Use the Divergence Theorem to compute the net outward flux of:

$$ F = \langle x^2, y^2, z^2 \rangle $$ $S$ is the sphere: $$ \{(x,y,z): x^2 + y^2 + z^2 = 25\} $$

First, I took: $$ \nabla \cdot F = 2x + 2y + 2z $$

Then, I tried setting up the triple integral with spherical coordinates, but it is just not working out for me.

  • $\begingroup$ $\nabla \cdot F$ is an odd function and your integration region is symmetric. $\endgroup$ – achille hui Aug 7 '15 at 1:46
  • $\begingroup$ @achillehui. bingo you beat me to it $\endgroup$ – Matematleta Aug 7 '15 at 1:46
  • $\begingroup$ I changed $< x^2, y^2, z^2 >$ to $\langle x^2, y^2, z^2 \rangle$ and $\nabla * F$ to $\nabla\cdot F$, and inserted the $\{\text{curly braces}\}$ where they belong. All standard usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 7 '15 at 4:11

Converting to spherical coordinates, we have

$$\nabla \cdot \vec F=2(x+y+z)=2(r\sin \theta \cos \phi+r\sin \theta \sin \phi +r\cos \theta)$$


$$\int_V \nabla \cdot \vec F dV=\int_0^{2\pi}\int_0^{\pi}\int_0^5 2(r\sin \theta \cos \phi+r\sin \theta \sin \phi +r\cos \theta)\,r^2\sin \theta \,dr\,d\theta\,d\phi=0$$

since $\int_0^{2\pi}\sin \phi d\phi=\int_0^{2\pi}\cos \phi d\phi=0$ and $\int_0^{\pi}\cos \theta \sin \theta d\theta=0$

  • $\begingroup$ So you just simply had to substitute the x, y, and z spherical coordinates.. It seems relatively simple, yet I'm always struggling with this section. Thank you. $\endgroup$ – dendritic Aug 7 '15 at 1:46
  • $\begingroup$ You're very welcome. It was my pleasure. As a suggested exercise, see if you can recover the result using the surface integral. $\endgroup$ – Mark Viola Aug 7 '15 at 1:47
  • $\begingroup$ Sure, will definitely try that out! $\endgroup$ – dendritic Aug 7 '15 at 1:48

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