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Math Help - Volume of Solid

  1. #1
    euler
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    Volume of Solid

    Find the volume of the solid above the cone z=\sqrt{x^2+y^2} and below the sphere x^2+y^2+z^2=1

    Obviously use polar coordinates here.

    So the two equations are z=r and z=\sqrt{1-r^2}.

    I'm trying to draw the region of integration in the x-y plane and getting stuck. I think the cone comes down to a single point at (0,0), so maybe that can be disregarded. The sphere has a projection of a unit circle. So is the region of integration the unit circle?

    Also what are the bounds for the integral?
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  2. #2
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    ok so we have V=\iiint_{E}dV

    E is the region that we want.
    so we convert to spherical coordinates.
    it is obvious that 0\le p\le 1 and 0\le \theta\le 2\pi

    to find the values for \phi we have

    z=p\cos(\phi)=\sqrt{(p\sin(\phi)\cos(\theta))^2+(p  \sin(\phi)\sin(\theta))^2}
    which is

    \cos(\phi)=\sin(\phi)\Longrightarrow \phi=\frac{\pi}{4}

    so we have

    \int_0^{2\pi}\int_0^{\frac{\pi}{4}}\int_0^1 p^2\sin(\phi)dp d\phi d\theta and i am sure that you can solve it from here.
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  3. #3
    Eater of Worlds
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    That spherical method is just what I was gonna post. That's what I arrived at also.

    Oh well, here's another way, but equivalent in rectangular.

    \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}dzdydx

    Now, do it in cylindrical coordinates.

    \int_{0}^{2{\pi}} \int_{0}^{\frac{1}{\sqrt{2}}} \int_{r}^{\sqrt{1-r^{2}}}rdzdrd{\theta}

    Thanks for the fix PH. Everytime I tried to edit last night I kept getting a "this page can't be displayed" window, so I just gave up.
    Last edited by galactus; November 30th 2006 at 03:58 AM.
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