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Math Help - Volume of a soild using double integralsand polar coordinates

  1. #1
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    Volume of a soild using double integralsand polar coordinates

    Hello

    I am trying to find the volume the region above the cone z=\sqrt{x^2+y^2} and below the sphere x^2+y^2+z^2=1

    I was told to consider the difference z=\sqrt{x^2+y^2} - \sqrt{1-(x^2+y^2)}

    and to set the two equations equal to each other to get bounds

    \sqrt{x^2+y^2}=\sqrt{1-(x^2+y^2)}

    simplifies to

    x^2+y^2 = \frac{1}{2}

    so
    0 \leq \theta \leq 2\pi

    and
    0 \leq r \leq \frac{1}{\sqrt{2}}

    I can easily set up the integral from here. But I don't understand the geometry behind these steps. Could someone explain to me please?
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  2. #2
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    Re: Volume of a soild using double integralsand polar coordinates

    Reverse the difference...

    \sqrt{1 - (x^2 + y^2)} - \sqrt{x^2 + y^2}

    \text{i.e. }\sqrt{1 - r^2} - r

    and see it as the range of the inner-most integral of three...

    \displaystyle{\int_0^{2 \theta} \int_0^{\frac{1}{\sqrt{2}}} \int_r^{\sqrt{1 - r^2}}\ r\ dz\ dr\ d\theta

    ... i.e. at every step along every 1/root2 radius about the origin in the z=0 plane, you are measuring the distance from the point on the cone directly above you (where z = r) up to as far as the point on the sphere directly above that (where z = root(1 - r^2)).
    Last edited by tom@ballooncalculus; April 26th 2012 at 04:58 AM.
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