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Triple integral in spherical coordinates

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data

    [tex]\int \int \int _R (x^2+y^2+z^2)dV[/tex]

    where R is the cylinder
    [itex]0\leq x^2+y^2\leq a^2[/itex],
    [itex]0\leq z\leq h[/itex]

    2. Relevant equations

    [tex]x = Rsin\phi cos\theta[/tex]
    [tex]y = Rsin\phi sin\theta[/tex]
    [tex]z = Rcos\phi[/tex]

    3. The attempt at a solution

    [tex]2*\int_{0}^{\pi/2}d\phi \int_{0}^{2\pi}d\theta \int_{0}^{h/cos\phi}dR R^4sin\phi [/tex]

    Are the bounds correct?
    Last edited: May 16, 2015
  2. jcsd
  3. May 16, 2015 #2
    Do you have to use spherical co-ordinates, why not cylindrical ones ?
  4. May 16, 2015 #3


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    This is not set properly if you insist on using spherical coordinates. You would have to break it up into two integrals, one where ##\rho## goes from ##0## to the side of the cylinder and another where it goes from ##0## to the top. A much better choice is cylindrical coordinates. Try setting it up that way and check back with us.
  5. May 16, 2015 #4
    Yes I realized at one time it would be easier if I used cylindrical coordinates, since the region itself is a cylinder. But I just wanted to know how to proceed with spherical coordinates.
  6. May 17, 2015 #5


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    OK, so give it a go. First set up$$
    \int_{\theta \text{ limits}}\int_{\phi \text{ limits}}\int_0^{\rho \text{ on the sides in terms of } \phi} \text{Integrand }
    \rho^2\sin\phi~d\rho d\phi d\theta$$then add$$
    \int_{\theta \text{ limits}}\int_{\phi \text{ limits}}\int_0^{\rho \text{ on the top in terms of } \phi} \text{Integrand }\rho^2\sin\phi~d\rho d\phi d\theta$$
  7. May 17, 2015 #6
    EDIT : Sorry, i thaught they were cylindrical
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