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CaptainSparkles

  • 2 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.

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  1. experimentX
    • 2 years ago
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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

  2. colorful
    • 2 years ago
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    are you sure that's the right dV ?

  3. colorful
    • 2 years ago
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    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
    • 2 years ago
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    @CaptainSparkles still with me?

  5. CaptainSparkles
    • 2 years ago
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    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
    • 2 years ago
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    good, so what are the bounds on phi and theta?

  7. CaptainSparkles
    • 2 years ago
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    Sorry experiment, I wish I could give you best answer also.

  8. colorful
    • 2 years ago
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    I did for you :)

  9. CaptainSparkles
    • 2 years ago
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    theta would be 0 to 2pi correct?

  10. colorful
    • 2 years ago
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    yes

  11. CaptainSparkles
    • 2 years ago
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    and phi is 0 to pi/2?

  12. colorful
    • 2 years ago
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    you got it :) ok I think you can take it from here, good job.

  13. CaptainSparkles
    • 2 years ago
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    Thanks.

  14. colorful
    • 2 years ago
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    welcome

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