Let $f(x,y,z) = xz-y^2+xyz$. We wish to calculate $\displaystyle \iint_{S} \nabla f \cdot \overline{n}d\sigma$ where $d\sigma$ is the area element of $S$ , and $\overline{n}$ is the outward pointing unit normal to $S$ at $(x,y,z)$. Applying the Divergence theorem, we get:

$$\displaystyle \iint_{S} \nabla f \cdot \overline{n}d\sigma = -2\iiint_{V} d\tau$$

My question is: where does the idea that this is the volume of the unit sphere come from?

  • $\begingroup$ If V is the unit sphere and $d\tau$ is the volume element of V, then the integration of the volume element gives the total volume. Here it will give the volume of V which is the volume of a unit sphere $\endgroup$ – Triatticus Mar 24 '18 at 6:00
  • $\begingroup$ @Triatticus Thanks for the comment. Do you mind expanding on your first sentence a bit? $\endgroup$ – Ryan Mar 24 '18 at 6:18
  • $\begingroup$ All you need to do is expand to spherical coordinates to prove it to yourself recall that $d\tau=dxdydz=r^2\sin{\phi}dr d\phi d\theta$ and the limits of integration are $0\leq r \leq 1$, $0\leq \phi \leq \pi$ and $0\leq \theta < 2\pi$. That is of you use $\phi$ as the polar angle, switch $\theta$ and $\phi$ for other conventions $\endgroup$ – Triatticus Mar 24 '18 at 6:41
  • $\begingroup$ @Triatticus I understand how to evaluate the final integral (provided that V is the unit sphere). My problem is understanding how we went from $S$ to the unit sphere. $\endgroup$ – Ryan Mar 24 '18 at 6:46
  • $\begingroup$ @Triatticus Ignore that, found where they stated that "S is the unit sphere." Thanks for your replies. $\endgroup$ – Ryan Mar 24 '18 at 6:50

Start with $\nabla f = (z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k}$.

Lets apply the divergence theorem and see what we have to work with.

$$ \iint_S \left[(z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k} \right] \cdot \bar{n}d\sigma = $$ $$ \iiint_V \nabla \cdot \left[ (z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k} \right] dV $$ $$ = \iiint_V \left[ 0\hat{i} -2 \hat{j} + 0\hat{k}\right] dV $$ $$ =-2 \iiint_V dV $$

This is the result you show. It states that the flux through your surface is -2 times the volume enclosed in the surface so you have a sink. Now the volume enclosed by the surface is whatever you want it to be. Usually it would be stated in the problem or it is a volume of interest by you.

Say we want to know the flux through the surface of a unit sphere. The volume is $\frac{4\pi}{3}$ so the flux through this surface is $-\frac{8\pi}{3}$.

In your question you never state what the surface, S, is.

  • $\begingroup$ Oh, okay! Thanks, I see. Another part of the question said that $S$ was the unit sphere. I missed that bit and was baffled by where the volume of the unit sphere came from! $\endgroup$ – Ryan Mar 24 '18 at 6:49

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