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.
Best Answer
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 $\int\int_{x^2+y^2=a} (A|_{z=0}\cdot-\hat{z})dA=-3\pi a^2$.