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Volume of solid

  1. Dec 1, 2006 #1
    volume of solid:(

    hi..I tried to solve it but ı couldnt .book says the answer is pi /6...please help me.

    question is; y=x and y=x^1/2 about y =1
     
  2. jcsd
  3. Dec 1, 2006 #2

    arildno

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    Pinpoint your problem.
     
  4. Dec 1, 2006 #3
    ok I showed my work..


    I used this formula
    integral Pı(outer r)^2-Pı(inner)^2



    inside integral from y=0 to y=1 pi(1-y)^2 - pi(1-y^2)^2

    my result is 7/6 pi..:( where am I wrong??
     
  5. Dec 1, 2006 #4
    integrate it with respect to x. The inner radius is [tex] 1-\sqrt{x} [/tex] and the outer radius is [tex] 1- x [/tex].

    So you have [tex] \int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx [/tex]
     
    Last edited: Dec 1, 2006
  6. Dec 1, 2006 #5

    arildno

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    Well, your integrand is utterly wrong!
    You are to rotate the VERTICAL strip [itex]\sqrt{x}\leq{y}\leq{x}[/itex] around y=1.
    That gives you the integral:
    [tex]V=\int_{0}^{1}\pi((1-x)^{2}-(1-\sqrt{x})^{2})dx=\pi\int_{0}^{1}(x^{2}-3x+2\sqrt{x})dx[/tex]
     
  7. Dec 1, 2006 #6

    arildno

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    Apart from being wrongly signed, you have rotated your figure around the line x=1, rather than around y=1.
    That accounts for your discrepancy.
    (Note, for example, that your INNER radius is the actual OUTER radius)
     
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