## CaptainSparkles Group Title 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. 2 years ago 2 years ago

1. experimentX Group Title

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

are you sure that's the right dV ?

3. colorful Group Title

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. colorful Group Title

@CaptainSparkles still with me?

5. CaptainSparkles Group Title

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. colorful Group Title

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

7. CaptainSparkles Group Title

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

8. colorful Group Title

I did for you :)

9. CaptainSparkles Group Title

theta would be 0 to 2pi correct?

10. colorful Group Title

yes

11. CaptainSparkles Group Title

and phi is 0 to pi/2?

12. colorful Group Title

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

13. CaptainSparkles Group Title

Thanks.

14. colorful Group Title

welcome