Flux: Surface integral of a sphere.


by Kizaru
Tags: flux, integral, sphere, surface
Kizaru
Kizaru is offline
#1
Sep15-09, 02:17 PM
P: 45
1. The problem statement, all variables and given/known data
Find the surface integral of [tex]\vec{r}[/tex] over a surface of a sphere of radius a and center at the origin. Also find the volume integral of [tex]\nabla[/tex] [tex]\bullet[/tex] [tex]\vec{r}[/tex].


2. Relevant equations
Divergence theorem.


3. The attempt at a solution
First I did the volume integral part of the divergence theorem. I obtained [tex]\nabla[/tex] [tex]\bullet[/tex] [tex]\vec{r}[/tex] = 1 + 1 + 1 = 3. So I figured, the answer must be 3*volume = 4[tex]\pi[/tex]r[tex]^{3}[/tex] (I don't know why the pi looks like an exponent, but it's 4 pi r^3)

This answer seems like a correct one.

Now the surface integral I'm having trouble with. Knowing that the equation of the sphere is
x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]=a[tex]^{2}[/tex], I found [tex]\nabla[/tex] [tex]\bullet[/tex] (x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]) to obtain the normal. The [tex]\vec{r}[/tex] [tex]\bullet[/tex] [tex]\vec{n}[/tex] = 2x[tex]^{2}[/tex] + 2y[tex]^{2}[/tex] + 2z[tex]^{2}[/tex].

So I would integrate this over the surface in Cartesian coordinates, or convert to spherical and integrate? Is the normal suppose to be the normal unit vector? I appear to be obtaining the wrong answer no matter which way I am doing this. What exactly would the integral in cartesian coordinates contain for boundaries?

Thanks. Sorry if the latex syntax is not perfect.
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LCKurtz
LCKurtz is offline
#2
Sep15-09, 02:55 PM
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For the first integral your answer should be 4 pi a^3, not 4 pi r^3.

You probably aren't asked to find "the surface integral of [tex]\vec{r}[/tex] over a surface of a sphere". I'm guessing you are asked to find the flux integral for the vector field [tex]\vec{r}[/tex]. In other words you are to calculate

[tex]\int\int_S \vec r \cdot \hat n\, dS[/tex]

where [tex]\hat n[/tex] is the unit outward normal. In the case of your sphere your unit outward normal is:

[tex] \hat n = \frac {\vec r}{|\vec r |}[/tex]

Now the natural way to do such an integral would be spherical coordinates. But when you evaluate [tex] \vec r \cdot \hat n[/tex] on the surface of the sphere you should see a shortcut.
Kizaru
Kizaru is offline
#3
Sep15-09, 03:39 PM
P: 45
Thanks!


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