# Gauss Theorem example

Verify the Gauss theorem for the vector field $F(x)=\frac{x }{\|x\|},$ where $x \in W \subset \mathbb{R}^3$ and $$W=\left\{(x,y,z) \in \mathbb{R}^3 \left/ a^2\right.\leqslant x^2 + y^2 + z^2 \leqslant b^2\right\}$$

Thanks you very much!

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Normally if I saw $x^\wedge 2$, I would think that what was intended was $x^2$. But you seem to have gone to some trouble to avoid writing $x^2$. So what does it mean? –  Michael Hardy Jun 25 '13 at 23:07
@MichaelHardy Thanks. I made that on Wolfram alpha, and copy paste goes wrong sometimes.. –  Argentino2013 Jun 25 '13 at 23:13
So what's giving you trouble? –  Muphrid Jun 25 '13 at 23:21
I try to prove it, but i can't do it. I'm not very good with this, for that i came here –  Argentino2013 Jun 25 '13 at 23:39
It does help to know what you did try. It could be you just had a simple mistake. –  Muphrid Jun 25 '13 at 23:42

Gauss's Theorem states $$\iiint_{\Omega} (\nabla \cdot \overrightarrow{F}) \space dV = \iint_{\partial \Omega} (\overrightarrow{F} \cdot \hat n) \space dS.$$

Verifying the theorem for the given function is just a matter of calculating the LHS integral and the RHS integral and checking that LHS = RHS.

\begin{align} \nabla \cdot \overrightarrow{F} &= \nabla \cdot (\frac{\overrightarrow{x}}{\|x\|}) \\& = \frac{1}{\|x\|} (\nabla \cdot \overrightarrow{x}) + \overrightarrow{x} \cdot(\nabla \frac{1}{\|x\|}) \\&= \frac{3}{\|x\|} + \overrightarrow{x} \cdot (\frac{-\overrightarrow{x}}{{\|x\|}^3}) \\&= \frac{2}{\|x\|} \end{align}

\begin{align} \iiint_{W} (\nabla \cdot \overrightarrow{F}) \space dV &= \iiint_{W} \frac{2}{\|x\|} \space dV \\&= \int_0^{2 \pi} \int_0^{\pi} \int_a^b (\frac{2}{r})(r^2 \sin{\theta} \space dr d\theta d\phi) \\&= \int_0^{2 \pi} \int_0^{\pi} \int_a^b 2r \sin{\theta} \space dr d\theta d\phi \\&= (2\pi)( 2)(b^2 - a^2) \\&=4\pi(b^2 -a^2) \end{align}

The surface integral is much less tedious since the unit normal vector to the surface of integration is $\hat n = \frac{\overrightarrow{x}}{\|x\|} = \overrightarrow{F}$ on the outer spherical surface and $\hat n = -\frac{\overrightarrow{x}}{\|x\|} = -\overrightarrow{F}$ on the inner surface. The integral is seen to reduce to just the difference of surface areas of the outer and inner spheres. Thus,

$$\iint_{\partial W} (\overrightarrow{F} \cdot \hat n) \space dS = \iint_{\partial W}dS = 4\pi b^2 - 4\pi a^2 = 4\pi (b^2 - a^2).$$

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Thanks you david! Was very nice explained –  Argentino2013 Jun 26 '13 at 0:52

I made it in spherical coordinates I use (r,v,u) for calculus:

$\ \int _a^b\int _0^{2\text{Pi}}\int _0^{\text{Pi}}(2)r*\text{Sin}[u]dudvdr$

$\ 4 \left(-a^2+b^2\right) \pi$

And then:

$\ \int _0^{\text{Pi}}\int _0^{2\text{Pi}}(\text{Sin}[u]*\text{Cos}[v]*\text{Sin}[u]*\text{Cos}[v]*b{}^{\wedge}2*\text{Sin}[u] +\text{Sin}[u]\text{Sin}[v]* \text{Sin}[u]\text{Sin}[v]*b{}^{\wedge}2*\text{Sin}[u]+\text{Cos}[u]\text{Cos}[u]*b{}^{\wedge}2*\text{Sin}[u])dvdu$

$\ = 4 b^2 \pi$

$\ \int _0^{\text{Pi}}\int _0^{2\text{Pi}}(-\text{Sin}[u]*\text{Cos}[v]*\text{Sin}[u]*\text{Cos}[v]*a{}^{\wedge}2*\text{Sin}[u] -\text{Sin}[u]\text{Sin}[v]* \text{Sin}[u]\text{Sin}[v]*a{}^{\wedge}2*\text{Sin}[u]-\text{Cos}[u]\text{Cos}[u]*a{}^{\wedge}2*\text{Sin}[u])dvdu$

$\ = -4 a^2 \pi$

Then:

$\ = 4 b^2 \pi -4 a^2 \pi$

Thanks to Davi and Michael

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