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Find $\iint\limits_\Sigma(x\sqrt{x^2+y^2+z^2}\hat{\imath}+y\sqrt{x^2+y^2+z^2}\hat{\jmath}+z\sqrt{x^2+y^2+z^2}\hat{k})\cdot\hat{n}dS$ Where $\Sigma$ is the torus $\rho=\sin\phi$ Oriented by the normal directed outward.

I got this as a study question for an upcoming exam of mine, and I've been having some trouble with it.

What I've been able to do so far:

Using the Divergence Theorem, which states that $\iint\limits_\Sigma\vec{F}\cdot\hat{n}dS = \iiint\limits_E div(\vec{F})dV$ $$ div(\vec{F}(x,y,z)=P\hat{\imath}+Q\hat{\jmath}+R\hat{k})=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$ $$ \implies div\left(x\sqrt{x^2+y^2+z^2}\hat{\imath}+y\sqrt{x^2+y^2+z^2}\hat{\jmath}+z\sqrt{x^2+y^2+z^2}\hat{k}\right) $$ $$ = \frac{2x^2+y^2+z^2}{\sqrt{x^2+y^2+z^2}}+ \frac{x^2+2y^2+z^2}{\sqrt{x^2+y^2+z^2}}+\frac{x^2+y^2+2z^2}{\sqrt{x^2+y^2+z^2}} = \frac{4x^2+4y^2+4z^2}{\sqrt{x^2+y^2+z^2}} = 4(x^2+y^2+z^2)^{3/2} $$ $$ \implies \iint\limits_\Sigma\vec{F}\cdot\hat{n}dS = 4\iiint\limits_E(x^2+y^2+z^2)^{3/2}dV $$

What I still can't figure out:

How to get the bounds. I now have this triple integral, which needs bounds. Most of the problems I've done in the past have come out to be the triple integral of some constant, which is just the constant multiplied by the volume of the object, which is usually something simple like a cylindar or sphere. This one seems to actually require evaluating. How do I find the bounds to my new triple integral?

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The equation for the torus is clearly directing you to switch to spherical coordinates $(\rho, \phi, \theta)$. The bounds are then $0 \le \rho \le \sin \phi$, $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$

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  • $\begingroup$ I was hoping for the process, not the answer. Once I switch to sphereical coordinates, how do I find that those are the bounds? $\endgroup$ – user198780 Dec 9 '14 at 5:39
  • $\begingroup$ Intuition. You want to integrate over the entire function, so $\sin\phi$ is the upper bound for $\rho$. There are no other conditions so $\phi$ and $\theta$ just go all the way. $\endgroup$ – Dylan Dec 9 '14 at 5:46

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