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March 28, 2017

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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = 10 x and y = 5 x^2 about y =0

Find the volume of the solid obtained by rotating the region bounded by y=8 x^2, x = 1, and y = 0 , about the x-axis.

I saw some other people were having trouble too and looked at those, but it makes no sense to me. Help? :(

  • Calculus I don't understand - ,

    ok. for volume generated by rotating around y=0, you can use discs or shells. The volume of a disc, or washer with a hole in it, is
    v=π(R^2-r^2)h
    where the height is dx and the radii are the distance from the x-axis to the two curves.

    The upper curve is y=10x and the lower curve is y=5x^2. If you take a thin slice of the area and rotate it around the x-axis, you will get a washer.

    Since the two curves intersect at (0,0) and (2,20), we integrate along x from 0 to 2.

    v = ∫[0,2] π(R^2-r^2) dx
    = π∫[0,2] (10x)^2 - (5x^2)^2 dx
    = π∫[0,2] 100x^2 - 25x^4 dx
    = π(100/3 x^3 - 5x^5) [0,2]
    = π(800/3 - 160)
    = 320/3 π

    Now, you can also use shells. Recall that the volume of a thin cylinder of thickness dy is just

    v = 2πrh dy

    Now, in this case, the radius is just y, since we are rotating around the x-axis. The height h is the distance between the two curves

    x = √(y/5) and
    x = y/10

    So, our shell volume is

    v = ∫[0,20] 2πy(√(y/5)-y/10)
    = 2π∫[0,20] 1/√5 y^(3/2) - y^2/10 dy
    = 2π(2/5√5 y^(5/2) - y^3/30) [0,20]
    = 2π(2/5√5 * 400√20 - 800/3)
    = 2π(320 - 800/3)
    = 320/3 π


    Now see whether you can set up the other one. As you can probably see, using discs is easier in this case.

  • Calculus I don't understand - ,

    Thank you so much that makes sense more sense to me! :)

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