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Math Help - Finding volume of a function rotating around the x-axis

  1. #1
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    Finding volume of a function rotating around the x-axis

    Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by and about the -axis.

    So I know it is pi[ outer^2-inner^2]

    the boundaries are 0 to 2. and I know that y=8x is on top.

    But I can't seem to get the right answer, I would get a negative number...
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  2. #2
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    Does the integral look like \displaystyle \pi \int_0^2 [(8x)^2 - (x^4)^2] \, dx? If so, then the answer is \dfrac{1024\pi}{9}.

    \displaystyle \pi \int_0^2 [(8x)^2 - (x^4)^2] \, dx

    \displaystyle = \pi \int_0^2 (64x^2 - x^8) \, dx

    \displaystyle = \pi [\dfrac{64}{3}x^3 - \dfrac{1}{9}x^9]_0^2

    \displaystyle = \pi (\dfrac{64}{3}2^3 - \dfrac{1}{9}2^9)

    \dfrac{1024\pi}{9}
    Last edited by NOX Andrew; March 6th 2011 at 01:52 PM.
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