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Disk Method

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    The curve x=y^(2) and the line x=4 is rotated about the x axis.

    2. Relevant equations

    pi* integral from a to b of Radius^(2)

    3. The attempt at a solution

    pi* integral from 0 to 4 of (square root of x)^(2) dx.

    My teacher has this answer as 8pi but I think that that is wrong. Shouldn't it be 16 pi because you have to account for symmetry?
     
  2. jcsd
  3. Apr 9, 2012 #2

    LCKurtz

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    Your formula for the volume of the little disk is for the whole disk: ##\pi r^2 dx##. That's the area of the circle times the thickness. Your teacher is correct.
     
  4. Apr 9, 2012 #3
    I see!!! How would you do this if you are doing the shell method? I am stuck at this part:

    2pi* integral from -2 to 2 of y[4-y^(2)] dy. I am stuck there lol.
     
  5. Apr 10, 2012 #4

    LCKurtz

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    You almost have it correct. But remember that the ##2\pi y## accounts for revolving the area. You only revolve the area above the ##x## axis. Your original statement of the problem is not well worded in that regard. So ##y## would only vary from 0 to 2. Otherwise the volume is given twice.
     
  6. Apr 10, 2012 #5
    I see and thanks. However, how come for this problem: y=x^(2), y=2-x^(2), about x=1 the limits are from -1 to 1.?
     
  7. Apr 10, 2012 #6

    LCKurtz

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    Remember, the limits are always for the area that is being revolved. Think about what limits you would use if you were just calculating the area.
     
  8. Apr 10, 2012 #7
    Thank you!!
     
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