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Finding volume though revolution

  1. May 6, 2009 #1
    Write a definite integral that gives the volume of the solid of revolution formed by revolving the region
    bounded by the graph of x = e^(−y^2) and the y-axis between y=0 and y=1, about the x-axis.

    Not sure how to set up the problem
     
  2. jcsd
  3. May 6, 2009 #2
    Shells look easier than discs. Do you know shells?
     
  4. May 6, 2009 #3
    Yes. 2*pi*r*f(x) dx
     
  5. May 6, 2009 #4
    In this case, the shells are "sideways" so it's 2*pi*r*f(y) dy. And the r is y.
     
  6. May 6, 2009 #5
    Then if f(x) = e^(-y^2)

    then let y = e^(-y^2)

    ln(y) = ln( e^(-y^2))

    ln(y) = -x^2

    -sqrt( ln(y) ) = x

    - ln(y)^.5 = x

    so its

    V = 2*pi*y*-ln(y^.5)

    1
    INT [V]
    0

    ?????
     
  7. May 6, 2009 #6
    No, much easier than that. (Did you draw the graph? I hope so!) Just f(y)=e^(-y^2).

    No solving for anything required. This is a very quick problem.
     
  8. May 6, 2009 #7
    2pi*y*f(y)
     
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