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Calculate the following double integral

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the double integral:

    [tex]\iint\limits_D x^{5}y^{6}dxdy[/tex]
    where D = {(x,y): x9 ≤ y ≤ x1/9}

    2. Relevant equations



    3. The attempt at a solution

    I didn't think this problem would be too hard, but it seems I'm really not good with double integrals.

    Anyway, I first tried to find the right interval. The curve y=x9 is below y=x1/9 in two intervals, first of all 0 ≤ x ≤ 1 and -∞ ≤ x ≤ -1. I was wondering about this for some time, but then I thought that I probably needed an enclosed D, so I discarded the second interval.

    When x=0, y=0, and when x=1, y=1 in both curves.

    [tex]\iint\limits_D x^{5}y^{6}dxdy = \int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy[/tex]

    I tried that, because that was something they explained to us during the lecture and well.. it seemed to apply. And those two integrals are really very easy:

    [tex]\int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy = \left[ \frac{1}{6} x^6 \right] ^{1}_{0} \left[ \frac{1}{7} y^7 \right]^{1}_{0} = ( \frac{1}{6})( \frac{1}{7}) = \frac{1}{42}[/tex]

    And this is wrong.

    Help?
     
  2. jcsd
  3. Nov 19, 2009 #2

    Mark44

    Staff: Mentor

    The two curves intersect at (-1, -1), (0, 0), and (1, 1). When 0 <= x <= 1, x^(1/9) >= x^9, and when -1 <= x <= 0, x^9 >= x^(1/9). So that means you'll need to split your double integral into a pair of iterated integrals, one for each of the two intervals.
    Hope that helps.
     
  4. Nov 19, 2009 #3
    You must integrate over y first because the limits of the y integral contain x values. After the y integral, you do the x integral. You should end up with an expression with only x's.
     
  5. Nov 19, 2009 #4

    Mark44

    Staff: Mentor

    Also, I don't think there's a justification for splitting up the iterated integrals as you did. Flatmaster's advice of integrating with respect to y first, and then with respect to x, is good.
     
  6. Nov 19, 2009 #5
    Oh thank you! Talk about change in thinking.

    And anyway, I didn't use that second interval, because it didn't fit with the area that I was given, since it stated that x^(1/9) needed to be >= x^9. It gave me the right answer, so I'm very happy now!
     
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