발산 정리

(a) 구면좌표를 사용하자. $S$ 는 원점을 중심으로 하는 반지름 $a$ 인 구면이므로 다음과 같이 매개화할 수 있다.

$$\begin{aligned} \mathbf{r}(\theta,\varphi) &=a\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} + a\cos\varphi\,\hat{\mathbf{k}} \\ \frac{\partial \mathbf{r}}{\partial \theta} &=-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}} \\ \frac{\partial \mathbf{r}}{\partial \varphi} &= \phantom{-}a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \\ \hat{\mathbf{n}}\,\mathrm{d}S &=\pm\frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \varphi} \ \mathrm{d}\theta\, d\varphi\\ &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ -a\sin\varphi\sin\theta & a\sin\varphi\cos\theta & 0 \\ a\cos\varphi\cos\theta & a\cos\varphi\sin\theta & -a\sin\varphi \end{matrix}\right] \mathrm{d}\theta\, d\varphi\\ &=\pm\Big(-a^2\sin^2\varphi\cos\theta\,\hat{\boldsymbol{\imath}} -a^2\sin^2\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a^2\sin\varphi\cos\varphi\,\hat{\mathbf{k}}\Big)\ \mathrm{d}\theta\, d\varphi \\ &=\mp a^2\sin\varphi \big(\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +\cos\varphi\,\hat{\mathbf{k}}\Big)\ \mathrm{d}\theta\, d\varphi \end{aligned}$$

바깥쪽 법선이 되려면 $+$ 부호를 택해야 하므로

$$\begin{aligned} \hat{\mathbf{n}}\,\mathrm{d}S &= a^2\sin\varphi \big(\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +\cos\varphi\,\hat{\mathbf{k}}\Big)\ \mathrm{d}\theta\, d\varphi \\ \mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S &= z(\theta,\varphi)\,\hat{\mathbf{k}}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = a^3 \sin\varphi\cos^2\varphi\ \mathrm{d}\theta\,d\varphi \end{aligned}$$

이고,

$$\begin{aligned} \iint _S\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S &= a^3\int_0^\pi d\varphi\int_0^{2\pi}\mathrm{d}\theta\ \sin\varphi\,\cos^2\varphi\\ &=2\pi a^3 \int_0^\pi d\varphi \ \sin\varphi\,\cos^2\varphi =2\pi a^3\left[-\frac{1}{3}\cos^3\varphi\right]_0^\pi \\ &=\frac{4}{3}\pi a^3 \end{aligned}$$

이다.

(b) 입체 $x^2+y^2+z^2\le a^2$ 을 $V$ 라 하자.

$${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(0) +\frac{\partial }{\partial y}(0) +\frac{\partial }{\partial z}(z) =1$$

이므로 발산 정리에 의해

$$\begin{aligned} \iint _{\mathcal{S}}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iiint _V{ \boldsymbol{\nabla} }\cdot\mathbf{F}\ \mathrm{d}V =\iiint _V \ \mathrm{d}V =\text{부피}(V) =\frac{4}{3}\pi a^3 \end{aligned}$$

이다.