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Thread: volume enclosed by spheres

  1. #1
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    volume enclosed by spheres

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

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

    --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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