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Triple integral spherical

  1. Jul 29, 2014 #1
    1. The problem statement, all variables and given/known data

    ##\iiint_W (x^2+y^2+z^2)^{5/2}## W is the ball ##x^2+y^2+z^2 \le 1##





    3. The attempt at a solution

    changing to spherical

    ##0 \le \theta \le 2\pi ; 0 \le \phi \le \pi ; 0 \le \rho \le 1##


    ##(x^2 + y^2 + z^2)^{5/2} \Rightarrow ((\rho \sin \phi \cos \theta)^2 + (\rho sin \phi \sin \theta)^2 + (\rho \cos \phi)^2)^{5/2} = \rho^5##


    ##\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^5(\rho^2\sin\phi) d\rho d\phi d\theta##


    Is that a correct setup?
     
  2. jcsd
  3. Jul 29, 2014 #2

    Matterwave

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    Looks good to me. :)
     
  4. Jul 29, 2014 #3
    ##\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^5(\rho^2\sin\phi) d\rho d\phi d\theta##


    ##\int_{0}^{2\pi}\int_{0}^{\pi}\rho^7\sin\phi d\rho d\phi d\theta##

    ##\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{8}\sin\phi d\phi d\theta##

    ##\int_{0}^{2\pi} \frac{1}{4} d\theta##

    ##\frac{\pi}{2}##
     
  5. Jul 29, 2014 #4

    Matterwave

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  6. Jul 29, 2014 #5

    HallsofIvy

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    Very good. Also note that because the limits of integration on each integral are constants, not depending on the other variables, this is the same as
    [tex]\left(\int_0^{2\pi} d\theta\right)\left(\int_0^\pi sin(\phi) d\phi\right)\left(\int_0^1\rho^7 d\rho\right)[/tex]
     
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