Use spherical coordinates to evaluate \int_Q (2+x^2+y^2+z^2)^(-1/2) dV where Q is the region above the xy-plane outside x^2+y^2+z^2=1 and inside x^2+y^2+z^2=16. Any help regarding setting up the limits would be appreciated.

Question

Use spherical coordinates to evaluate \int_Q (2+x^2+y^2+z^2)^(-1/2) dV where Q is the region above the xy-plane outside x^2+y^2+z^2=1 and inside x^2+y^2+z^2=16.  Any help regarding setting up the limits would be appreciated.

experimentx · Answer

$$ \int_Q (2+x^2+y^2+z^2)^{-1/2} dV $$ Change $ x^2 + y^2 + z^2 = r^2$ and $ dV = r^2 d\phi d\theta dr$ Rest should be easy, For reference: http://mathworld.wolfram.com/SphericalCoordinates.html

anonymous · Answer

are you sure that's the right dV ?

anonymous · Answer

oh$$dV=\rho^2\sin\phi d\rho d\phi d\theta$$anyway, for the bounds we have the region lying between two spheres$$x^2+y^2+z^2=\rho^2$$and between the two equations we have$$1\le\rho\le4$$does that make sense to you?

anonymous · Answer

@CaptainSparkles still with me?

anonymous · Answer

Ahh.. yes.  Thanks, its just hard for me to get a hand of these but once I see someone else do it I understand.

anonymous · Answer

good, so what are the bounds on phi and theta?

anonymous · Answer

Sorry experiment, I wish I could give you best answer also.

anonymous · Answer

I did for you :)

anonymous · Answer

theta would be 0 to 2pi correct?

anonymous · Answer

yes

anonymous · Answer

and phi is 0 to pi/2?

anonymous · Answer

you got it :)
ok I think you can take it from here, good job.

anonymous · Answer

Thanks.

anonymous · Answer

welcome