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Math Help - Volume enclosed by a cone in spherical coordinates

  1. #1
    s7b
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    Volume enclosed by a cone in spherical coordinates

    Hi,

    I really don't understand how to set up this integral;

    Find the volume of the solid enclosed by the cone z=sqrt(x^2 + y^2) between the planes z=1 and z=2

    The only thing I did so far was convert the first z equation into phi=pi/4
    I'm not sure where to go from here.
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  2. #2
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    Quote Originally Posted by s7b View Post
    Hi,

    I really don't understand how to set up this integral;

    Find the volume of the solid enclosed by the cone z=sqrt(x^2 + y^2) between the planes z=1 and z=2

    The only thing I did so far was convert the first z equation into phi=pi/4
    I'm not sure where to go from here.
    The \phi angle opens up from 0 to \tfrac{\pi}{4}.
    The \theta angles travels a complete revolution.
    What is trickier is finding the bounds on \rho.

    If z=1 then the plane cut is 1 unit above the xy plane, but the side of the cone makes an angle of \tfrac{\pi}{4}, therefore the length of the side of the cone is \cos \tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}. For z=2 by similar argument we get \rho = 2\cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}. Therefore, the integral becomes,
    \int_{\sqrt{2}/2}^{\sqrt{2}} \int_0^{2\pi} \int_0^{\pi/4}\rho^2 \sin \phi ~ d\phi ~ d\theta ~ d\rho
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    The \phi angle opens up from 0 to \tfrac{\pi}{4}.
    The \theta angles travels a complete revolution.
    What is trickier is finding the bounds on \rho.

    If z=1 then the plane cut is 1 unit above the xy plane, but the side of the cone makes an angle of \tfrac{\pi}{4}, therefore the length of the side of the cone is \cos \tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}. For z=2 by similar argument we get \rho = 2\cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}. Therefore, the integral becomes,
    \int_{\sqrt{2}/2}^{\sqrt{2}} \int_0^{2\pi} \int_0^{\pi/4}\rho^2 \sin \phi ~ d\phi ~ d\theta ~ d\rho
    Surely the limits for r would be from r=\sqrt{2} to r=2\sqrt{2}. No? Using the lengths as \tfrac{1}{\cos \tfrac{\pi}{4}} and \tfrac{2}{\cos \tfrac{\pi}{4}}.
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