Triple integral spherical

  • #1
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Homework Statement



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





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?
 

Answers and Replies

  • #2
Matterwave
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Looks good to me. :)
 
  • #3
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##\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}##
 
  • #4
Matterwave
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Yep.
 
  • #5
HallsofIvy
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##\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}##
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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