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Math Help - Volume of an Ice Cream cone.

  1. #1
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    Volume of an Ice Cream cone.

    I am quite perplexed by this problem and I have already graphed it within WinPlot. Would a good step be finding the plane of intersection?

    "Find the volume of an ice cream cone bounded by the hemisphere
    z = (8- x^2 - y^2)^(1/2) and the cone z = (x^2 + y^2) ^ (1/2)."
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  2. #2
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    Quote Originally Posted by EuptothePiI1 View Post
    I am quite perplexed by this problem and I have already graphed it within WinPlot. Would a good step be finding the plane of intersection?

    "Find the volume of an ice cream cone bounded by the hemisphere
    z = (8- x^2 - y^2)^(1/2) and the cone z = (x^2 + y^2) ^ (1/2)."
    Disclaimer I am using the spherical coordinates in the following form

    (\rho,\theta,\phi) where \theta is the measure of the angle from the positive z axis and \phi is the plane angle.
    Alot of calculus books reverse the two angles.
    First note that you will want to use spherical coordinates setting the equations equal gives

    8-x^2-y^2=x^2+y^2 \iff x^2+y^2=4 plugging this into either equation gives z=2

    So the circle of intersection of the two objects is described above

    Using the above information we can find the angle from the positive z-axis to be

    \displaystyle \theta=\tan^{-2}\left( \frac{2}{2}\right)=\frac{\pi}{4}

    So the radius \rho=\sqrt{8} and the plane angle \phi needs to sweep out the whole circles so it goes from 0 \text{ to } 2\pi

    This gives

    \displaystyle \iiint_{V}dV=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{8}}\rho^2 \sin(\theta)\,d\rho \, d\theta \, d\phi
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  3. #3
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    I am going to attempt to follow what you showed me on paper now. I will edit this post once I come to a conclusion. Thank you very much for your time.

    (15 minutes later, View Below)

    Huzzah! Success! Thank you *very* much.
    Last edited by EuptothePiI1; December 6th 2010 at 06:35 PM.
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