발산 정리

We will be using the divergence theorem in both parts (a) and $b$. So as a prelimary calculation, let's find the divergence of $\mathbf{H}(x,y,z)=\frac{(x,y,z)-(a,b,c)}{{[(x-a)^2+(y-b)^2+(z-b)^2]}^{3/2}}$ for any $(a,b,c)$. If $(x,y,z)\ne(a,b,c)$,

$$\begin{aligned} &{ \boldsymbol{\nabla} }\cdot\mathbf{H}(x,y,z) \\ & =\frac{\partial }{\partial x}\frac{x-a} {{[(x-a)^2+(y-b)^2+(z-c)^2]}^{3/2}} +\frac{\partial }{\partial y}\frac{y-b} {{[(x-a)^2+(y-b)^2+(z-c)^2]}^{3/2}} \\& +\frac{\partial }{\partial z}\frac{z-c} {{[(x-a)^2+(y-b)^2+(z-c)^2]}^{3/2}} \\ & =\frac{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]\ -\ (x-a)\ \frac{3}{2}(2(x-a))} {{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]}^{5/2}} \\& +\frac{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]\ -\ (y-b)\ \frac{3}{2}(2y)} {{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]}^{5/2}} \\& +\frac{\big[(x-a)^2+(y-b)^2+z^2\big]\ -\ (z-c)\ \frac{3}{2}(2(z-c))} {{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]}^{5/2}} \\ & =\frac{3\big[(x-a)^2+(y-b)^2+(z-c)^2\big]\ -\ 3(x-a)^2-3(y-b)^2-3(z-c)^2} {{\big[(x-a)^2+(y-b)^2+(z-c)^2\big]}^{5/2}} \\ &=0 \end{aligned}$$

If $(x,y,z)=(a,b,c)$, $\mathbf{H}(x,y,z)$ is not defined and hence ${ \boldsymbol{\nabla} }\cdot\mathbf{H}(x,y,z)$ is also not defined.

(b) By the above preliminary computation with $(a,b,c)=(3,2,2)$, ${ \boldsymbol{\nabla} }\cdot\mathbf{G}$ is defined and zero for all $(x,y,z)\ne (3,2,2)$, and, in particular for all $(x,y,z)$ in

$$V=\big\{ \ (x,y,z)\ \big|\ x^2+2y^2+3z^2\le 16\ \big\}$$

So, by the divergence theorem,

$$\begin{aligned} \iint _S \mathbf{G}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iiint _V{ \boldsymbol{\nabla} }\cdot\mathbf{G}\,\mathrm{d}V =0 \end{aligned}$$

(a) Because $(1,1,2)$ is inside $V$, we cannot use the argument of part (b), to conclude that the integral is zero. Let $\varepsilon >0$ be small enough that

$$S_\varepsilon = \big\{ \ (x,y,z)\ \big|\ (x-2)^2 + (y-1)^2 +(z-1)^2=\varepsilon ^2\ \big\}$$

is completely contained inside $V$, as in the sketch below.

Set

$$V_\varepsilon =\big\{ \ (x,y,z)\ \big|\ x^2+2y^2+3z^2\le 16,\ (x-2)^2 + (y-1)^2 +(z-1)^2\ge\varepsilon ^2\ \big\}$$

The boundary, $\partial V_\varepsilon$, of $V$ consists of two parts — $S$ and $S_\varepsilon$, with the normals as in the figure above. The divergence ${ \boldsymbol{\nabla} }\cdot\mathbf{F}$ of $\mathbf{F}$ is well-defined and zero throughout $V_\varepsilon$. Consequently, the divergence theorem gives

$$\begin{aligned} 0 &= \iiint _V{ \boldsymbol{\nabla} }\cdot\mathbf{F}\,\mathrm{d}V = \iint _S \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S + \iint _{S_\varepsilon } \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \end{aligned}$$

So

$$\iint _S \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S = -\iint _{S_\varepsilon } \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S$$

The unit normal to $S_\varepsilon$ at the point $(x,y,z)$ on $S_\varepsilon$ is

$$\hat{\mathbf{n}} =-\frac{1}{\varepsilon }\big[(x-2)\,\hat{\boldsymbol{\imath}} + (y-1)\,\hat{\boldsymbol{\jmath}} +(z-1)\,\hat{\mathbf{k}}\big]$$

(Recall that $|(2-1)\,\hat{\boldsymbol{\imath}} + (y-1)\,\hat{\boldsymbol{\jmath}} +(z-1)\,\hat{\mathbf{k}}|=\varepsilon$ on $S_\varepsilon$. So, on $S_\varepsilon$,

$$\begin{aligned} \mathbf{F}\cdot\hat{\mathbf{n}} &= -\frac{1}{\varepsilon } \left(\frac{(x,y,z)-(2,1,1)}{{\big[(x-2)^2+(y-1)^2+(z-1)^2\big]}^{3/2}}\right) \cdot\big[(x-2)\,\hat{\boldsymbol{\imath}} + (y-1)\,\hat{\boldsymbol{\jmath}} +(z-1)\,\hat{\mathbf{k}}\big] \\ &=-\frac{1}{\varepsilon } \left(\frac{(x-2)^2+(y-1)^2+(z-1)^2}{{\big[(x-2)^2+(y-1)^2+(z-1)^2\big] }^{3/2}}\right)\\ &= -\frac{1}{\varepsilon ^2} \end{aligned}$$

Hence

$$\begin{aligned} \iint _S \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S = -\iint _{S_\varepsilon } \left(-\frac{1}{\varepsilon ^2}\right)\,\mathrm{d}S =\frac{1}{\varepsilon ^2} (4\pi\varepsilon ^2) =4\pi \end{aligned}$$