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Thread: another triple integral problem..

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face another triple integral problem..

    Calculate the volume of the solid that lies within $\displaystyle x^2+y^2+z^2=49$ above the xy=plane and below $\displaystyle z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$

    Having trouble visualizing what that cone would look like.
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  2. #2
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    It should look something like this:



    I would convert to polar co-ordinates.

    EDIT: Scratch that.
    Last edited by ANDS!; Jun 2nd 2010 at 08:29 PM. Reason: Wrong strategy. . .
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by ANDS! View Post
    It should look something like this:



    I would convert to polar co-ordinates.

    EDIT: Scratch that.
    Thanks though. Yeah I was considering spherical, my trouble is obtaining the angle phi for the cone.
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  4. #4
    Super Member 11rdc11's Avatar
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    It resembles a snowball.

    $\displaystyle z^2 = \frac{1}{3}(x^2 + y^2)$

    $\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}(\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta})$

    $\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}\rho^2\sin^2{\phi}$

    $\displaystyle 3= \tan^2{\phi}$

    you can finish up from here to find phi?
    Last edited by 11rdc11; Jun 2nd 2010 at 10:12 PM.
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  5. #5
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    It resembles a snowball.

    $\displaystyle z^2 = \frac{1}{3}(x^2 + y^2)$

    $\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}(\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta})$

    $\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}\rho^2\sin^2{\phi}$

    $\displaystyle 3= \tan^2{\phi}$

    you can finish up from here to find phi?
    Sorry I am not sure how to solve that last line.
    With phi solved would the volume then be described by this:
    $\displaystyle \int_?^{\frac{\pi}{2}}\int_0^{2\pi}\int_0^7p^2sin{ \phi} dpd{\theta}d{\phi}$
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    Sorry I am not sure how to solve that last line.
    With phi solved would the volume then be described by this:
    $\displaystyle \int_?^{\frac{\pi}{2}}\int_0^{2\pi}\int_0^7p^2sin{ \phi} dpd{\theta}d{\phi}$
    $\displaystyle 3 = \tan^2{\phi}$

    $\displaystyle \sqrt{3} = \tan{\phi}$

    $\displaystyle \arctan{\sqrt{3}} = \phi$

    $\displaystyle \phi = \frac{\pi}{3}$

    $\displaystyle \int_{0}^{\frac{\pi}{3}}\int_0^{2\pi}\int_0^7p^2si n{\phi} dpd{\theta}d{\phi}$
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  7. #7
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    Quote Originally Posted by Em Yeu Anh View Post
    Calculate the volume of the solid that lies within $\displaystyle x^2+y^2+z^2=49$ above the xy=plane and below $\displaystyle z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$

    Having trouble visualizing what that cone would look like.
    $\displaystyle z= \sqrt{\frac{1}{3}x^2+ \frac{1}{3}y^2}$, restricted to the xz-plane so that y= 0, is $\displaystyle z= \pm\sqrt{\frac{1}{3}}x$, two lines with slopes $\displaystyle \pm\sqrt{\frac{1}{3}}$.

    In spherical coordinates, then, $\displaystyle \phi= cot^{-1}\left(\sqrt{\frac{1}{3}}\right)= \frac{\pi}{3}$.


    Blast! 11rdc11 beat me again!
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