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$\newcommand{\mbf}{\mathbf}$ Evaluate $$ \iint \limits_{S} \mbf{F} \cdot d \mbf{S} $$ where $\mbf{F} = 3xy^2 \mbf{i} + 3x^2y \mbf{j} + z^3 \mbf{k}$ and $S$ is the surface of the unit sphere.

I have written my solution below -- is it correct? Also, how does one know when to use Green/Stokes/Divergence theorem to evaluate a surface/line integral?

Note that evaluating this surface integral directly yields an ugly integral of $\int \sin^5(x)\, dx$. We will resort to Gauss' Divergence Theorem to simplify matters.

Indeed, by Gauss' Divergence Theorem, we have $$ \iint \limits_{S} \mbf{F} \cdot d \mbf{S} = \iiint \limits_{S_{\text{int}}} (\nabla \cdot \mbf{F}) \, dV. $$ where $S_{\text{int}}$ denotes the interior of the sphere. Now, $\nabla \cdot \mathbf{F} = 3(x^2+y^2+z^2)$. Now, we are just integrating $$ \iiint \limits_{S_{\text{int}}} 3(x^2+y^2+z^2) \, dV. $$ Let's transform this to spherical coordinates. We obtain the equivalent integral $$ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} 3 \rho^4 \sin \phi \, d \rho \, d \phi \, d \theta. $$ $$ = \frac{3}{5} \int_{0}^{2 \pi} \int_{0}^{\pi} \sin \phi \, d \phi \, d \theta $$ $$ = \frac{12 \pi}{5}. $$

WeierstrassSauce
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    Your choice of theorem to apply is correct, but your final simplication of $\nabla\cdot\mathbf F$ is incorrect. Remember that you're integrating over the inside of the unit sphere when you do the volume integral. – Ted Shifrin Dec 15 '13 at 18:44
  • You need to integrate $\nabla \cdot \mathbf{F} = 3(x^2+y^2+z^2)$ over $x^2+y^2+z^2<1$ – yoyo Dec 15 '13 at 19:01
  • Oh, of course, silly error! I will edit my current solution. – WeierstrassSauce Dec 15 '13 at 19:10
  • I have fixed the solution. – WeierstrassSauce Dec 15 '13 at 19:16
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    Pay attention when you switch to polar coordinates. You wrote $dV=d\rho d\theta d\phi$, which is wrong. – Giuseppe Negro Dec 15 '13 at 19:24
  • Depending on what country you're from, WeierstraßSauce, you might call this spherical coordinates. But, regardless, as @GiuseppeNegro indicated, you need to the correct formula for $dV$, incorporating what I like to call the correct "fudge factor." – Ted Shifrin Dec 15 '13 at 19:28
  • Ah, that's right -- I need to multiply by the magnitude of the Jacobian of the transformation, $\rho^2 \sin \phi$. I believe I now have a correct solution. – WeierstrassSauce Dec 15 '13 at 19:41

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