal $\pm\frac{1}{\sqrt{2}}(1,0,-1)$, the unit normal used in Stokes' theorem is $\hat{\mathbf{n}}=\frac{1}{\sqrt{2}}(-1,0,1)$. By Stokes' theorem

$$\begin{aligned} \oint_C \mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _D\nabla\times\mathbf{F}\cdot\hat{\mathbf{n}}\ \mathrm{d}S =\frac{1}{\sqrt{2}}\iint _D(3y^2z^3,1,1)\cdot(-1,0,1)\ \mathrm{d}S \\ &=\frac{1}{\sqrt{2}}\iint _D(1-3y^2z^3)\ \mathrm{d}S \end{aligned}$$

The disk $D$ is invariant under the reflection $(x,y,z)\rightarrow(-x,y,-z)$. Since $y^2z^3$ is odd under this reflection, $\iint _D y^2z^3\ \mathrm{d}s=0$ and

$$\oint \mathbf{F}\cdot \mathrm{d}\mathbf{r} =\frac{1}{\sqrt{2}}\iint _D \ \mathrm{d}S =\frac{1}{\sqrt{2}}{\rm Area}(D)$$

Because the centre of the ball $x^2+y^2+z^2\le 2$ (namely $(0,0,0)$) is contained in the plane $x=z$, the radius of the disk $D$ is the same as the radius of the sphere $x^2+y^2+z^2=2$. So $D$ has radius $\sqrt{2}$ and

$$\oint \mathbf{F}\cdot \mathrm{d}\mathbf{r} =\frac{1}{\sqrt{2}}{\rm Area}(D) =\frac{1}{\sqrt{2}}\pi{\big(\,\sqrt{2}\,\big)}^2 =\sqrt{2}\pi$$