면적분과 플럭스

We are to find the surface area of part of a hemisphere. On the hemisphere

$$\begin{aligned} z=f(x,y)=\sqrt{a^2-x^2-y^2}\qquad f_x(x,y)=-\frac{x}{\sqrt{a^2-x^2-y^2}}\qquad f_y(x,y)=-\frac{y}{\sqrt{a^2-x^2-y^2}} \end{aligned}$$

so that

$$\begin{aligned} \mathrm{d}S&=\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,\mathrm{d}x\,\mathrm{d}y =\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \\ &=\sqrt{\frac{a^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$$

In polar coordinates, this is $\mathrm{d}S=\frac{a}{\sqrt{a^2-r^2}}\,r\,\mathrm{d}r\,\mathrm{d}\theta$. We are to find the surface area of the part of the hemisphere that is inside the cylinder, $x^2-ax+y^2=0$, which in polar coordinates becomes $r^2-ar\cos\theta=0$ or $r=a\cos\theta$. The top half of the domain of integration is sketched below.

So the

$$\begin{aligned} {\rm Surface\ Area} &= 2\int_0^{\pi/2}\mathrm{d}\theta\int_0^{a\cos\theta}\mathrm{d}r\ r \frac{a}{\sqrt{a^2-r^2}} = 2a\int_0^{\pi/2}\mathrm{d}\theta\ \Big[-\sqrt{a^2-r^2}\,\Big]_0^{a\cos\theta} \\ &= 2a\int_0^{\pi/2}\mathrm{d}\theta\ \big[a-a\sin\theta\big] \\ &= 2a^2\Big[\theta+\cos\theta\Big]_0^{\pi/2} =a^2[\pi-2] \end{aligned}$$

In cylindrical coordinates

$$x=r\cos\theta\qquad y=r\sin\theta\qquad z=z$$

In these coordinates the equation, $x^2+y^2=2ay$, of the cylinder becomes

$$r^2=2ar\sin\theta\qquad\text{or}\qquad r=2a\sin\theta$$

That is, $r=f(\theta)$ with $f(\theta) =2a\sin\theta$. Parametrize the cylinder by $\mathbf{r}(\theta,z) = x(\theta,z)\,\hat{\boldsymbol{\imath}} + y(\theta,z)\,\hat{\boldsymbol{\jmath}} + z(\theta,z)\,\hat{\mathbf{k}}$ with

$$\begin{aligned} x(\theta,z)&=f(\theta)\cos\theta=2a\sin\theta\cos\theta=a\sin2\theta\\ y(\theta,z)&=f(\theta)\sin\theta=2a\sin\theta\sin\theta=a(1-\cos2\theta)\\ z(\theta,z)&=z \end{aligned}$$

Under this parametrization,

$$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}=2a\cos2\theta\,\hat{\boldsymbol{\imath}}+2a\sin2\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial z}=\hat{\mathbf{k}} &\implies \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z} =-2a\cos2\theta\,\hat{\boldsymbol{\jmath}}+2a\sin2\theta\,\hat{\boldsymbol{\imath}} \\ &\implies \mathrm{d}S=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z}\right|\, \mathrm{d}\theta\,\mathrm{d}z =2a\,\mathrm{d}\theta\,\mathrm{d}z \end{aligned}$$

We still have to determine the limits of integration. The figure on the left below provides a top view of the cylinder.

From it we see that $0\le\theta\le\pi$. The cone $z^2=x^2+y^2=r^2$ (i.e. $z=\pm r$) and the cylinder $r=2a\sin\theta$ intersect at $z^2=r^2=4a^2\sin^2\theta$. So, for each fixed $\theta$, $z$ runs from $-2a\sin\theta$ to $z=+2a\sin\theta$. (See the figure on the right above. It shows a constant $\theta$ cross-section.) Finally,

$$\begin{aligned} {\rm Area} &= \int_{|z|\le 2a\sin\theta}2a\,\mathrm{d}\theta\,\mathrm{d}z =2a\int_0^\pi \!\!\mathrm{d}\theta\int_{-2a\sin\theta}^{2a\sin\theta} \!\!\!\mathrm{d}z =8a^2\int_0^\pi \!\!\mathrm{d}\theta\ \sin\theta =8a^2\Big[-\cos\theta\Big]_0^\pi \\ &=16a^2 \end{aligned}$$