12. 선적분의 성질과 그린 정리Properties of Line Integrals and Green's Theorem

Solution. The function $f(x,y,z)=-\dfrac{1}{x^2+y^2+z^2}$ is $C^1$ on $\mathbb{R}^3\setminus\{\mathbf{0}\}$, and since $C$ does not pass through the origin, $C$ lies entirely within the domain of $f$.

Although the origin is removed from the domain, simple connectedness of the domain is not needed to apply the Fundamental Theorem of Line Integrals — a potential function $f$ with $\mathbf{F}=\nabla f$ is already given on a domain containing $C$, so it suffices that $C$ be a smooth curve inside that domain. (Simple connectedness is only needed for the converse direction, to conclude that a closed vector field is conservative.) Hence by the Fundamental Theorem of Line Integrals,

$$\int_C \mathbf{F}\cdot d\mathbf{r}=f(0,0,2)-f(1,0,0)$$

holds.

Since $f(0,0,2)=-\dfrac{1}{4}$ and $f(1,0,0)=-1$,

$$\int_C \mathbf{F}\cdot d\mathbf{r}=-\frac{1}{4}-(-1)=\frac{3}{4}.$$

Solution. Since $\mathbf{F}=\nabla f$ with $f(x,y,z)=xyz$, the Fundamental Theorem of Line Integrals gives, regardless of the path taken,

$$W=\int_C \mathbf{F}\cdot d\mathbf{r}=f(B)-f(A).$$

Since $f(B)=f(1,6,-4)=1\cdot 6\cdot(-4)=-24$ and $f(A)=f(-1,3,9)=(-1)\cdot 3\cdot 9=-27$,

$$W=-24-(-27)=3.$$

Proof. Write $\mathbf{F}=M\mathbf{i}+N\mathbf{j}+P\mathbf{k}$ with $M=e^x\cos y+yz$, $N=xz-e^x\sin y$, $P=xy+z$. The component functions and their partial derivatives are continuous on all of $\mathbb{R}^3$, and $\mathbb{R}^3$ is simply connected, so it suffices to show $\nabla\times\mathbf{F}=\mathbf{0}$ in order to conclude that $\mathbf{F}$ is conservative.

Computing,

$$\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}=x-x=0,\qquad \frac{\partial M}{\partial z}-\frac{\partial P}{\partial x}=y-y=0,\qquad \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=(z-e^x\sin y)-(z-e^x\sin y)=0,$$

so $\nabla\times\mathbf{F}=\mathbf{0}$. Hence $\mathbf{F}$ is a conservative vector field.

To find a potential function $f$: since $\dfrac{\partial f}{\partial x}=M=e^x\cos y+yz$, integrating with respect to $x$ gives

$$f(x,y,z)=e^x\cos y+xyz+g(y,z).$$

Differentiating with respect to $y$, $\dfrac{\partial f}{\partial y}=-e^x\sin y+xz+g_y(y,z)$, which must equal $N=xz-e^x\sin y$, so $g_y(y,z)=0$, i.e., $g=g(z)$. Finally $\dfrac{\partial f}{\partial z}=xy+g'(z)$ must equal $P=xy+z$, so $g'(z)=z$, i.e., $g(z)=\dfrac{z^2}{2}+C$. Therefore

$$f(x,y,z)=e^x\cos y+xyz+\frac{z^2}{2}$$

is a potential function for $\mathbf{F}$.

Proof. Write $\mathbf{F}=M\mathbf{i}+N\mathbf{j}+P\mathbf{k}$ with $M=2x-3$, $N=-z$, $P=\cos z$. Since $\mathbf{F}$ is smooth on all of $\mathbb{R}^3$, a necessary condition for $\mathbf{F}$ to be conservative is $\nabla\times\mathbf{F}=\mathbf{0}$.

Computing,

$$\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}=0-(-1)=1\neq 0,$$

so $\nabla\times\mathbf{F}\neq\mathbf{0}$. Hence $\mathbf{F}$ is not a conservative vector field.

