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Homework Help: Plane intergration

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine [tex] \int \int_D (x^2+y^2)dA[/tex], where D is the area in the first quadrant of the xy-plane restricted by y=0, y=x, xy=1 and x^2-y^2=1, using variable change.

    2. Relevant equations

    Here's a picture of the situation:
    http://www.aijaa.com/img/b/00050/3858610.jpg [Broken]

    3. The attempt at a solution

    I'm out of ideas here, the only thing I came up with is to use polar coordinates. That would give me [tex] \int_0^{ \pi /4} \int_0^? r^3drd \theta[/tex]. As you can probably see, I have no clue as to what the limits are for r...
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 22, 2009 #2
    Stick with the original integral, and break it down into two segments. The first segment is the area bounded by:

    y = 0
    y = x
    x = 1

    The integral over this segment is:
    [tex] \int_0^1\left(\int_0^x (x^2+y^2) dy\right)dx[/tex]

    You should have no trouble evaluating this integral.

    What remains is the integral over the area bounded by the three curves:
    x = 1 (lower bound for x)
    y=1/x (upper bound for y)
    x^2 - y^2 = 1, e.g. [tex]y = +\sqrt{x^2-1}[/tex] (lower bound for y). Note that x>1.

    Note that [tex]y = +\sqrt{x^2-1}[/tex] and y=1/x intersect at the point where

    [tex] \frac{1}{x_{\textrm{max}}} = \sqrt{x^2_{\textrm{max}}-1}[/tex]
    Solve this equation to get the upperbound for x( [tex]\equiv x_{\textrm{max}}[/tex])

    What remains is the integral with all bounds included:
    [tex]\int_1^{x_{\textrm{max}}} \left[\int_{\sqrt{x^2-1}}^{1/x}(x^2 + y^2)dy \right]dx[/tex]

    Which you should be able to compute. Don't forget to add both integrals for the final answer.
  4. Mar 22, 2009 #3
    Thanks for the reply, xepma. But the problem states that we have to solve this using variable change. So you're suggestion is of no use to me, though it would work.
  5. Mar 23, 2009 #4
    Okay, I've found two possible ways to do this. If I let u=xy and v=x^2-y^2 I get an integral that looks like [tex] \int_0^1 \int_0^1 - \frac{1}{2} dvdu[/tex], which gives me [tex]- \frac{1}{2} [/tex], but the correct answer is [tex] \frac{1}{2} [/tex].

    If I choose u=x^2-y^2 and v=xy, I get almost the same calculation: [tex] \int_0^1 \int_0^1 \frac{1}{2} dvdu[/tex]. So the Jacobian determinant gives me [tex] \frac{1}{2} [/tex] instead of [tex]- \frac{1}{2} [/tex].

    I can get the right answer by altering one of the bounds in case 1. I'm just wondering, which one should I alter and why?
  6. Mar 23, 2009 #5


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    Science Advisor

    The sign of a determinant depends on the order of the rows or columns which is not relevant here.
    The differential is the absolute value of the Jacobian determinant so you get 1/2 in either case.
  7. Mar 23, 2009 #6
    Thanks a lot, HallsofIvy! You really saved my butt. Our teacher uses the same method of denoting absolute value and determinant, so it's easy to confuse them.
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