그린 정리

(a) Expressing the left hand side as an iterated integral, with $y$ as the inner integration variable, we have

$$\begin{aligned} \iint _R\frac{\partial f}{\partial y}(x,y)\ \mathrm{d}x\,\mathrm{d}y &=\int_0^1\mathrm{d}x\left[\int_0^1\mathrm{d}y\ \frac{\partial f}{\partial y}(x,y)\right] \\ &=\int_0^1\mathrm{d}x\ \big[f(x,1)-f(x,0)\big] \\ &\text{by the fundamental theorem of calculus} \\ &=\int_0^1 f(x,1)\ \mathrm{d}x - \int_0^1 f(x,0)\ \mathrm{d}x \end{aligned}$$

(b) Define $F_1(x,y)=f(x,y)$ and $F_2(x,y)=0$. Then. by Green's theorem

$$\begin{aligned} \iint _R\frac{\partial f}{\partial y}(x,y)\ \mathrm{d}x\,\mathrm{d}y &=-\iint _R\Big[\frac{\partial F_2}{\partial x}(x,y) -\frac{\partial F_1}{\partial y}(x,y)\Big]\ \mathrm{d}x\,\mathrm{d}y \\ &=-\int_{\partial R}\big[F_1(x,y)\,\mathrm{d}x +F_2(x,y)\,\mathrm{d}y\big] \\ &=-\int_{\partial R}f(x,y)\,\mathrm{d}x \end{aligned}$$

The boundary of $R$, oriented counterclockwise, is the union of four line segments.

$$\begin{aligned} \\[-0.1in] &\text{$C_1$ from $(0,0)$ to $(1,0)$} \qquad\smash{  }\\ &\text{$C_2$ from $(1,0)$ to $(1,1)$} \\ &\text{$C_3$ from $(1,1)$ to $(0,1)$} \\ &\text{$C_4$ from $(0,1)$ to $(0,0)$} \\[-0.1in] \end{aligned}$$

Now $x$ is constant on $C_2$ and $C_4$ so that

$$\int_{C_2}f(x,y)\,\mathrm{d}x = \int_{C_4}f(x,y)\,\mathrm{d}x =0$$

So, using $-C_3$ to denote the line segment from $(0,1)$ to $(1,1)$

$$\begin{aligned} \iint _R\frac{\partial f}{\partial y}(x,y)\ \mathrm{d}x\,\mathrm{d}y &=-\left[\int_{C_1}f(x,y)\,\mathrm{d}x+\int_{C_3}f(x,y)\,\mathrm{d}x\right]\\ &=\int_{-C_3}f(x,y)\,\mathrm{d}x-\int_{C_1}f(x,y)\,\mathrm{d}x \\ &=\int_0^1 f(x,1)\ \mathrm{d}x - \int_0^1 f(x,0)\ \mathrm{d}x \end{aligned}$$

Let $\mathbf{r}(s)=x(s)\,\hat{\boldsymbol{\imath}}+y(s)\,\hat{\boldsymbol{\jmath}}$ be a counterclockwise parametrization of $C$ by arc length. Then $\hat{\mathbf{T}}(s) = \mathbf{r}'(s) = x'(s)\,\hat{\boldsymbol{\imath}}+y'(s)\,\hat{\boldsymbol{\jmath}}$ is the forward pointing unit tangent vector to $C$ at $\mathbf{r}(s)$ and $\hat{\mathbf{n}}(s) = \mathbf{r}'(s)\times\hat{\mathbf{k}}=y'(s)\,\hat{\boldsymbol{\imath}}-x'(s)\,\hat{\boldsymbol{\jmath}}$. To see that $\mathbf{r}'(s)\times\hat{\mathbf{k}}$ really is $\hat{\mathbf{n}}(s)$, note that $y'(s)\,\hat{\boldsymbol{\imath}}-x'(s)\,\hat{\boldsymbol{\jmath}}$

• has the same length, namely $1$, as $\mathbf{r}'(s)$ (recall that $\mathbf{r}(s)$ is a parametrization by arc length),
• lies in the $xy$-plane and
• is perpendicular to $\mathbf{r}'(s)$. (Check that $\mathbf{r}'(s)\cdot\big[y'(s)\,\hat{\boldsymbol{\imath}}-x'(s)\,\hat{\boldsymbol{\jmath}}\big]=0$.)
• Use the right hand rule to check that $\mathbf{r}'(s)\times\hat{\mathbf{k}}$ is $\hat{\mathbf{n}}$ rather than $-\hat{\mathbf{n}}$.

