벡터 미적분 전체 문제

(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.