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Finding volume using integrals

  1. Nov 16, 2006 #1
    So here is the question

    find the volume of the solid generated by revolving the region bounded by the curve y = sqrt(16 - x^2) and the line y = 0 about the x axis.

    this is how I solved it

    [tex] \int_{0}^{4} \Pi (16 - y^{2}) \; dy [/tex]
    [tex] \Pi \int_{0}^{4} (16 - y^{2}) \; dy [/tex]

    pi(16*y - y^3/3) from 0 to 4
    pi(16*4 - 64/3 - 0)

    so the answer I got is (128 pi)/3

    is this correct.
     
  2. jcsd
  3. Nov 17, 2006 #2

    radou

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    Unless I'm missing something, you should multiply your whole integral with 2, since your function is defined from -4 to 4 and symmetric.
     
  4. Nov 17, 2006 #3

    arildno

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    Well, why haven't you used -4 as your lower limit?
    Note that this would double the volume of your object.
     
  5. Nov 17, 2006 #4

    HallsofIvy

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    Another check is to recognize that [itex]y= \sqrt{16- x^2}[/itex] is the upper half of a circle and so the figure formed is a sphere of radius 4. Its volume is [itex](4/3)\pi (4)^3[/itex], twice your answer.
     
  6. Nov 17, 2006 #5
    thanks guys i completely missed that. you are right it should be -4 to 4... other than that does everything else look ok.
     
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