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March 25, 2017

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Find the volume of the solid obtained by rotating the region bounded by y= x^2 and x=y^2 about the x-axis.

  • calculus - ,

    The bounded region is x =0 to x = 1, so those are your limits of integration.

    The integral is of
    pi*(x^2 - sqrt x)^2 dx.

    Visualize it as a stack of washers with holes in the middle.

  • calculus - ,

    I think that
    the volume=pi*Int(from 0 to 1) (sqrt(x))^2dx-pi*Int(0 to 1)(x^2)^2dx=
    pi(1/2-1/5)

  • calculus - ,

    Mgraph is correct. I should have written
    Integral of pi[(sqrtx)^2 - (x^2)^2] dx
    0 to 1

    which is pi*(1/2 - 1/5)

  • calculus - ,

    Thank you so much that makes alot of since.

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