Proof. $\mathbf{F}$ is defined on $\mathbb{R}^3$ with the $z$-axis removed. Let $M=\dfrac{-y}{x^2+y^2}$, $N=\dfrac{x}{x^2+y^2}$, $P=0$.

(Showing $\mathbf{F}$ is closed.) Computing,

$$\frac{\partial N}{\partial x}=\frac{(x^2+y^2)-x(2x)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2},\qquad \frac{\partial M}{\partial y}=\frac{-(x^2+y^2)+y(2y)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2},$$

so $\dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}$. Since $P=0$, the remaining conditions $\dfrac{\partial P}{\partial y}-\dfrac{\partial N}{\partial z}=0$ and $\dfrac{\partial M}{\partial z}-\dfrac{\partial P}{\partial x}=0$ hold trivially. Hence $\nabla\times\mathbf{F}=\mathbf{0}$ throughout the domain, so $\mathbf{F}$ is a closed vector field.

(Showing $\mathbf{F}$ is not conservative.) Along the unit circle $C:\mathbf{r}(t)=(\cos t)\mathbf{i}+(\sin t)\mathbf{j}$ $(0\le t\le 2\pi)$ centered at the origin, $dx=-\sin t\,dt$, $dy=\cos t\,dt$, and the circulation is

$$\oint_C \mathbf{F}\cdot d\mathbf{r}=\int_0^{2\pi} \Big[(-\sin t)(-\sin t)+(\cos t)(\cos t)\Big]\,dt=\int_0^{2\pi} 1\,dt=2\pi\neq 0.$$

If $\mathbf{F}$ were conservative, the line integral around every closed curve would be $0$; since this fails, $\mathbf{F}$ is not a conservative vector field.

This example shows that $\nabla\times\mathbf{F}=\mathbf{0}$ (i.e., $\mathbf{F}$ closed) does not imply $\mathbf{F}$ is conservative when the domain is not simply connected (here, the $z$-axis has been removed).

Solution. Let $M=y^2+2czx$, $N=y(bx+cz)=bxy+cyz$, $P=y^2+cx^2$. Since $\mathbf{F}$ is smooth on all of $\mathbb{R}^3$, $\mathbf{F}$ is conservative if and only if $\nabla\times\mathbf{F}=\mathbf{0}$.

Computing the three conditions,

$$\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}=2y-cy=(2-c)y,\qquad \frac{\partial M}{\partial z}-\frac{\partial P}{\partial x}=2cx-2cx=0,\qquad \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=by-2y=(b-2)y.$$

These must vanish for all $x,y,z$, so $(2-c)y=0$ and $(b-2)y=0$ must hold identically, which forces $c=2$ and $b=2$.

Hence $b=2$, $c=2$.

Proof.

(1) Let $M=y$, $N=x$, $P=4$, so $\mathbf{F}=M\mathbf{i}+N\mathbf{j}+P\mathbf{k}$ is smooth on all of $\mathbb{R}^3$. Computing,

$$\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z}=0-0=0,\qquad \frac{\partial M}{\partial z}-\frac{\partial P}{\partial x}=0-0=0,\qquad \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=1-1=0,$$

so $\nabla\times\mathbf{F}=\mathbf{0}$; since $\mathbb{R}^3$ is simply connected, $y\,dx+x\,dy+4\,dz$ is an exact form.

(2) To find a potential function $f$: since $\dfrac{\partial f}{\partial x}=y$, $f=xy+g(y,z)$, and $\dfrac{\partial f}{\partial y}=x+g_y=x$ gives $g_y=0$, i.e., $g=g(z)$. Then $\dfrac{\partial f}{\partial z}=g'(z)=4$ gives $g(z)=4z+C$. Hence $f(x,y,z)=xy+4z$ is a potential function, and

$$\int_{(1,1,1)}^{(2,3,-1)} y\,dx+x\,dy+4\,dz=f(2,3,-1)-f(1,1,1)=\big(6-4\big)-\big(1+4\big)=2-5=-3.$$

Solution.

