발산 정리

Solution 1 (divergence theorem): Set $\mathbf{F} = (x + y, x + z, y + z)$. Then ${ \boldsymbol{\nabla} }\cdot\mathbf{F} = 2$. That's really simple. So let's try using the divergence theorem.

• Set $S=\big\{ \ (x,y,z)\ \big|\ x^2+z^2=4,\ 0\le y\le 3\ \big\}$. We are to compute $\iint _S\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S$, with $\hat{\mathbf{n}}$ denoting the outward normal to $S$. $S$ is not the boundary of a solid, so we cannot compute $\iint _S\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S$ by applying the divergence theorem directly. The figure on the left below shows the part of $S$ that is in the first octant.

• On the other hand $S$, is “almost” the boundary of

$$V = \big\{ \ (x,y,z)\ \big|\ x^2+z^2\le 4,\ 0\le y\le 3\ \big\}$$

The boundary, $\partial V$ of $V$ consists of three pieces — $S$ and the two disks

$$D_l = \big\{ \ (x,y,z)\ \big|\ x^2+z^2\le 4,\ y=0\ \big\}\qquad D_r = \big\{ \ (x,y,z)\ \big|\ x^2+z^2\le 4,\ y=3\ \big\}$$

The figure on the right above shows the parts of $S$, $V$, $D_l$ and $D_r$ that are in the first octant.

The outward normal to $D_r$ is $\hat{\boldsymbol{\jmath}}$ and the outward normal to $D_l$ is $-\hat{\boldsymbol{\jmath}}$, to the divergence theorem gives

$$\begin{aligned} \iiint _V { \boldsymbol{\nabla} }\cdot \mathbf{F}\,\mathrm{d}V &=\iint _{\partial V}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \\ &=\iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S + \iint _{D_r}\mathbf{F}\cdot\hat{\boldsymbol{\jmath}}\,\mathrm{d}S + \iint _{D_l}\mathbf{F}\cdot(-\hat{\boldsymbol{\jmath}})\,\mathrm{d}S \end{aligned}$$

Since ${ \boldsymbol{\nabla} }\cdot\mathbf{F} = 2$ and $\mathbf{F}\cdot\hat{\boldsymbol{\jmath}} = x+z$,

$$\begin{aligned} \iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &= \iiint _V 2\,\mathrm{d}V -\iint _{x^2+z^2\le 4}(x+z)\,\mathrm{d}x\mathrm{d}z - \iint _{x^2+z^2\le 4}(-x-z)\,\mathrm{d}x\mathrm{d}z \\ &= \iiint _V 2\,\mathrm{d}V \\ &= 2\ \text{volume}(V) =2(\pi 2^2) 3 =24\pi \end{aligned}$$

Solution 2 (direct evaluation): Let's parametrize the surface by

$$\begin{aligned} \mathbf{r}(\theta,y) &= 2\cos\theta\,\hat{\boldsymbol{\imath}} +y\,\hat{\boldsymbol{\jmath}} + 2\sin\theta\,\hat{\mathbf{k}}\qquad 0\le\theta<2\pi,\ 0\le y\le 3 \end{aligned}$$

Then

$$\begin{aligned} \tfrac{\partial\mathbf{r}}{\partial\theta} &=\big(-2\sin\theta\,,\, 0\,,\, 2\cos\theta \big) \\ \tfrac{\partial\mathbf{r}}{\partial y} &=\big(0\,,\, 1\,,\, 0\big) \\ \hat{\mathbf{n}}\mathrm{d}S &=\pm\tfrac{\partial\mathbf{r}}{\partial\theta} \times \tfrac{\partial\mathbf{r}}{\partial y} \mathrm{d}\theta\mathrm{d}y =\pm \big(-2\cos\theta\,,\, 0\,,\, -2\sin\theta\big)\mathrm{d}\theta\mathrm{d}y \end{aligned}$$

To get the outward normal, we want the minus sign. So

$$\hat{\mathbf{n}}\mathrm{d}S =\big(2\cos\theta\,,\, 0\,,\, 2\sin\theta\big)\mathrm{d}\theta\mathrm{d}y$$

and, since

$$\mathbf{F}\big(\mathbf{r}(\theta,y)\big) =\big(2\cos\theta + y\,,\,2\cos\theta+2\sin\theta\,,\,y+2\sin\theta\big)$$

the specified flux is

$$\begin{aligned} \iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &= \int_0^{2\pi}\mathrm{d}\theta\int_0^3\mathrm{d}y \ \big(2\cos\theta + y\,,\,2\cos\theta+2\sin\theta\,,\,y+2\sin\theta\big)\cdot \big(2\cos\theta\,,\,0\,,\, 2\sin\theta\big) \\ &= \int_0^{2\pi}\mathrm{d}\theta\int_0^3\mathrm{d}y \ \big(4\cos^2\theta + 2y\cos\theta +2y\sin\theta+4\sin^2\theta\big) \\ &= \int_0^{2\pi}\mathrm{d}\theta\int_0^3\mathrm{d}y \ \big(4 + 2y\cos\theta +2y\sin\theta\big) \end{aligned}$$

Since $\int_0^{2\pi}\mathrm{d}\theta\ \cos\theta = \int_0^{2\pi}\mathrm{d}\theta\ \sin\theta = 0$,

$$\begin{aligned} \iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &= 4\int_0^{2\pi}\mathrm{d}\theta\int_0^3\mathrm{d}y =4(2\pi)3 =24\pi \end{aligned}$$