Flux: Surface integral of a sphere.

[Kizaru]

Homework Statement

Find the surface integral of \vec{r} over a surface of a sphere of radius a and center at the origin. Also find the volume integral of \nabla \bullet \vec{r}.

Homework Equations

Divergence theorem.

The Attempt at a Solution

First I did the volume integral part of the divergence theorem. I obtained \nabla \bullet \vec{r} = 1 + 1 + 1 = 3. So I figured, the answer must be 3*volume = 4\pir^{3} (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^{2}+y^{2}+z^{2}=a^{2}, I found \nabla \bullet (x^{2}+y^{2}+z^{2}) to obtain the normal. The \vec{r} \bullet \vec{n} = 2x^{2} + 2y^{2} + 2z^{2}.

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.

 

[LCKurtz]

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 \vec{r} over a surface of a sphere". I'm guessing you are asked to find the flux integral for the vector field \vec{r}. In other words you are to calculate

\int\int_S \vec r \cdot \hat n\, dS

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

\hat n = \frac {\vec r}{|\vec r |}

Now the natural way to do such an integral would be spherical coordinates. But when you evaluate \vec r \cdot \hat n on the surface of the sphere you should see a shortcut.

 

[Kizaru]

Thanks!