Substituting spherical coordinates to evaluate an integral

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
HMPARTICLE
Messages
95
Reaction score
0
I have to evaluate

$$\int^1_{-1} \int^{ \sqrt {1-x^2}}_{ - \sqrt {1-x^2}} \int^1_{-\sqrt{x^2+y^2}}dzdydx$$

using spherical coordinates.

This is what I have come up with

$$\int^1_{0} \int^{ 2\pi}_0 \int^{3\pi/4}_{0}r^2\sin\theta d\phi d\theta dr$$

by a combination of sketching and substituting spherical coordinates.

After evaluating I obtain this integral to equal 3.57.

where as the first one evaluates to 5.236.

These are so difficult :(
 
Physics news on Phys.org
I am not convinced you did that right.
The volume of integration appears to be a sphere for z<0 and a cylinder for z>0. Your spherical integral doesn't look like that.

Are you allowed to use spherical integral for the lower half and cylindrical integral for the upper half? Or maybe just geometry...1/2 volume of unit sphere + volume of unit cylinder = 5.236.