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Washer method problem

  1. Nov 19, 2008 #1
    1. Find the volume generated by revolving the area bounded by x=y[tex]^{2}[/tex] and x=4 about the line y=2.


    Is it [tex]\pi\int(4-(2-\sqrt{x})^{2}dx[/tex] from 0 to 4?
     
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  3. Nov 19, 2008 #2

    Dick

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    No. The inner radius is (2-sqrt(x)). What's the outer radius? It doesn't lie on the x-axis.
     
  4. Nov 19, 2008 #3
    oh whoops forgot to add a bracket at the end. Thanks for your help.
     
  5. Nov 19, 2008 #4

    Dick

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    Hold it. I gave it a second thought and edited my answer. The area enclosed by x=4 and y^2=x has pieces both about and below the x-axis.
     
    Last edited: Nov 19, 2008
  6. Nov 19, 2008 #5
    So it's 4 instead of 2?
     
  7. Nov 19, 2008 #6

    Dick

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    No, y^2=x gives y=sqrt(x) or y=-sqrt(x). Those are the boundaries of the region. Your original answer would be ok if they had specified y=0 as a boundary. But they didn't.
     
  8. Nov 19, 2008 #7
    oh okay. So would y=2 factor into the outer radius too? Meaning r = [2 - (-sqrt(x))]?
     
  9. Nov 19, 2008 #8

    Dick

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    Exactly. The outer radius is 2+sqrt(x).
     
  10. Nov 19, 2008 #9
    awesome. Thanks again.
     
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