Evaluating Surface Integral with Divergence Theorem

[jcertel]

1.The problem asks " use the divergence theorem to evaluate the surface integral $$\int\int$$ F.ds
for F(x,y,z) = <x3y,x2y2,−x2yz>


where S is the solid bounded by the hyperboloid x^2 + y^2 - z^2 =1 and the planes z = -2 and z=2.

i know that the
$$\int\int$$ F.ds = $$\int\int\int$$ divFdv
but in not sure on the limits of integration.

when i found the divF i got 4x^2y where do i gor from here i turned it into polar with x=[rcos]\theta[/tex]
y=rsin$$\theta$$
but I am not sure where to go from here?

 

[ideasrule]

i know that the
$$\int\int$$ F.ds = $$\int\int\int$$ divFdv
but in not sure on the limits of integration.

You want to integrate over all of the volume bound by the surface S, so the limits are whatever you need to include all of the volume.

when i found the divF i got 4x^2y where do i gor from here i turned it into polar with x=[rcos]\theta[/tex]
y=rsin$$\theta$$
but I am not sure where to go from here?

OK, that's good. Now you integrate 4x^2y over the entire volume of the hyperboloid that's bound by z=2 and z=-2. I would personally do this:

1. For a specific z value and specific r value, integrate around a ring of radius r. This should get rid of theta.
2. Integrate from r=0 to the largest possible r value for this specific z.
3. Integrate from z=-2 to z=2.

But that's just my personal preference.

 

[jcertel]

ok i that with the limits

-2<Z<2
0<$$\theta$$<$$\pi$$
0<r<4

now do i need to change 4(x^2)y to 4(r^2)(cos^2$$\theta$$)sin$$\theta$$? when i integrate that get 0? I am not sure if that's correct.