- #1
Kizaru
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Homework Statement
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].
Homework Equations
Divergence theorem.
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.