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Double Integral - Volume question

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid bounded by z = 0 and z = 2xy, lying in the first quadrant and bounded by the curves y = x^2 and x+y = 2

    2. Relevant equations


    3. The attempt at a solution
    I have an answer, but just asking if I've done it correctly, since we arent given the solution:

    Intersection of x^2 and 2-x -> x = 1 or x = -2

    Limits
    x^2 <= y <= 2-x
    0 <= x <= 1
    0 <= z <= 2xy

    [tex] Volume = \int ^{1}_{0}\int^{2-x}_{x^2} (2xy - 0) dy dx [/tex]
    [tex] = \int ^{1}_{0} 4x - 4x^2 dx [/tex]
    [tex] = 2/3 [/tex]
     
    Last edited by a moderator: Jun 4, 2009
  2. jcsd
  3. Jun 4, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, this is the correct integral

    How did you get this? The integral of y will involve y2 and letting y= x2 for the lower limit will give x4. Multiplying by the x in the original integral, this integrand should have an x5 in it!

     
  4. Jun 4, 2009 #3
    Right -> I mistook the x^2 for an x :p. Trying to do too many steps in your head inevitably leads to errors.
    So the new integral then is

    [tex] \int^{1}_{0} x^3 - 4x^2 + 4x - x^5 dx [/tex]

    3/4

    Should be right :p
    Thanks man
     
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