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Math Help - spherical coordinates

  1. #1
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    spherical coordinates

    Use spherical coordinates to show that \int\int\int_{E} z dV = \frac{15\pi}{16}, where E lies between the spheres x^2 + y^2 + z^2 = 1 and x^2 + y^2 + z^2 = 4 in the first octant (i.e. x, y, z \geq 0).
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    Use spherical coordinates to show that \int\int\int_{E} z dV = \frac{15\pi}{16}, where E lies between the spheres x^2 + y^2 + z^2 = 1 and x^2 + y^2 + z^2 = 4 in the first octant (i.e. x, y, z \geq 0).
    z=\rho\cos\phi
    \rho^2 = x^2+y^2+z^2
    dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta

    So the bounds are:

    \rho: 1\to 2
    \phi: 0\to\pi/2
    \theta: 0\to\pi/2

    This makes the integrand:

    \rho\cos\phi*\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = \rho^3\sin(\phi)\cos(\phi)\,d\rho\,d\phi\,d\theta

    The integral is:

    \int_0^{\pi/2}\int_0^{\pi/2}\int_1^2 \rho^3\sin(\phi)\cos(\phi)\,d\rho\,d\phi\,d\theta

    Here's the evaluation of the integral, if you need it:
    Spoiler:

    Evaluating the innermost integral gives you \left(\frac{\rho^4}{4}\sin(\phi)\cos(\phi)\right)\  bigg|_1^2 = \frac{15}{4}\sin(\phi)\cos(\phi)

    Noticing that \sin(\phi)\cos(\phi)=\frac{1}{2}\sin(2\phi), our integral is now:

    \frac{15}{8}\int_0^{\pi/2}\int_0^{\pi/2}\sin(2\phi)\,d\phi\,d\theta

    Evaluating the next integral, we have \left(-\frac{1}{2}\cos(2\phi)\right)\bigg|_0^{\pi/2} = -\frac{1}{2}(-1-1) = 1

    Now the integral is \frac{15}{8}\int_0^{\pi/2}\,d\theta = \left(\frac{15}{8}\theta\right)\bigg|_0^{\pi/2} = \frac{15\pi}{16}

    So the integral is \frac{15\pi}{16}.
    Last edited by redsoxfan325; April 23rd 2009 at 08:40 AM.
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