발산 정리

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

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

(b) Define the vector field $\mathbf{F}(x,y,z) = f(x,y,z)\,\hat{\mathbf{k}}$. Then the divergence of $\mathbf{F}$ is ${ \boldsymbol{\nabla} }\cdot\mathbf{F}(x,y,z) = \frac{\partial f}{\partial z}(x,y,z)$. The boundary of the cube $V$ is the union of six faces

$$\begin{aligned} S_1&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le y\le 1,\ z=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\mathbf{k}}$} \\ S_2&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le y\le 1,\ z=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\mathbf{k}}$} \\ S_3&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le z\le 1,\ y=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\boldsymbol{\jmath}}$} \\ S_4&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le z\le 1,\ y=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\boldsymbol{\jmath}}$} \\ S_5&=\big\{ \ (x,y,z)\ \big|\ 0\le y\le 1,\ 0\le z\le 1,\ x=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\boldsymbol{\imath}}$} \\ S_6&=\big\{ \ (x,y,z)\ \big|\ 0\le y\le 1,\ 0\le z\le 1,\ x=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\boldsymbol{\imath}}$} \end{aligned}$$

Observe that

$$\begin{aligned} \mathbf{F}\cdot\hat{\mathbf{n}} = f\,\hat{\mathbf{k}}\cdot\hat{\mathbf{n}} = \begin{cases} +f &\text{on $S_1$} \\ -f &\text{on $S_2$} \\ 0 &\text{on $S_3$, $S_4$, $S_5$, $S_6$} \end{cases} \end{aligned}$$

So the divergence theorem gives

$$\begin{aligned} \iiint _V\frac{\partial f}{\partial z}(x,y,z)\ \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z &=\iiint _V { \boldsymbol{\nabla} }\cdot\mathbf{F}(x,y,z)\ \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\ &= \iint _{\partial V} \mathbf{F}\cdot\hat{\mathbf{n}}\ \mathrm{d}S =\sum_{j=1}^6 \iint _{S_j} \mathbf{F}\cdot\hat{\mathbf{n}}\ \mathrm{d}S\\ &=\iint _{S_1} f\ \mathrm{d}S -\iint _{S_2} f\ \mathrm{d}S \\ &=\iint _R f(x,y,1)\ \mathrm{d}x\,\mathrm{d}y - \iint _R f(x,y,0)\ \mathrm{d}x\,\mathrm{d}y \end{aligned}$$

(a) The divergence of $\phi\,\mathbf{a}$ is ${ \boldsymbol{\nabla} }\cdot(\phi\,\mathbf{a}) ={ \boldsymbol{\nabla} }\phi\cdot\mathbf{a}+\phi{ \boldsymbol{\nabla} }\cdot\mathbf{a}={ \boldsymbol{\nabla} }\phi\cdot\mathbf{a}$, since $\mathbf{a}$ is constant. So, by the divergence theorem,

$$\iint _{\partial V}\phi\,\mathbf{a}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =\iiint _V{ \boldsymbol{\nabla} }\cdot(\phi\,\mathbf{a})\ \mathrm{d}V =\iiint _V { \boldsymbol{\nabla} }\phi\cdot\mathbf{a}\ \mathrm{d}V \implies \left[\iint _{\partial V}\phi\hat{\mathbf{n}}\,\mathrm{d}S -\iiint _V { \boldsymbol{\nabla} }\phi\,\mathrm{d}V\right]\cdot\mathbf{a}=0$$

This is true for all vectors $\mathbf{a}$. In particular, applying this with $\mathbf{a}=\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\mathbf{k}}$, we have that all three components of $\left[\iint _{\partial V}\phi\hat{\mathbf{n}}\,\mathrm{d}S -\iiint _V { \boldsymbol{\nabla} }\phi\,\mathrm{d}V\right]$ are zero. So

$$\iint _{\partial V}\phi\,\hat{\mathbf{n}}\,\mathrm{d}S-\iiint _V { \boldsymbol{\nabla} }\phi\,\mathrm{d}V=0$$

(b) By part (a), with $\phi=x^2+y^2+z^2$ and ${ \boldsymbol{\nabla} }\phi = 2x\,\hat{\boldsymbol{\imath}}+2y\,\hat{\boldsymbol{\jmath}}+2z\,\hat{\mathbf{k}}$,

$$\frac{1}{2|V|}\iint _{\partial V} (x^2+y^2+z^2)\,\hat{\mathbf{n}}\,\mathrm{d}S =\frac{1}{2|V|} \iiint _V (2x\,\hat{\boldsymbol{\imath}}+2y\,\hat{\boldsymbol{\jmath}}+2z\,\hat{\mathbf{k}})\,\mathrm{d}V =(\bar x,\bar y,\bar z)$$