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Volume of Solid Revolution

  1. Jul 26, 2011 #1
    Hi all. I've just hit a block in the following question:

    [Find the volume of the solid...] "The region in the first quadrant bounded by the curve y = x^2, below by the x-axis, and on the right by the line x = 1, revolved around the axis x = -1."

    I've tried nearly 2 hours figuring the question out but can't seem to find the correct fomula for integration due the empty space between x = 0 and the axis x = -1.

    Any help will be appreciated.
     
  2. jcsd
  3. Jul 26, 2011 #2
    Divide the x-y curve in small rectangles of base dx and height x^2.

    Now revolve each of theese rectangles around x=-1, obtaining a cylindrical shell of internal radius
    \pi (1+x)^2 and external radius \pi (i + x + dx)^2.

    The base area of this shell is

    dA(x) = \pi (1+x+dx)^2 - \pi (1+x)^2 = 2\pi(1+x)dx

    and the volume of the shell is dV(x) = x^2 dA(x)

    Now integrate from x = 0 to x = 1.
     
  4. Jul 26, 2011 #3
    I understand that the Shell method is easiest way to do this.

    However, this exercise is supposed to be done before the lesson on Shell revolution.

    Is is possible to use the Slice and Washer method to solve this?
     
  5. Jul 26, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, it is. Do it as two separate problems. First do the problem of the volume you get rotating the region from x=-1 to the vertical line x= 1, with height 1. That. of course, is just the volume of the cylinder of radius 2 and height 1. Then do the problem of the volume of the region from x= -1 to x= sqrt(y) rotating around x= -1 again with y up to 1. Finally, subtract the second from the first (that is the same as the "washer" method).
     
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