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Math Help - shell method #2

  1. #1
    Junior Member
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    shell method #2

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

    y=x^2
    x=y^2

    about the line x= -1

    I'm having trouble deciphering on what to do. When I tried to solve it I got 3pi/10. But, I checked it and the answer was incorrect. Can someone show me the steps on how to setup the problem?
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  2. #2
    Senior Member
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    First, we note that y=x^2 defines a parabola facing up, while x=y^2 defines a parabola facing right. These two curves intersect at (0,0) and (1,1), defining a small region between these two points.

    To find the volume of revolution about the line x=-1, we may use the Disc Method:

    \int_0^1(\pi(\sqrt{y}+1)^2-\pi(y^2+1)^2)\,dy.\;\;\;\;\;\;\;\;\;\;(\mbox{formu  la used: }(\pi {r_2}^2-\pi {r_1}^2)\,dy)

    Or, alternatively, we may use the Shell Method:

    \int_0^1 2\pi(x+1)(\sqrt{x}-x^2)\,dx.\;\;\;\;\;\;\;\;\;\;(\mbox{formula used: }2\pi rh\,dx)
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  3. #3
    MHF Contributor
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    (Scrubbed)

    Sorry, yes, rotation unnecessary... I read x = -1 as y = -x!
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