Divergence theorem

Tyst

1. The problem statement, all variables and given/known data
Here is a link to the problem:
http://www.brainmass.com/homework-he...c-theory/68800


3. The attempt at a solution
To find the divergence

1/r^2*d(r)*(r^2*r^2*cos(theta))
+[1/r*sin(theta)]*d(theta)*(sin(theta)*r^2*cos(phi))
-[1/r*sin(theta)]*d(phi)*(r^2*cos(theta)*sin(phi))

Which gives

1/r^2*4*r^3*cos(theta)
+[1/r*sin(theta)]*(cos(theta)*r^2*cos(phi))
-[1/r*sin(theta)]*(r^2*cos(theta)*cos(phi))

Is this correct?[quote]
Looks correct to this point

What do you mean by "following this"? How did you get that and for what?

It would help if you used parenthes:
div v= (1/r^2)D_r(r²v_r)+ (1/r sin theta)D_theta(sin theta v_theta)+ (1/r sin theta) D_phi(v_phi)

Yes, the derivative only applies to the expression immediately following. Usually it is in the derivative symbol or in parentheses to indicate that.

Defennder

I'm a little confused as to the notational convention for [tex]\phi \ \mbox{and} \ \theta[/tex]. Which one in this question is the azimuthal angle to the xy plane?

Tyst

Phi is the azimuthal angle

Tyst

4rcos(theta) came from simplifying the expressions above it, it is the divergence.

And thank you - you answered my question!