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Homework Help: Volume Problem

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the region bounded by the curves y= lnx and y=( x-3)^2

    Find the volume of the solid obtained by rotating the region about the y-axis

    2. Relevant equations

    3. The attempt at a solution

    For this I solve for the x so i got x= e^y and x= (y)^(1/2) +3

    and for volume equation i got :
    ∫ ((y)^(1/2) +3)^2 - (e^y)^2 = 12.481(pi). My answer is

    different from the book. So can someone point out wat i done wrong and help me to get the right answer

    Thank you
  2. jcsd
  3. Apr 11, 2013 #2

    Simon Bridge

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    I cannot see what you did wrong because you did not show your working - you went right from the integral to the final numerical value ... and the integration you show does not look like the one for volume of rotation.

    So I cannot tell what you were doing.
    I suspect you need to sketch the two curves in the x-y plane and shade in the region bounded by them. The curve ##y=(x-3)^2## intersects the curve ##x=e^y## in two places. ##x=\sqrt{y}+3## only gives you part of the curve you need.
    Last edited: Apr 11, 2013
  4. Apr 11, 2013 #3


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    You appear to be using the 'washer' method, i.e. a stack of elemental discs. This isn't going to work as a single range of integration because there's a sudden change of behaviour at the lower solution of ln(x) = (x-3)2. (Try drawing the region, as Simon advises.) I suggest the cylinder method.
  5. Apr 11, 2013 #4

    Is this the right set up for the shell method

    radius = x and height = lnx- (x-3)^2 so i get

    A= 2(pi)x((x)(lnx-(x-3)^2) = 2(pi)(xlnx - (x)(x-3)^2 )

    so volume =

    2(pi) ∫ (xlnx - (x)(x-3)^2

    Last edited: Apr 11, 2013
  6. Apr 11, 2013 #5


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    Yes, except that I get slightly different bounds: 2.130, 4.198.
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