# Surface integral over an inconvenient surface

I was looking at U Penn's sample prelim (http://www.math.upenn.edu/grad/sampleprelim.pdf) and most of the problems seem fairly trivial. However, I'm having a lot of trouble with a surface integral. The problem reads:

Let $$\vec{F}$$ be a vector field defined in $$\mathbb{R}^3$$ minus the origin by

$${\displaystyle\vec{F}=\frac{\vec{r}}{\|\vec{r}\|^3}=\frac{x\vec{i}+y\vec{j}+z\vec{k}}{(x^2+y^2+z^2)^{3/2}}}$$ for $$\vec{r}\neq 0$$.

a) Compute $$div\vec{F}$$.

b) Let $$S$$ be the sphere of radius 1 centered at $$(x,y,z) = (2,0,0)$$. Compute

$$\iint_S \vec{F}\cdot\vec{n}\ dS}$$.

Part a) is no problem. I'm a little confused about part b). I tried using the divergence theorem, but the computation required seems hard too hard; converting to spherical coordinates or quasi-spherical coordinates (i.e. let $$x=\rho\sin\phi\cos\theta+2$$ and the others same as normal) doesn't seem to help because it'll make either the limits or the integral hard. I didn't have much luck with a direct computation of the surface integral either.

The only vector Calculus that I know is from Calc 3, which was awhile ago. So I suspect there's something I'm not seeing. Thanks for your help.

• From part a) you know $\nabla \cdot \mathbf F = 0$ everywhere except the origin. The origin is outside the sphere of integration. The surface integral in b) is then $\iiint_V 0 dV = 0$. – David H Sep 17 '13 at 17:05
• Well, that seems more in-line with the difficulty of the rest of the sample test, but isn't $div\vec{F}=\frac{\partial }{\partial x}\left( \frac{x}{(x^2+y^2+z^2)^{3/2}}\right)+\frac{\partial }{\partial y}\left( \frac{y}{(x^2+y^2+z^2)^{3/2}}\right)+\frac{\partial }{\partial z}\left( \frac{z}{(x^2+y^2+z^2)^{3/2}}\right)\\ =\frac{3(x^2+y^2+z^2)-9}{(x^2+y^2+z^2)^{3/2}}$? – Charles Sep 17 '13 at 19:11
• Lol. Silly me... I take it back. I guess I was too confident about computing the divergence. – Charles Sep 17 '13 at 19:16

$\frac{\partial }{\partial x}\left( \frac{x}{(x^2+y^2+z^2)^{3/2}}\right) = \frac{1}{(x^2+y^2+z^2)^{3/2}} \frac{\partial }{\partial x} (x) + x \frac{\partial }{\partial x} \frac{1}{(x^2+y^2+z^2)^{3/2}}$
$= \frac{1}{(x^2+y^2+z^2)^{3/2}} + x \left( -\frac32 \frac{2x}{(x^2+y^2+z^2)^{5/2}} \right)$
$= \frac{1}{(x^2+y^2+z^2)^{3/2}} - \frac{3x^2}{(x^2+y^2+z^2)^{5/2}}$, and similarly for the $y$ and $z$ partials. Adding them up, the divergence is,
$\nabla \cdot \mathbf F = \frac{3}{(x^2+y^2+z^2)^{3/2}} - \frac{3(x^2 + y^2 + z^2)}{(x^2+y^2+z^2)(x^2+y^2+z^2)^{3/2}} = 0$, except at $x=0$.
Answer to part b). According to part a), the divergence of ${\bf F}$ is 0 everywhere except at $x=0$. But $0$ is not inside $S$, the divergence theorem gives $\iiint 0 dV = 0$. It is a very convenient surface.