I am trying to calculate the volumn of $$\Omega=\{(x,y,z)\in\mathbb R^3|(\frac{x}{2})^2+y^2+z^2\le 1\}$$ A parametrization of the boundary of $\Omega$ is $$\gamma:[0,2\pi]^2\to\mathbb R^3$$ $$\gamma(\phi,\psi)=(2\cos(\phi)\cos(\psi),\sin(\phi)\cos(\psi),\sin(\psi))$$
So with Stokes we have $$\int_\Omega d\mathbb R^3=\int_{\gamma} x\ \text{d}y\wedge\text{d}z$$
$\gamma^*(x\ \text{d}y\wedge\text{d}z)=2\cos(\phi)\cos(\psi)[\cos(\phi)\cos(\psi)\ \text{d}\phi-\sin(\phi)\sin(\psi)\ \text{d}\psi]\wedge[\cos(\psi)\ \text{d}\psi] =$ $$2\cos^2(\phi)\cos^3(\psi)\ \text{d}\phi\wedge\text{d}\psi$$
$\implies$ $$\int_{\gamma} x\ \text{d}y\wedge\text{d}z=\int_0^{2\pi}\Big(\int_0^{2\pi}2\cos^2(\phi)\cos^3(\psi)\text{d}\phi\Big)\ \text{d}\psi$$ = $$2\pi\int_0^{2\pi}\cos^3(\psi)\ \text{d}\psi $$ = $$2\pi\cdot 0=0$$
But it is supposed to be $\frac{8}{3}\pi$. I know that there are other, easier ways to calculate this, but I wanted to try it that way to practise line integrals. Any help is greatly appreciated!
I know the mistake:
The right parametrization of the boundary of $\Omega$ is $$\gamma:[0,2\pi]\times[\frac{-\pi}{2},\frac{\pi}{2}]\to\mathbb R^3$$ $$\gamma(\phi,\psi)=(2\cos(\phi)\cos(\psi),\sin(\phi)\cos(\psi),\sin(\psi))$$
$$\int_{\gamma} x\ \text{d}y\wedge\text{d}z=\int_\frac{-\pi}{2}^\frac{\pi}{2}\Big(\int_0^{2\pi}2\cos^2(\phi)\cos^3(\psi)\text{d}\phi\Big)\ \text{d}\psi$$ = $$2\pi\int_\frac{-\pi}{2}^\frac{\pi}{2}\cos^3(\psi)\ \text{d}\psi $$ = $$2\pi\cdot \frac{4}{3}=\frac{8}{3}\pi$$