What is the solution to a multiple integral problem using polar coordinates?

[kidsmoker]

Homework Statement

By transforming to polar coordinates, show that

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

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

Homework Equations

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

where J is the Jacobian.

The Attempt at a Solution

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

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

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!

 

[CompuChip]

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 \sqrt{2}: it always starts at 0 but it only runs to \sqrt{2} for \theta = \pi / 4.

It may help if you draw the triangle.

 

[kidsmoker]

Sorry I've just realized 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 \theta = 0 and r=\sqrt{2} at \theta = \pi / 4. So these are my boundaries? Or am I misunderstanding it?

Thanks.

 

[CompuChip]

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

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 \theta = 0 then r should run from 0 to 1, and if \theta = \pi / 4 then r should run from 0 to r=\sqrt{2}. If \theta = \pi / 8 then r runs from 0 to ... ?

It is precisely analagous to the expression
I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)} dy dx
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?

 

[kidsmoker]

Ah okay... I think i see now. So I should have wrote

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 ... ?

 

[CompuChip]

No, the point I was trying to make is that instead of \sqrt{2}, you should have something which depends on \theta - just like you have an x as integration boundary in the first form.

 

[kidsmoker]

Ahh, i see! Right so it should be

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

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

Thanks!

 

[CompuChip]

Ahh, i see! Right so it should be

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

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

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

Thanks!

Yes, that was a bit of a problem for me too... the easiest way I see would be something like this:
First substitute x = r^2 \cos\theta \sin\theta and write the integral in the form
\int_0^{\cdots} \frac{1}{x^2 + a x + b} \, dx = \int_0^{\cdots} \frac{1}{(x - x_1)(x - x_2)} \, dx
then note that this integral is equal to
\frac{\log(x + a) - \log(x + b)}{b - a}.
In principle it looks easy, but prepare your trig identities... you will need them :sad:

 

[kidsmoker]

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!