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Multiple integral problem

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    By transforming to polar coordinates, show that

    [tex]I = \int\int_{T}\frac{1}{(1+x^2)(1+y^2)}dxdy = \int^{\pi/4}_{0}\frac{log(\sqrt{2}cos(\theta))}{cos(2\theta)}d\theta[/tex]

    where T is the triangle with successive vertices (0,0),(1,0),(1,1).


    2. Relevant equations

    [tex]I = \int\int_{K} f(x,y)dxdy = \int\int_{K'} g(u,v)*J*dudv[/tex]

    where J is the Jacobian.

    3. The attempt at a solution

    The Jacobian is r, as always with a transformation to polar coordinates, so we get that

    [tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dydx = \int^{\pi/4}_{0}\int^{\sqrt2}_{1}\frac{r}{(1+r^2cos^2(\theta))(1+r^2sin^2(\theta))}drd\theta[/tex]

    Firstly, is this correct? Secondly, if it is, could you give me a hint as to how to solve it to get the answer given? The obvious thing seems to be to split it into partial fractions, but I did try this once and didn't seem to get anywhere?

    Thanks!
     
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2

    CompuChip

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    The transformation of coordinates looks correct, but you may want to check your integration boundaries. The triangle is "standing on it's tip" so for x = 0 you want to integrate y from 0 to 1, etc. In polar coordinates then, r doesn't run from 1 to [itex]\sqrt{2}[/itex]: it always starts at 0 but it only runs to [itex]\sqrt{2}[/itex] for [itex]\theta = \pi / 4[/itex].

    It may help if you draw the triangle.
     
  4. Jan 24, 2009 #3
    Sorry i've just realised i got the coordinates of one of the vertices of the triangle muddled up. It should be (0,0),(1,0),(1,1) not (0,0),(0,1),(1,1) which i originally wrote.

    The integration boundaries in polar coordinates do confuse me. With the correct vertices, surely r=1 at [itex]\theta = 0[/itex] and [itex]r=\sqrt{2}[/itex] at [itex]\theta = \pi / 4[/itex]. So these are my boundaries? Or am I misunderstanding it?

    Thanks.
     
  5. Jan 24, 2009 #4

    CompuChip

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    Yes, you are correct about those values. But let's suppose that the theta integration is the outer one and the r-integration the inner one. That means that for every value of theta you will have to do the r-integral. So if [itex]\theta = 0[/itex] then r should run from 0 to 1, and if [itex]\theta = \pi / 4[/itex] then r should run from 0 to [itex]r=\sqrt{2}[/itex]. If [itex]\theta = \pi / 8[/itex] then r runs from 0 to ... ?

    It is precisely analagous to the expression
    [tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)} dy dx[/tex]
    you gave: when you do the inner integration over y, your x is fixed. You can pretend that you are walking along the x-axis, and for each point there doing the y-integral. However the upper boundary for y depends on what the "current" value of x is, so you have to integrate y from 0 to x and not from 0 to 1, despite the fact that y runs to 0 for x = 0 and y runs all the way to 1 for x = 1.

    Do you understand?
     
  6. Jan 24, 2009 #5
    Ah okay... I think i see now. So I should have wrote

    [tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{\sqrt2}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta[/tex] ... ?
     
  7. Jan 24, 2009 #6

    CompuChip

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    No, the point I was trying to make is that instead of [itex]\sqrt{2}[/itex], you should have something which depends on [itex]\theta[/itex] - just like you have an x as integration boundary in the first form.
     
  8. Jan 25, 2009 #7
    Ahh, i see! Right so it should be

    [tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{1/cos\theta}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta[/tex]

    I'm still left unable to solve the integral though :-( Any tips for the best method to use?

    Thanks!
     
  9. Jan 26, 2009 #8

    CompuChip

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    Very good, that's what I had in mind too :smile:

    Yes, that was a bit of a problem for me too... the easiest way I see would be something like this:
    First substitute [tex]x = r^2 \cos\theta \sin\theta[/tex] and write the integral in the form
    [tex]\int_0^{\cdots} \frac{1}{x^2 + a x + b} \, dx = \int_0^{\cdots} \frac{1}{(x - x_1)(x - x_2)} \, dx[/tex]
    then note that this integral is equal to
    [tex]\frac{\log(x + a) - \log(x + b)}{b - a}[/tex].
    In principle it looks easy, but prepare your trig identities... you will need them :sad:
     
  10. Jan 26, 2009 #9
    Ah okay cool. I managed to do it using the substitution t=r^2 and then splitting it into partial fractions. Was a bit long-winded but got the right answer.

    Thanks!
     
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