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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.

Mathematics
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\[ \int_Q (2+x^2+y^2+z^2)^{-1/2} dV \] 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
are you sure that's the right dV ?
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?

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Other answers:

@CaptainSparkles still with me?
Ahh.. yes. Thanks, its just hard for me to get a hand of these but once I see someone else do it I understand.
good, so what are the bounds on phi and theta?
Sorry experiment, I wish I could give you best answer also.
I did for you :)
theta would be 0 to 2pi correct?
yes
and phi is 0 to pi/2?
you got it :) ok I think you can take it from here, good job.
Thanks.
welcome

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