Divergence Theorem and flux

Question

Given a vector field $A=:(xz,x^2,2xyz+z^2+3)$, and a region bounded below by $z=0$, enclosed by $x^2+y^2+4z^2=a^2$ with $a$ positive.

Can anyone help me to find the flux of it by divergence Theorem and by surface integral separately?

$\nabla A = 3z+2xy$, so by divergence theorem, $\iiint \nabla A dV = \iiint 3z+2xy dV$, where V is the region given above. I tried to use spherical coordinates, but I don't know how to bound it.

For $\iint A\cdot N dS$, I use the gradient of the surface $x^2+y^2+4z^2=a^2$, and dot it by the vector field. Still, it is too complex and I doubt if I can integrate it.

Answer 1

Recall that for a surface with parameterization $\vec{x}(\phi,\theta)$ we can compute $\vec{N}$ as $-\partial_\phi\vec{x} \times \partial_\theta\vec{x}$. Parameterize the surface by $\theta\in [0,2\pi)$ and $\phi\in [0,\frac{\pi}{2})$ as

$$x=a\cos(\theta)\sin(\phi)\\y=a\sin(\theta)\sin(\phi)\\z=\frac{a}{2}\cos(\phi)$$

$$\partial_{\theta }\vec{x}=\begin{pmatrix}-a\sin \left(\theta \right)\sin \left(\phi \right)\\ a\cos \left(\theta \right)\sin \left(\phi \right)\\ 0\end{pmatrix},\:\partial _{\phi }\vec{x}=\begin{pmatrix}a\cos \left(\theta \right)\cos \left(\phi \right)\\ a\sin \left(\theta \right)\cos \left(\phi \right)\\ -\frac{a}{2}\sin\left(\phi \right)\end{pmatrix}$$ $$\partial_\phi\vec{x} \times \partial_\theta\vec{x}=\begin{pmatrix}\frac{-a^2}{2}\cos(\theta)\sin^2(\phi) \\ \frac{-a^2}{2}\sin(\theta)\sin^2(\phi)\\ -a^2\sin(\phi)\cos(\phi)\end{pmatrix} $$

$$\vec{A}=\begin{pmatrix}\frac{a^2}{2}\cos \left(\phi \right)\sin \left(\phi \right)\cos \left(\theta \right)\\ a^2\cos ^2\left(\theta \right)\sin ^2\left(\phi \right)\\ a^3\cos \left(\theta \right)\sin \left(\theta \right)\sin^2 \left(\phi \right)\cos\left(\phi \right)+\frac{a^2}{4}\cos ^2\left(\phi \right)+3\:\:\end{pmatrix}$$

$\vec{A}\cdot \vec{N}dS =(\frac{a^4}{4}\cos(\phi)\cos^2(\theta)\sin^3(\phi)+\frac{a^4}{2}\sin(\theta)\sin^4(\phi)\cos^2(\theta)+a^5\cos(\theta)\sin(\theta)\sin^3(\phi)\cos^2(\phi)+\frac{a^4}{4}\cos^3(\phi)\sin(\phi)+3a^2\sin(\phi)\cos(\phi))d\phi d\theta $.

This looks bad, I know! But realize we're integrating from $0$ to $2\pi$ in $\theta$ so we can get rid of all the terms with odd powers of $\cos(\theta)$ or $\sin(\theta)$, and the integrals are easy:

$$\int_0^{2\pi}\int_0^\frac{\pi}{2}\left(\frac{a^4}{4}\cos^2(\theta)\cos(\phi)\sin^3(\phi)+\frac{a^4}{4}\cos^3(\phi)\sin(\phi)+3a^2\cos(\phi)\sin(\phi)\right)d\phi d\theta$$ $$=\frac{a^4\pi}{16} + \frac{a^4\pi}{8} + 3a^2\pi$$

Again, this is only the flux through the upper surface of the spheroid. The bottom surface (at z=0) has a negative flux of $\int\int_{x^2+y^2=a} (A|_{z=0}\cdot-\hat{z})dA=-3\pi a^2$.

Answer 2

Since this has gone unanswered for a bit I'll go ahead: Drawing a picture of the region will help! You're right, spherical coordinates might be good. But I don't know they make it any easier in this case, so let's work with Cartesian for now and then I'll address other coordinates. We are looking for the flux of A through the upper half of this oblate spheroid.

