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Math Help - Volume

  1. #1
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    Volume

    Find the volume of the given solid: above the cone
    z = ((x^2) + (y^2))^0.5 and below the sphere (x^2) + (y^2) + (z^2)=1.

    Thanks for any help.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by eigenvector11 View Post
    Find the volume of the given solid: above the cone
    z = ((x^2) + (y^2))^0.5 and below the sphere (x^2) + (y^2) + (z^2)=1.

    Thanks for any help.
    As a triple integral, you can use cylindrical coordinates.

    Note that r\leqslant z\leqslant \sqrt{1-r^2} and 0\leqslant\theta\leqslant2\pi

    Now to find r, find where the surfaces intersect.

    This would be where z=\sqrt{r^2}\implies r^2+(\sqrt{r})^2=1\implies r^2=\tfrac{1}{2}\implies r=\frac{\sqrt{2}}{2}. Thus, 0\leqslant r\leqslant \frac{\sqrt{2}}{2}

    Therefore, your triple integral would be \int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}}\int_r^{\s  qrt{1-r^2}}r\,dz\,dr\,d\theta

    Does this make sense?
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  3. #3
    Member Mentia's Avatar
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    Once again we use cylindrical coordinates:

    For the cone we have, in cylindrical coordinates: r = z,

    For the sphere we have: r = \sqrt[]{1-z^2}

    Note that the cone is inverted, so the volume makes an ice-cream cone shape. Where exactly does the cone run into the sphere?

    Thats when z = \sqrt[]{1-z^2}, or z = \pm \, \sqrt[]{\frac{1}{2}}. This is not quite true since, as it is defined, the cone doesn't extend into the negative z range. So we stick with positive \sqrt[]{\frac{1}{2}}.

    This time we break it up into two integrals. We have to change over from the cone to the sphere when we reach z = \sqrt[]{\frac{1}{2}}

    Volume =  \int_{0 }^{ \sqrt[ ]{ \frac{ 1}{2 }  }  }  \int_{0 }^{2\pi }  \int_{ 0}^{ z} rdrd \theta dz +  \int_{ \sqrt[ ]{ \frac{1 }{2 }  }  }^{ 1}  \int_{0 }^{2 \pi }  \int_{0 }^{ \sqrt[ ]{ 1-z^2}  } rdrd \theta dz

    Note how in the first integral we integrate r from 0 to z, corresponding to the cone. In the second we integrate corresponding to the sphere.

    I got -\frac{1}{3} \left(-2+\sqrt{2}\right) \pi \approx .613 for the volume.
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  4. #4
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    I would recommend spherical coordinates because I see much more spherical symmetry than cylindrical symmetry!

    The sphere x^2+ y^2+ z^2= 1 in spherical coordinates is just \rho= 1.

    The cone z= \sqrt{x^2+ y^2} is \rho cos(\phi)= \rho sin(\phi) or tan(\phi)= 1 so \phi= \pi/4.

    The volume is given by
    \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/4}\int_{\rho= 0}^1 \rho^2 sin(\phi)d\rho d\phi d\theta = \left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi/4} sin(\phi)d\phi\right)\left(\int_0^1 \rho^2 d\rho\right)
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