Let $x^2+y^2+z^2=R^2$ be the equation of the sphere we want to calculate the flux through, and $\vec{r}=x\hat i+y\hat j+z\hat k$ be the position vector field. We can compute it via divergence theorem:
$$\Phi=\iiint_V(\nabla\cdot \vec r)dV=3\iiint_V dV=3\dfrac{4}{3}\pi R^3=4\pi R^3.$$
Or like this too:
$$\Phi=\oint_S \vec r\cdot d\vec S=\oint_S R(\hat r\cdot\hat r)dS=R\oint_S dS=4\pi R^3.$$
But when I try to calculate it using the fact that $dS=||\partial_x \vec r\times \partial_y \vec r||dxdy$ ($\vec r(x,y)$ being the parametrized surface), I fail:
Let's parametrize the surface like
$$\vec r(x,y)= \left(x,y,\pm\sqrt{R^2-x^2-y^2}\right).$$
Now we can calculate the cross product,
$$\partial_x \vec r\times \partial_y \vec r=\left|\begin{array}
\hat{i} & \hat j& \hat k\\
1 & 0 & -\frac{x}{z}\\
0 & 1 & -\frac{y}{z}
\end{array}\right|=\dfrac{x}{z}\hat i+\dfrac{y}{z}\hat j+\hat k,$$
and thus we'll get the norm:
$$\Longrightarrow ||\partial_x \vec r\times \partial_y \vec r||=\sqrt{1+\dfrac{x^2}{z^2}+\dfrac{y^2}{z^2}}=\sqrt{\dfrac{x^2+y^2+z^2}{z^2}}=\dfrac{R}{|z|}.$$
We can now apply these equations to the surface integral:
$$\Longrightarrow \Phi=\oint_S R(\hat r\cdot \hat r)dS=R\oint_S \dfrac{R}{|z|}dxdy.$$
And changing into spherical coordinates we'd get that
$$R^2\int_0^{2\pi}\int_0^\pi\dfrac{R^2\sin\theta}{R|\cos\theta|} d\theta d\Phi=2\pi R^3 I,$$
and $I$ does not converge…
Best Answer
First, why not parametrize in spherical coordinates? :)
Switching an integral in $(x,y)$ over the disk to spherical coordinates is, of course, going to involve $\rho$ and $\phi$. But it’s silly. Switching to polar coordinates is what you want. So $$\iint_D \frac{R^2}{|z|}\,dx\,dy = R^2 \int_0^{2\pi}\int_0^R \frac r{\sqrt{R^2-r^2}}dr\,d\theta = 2\pi R^2\Big[{-}\sqrt{R^2-r^2}\Big]_0^R = 2\pi R^3.$$ And you get another for the integral over the lower hemisphere.