I want to calculate:

$$\iiint_V div (\overrightarrow F \cdot \space dV) $$

with $\overrightarrow F=x^3\hat i+ y^3 \hat j+z^3\hat k$ and Surface of sphere given as $x^2+y^2+z^2=r^2$

So, first I calculate $div.\overrightarrow F=3(x^2+y^2+z^2)$

Then using spherical co-ordinates, we know:

$x= rsin\theta \cdot cos\phi$
$y= rsin\theta \cdot sin\phi$
$z= rcos\phi$

Now what I don't understand is the following given according the textbook:

$dxdydz=r^2sin\theta \cdot dr \cdot d\theta \cdot d\phi$

Where does $dxdydz$ come from?

and How are these limits are calculated?

  • For $r$, the limit is $0$ to a
  • For $\theta$ the limit is $0$ to $\pi$
  • For $\phi$ the limit is $0$ to $2\pi$

Using these limits, the result can be calculated using integration.

  • $\begingroup$ Look up basically any calculus reference; read the section about triple integrals and then the section about integration in spherical coordinates. (Also, with that choice of coordinates, your limits are wrong: $\theta$ should run from $0$ to $2 \pi$ while $\phi$ runs from $0$ to $\pi$.) $\endgroup$ – Ian Apr 26 '16 at 11:55
  • $\begingroup$ Another hint: $dV=dxdydz$ $\endgroup$ – Candidate Apr 26 '16 at 11:58
  • $\begingroup$ I don't have any old reference book. Any reference link might help. $\endgroup$ – user963241 Apr 26 '16 at 12:17
  • $\begingroup$ @Ian The coordinates $(r,\theta,\phi)$ follow the right handed (physics, spherical harmonics) convention where $\phi$ is the azimuthal angle and $\theta$ is the zenith angle. $\endgroup$ – for Monica Apr 26 '16 at 13:03
  • $\begingroup$ @lastresort Oh, yes, I see how that would go, and the volume element is modified accordingly. OK, yes, disregard that remark. $\endgroup$ – Ian Apr 26 '16 at 13:05

Not exactly an answer to your question, but a bit long for a comment:

Since $r$ denotes the radius of your sphere, it's best to use a different symbol, such as $\rho$, to denote the "radius function" (i.e., the distance to the origin).

Be that as it may, you can avoid the nuisance of integrating the angular variables in your question because $\nabla \cdot F = 3\rho^{2}$ is constant on each spherical shell $S_{\rho}$. Consequently, $$ \iiint_{V} (\nabla \cdot F)\, dV = \int_{0}^{r} \left(\iint_{S_{\rho}} 3\rho^{2}\, dA\right) d\rho = \int_{0}^{r} 3\rho^{2} \left(\iint_{S_{\rho}} dA\right) d\rho = 3(4\pi)\int_{0}^{r} \rho^{4}\, d\rho. $$


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.