발산 정리

${\iint }_{S}\text{F}\cdot d\text{S}=72$

${e}^{-1}-1$

(a) $\operatorname{curl}\vec F = 1$. (b) $\operatorname{curl}\vec F\cdot \vec n = 1$, where $\vec n$ is a unit vector normal to $\mathcal{S}$.

(a) Let's use spherical coordinates. As $S$ is the sphere of radius $a$ centred on the origin, we can parametrize it by

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

For the outward normal, we want the $+$ sign, so

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

and

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

(b) Call the solid $x^2+y^2+z^2\le a^2$, $V$. As

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

the divergence theorem gives

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

The circle $x^2+y^2=4y$, or equivalently, $x^2+(y-2)^2=4$, has radius $2$ and centre $(0,2)$. On the bottom surface, $z=0$ and the outward normal is $-\hat{\mathbf{k}}$, so that $$\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =-\iint _{D}\mathbf{F}\cdot\hat{\mathbf{k}}\,\mathrm{d}x\mathrm{d}y =-\iint _D(2x+3y)\,\mathrm{d}x\mathrm{d}y$$ By symmetry, the centre of mass, $(\bar x,\bar y)$, of the circle is $(0,2)$. Here $\bar x$ and $\bar y$ are the average values

$$\begin{aligned} \bar x = \frac{\iint _D x\,\mathrm{d}x\mathrm{d}y}{\iint _D \mathrm{d}x\mathrm{d}y}\qquad \bar y = \frac{\iint _D y\,\mathrm{d}x\mathrm{d}y}{\iint _D \mathrm{d}x\mathrm{d}y} \end{aligned}$$

of $x$ and $y$ over $D$. As the disk $D$ has area $4\pi$,

$$\iint _D x\,\mathrm{d}x\mathrm{d}y =4\pi\bar x =0\qquad \iint _D y\,\mathrm{d}x\mathrm{d}y =4\pi\bar y =8\pi$$

and $$\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =-4\pi(2\bar x+3\bar y) =-4\pi(2\times 0+3\times 2) =-24\pi$$ As

$$\begin{aligned} { \boldsymbol{\nabla} }\cdot\mathbf{F} &=\frac{\partial }{\partial x}(x+x^2y) +\frac{\partial }{\partial y}(y-xy^2) +\frac{\partial }{\partial z}(z+2x+3y) =(1+2xy)+(1-2xy)+(1)\\ &=3 \end{aligned}$$

the divergence theorem gives

$$\begin{aligned} \iint _{\mathcal{S}}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iiint _R{ \boldsymbol{\nabla} }\cdot\mathbf{F}\ \mathrm{d}V -\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S\\ &=\iiint _R3\ \mathrm{d}V -(-24\pi) =3\text{Vol}(R)+24\pi =3\times10+24\pi \\ &=30+24\pi \end{aligned}$$