## anonymous 4 years ago 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.

1. experimentX

$\int_Q (2+x^2+y^2+z^2)^{-1/2} dV$ 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

2. anonymous

are you sure that's the right dV ?

3. anonymous

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?

4. anonymous

@CaptainSparkles still with me?

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

I did for you :)

9. anonymous

theta would be 0 to 2pi correct?

10. anonymous

yes

11. anonymous

and phi is 0 to pi/2?

12. anonymous

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

13. anonymous

Thanks.

14. anonymous

welcome