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Volumes of Disks and Washers

  1. Apr 28, 2008 #1
    Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
    y=x^2
    y=2x
    about the x-axis.
    So i used intergral 0 to 2 (2x-x^2)^2 dx
    this is wat i got....
    pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))
    its wrong so wat did i do wrong?:confused:
     
  2. jcsd
  3. Apr 28, 2008 #2
    That region will make a washer. Your interval [0,2] looks right.

    However it might be beneficial to find a small piece of volume of the actual shape.

    V=A*h so an infinitely small piece of volume would be dV = A*dh, or more approately in our case:

    dV=A*dx

    The area is pi*r^2 since it forms a washer, so [tex]A=\pi (r_1^2-r_2^2)[/tex]

    Where r1 and r2 are the distances from the x-axis where r1>r2

    The Area portion is what you did wrong, do you see where to go from there?

    Next time you should post questions like this (non-conceptual) questions in the homework help.
     
    Last edited: Apr 28, 2008
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