(1) Let $M=\dfrac{2x}{x^2+y^2+z^2}$, $N=\dfrac{2y}{x^2+y^2+z^2}$, $P=\dfrac{2z}{x^2+y^2+z^2}$. Direct computation confirms $\nabla f=M\mathbf{i}+N\mathbf{j}+P\mathbf{k}$ for $f(x,y,z)=\ln(x^2+y^2+z^2)$ (away from the origin). Assuming the path does not pass through the origin, the Fundamental Theorem of Line Integrals gives

$$\int_{(-1,-1,-1)}^{(2,2,2)} \frac{2x\,dx+2y\,dy+2z\,dz}{x^2+y^2+z^2}=f(2,2,2)-f(-1,-1,-1).$$

Since $f(2,2,2)=\ln 12$ and $f(-1,-1,-1)=\ln 3$, the value is

$$\ln 12-\ln 3=\ln 4.$$

(2) Let $M=\dfrac{1}{y}$, $N=\dfrac{1}{z}-\dfrac{x}{y^2}$, $P=-\dfrac{y}{z^2}$. Since $\dfrac{\partial M}{\partial y}=-\dfrac{1}{y^2}=\dfrac{\partial N}{\partial x}$, $\dfrac{\partial M}{\partial z}=0=\dfrac{\partial P}{\partial x}$, and $\dfrac{\partial N}{\partial z}=-\dfrac{1}{z^2}=\dfrac{\partial P}{\partial y}$, this is an exact form. For the potential function, $\dfrac{\partial f}{\partial x}=\dfrac{1}{y}$ gives $f=\dfrac{x}{y}+g(y,z)$, and $\dfrac{\partial f}{\partial y}=-\dfrac{x}{y^2}+g_y=\dfrac{1}{z}-\dfrac{x}{y^2}$ gives $g_y=\dfrac{1}{z}$, i.e., $g=\dfrac{y}{z}+h(z)$. Finally $\dfrac{\partial f}{\partial z}=-\dfrac{y}{z^2}+h'(z)=-\dfrac{y}{z^2}$ gives $h'(z)=0$. Hence $f(x,y,z)=\dfrac{x}{y}+\dfrac{y}{z}$, and

$$\int_{(1,1,1)}^{(2,2,2)} \frac{1}{y}\,dx+\left(\frac{1}{z}-\frac{x}{y^2}\right)dy-\frac{y}{z^2}\,dz=f(2,2,2)-f(1,1,1)=(1+1)-(1+1)=0.$$

Solution. Let $M=x^2$, $N=yz$, $P=\dfrac{y^2}{2}$. Since $\dfrac{\partial M}{\partial y}=0=\dfrac{\partial N}{\partial x}$, $\dfrac{\partial M}{\partial z}=0=\dfrac{\partial P}{\partial x}$, and $\dfrac{\partial N}{\partial z}=y=\dfrac{\partial P}{\partial y}$, this is an exact form, and since $\mathbb{R}^3$ is simply connected, the value of the line integral is the same for any path with the same endpoints as $C$.

For the potential function, $\dfrac{\partial f}{\partial x}=x^2$ gives $f=\dfrac{x^3}{3}+g(y,z)$; $\dfrac{\partial f}{\partial y}=g_y=yz$ gives $g=\dfrac{y^2z}{2}+h(z)$; and $\dfrac{\partial f}{\partial z}=\dfrac{y^2}{2}+h'(z)=\dfrac{y^2}{2}$ gives $h'(z)=0$. Hence

$$f(x,y,z)=\frac{x^3}{3}+\frac{y^2z}{2},$$

and by the Fundamental Theorem of Line Integrals,

$$\int_C x^2\,dx+yz\,dy+\frac{y^2}{2}\,dz=f(0,3,4)-f(0,0,0)=\left(0+\frac{9\cdot 4}{2}\right)-0=18.$$

Solution. For a planar vector field $\mathbf{F}=M\mathbf{i}+N\mathbf{j}$, the circulation density is $\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}$, and the flux density is $\dfrac{\partial M}{\partial x}+\dfrac{\partial N}{\partial y}$.

