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Double Integral Over General Region

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

    1. Find the volume of the solid which is under the surface z = 2x + y2 and above the region bounded by x = y^2 and x = y^3.

    2. Relevant equations



    3. The attempt at a solution

    So first I graphed x=y^3 and x=y^2. (http://h.imagehost.org/view/0716/Math_Problem [Broken])
    I found their points of intersection (y=1 or y =0).
    Set up double integral as Integral from 0 to 1 Integral from y^2 to y^3 of (2x+y^2) dx dy
    where y^2<x<y^3 and 0<y<1

    I calculated the integral and got 1/7 plus 1/6 minus 2/5


    Is my work correct?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 4, 2009 #2

    Mark44

    Staff: Mentor

    Your answer is almost right; your sign is wrong. It should be 2/5 - 1/7 - 1/6 = 19/210
    For each horizontal strip, the left boundary is x = y^3 and the right boundary is x = y^2. You have them reversed in your inner integral, which gives you the opposite sign.
     
  4. Nov 4, 2009 #3
    Based on your picture, shouldn't it be [tex]y=x^3[/tex] and [tex] x=y^2[/tex]?

    However, if you did write the equations correctly, then you've drawn the region wrong.
     
  5. Nov 4, 2009 #4

    Mark44

    Staff: Mentor

    vandyboy's graph for x = y^3 is incorrect. He has actually drawn the graph of y = x^3.
     
  6. Nov 4, 2009 #5
    Oops. Yeah I just noticed that. thanks guys
     
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