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Double Integration (Stuck at square root step) (Solution Included).

  1. Mar 20, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and solution are included.


    2. Relevant equations
    Double integration.


    3. The attempt at a solution
    Firstly, I'd like to mention that the additional ρ under the square root is there accidentally and that it should be outside of the square root such that it forms the ending ρ dρ dθ.

    I'm stuck at the sqrt(4a^2 - ρ^2) ρ dρ step. I have a feeling that I need to use trigonometry with ρ = 2acosθ or ρ = 2asinθ but firstly, which of the two would I choose and secondly given that I then have to integrate with respect to θ, must I replace ρ with a variable other than θ? Why or why not?

    Any help would be greatly appreciated!
    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2012 #2
    I did it with [itex]\rho=2a \sin t[/itex].
     
  4. Mar 21, 2012 #3

    s3a

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    Thanks for saying but I also just realized that I don't know why z = √(4a^2 - ρ^2). Could you explain that to me please?
     
  5. Mar 21, 2012 #4

    HallsofIvy

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    Staff Emeritus
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    The "sphere of radius 2a", which forms part of the boundary, has equation [itex]x^2+ y^2+ z^2= 4a^2[/itex]. In polar or cylindrical coordinates (not spherical coordinates), [itex]\rho=\sqrt{x^2+ y^2}[/itex] so that equation becomes [itex]\rho^2+ z^2= 4a^2[/itex] so that [itex]z^2= 4a^2- \rho^2[/itex] and [itex]z= \pm\sqrt{4a^2- \rho^2}[/itex].
     
  6. Mar 21, 2012 #5

    s3a

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    How did you go from ρ = √(x^2 + y^2) to ρ^2 + z^2 = 4a^2?
     
  7. Mar 22, 2012 #6
    Change [itex]\rho=\sqrt{x^2+y^2}[/itex] in [itex]x^2+y^2+z^2=4a^2[/itex]
     
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