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Just got done with a test

  1. Apr 13, 2005 #1
    One question on it was reverse the following integral and then evaluate it:

    [tex]\int_{0}^{\pi}\int_{x^2}^{\pi^2} x^3y dydx[/tex]

    So I got the following reversed integral:
    [tex]\int_{0}^{\pi^2}\int_{-\sqrt{y}}^{\sqrt{y}} x^3y dxdy[/tex]

    When you first evaluate in respect to x you get 0, am I doing something wrong?
  2. jcsd
  3. Apr 13, 2005 #2
    minor mistake really... the second integral should be
    [tex]\int_0^{\pi^2}\int_0^{\sqrt{y}} x^3 y \partial x \partial y [/tex]
    I really doubt you would lose a lot of marks

    unless I've made a mistake which is possible
  4. Apr 13, 2005 #3
    Yeah, as always, drawing out your region really helps. This is an interesting part of calculus where you can actually visualize things... unlike many other math courses.
  5. Apr 13, 2005 #4
    Damn it, I put -sqrt(y) because in the book, the same problem (with x^2 in the lower bound of the first integral) like it put -sqrt(y) in the lower bound for it's solution. I would get lot of points taken off... it's a 15 point problem, the professors are harsh, i would get taken off -10 or something like that
  6. Apr 13, 2005 #5
    The book problem...

    The book problem was:
    [tex]\int_{-2}^{2}\int_{x^2}^{4} x^2y dydx[/tex]

    Their answer was:
    [tex]\int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}} x^2y dydx[/tex]

    Why was there a -sqrt(y) in this problem?
  7. Apr 13, 2005 #6
    The reason for that comes from the fact that part of your domain for the region extends into the dreaded negative x.
  8. Apr 13, 2005 #7
    and might i ask, did you mean to interchange the differential elements in the second integrand?

    And also, NIN rocks!!!
    Last edited: Apr 13, 2005
  9. Apr 13, 2005 #8
    Yes, I meant to do that. Damn it! I should have changed there right then because the problem would have never come out to zero. Oh well. Thanks.
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