(1) With $M=cx$, $N=cy$, the circulation density is $0-0=0$ and the flux density is $c+c=2c$.

(2) With $M=-cy$, $N=cx$, the circulation density is $c-(-c)=2c$ and the flux density is $0+0=0$.

(3) With $M=y$, $N=0$, the circulation density is $0-1=-1$ and the flux density is $0+0=0$.

(4) With $M=\dfrac{-y}{x^2+y^2}$, $N=\dfrac{x}{x^2+y^2}$ ($(x,y)\neq(0,0)$), we have $\dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$, so the circulation density is $0$. Also

$$\frac{\partial M}{\partial x}=\frac{2xy}{(x^2+y^2)^2},\qquad \frac{\partial N}{\partial y}=\frac{-2xy}{(x^2+y^2)^2},$$

so the flux density is $\dfrac{2xy}{(x^2+y^2)^2}-\dfrac{2xy}{(x^2+y^2)^2}=0$.

Solution. If $\mathbf{F}=M\mathbf{i}+N\mathbf{j}$ is conservative, there is a potential function $f$ with $M=f_x$, $N=f_y$. Since $\mathbf{F}$ is smooth, $f$ is $C^2$, and by the equality of mixed partial derivatives (Clairaut's theorem), $f_{xy}=f_{yx}$. Hence the circulation density is

$$\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=f_{yx}-f_{xy}=0.$$

That is, the circulation density of a conservative vector field is always $0$.

Proof. Here $M=x-y$, $N=x$, and the region $R$ enclosed by $C$ is the unit disk. Since $C$ is $x=\cos t$, $y=\sin t$ $(0\le t\le 2\pi)$, we have $dx=-\sin t\,dt$, $dy=\cos t\,dt$.

(1) Since $\displaystyle\oint_C \mathbf{F}\cdot\mathbf{T}\,ds=\oint_C M\,dx+N\,dy$, computing the left side directly,

$$\oint_C M\,dx+N\,dy=\int_0^{2\pi}\Big[(\cos t-\sin t)(-\sin t)+(\cos t)(\cos t)\Big]dt=\int_0^{2\pi}\Big[1-\sin t\cos t\Big]dt.$$

Since $\displaystyle\int_0^{2\pi}\sin t\cos t\,dt=\frac{1}{2}\int_0^{2\pi}\sin 2t\,dt=0$, the left side equals $2\pi$.

For the right side, $\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=1-(-1)=2$, so

$$\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=\iint_R 2\,dx\,dy=2\cdot(\text{area of the unit disk})=2\pi.$$

Both sides equal $2\pi$, so the identity holds.

(2) Since $\displaystyle\oint_C \mathbf{F}\cdot\mathbf{n}\,ds=\oint_C M\,dy-N\,dx$, computing the left side directly,

$$\oint_C M\,dy-N\,dx=\int_0^{2\pi}\Big[(\cos t-\sin t)(\cos t)-(\cos t)(-\sin t)\Big]dt=\int_0^{2\pi}\cos^2 t\,dt=\pi$$

(the last equality holds since the average of $\cos^2t$ over one period is $\tfrac12$, so $\displaystyle\int_0^{2\pi}\cos^2t\,dt=\pi$).

For the right side, $\dfrac{\partial M}{\partial x}+\dfrac{\partial N}{\partial y}=1+0=1$, so

$$\iint_R\left(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}\right)dx\,dy=\iint_R 1\,dx\,dy=\text{area of the unit disk}=\pi.$$

Both sides equal $\pi$, so the identity holds.

