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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

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

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.

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  • $\begingroup$ 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 $\iiint_V 0 dV = 0$. $\endgroup$ – David H Sep 17 '13 at 17:05
  • $\begingroup$ Well, that seems more in-line with the difficulty of the rest of the sample test, but isn't $div\vec{F}=\frac{\partial }{\partial x}\left( \frac{x}{(x^2+y^2+z^2)^{3/2}}\right)+\frac{\partial }{\partial y}\left( \frac{y}{(x^2+y^2+z^2)^{3/2}}\right)+\frac{\partial }{\partial z}\left( \frac{z}{(x^2+y^2+z^2)^{3/2}}\right)\\ =\frac{3(x^2+y^2+z^2)-9}{(x^2+y^2+z^2)^{3/2}}$? $\endgroup$ – Charles Sep 17 '13 at 19:11
  • $\begingroup$ Lol. Silly me... I take it back. I guess I was too confident about computing the divergence. $\endgroup$ – Charles Sep 17 '13 at 19:16
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$ \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 $x=0$.

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  • $\begingroup$ I was so confident about (a) that I didn't think to check my work! $\endgroup$ – Charles Sep 18 '13 at 5:33
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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.

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