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Math Help - Finding the volume revolved around the given axis?

  1. #1
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    Finding the volume revolved around the given axis?

    y = x^2
    y = 2x

    What is the volume of the region bounded by y = x^2 and y = 2x around y = 3?
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  2. #2
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    Quote Originally Posted by AlphaRock View Post
    y = x^2
    y = 2x

    What is the volume of the region bounded by y = x^2 and y = 2x around y = 3?
    I guess the basics steps for solving this would be:

    find the area of the half-ellipse [formed by y=x^2 around y=3] = 4/3 \piabc

    then subtract the area of the internal cone formed by the revolution of y=2x from the point of origin following the radius of x=3/2.

    so the area of the cone is 1/3 \pi(1.5)^2(3) = 2.25 \pi

    now just find the area of the half-ellipse formed by y=x^2 revolved around the y axis with a bound at {y=0 and y=3}.

    then subtract:

    Area of half an Ellipse - 2.25 \pi

    Area of the Ellipse = (4/3) \pi(1.73)(3)(1.73) = ((4/3)(9) \pi)/2 <--- remember its only half a ellipse

    so the answer is 6 \pi - 2.25 \pi = 3.75 \pi... i think.. O and don't forget to include unit^3 in there.
    Last edited by Warrenx; April 19th 2010 at 06:57 PM.
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