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Math Help - Finding Volume using spherical coordinates

  1. #1
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    Finding Volume using spherical coordinates

    Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere \rho=4 cos\phi and the hemisphere \rho=6, z\geq0. Then evaluate the integral.

    I found the integral with the correct limits for the solid in spherical coordinates as the following: V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta

    I would like to check my volume, the answer which i got is \frac{26\pi}{3}.
    Last edited by Belowzero78; March 26th 2010 at 06:11 PM.
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  2. #2
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    As long as all your bounds are correct for the figure, your answer is off. I obtained 400*pi/3 for that integral as is.
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    Alright, do you mind to show a few steps?
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    If you show me your steps, it will be quicker for me to find your error.
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    Quote Originally Posted by Belowzero78 View Post
    Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere \rho=4 cos\phi and the hemisphere \rho=6, z\geq0. Then evaluate the integral.

    I found the integral with the correct limits for the solid in spherical coordinates as the following: V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta

    I would like to check my volume, the answer which i got is \frac{26\pi}{3}.

    2\pi\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi

    = 2\pi\int_{0}^{\pi/2} \frac{\rho^3}{3}\sin\phi \bigg|^{6}_{4\cos\phi} d\phi

    = 2\pi\int_{0}^{\pi/2} 72\sin\phi -\frac{64\cos^3\phi\sin\phi}{3}\ d\phi

    = 2\pi( -72\cos\phi +\frac{64\cos^4\phi}{12}) \bigg|^\frac{\pi}{2}_{0}

    = 2\pi(72-\frac{16}{3}) = \frac{400\pi}{3}
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