Solution. $C$ is the boundary of the square $[0,1]\times[0,1]$. Let $M=-y^2$, $N=xy$; then $\dfrac{\partial N}{\partial x}=y$, $\dfrac{\partial M}{\partial y}=-2y$, so by Green's Theorem,

$$\oint_C xy\,dy-y^2\,dx=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=\iint_R 3y\,dx\,dy=3\int_0^1\int_0^1 y\,dx\,dy=3\cdot\frac{1}{2}=\frac{3}{2}.$$

Solution.

(1) Let $M=4x^3y$, $N=x^4$; then $\dfrac{\partial N}{\partial x}=4x^3=\dfrac{\partial M}{\partial y}$, so by Green's Theorem,

$$\oint_C 4x^3y\,dx+x^4\,dy=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=\iint_R 0\,dx\,dy=0.$$

(2) Let $M=x^3$, $N=-y^3$; then $\dfrac{\partial N}{\partial x}=0=\dfrac{\partial M}{\partial y}$, so likewise

$$\oint_C -y^3\,dy+x^3\,dx=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=0.$$

Solution. Let $R=[-1,1]\times[-1,1]$. With $M=2e^{xy}$, $N=y^3$, we have $\dfrac{\partial M}{\partial x}=2ye^{xy}$, $\dfrac{\partial N}{\partial y}=3y^2$, so by the flux form of Green's Theorem,

$$\text{flux}=\oint_C M\,dy-N\,dx=\iint_R\left(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}\right)dx\,dy=\iint_R\big(2ye^{xy}+3y^2\big)\,dx\,dy.$$

First,

$$\int_{-1}^1\int_{-1}^1 2ye^{xy}\,dx\,dy=\int_{-1}^1 \Big[2e^{xy}\Big]_{x=-1}^{x=1}\,dy=\int_{-1}^1 2\big(e^{y}-e^{-y}\big)\,dy=\int_{-1}^1 4\sinh y\,dy=0$$

since $\sinh y$ is odd and the interval is symmetric. Also,

$$\int_{-1}^1\int_{-1}^1 3y^2\,dx\,dy=3\left(\int_{-1}^1 y^2\,dy\right)\left(\int_{-1}^1 dx\right)=3\cdot\frac{2}{3}\cdot 2=4.$$

Hence the flux is

$$0+4=4.$$

Solution. Let $M=3xy-\dfrac{x}{1+y^2}$, $N=e^x+\tan^{-1}y$. Then

$$\frac{\partial M}{\partial x}=3y-\frac{1}{1+y^2},\qquad \frac{\partial N}{\partial y}=\frac{1}{1+y^2},$$

so the divergence is $\dfrac{\partial M}{\partial x}+\dfrac{\partial N}{\partial y}=3y$. By the flux form of Green's Theorem, the flux is

$$\iint_R 3y\,dx\,dy$$

where $R$ is the region enclosed by the cardioid $r=a(1+\cos\theta)$.

The cardioid's value of $r$ is unchanged under $\theta\mapsto-\theta$, so it is symmetric about the $x$-axis; that is, $R$ is invariant under the reflection $(x,y)\mapsto(x,-y)$. Since the integrand $3y$ changes sign under this reflection,

$$\iint_R 3y\,dx\,dy=-\iint_R 3y\,dx\,dy,$$

so $\displaystyle\iint_R 3y\,dx\,dy=0$.

Hence the flux is $0$.

