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Math Help - Triple integral, spheric coordinates

  1. #1
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    Triple integral, spheric coordinates

    Hi there. I have some doubts about this exercise. It asks me to use the appropriated coordinates to find the volume of the region indicated below. The region is determined by the first octant under the sphere x^2+ y^2+ z^2=16 and inside the cylinder x^2+ y^2=4x
    The first thing I did was to complete the square with the cylinder: (x-2)^2+ y^2=4
    And then I thought of using spherical coordinates, this way:
    \displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}  \displaystyle\int_{0}^{\displaystyle\frac{\pi}{4}}  \displaystyle\int_{0}^{\sec(\theta)}\rho^2 \sin \phi d\rho d\theta d\phi  + \displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}  \displaystyle\int_{0}^{\displaystyle\frac{\pi}{4}}  \displaystyle\int_{\sec(\theta)}^{4\cos\theta}\rho  ^2 \sin \phi d\rho d\theta d\phi
    Is this okay?
    Bye there, thanks for anyhelp.

    Edit: mm, now I see it isn't. Unless I think its not, because it doesn't defines the region under the sphere. I think I should use cylindrical coordinates.
    Last edited by Ulysses; November 18th 2010 at 05:26 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I was going to suggest cylindrical

    z ranges from 0 to \sqrt{16-r^2}

    while the base is r=4\cos\theta

    Try \int_0^{\pi}\int_0^{4\cos\theta}\int_0^{\sqrt{16-r^2}}rdzdrd\theta
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