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Volumes of solids of revolutions

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    In the question your working with the region bounded by the two curves y=x^2 and y=x^3.

    In the first part of the question you had to revolve the region around the x-axis and find the area which I managed to do by subtracting two areas from each other.

    The second part of the question asked to revolve the region around the y-axis which I didn't know how to do since I couldn't find (x^3)dy
     
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  3. Jul 7, 2011 #2

    rock.freak667

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    Revolving around the y-axis would give:

    V= π∫ x2 dy

    so if y=x3, then x = y 1/3 and hence x2 = y2/3
     
  4. Jul 7, 2011 #3
    That's what I tried before but kept getting the answer wrong. I think my text book is wrong, did you get 1/10pi u^3?
     
  5. Jul 7, 2011 #4

    SammyS

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    Using the washer method, the volume would be: V = π ((x2)2- (x1)2)dy,

    where x2=y(1/3) and x1 = y(1/2).
     
  6. Jul 7, 2011 #5

    rock.freak667

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    I do

    It will end up as the same since OP will have to subtract one from the other while that way will make into one simple integral.
     
  7. Jul 7, 2011 #6

    SammyS

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    When OP mentioned "subtracting two areas from each other", I thought he literally meant areas, not volumes. Oh well !
     
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