Solution. Let $M=\tan^{-1}\dfrac{y}{x}$, $N=\ln(x^2+y^2)$. Computing,

$$\frac{\partial M}{\partial y}=\frac{x}{x^2+y^2},\qquad \frac{\partial N}{\partial x}=\frac{2x}{x^2+y^2},\qquad \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=\frac{x}{x^2+y^2}.$$

$M$ is discontinuous at $x=0$ (jumping between $\pm\dfrac{\pi}{2}$), and the interior of $R:1\le r\le 2,\ 0\le\theta\le\pi$ contains the segment $x=0$ ($\theta=\dfrac{\pi}{2}$, $1\le y\le 2$), so Green's Theorem cannot be applied to $R$ as a whole directly. However, splitting $R$ along this segment into $R_1$ ($0\lt\theta\lt\dfrac{\pi}{2}$) and $R_2$ ($\dfrac{\pi}{2}\lt\theta\lt\pi$), $\mathbf{F}$ is smooth on the closure of each. Traversing the boundary of each region positively, the shared segment is crossed twice in opposite directions; along it $dx=0$, so the $M\,dx$ term vanishes automatically regardless of the jump in $M$, while $N=\ln(x^2+y^2)$ is continuous at $x=0,\ y\gt 0$, so the two contributions of the $N\,dy$ term cancel. Hence

$$\oint_{\partial R_1} M\,dx+N\,dy+\oint_{\partial R_2} M\,dx+N\,dy=\oint_C M\,dx+N\,dy,$$

and applying Green's Theorem to each piece and adding shows that

$$\oint_C M\,dx+N\,dy=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=\iint_R \frac{x}{x^2+y^2}\,dx\,dy$$

still holds (i.e., the discontinuity of $M$ does not affect the result here).

Switching to polar coordinates with $x=r\cos\theta$, $\dfrac{x}{x^2+y^2}=\dfrac{\cos\theta}{r}$ and $dx\,dy=r\,dr\,d\theta$, so

$$\iint_R \frac{x}{x^2+y^2}\,dx\,dy=\int_0^\pi\int_1^2 \frac{\cos\theta}{r}\cdot r\,dr\,d\theta=\left(\int_0^\pi\cos\theta\,d\theta\right)\left(\int_1^2 dr\right)=\big[\sin\theta\big]_0^\pi\cdot 1=0.$$

Hence the circulation is $0$.

Solution. Since $x=\cos^3 t$, $y=\sin^3 t$, we have $dx=-3\cos^2t\sin t\,dt$, $dy=3\sin^2t\cos t\,dt$. By Green's Theorem, the area can be found as

$$A=\frac{1}{2}\oint_C x\,dy-y\,dx.$$

Computing,

$$x\,dy-y\,dx=\cos^3t\cdot 3\sin^2t\cos t\,dt-\sin^3t\cdot(-3\cos^2t\sin t)\,dt=3\cos^4t\sin^2t\,dt+3\sin^4t\cos^2t\,dt$$ $$=3\sin^2t\cos^2t(\cos^2t+\sin^2t)\,dt=3\sin^2t\cos^2t\,dt,$$

and since $\sin^2t\cos^2t=\dfrac{1}{4}\sin^2 2t=\dfrac{1-\cos 4t}{8}$,

$$A=\frac{1}{2}\int_0^{2\pi} 3\sin^2t\cos^2t\,dt=\frac{3}{2}\int_0^{2\pi}\frac{1-\cos 4t}{8}\,dt=\frac{3}{16}\Big[t-\frac{\sin 4t}{4}\Big]_0^{2\pi}=\frac{3}{16}\cdot 2\pi=\frac{3\pi}{8}.$$

Proof. Let $M=xy^2$, $N=x^2y+2x$; then $\dfrac{\partial N}{\partial x}=2xy+2$, $\dfrac{\partial M}{\partial y}=2xy$, so

$$\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=2$$

is a constant independent of position. If $C$ is the boundary of a square of side length $a$, then by Green's Theorem,

$$\oint_C xy^2\,dx+(x^2y+2x)\,dy=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dx\,dy=\iint_R 2\,dx\,dy=2\cdot(\text{area})=2a^2.$$

Since the integrand $2$ is constant, this value depends only on the area $a^2$ of the square, regardless of where the square is placed in the plane or how it is oriented. Hence the value of the line integral depends only on the area of the square.