3
$\begingroup$

Find flux of $ \vec{F} =<0,0,z>$ through the sphere of radius $a$ centered at the origin.

Using Gauss Divergence Theorem

$$ \nabla \cdot \vec{F} = 1 \\ \therefore \int_S \vec{F}\cdot \hat{n} ~dS= \iiint_R 1 ~dV =\frac{4}{3}\pi a^3 $$


Calculating the surface integral by using sperical co-ordinates $ (\rho, \phi, \theta )$

$$ \hat{n} = \frac{1}{a} <x,y,z> \\ \vec{F} \cdot \hat{n}=\frac{z^2}{a}$$

w.k.t

$$ dS= a^2 \sin(\theta) ~ d\phi d\theta \\ z= a cos(\phi)$$

substituting:

$$\int_s \vec{F}\cdot \hat{n} ~dS = \int_0^{2\pi} \int_0^\pi \frac{a^2 cos^2(\phi)}{a} a^2 \sin(\theta) ~ d\phi d\theta \\ $$

This integral will evaluate to zero, because $sin(\theta)$ will be integrated from $0$ to $2\pi$.


I'm getting two different answers. Where am I going wrong?

$\endgroup$
4
$\begingroup$

In spherical coordinates $(r,\theta,\phi)$, $z=r\cos(\theta)$ and $dS=r^2\sin(\theta)\,d\theta\,d\phi$ where $\theta \in[0,\pi]$ and $\phi\in[0,2\pi]$.

Then, note that

$$\int_0^{\pi}\cos^2(\theta)\sin(\theta)\,d\theta=\frac{2}{3}$$

so that

$$\oint_{|r|=a}\vec F(\vec r)\cdot \hat n\,dS=\int_0^{2\pi}\int_0^\pi \frac{a^2\cos^2(\theta)}{a}\,a^2\sin(\theta)\,d\theta\,d\phi=\frac{4\pi a^3}{3}$$

as expected!

$\endgroup$
  • $\begingroup$ I see where i went wrong, thanks! $\endgroup$ – jumpmonkey Jun 4 '17 at 13:51
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jun 4 '17 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.