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Math Help - volume enclosed by spheres

  1. #1
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    volume enclosed by spheres

    Find the volume enclosed by the spheres x^{2}+y^{2}+z^{2}=9 and x^{2}+y^{2}+(z-2)^{2}=4
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  2. #2
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    Spherical Coordinates

    If we use spherical coordinates (\rho,\theta,\phi), the spheres are thus \rho=3 and \rho=4\cos\theta. We see our volume is bounded by \rho=3 for 0\le\theta<\arccos\frac{3}{4} and \rho=4\cos\theta for \arccos\frac{3}{4}\le\theta\le\frac{\pi}{2}. Thus, we have:
    V=\int_{0}^{\arccos\frac{3}{4}}\int_{0}^{2\pi}\int  _{0}^{3}\rho^2\sin\theta\,d\rho\,d\phi\,d\theta+\i  nt_{\arccos\frac{3}{4}}^{\frac{\pi}{2}}\int_{0}^{2  \pi}\int_{0}^{4\cos\theta}\rho^2\sin\theta\,d\rho\  ,d\phi\,d\theta
    V=\int_{0}^{\arccos\frac{3}{4}}18\pi\sin\theta\,d\  theta+\int_{\arccos\frac{3}{4}}^{\frac{\pi}{2}}\fr  ac{128\pi}{3}\cos^3\theta\sin\theta\,d\theta
    V=\frac{9\pi}{2}+\frac{27\pi}{8}=\frac{63}{8}\pi

    --Kevin C.
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  3. #3
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    Cool. That is identical to how I set it up. Just verifying.
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