Surface Integral

For the surface integral bit, note that the top of the surface $S$ is given by $z=f(x,y)=\frac{1}{2}\sqrt{{a^2}-x^2-y^2}$, and the bottom, is of course just the circle on the $xy$ plane of radius $a$.

The normal vector field for the top of $S$ is just the gradient of $z=f(x,y)$.

$$\vec{N}=\begin{pmatrix}xz\\ yz\\ 1 \end{pmatrix} = \begin{pmatrix}\frac{x}{2\sqrt{a^2-x^2-y^2}}\\ \frac{y}{2\sqrt{a^2-x^2-y^2}}\\ 1\end{pmatrix}$$

Since $dS=\sqrt{1+(\partial_xf)^2+(\partial_yf)^2}=||\nabla f||$ I won't bother calculating the unit normal. Also note that I changed the sign on the first two components. There are two normal vector fields $(f_x,f_y,-1)$ and $(-f_x,-f_y,1)$ pointing in and out of the surface.

Now $\vec{A}\cdot \vec{N} = \frac{x^2z+x^2y}{2\sqrt{a^2-4x^2-4y^2}}+2xzy+z^2+3$ At this point, we're integrating over the circular region on the xy plane, so we'll go ahead and use polar coordinates. We'll also substitute in the $z=\frac{1}{2}\sqrt{a^2-r^2}$. This gives us

$$\frac{\frac{x^2}{2}\sqrt{a^2-r^2}+x^2y}{2\sqrt{a^2-r^2}}+xy\sqrt{a^2-r^2}+\frac{1}{4}\left(a^2-r^2\right)+3$$

$$\frac{x^2y}{2\sqrt{a^2-r^2}}+xy\sqrt{a^2-r^2}+\frac{1}{4}\left(a^2-y^2\right)+3$$ $$=\frac{r^2\cos ^2\left(\theta \right)\sin \left(\theta \right)}{2\sqrt{a^2-r^2}}+r^2\cos \left(\theta \right)\sin \theta \sqrt{a^2-r^2}+\frac{a^2}{4}-\frac{r^2}{4}\sin ^2\left(\theta \right)+3$$

The first two terms vanish because their integral over $\theta$ is $0$. So we just have

$$\int_0^{2\pi}\int_0^a\left(\frac{a^2r}{4}-\frac{r^3}{4}\sin^2(\theta)+3r\right)drd\theta $$ $$= \frac{a^4\pi}{4} - \frac{a^4\pi}{16}+3\pi a^2$$

Now flux through the bottom of the region (with normal vector $-\hat{z}$) is just the area $\pi a ^2$ times the $z-$component of $\vec{A}$ which is 3. The total flux then is just $\frac{3\pi a^4}{16}$.

Volume Integral

This is also not so bad. First we'll do the z-bounds. Write z as a function of x and y, as above, and integrate from 0 to whatever z(x,y) is. Then our cross-section is a circle in the xy plane. Write y(x) = \pm\sqrt{a^2-x^2}, and integrate from the negative to the positive square root. Finally, our cross section is just the line from -a to a on the x-axis. Hence, we have

$$\int _{-a}^a\:\int _{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\:\int _0^{\frac{1}{2}\sqrt{a^2-x^2-y^2}}\left(3z+2xy\right)dzdydx\:$$

$$\int _{-a}^a\:\int _{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\:\left(\frac{3\left(a^2-x^2-y^2\right)}{8}+xy\sqrt{a^2-x^2-y^2}\right)dydx$$

Again, it's perfectly fine to keep using Cartesian coordinates, but since we're integrating over a circle, polar coordinates are quicker, but remember the extra "r".

$$\int _0^{2\pi }\:\int _0^a\:\left(\frac{3\left(a^2-r^2\right)}{8}+r^2\sin \left(\theta \right)\cos \left(\theta \right)\sqrt{a^2-r^2}\right)rdrd\theta $$ $$=\frac{3\pi }{4}\left(\frac{a^4}{2}-\frac{a^4}{4}\right)=\frac{3\pi a^4}{16}$$