스토크스 정리

Given the form of $\mathbf{F}$, direct evaluation looks hard. So let's try Stokes' theorem, using as $S$ the part of the plane $G(x,y,z)=x+2y-z=7$ that is inside $x^2-2x+4y^2=15$. Then

$$\hat{\mathbf{n}}\,\mathrm{d}S = \pm \frac{{ \boldsymbol{\nabla} } G}{{ \boldsymbol{\nabla} } G\cdot\hat{\mathbf{k}}}\,\mathrm{d}x\,\mathrm{d}y = \pm\big(\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}-\hat{\mathbf{k}}\big)\,\mathrm{d}x\,\mathrm{d}y$$

As $C$ is oriented counterclockwise when viewed from high, Stokes' theorem specifies the upward pointing normal so that $\hat{\mathbf{n}}\,\mathrm{d}S = -\big(\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}-\hat{\mathbf{k}}\big)\,\mathrm{d}x\,\mathrm{d}y$.

From the observations that

$${ \boldsymbol{\nabla} }\times\mathbf{F}=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] e^{x^2}+yz & \cos(y^2)-x^2 & \sin(z^2)+xy \end{matrix} \right] =x\,\hat{\boldsymbol{\imath}}-(z+2x)\,\hat{\mathbf{k}}$$

and that we can rewrite $x^2-2x+4y^2=15$ as $(x-1)^2+4y^2=16$, we have

$$\begin{aligned} \oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S\\ &=\iint _{(x-1)^2+4y^2\le 16}\, [x\hat{\boldsymbol{\imath}}-(z+2x)\hat{\mathbf{k}}]\Big|_{z=-7+x+2y}\cdot(-1,-2,1)\,\mathrm{d}x\,\mathrm{d}y\\ &=\iint _{(x-1)^2+4y^2\le 16} [-x-(-7+x+2y+2x)]\,\mathrm{d}x\,\mathrm{d}y\\ &=\iint _{(x-1)^2+4y^2\le 16} [7-4x-2y]\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$$

To evaluate the integrals of $x$ and $y$ we use that, for any region $R$ in the $xy$–plane,

$$\bar x =\frac{\iint _R x\,\mathrm{d}x\,\mathrm{d}y}{\text{Area}(R)}\qquad \bar y =\frac{\iint _R y\,\mathrm{d}x\,\mathrm{d}y}{\text{Area}(R)}$$

Our ellipse is $\frac{(x-1)^2}{4^2}+\frac{y^2}{2^2}=1$ and so has area $\pi ab = \pi\times4\times 2=8\pi$ and centroid $(\bar x,\bar y) = (1,0)$. So, using $R=\big\{ \ (x,y)\ \big|\ \frac{(x-1)^2}{4^2}+\frac{y^2}{2^2}\le 1\ \big\}$,

$$\begin{aligned} \oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _R [7-4x-2y]\,\mathrm{d}x\,\mathrm{d}y\\ &=\text{Area}(R) \big\{7-4\bar x-2\bar y\}\\ &=8\pi[7-4\times 1-2\times 0]=24\pi \end{aligned}$$

(a) The curl is

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ 2+x^2+z & 0 & 3+x^2z \end{matrix} \right] =(1-2xz)\,\hat{\boldsymbol{\jmath}} \end{aligned}$$

(b) We are going to use Stokes' theorem. The specified curve $C$ is not closed and so is not the boundary of a surface. So we extend $C$ to a closed curve $\tilde C$ by appending to $C$ the line segment $L$ from $(2,0,0)$ to $(0,0,0)$. In the figure below, $C$ is the red curve and $\tilde C$ is $C$ plus the blue line segment.

The closed curve $\tilde C$ is boundary of the surface $S$ that is the union of

• the triangle $T_1$ in the $yz$-plane with vertices $(0,0,0)$, $(0,0,3)$ and $(0,1,0)$ and with normal vector $-\hat{\boldsymbol{\imath}}$ and
• the triangle $T_2$ in the $xy$-plane with vertices $(0,0,0)$, $(0,1,0)$ and $(2,0,0)$ and with normal vector $-\hat{\mathbf{k}}$.

So, by Stokes' theorem

$$\begin{aligned} \int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} + \int_L \mathbf{F}\cdot\mathrm{d}\mathbf{r} &= \int_{\partial S} \mathbf{F}\cdot\mathrm{d}\mathbf{r} = \iint _S { \boldsymbol{\nabla} }\times \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \\ &= \iint _{T_1} { \boldsymbol{\nabla} }\times \mathbf{F}\cdot(-\hat{\boldsymbol{\imath}})\,\mathrm{d}S + \iint _{T_2} { \boldsymbol{\nabla} }\times \mathbf{F}\cdot(-\hat{\mathbf{k}})\,\mathrm{d}S \\ &= \iint _{T_1} (1-2xz)\,\hat{\boldsymbol{\jmath}}\cdot(-\hat{\boldsymbol{\imath}})\,\mathrm{d}S + \iint _{T_2} (1-2xz)\,\hat{\boldsymbol{\jmath}}\cdot(-\hat{\mathbf{k}})\,\mathrm{d}S \\ &=0 \end{aligned}$$

Consequently the integral of interest

$$\begin{aligned} \int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} &=- \int_L \mathbf{F}\cdot\mathrm{d}\mathbf{r} =- \int_2^0 (2+x^2)\,\mathrm{d}x \qquad\text{since $\mathrm{d}y=\mathrm{d}z=z=0$ on $L$} \\ &=\int_0^2 (2+x^2)\,\mathrm{d}x ={\left[2x+\frac{x^3}{3}\right]}_0^2 =\frac{20}{3} \end{aligned}$$

(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.