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Washer Method

  1. May 7, 2005 #1
    R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis.

    How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the steps to do ? :frown:
     
  2. jcsd
  3. May 7, 2005 #2

    OlderDan

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    Your integral will be with respect to y, and you need to find the inner and outer radius as a function of y. That involves solving your equations for x.
     
  4. May 7, 2005 #3
    I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1

    I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12
     
  5. May 7, 2005 #4

    OlderDan

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    You mean x= sqrt(y)+1 and x=y/2 +1. Is that what you did? The answer given is correct (reduces to 16pi/3). When you take the difference between the lerger area and the smaller area of each washer, two terms cancel and two terms survive that need to be integrated.
     
  6. May 7, 2005 #5
    Thx man, I gotted the answer but I was too tired to realize it :rofl:
     
  7. May 8, 2005 #6
    Answer sheet = 64pie/15
    But I think the paper is wrong, cause it keeps giving me 64pi/12 (Unreduced)
     
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