Let $S$ be a surface given by $x^2+3y^2+z^2=1$ and the vector field
$F(x,y,z)=(x+y^2, -x, z-xy)$.Find $\iint _{S} F \cdot \vec{n}$
$dS$ with $\vec{n}$ pointing outside of the surface.
I've tried finding this using Gauss' theorem as the divergence of the vector field is an scalar ($div F = 2$).
So: $\iint_S F dS = \iiint_V 2 dV$
Now I made a change of variables with the surface, with $\begin{cases} x=u \\ y=\frac{v}{\sqrt3} \\ z=w \end{cases}$ then I have $u^2+v^2+w^2=1$, with the Jacobian $J=\frac{1}{\sqrt 3}$.
And then I made another change of variables but with spherical coordinates:
$\begin{cases} u=\rho \cos \theta \sin \phi \\ v = \rho \sin \theta \sin \phi \\ w = \rho \cos \phi \end{cases}$
with $\rho \in [0,1]$, $\theta \in [0, \pi]$, $\phi \in [0, 2\pi]$, and the Jacobian $J=\rho^2\sin \theta$.
The integral I got was $\int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta d \phi d\theta d\rho = 0$, and that's not the correct answer…
The correct answer is $\iint _{S} F \cdot \vec{n} = \frac{8\pi}{3 \sqrt 3}$
Best Answer
Your integral order is wrong.
$\displaystyle \int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta \, d \phi \, d\theta \, d\rho = 0$
Based on notation you have used, $\theta$ is the polar angle and $\phi$ is the azimuthal angle.
So, $0 \leq \theta \leq \pi \,$ and $0 \leq \phi \leq 2\pi \,$
Your integral should be
$\displaystyle \int _{0} ^{1} \frac{2}{\sqrt3} \rho^2 \bigg[ \int _{0} ^{2\pi} \bigg [\int _{0} ^{\pi} \sin \theta \, d \theta\bigg] \, d\phi \bigg] \, d\rho$
$\,$
You get $ \,\displaystyle \int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta \, \color{blue} {d \theta \, d\phi} \, d\rho = \frac{8 \pi}{3\sqrt3}$