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, 2016 at 11:55
  • $\begingroup$ Another hint: $dV=dxdydz$ $\endgroup$
    – Candidate
    Apr 26, 2016 at 11:58
  • $\begingroup$ I don't have any old reference book. Any reference link might help. $\endgroup$
    – user963241
    Apr 26, 2016 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. $\endgroup$
    – user332714
    Apr 26, 2016 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, 2016 at 13:05

1 Answer 1


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, $$ \iiint_{V} (\nabla \cdot F)\, dV = \int_{0}^{r} \left(\iint_{S_{\rho}} 3\rho^{2}\, dA\right) d\rho = \int_{0}^{r} 3\rho^{2} \left(\iint_{S_{\rho}} dA\right) d\rho = 3(4\pi)\int_{0}^{r} \rho^{4}\, d\rho. $$


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