If you're supposed to use the divergence theorem, then use the divergence theorem. You can essentially take a derivative inside the integral sign, and integrate over the volume instead.
Using the divergence thm, evaluate the following equation
where S is the surface of the ellipsoid
I am suck on this question because I have only done them for spheres.
I am not sure of how to get the normal and then what to do from there. I know it seems like I haven't done a lot on this question, but I am struggling to see how this question would be done.
Maybe... The divergence theorem states that
You're going to have to get the surface integral you started with to look like the RHS of the divergence theorem. So I think your original question still stands: how to get the normal vector.
How do you get the normal to the surface? The gradient is your answer. According to the wiki, if you have a surface defined implicitly: , then the normal at point is given by .
So, with your , the gradient vector is .
The formula in the divergence theorem, however, wants to see a unit normal vector. So, we take this result and divide by its length:
Next, we want to recover . That is, we need
(the RHS of this equation is your original integrand), or
It jumps out to me that it must be that . This is then the function of which you must take the divergence, which I think you'll agree is much easier. It had to be a vector function, in any case, right?
I think you'll find the integrand of the volume integral to be very simple indeed. You might even find the problem reduced to finding the volume of an ellipsoid.
Hope this helps.