So, by Green's theorem,

$$\begin{aligned} \oint_C\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}s &=\oint_C\left[F_1\,\frac{\mathrm{d}y}{\mathrm{d}s}-F_2\,\frac{\mathrm{d}x}{\mathrm{d}s}\right]\,\mathrm{d}s =\oint_C\left[-F_2\,\mathrm{d}x+F_1\,\mathrm{d}y\right] =\iint _R\left[\frac{\partial }{\partial x}F_1-\frac{\partial }{\partial y}(-F_2)\right]\ \mathrm{d}x\,\mathrm{d}y \\ &=\iint _R{ \boldsymbol{\nabla} }\cdot\mathbf{F}\ \mathrm{d}x\,\mathrm{d}y \end{aligned}$$

(a) Parametrize the circle by $x=a\cos\theta,\ y=a\sin\theta$, $0\le\theta\le 2\pi$. Then $\mathrm{d}x=-a\sin\theta\,\mathrm{d}\theta$ and $\mathrm{d}y=a\cos\theta\,\mathrm{d}\theta$ so that

$$\begin{aligned} \frac{1}{2\pi}\oint_C \frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2} =\frac{1}{2\pi}\int_0^{2\pi} \frac{a^2\cos^2\theta\,\mathrm{d}\theta+a^2\sin^2\theta\,\mathrm{d}\theta} {a^2\cos^2\theta+a^2\sin^2\theta} =\frac{1}{2\pi}\int_0^{2\pi}\mathrm{d}\theta=1 \end{aligned}$$

(b) The boundary of the square has four sides — one with $y=-1$, one with $x=1$, one with $y=1$ and one with $x=-1$.

To evaluate the integrals over the four sides

• parametrize the $y=-1$ part by $x$ so that $\mathbf{r}(x) = x\,\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}$, $\mathbf{r}'(x)=\hat{\boldsymbol{\imath}}$, with $x$ running from $-1$ to $1$,
• parametrize the $x=+1$ part by $y$ so that $\mathbf{r}(y) = \hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}$, $\mathbf{r}'(y)=\hat{\boldsymbol{\jmath}}$, with $y$ running from $-1$ to $1$,
• parametrize the $y=+1$ part by $x$ so that $\mathbf{r}(x) = x\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$, $\mathbf{r}'(x)=\hat{\boldsymbol{\imath}}$, with $x$ running from $1$ to $-1$, and
• parametrize the $x=-1$ part by $y$ so that $\mathbf{r}(y) = -\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}$, $\mathbf{r}'(y)=\hat{\boldsymbol{\jmath}}$, with $y$ running from $1$ to $-1$,

so that the integral

$$\begin{aligned} \frac{1}{2\pi}\oint_C \frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2} &=\frac{1}{2\pi}\overbrace{\int_{-1}^1 \frac{-(-1)\mathrm{d}x}{x^2+1}}^{y=-1{\rm\ part}} +\frac{1}{2\pi}\overbrace{\int_{-1}^1 \frac{(1)\mathrm{d}y}{1+y^2}}^{x=+1{\rm\ part}} +\frac{1}{2\pi}\overbrace{\int^{-1}_1 \frac{-(1)\mathrm{d}x}{x^2+1}}^{y=+1{\rm\ part}} +\frac{1}{2\pi}\overbrace{\int^{-1}_1 \frac{(-1)\mathrm{d}y}{1+y^2}}^{x=-1{\rm\ part}} \\ &=4\frac{1}{2\pi}\arctan x\bigg|_{-1}^1 =\frac{2}{\pi}\Big[\frac{\pi}{4}+\frac{\pi}{4}\Big] =1 \end{aligned}$$

(c) As in part (a) with $a=\sqrt{2}$, but with $\theta$ running from $0$ to $\pi$, the outer semicircle gives

$$\frac{1}{2\pi}\int_0^{\pi} \frac{a^2\cos^2\theta\,\mathrm{d}\theta+a^2\sin^2\theta\,\mathrm{d}\theta} {a^2\cos^2\theta+a^2\sin^2\theta} =\frac{1}{2\pi}\int_0^{\pi}\mathrm{d}\theta =\frac{1}{2}$$

As in part (a) with $a=1$, but with $\theta$ running from $\pi$ to $0$, the inner semicircle gives

$$\frac{1}{2\pi}\int^0_{\pi} \frac{a^2\cos^2\theta\,\mathrm{d}\theta+a^2\sin^2\theta\,\mathrm{d}\theta} {a^2\cos^2\theta+a^2\sin^2\theta} =\frac{1}{2\pi}\int^0_{\pi}\mathrm{d}\theta =-\frac{1}{2}$$

The two flat pieces each give zero, since on them $y=0$ and $\mathrm{d}y=0$. So

$$\frac{1}{2\pi}\oint_C \frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2} =\frac{1}{2}+0-\frac{1}{2}+0=0$$

An alternate solution, that uses Green's theorem, is given in the solution to the next question.