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CaptainSparkles
 3 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 xyplane 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.
CaptainSparkles
 3 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 xyplane 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.

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \int_Q (2+x^2+y^2+z^2)^{1/2} dV \] 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

colorful
 3 years ago
Best ResponseYou've already chosen the best response.2are you sure that's the right dV ?

colorful
 3 years ago
Best ResponseYou've already chosen the best response.2oh\[dV=\rho^2\sin\phi d\rho d\phi d\theta\]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?

colorful
 3 years ago
Best ResponseYou've already chosen the best response.2@CaptainSparkles still with me?

CaptainSparkles
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh.. yes. Thanks, its just hard for me to get a hand of these but once I see someone else do it I understand.

colorful
 3 years ago
Best ResponseYou've already chosen the best response.2good, so what are the bounds on phi and theta?

CaptainSparkles
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry experiment, I wish I could give you best answer also.

CaptainSparkles
 3 years ago
Best ResponseYou've already chosen the best response.0theta would be 0 to 2pi correct?

CaptainSparkles
 3 years ago
Best ResponseYou've already chosen the best response.0and phi is 0 to pi/2?

colorful
 3 years ago
Best ResponseYou've already chosen the best response.2you got it :) ok I think you can take it from here, good job.
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