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Evaluating triple integral with spherical coordinates

  1. Nov 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate the iterated integral ∫ (from 0 to 1) ∫ [from -sqrt(1-x^2) to sqrt(1-x^2) ] ∫ (from 0 to 2-x^2-y^2)

    the function given as √(x^2 + y^2) dz dy dx



    3. The attempt at a solution

    I changed the coordinates and I got the new limits as

    ∫(from 0 to pi) ∫(from (3pi)/2 to pi/2) ∫(from 0 to √2) √(x^2 + y^2) ρ^2 sin phi dρ dphi dθ

    What I'm having problems is with changing the function I need to integrate into spherical coordinates. Should I replace the values of x and y for it's spherical counterparts or is there an easier way via u-sub, etc?

    When i try to sub in x = ρ sin phi sin ρ and y = ρ sin phi cos ρ I get a mess. Can anyone nudge me in the right direction please?
     
  2. jcsd
  3. Nov 24, 2013 #2

    mfb

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    Staff: Mentor

    What do you get? This should give a nice result.
     
  4. Nov 24, 2013 #3
    Thanks for answering mfb,

    First I noticed I made a mistake, in spherical coordinates

    x= ρ sin[itex]\phi[/itex]cosθ
    y= ρ sin[itex]\phi[/itex]sinθ

    so now when I substitute that into the formula, I will get

    [itex]√[(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2[/itex]]

    What I get stuck with is I can't get rid of the square root. My process, so far, has been to expand everything, using the identity of cos/sin^2(x) to expand to either 1/2(1 +/- cos(2x) etc.

    At the end, before the first integration or even getting ρ^2 or sin phi involved, i get the answer below.

    [itex]√[ρ^2(\frac{1-cos(2 \phi)}{2})][/itex]

    I can probably take a picture of my work to post if that would help. Otherwise its a lot of writing. Please let me know what information you need to see where i made the mistake.
     
  5. Nov 24, 2013 #4

    LCKurtz

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    You're making it way too hard:$$
    \sqrt{(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2}=\sqrt{\rho^2\sin^2\phi(
    \cos^2\theta +\sin^2\theta)}$$Now do you see how to simplify it?
     
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