The azimuthal angle will range from to That you have correct. What's not correct is that the polar angle will also range from to That may change things a bit for you. I'm not seeing any other obvious problems. See what you get with this.
Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:
Evaluate where E is bounded by the xz-plane and the hemispheres and .
So my solution is this:
Using spherical coordinates, we have
Thus
But the answer is hmmm how did they get this?
Thanks!
Edit- I am incorrect. I did not see that we were bounded by the xz plane. OOPS!
The azimuthal will range from 0 to and theta will range from 0 to ....we want to cover the entire top half the hemisphere so there is no reason why theta would not be 360 degrees. Likewise pi over two corresponds with the top half (measuring from the Z-axis to the XY plane).
Reply to AllanCuz at Post # 5:
I think you're a bit confused. The polar angle in this problem is and inhabits the natural range in any spherical coordinates problem. The azimuthal angle in this problem is and inhabits the natural range in any spherical coordinates problem. However, in this problem, the shape described, which is that shape bounded by the plane, and the two hemispheres and , you have the artificial restriction There is no restriction on the polar angle.
Hence your limits, while producing the same answer for the integral as my limits, do not describe the shape in the original problem.
Draw a picture (as I did); that might help you see what I'm talking about.
[EDIT] Ah, I see you saw your mistake.
Your limits, actually, are a legitimate way of solving the problem if you think of the whole shape as rotated about the axis through radians such that becomes . So, if you believe that rotations preserve volumes and such, and since the integrand is insensitive to that rotation, your limits produce the same result.