# Multiple integral problem

1. Jan 24, 2009

### kidsmoker

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

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

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

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!

Last edited: Jan 24, 2009
2. Jan 24, 2009

### 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.

3. Jan 24, 2009

### kidsmoker

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

Thanks.

4. Jan 24, 2009

### CompuChip

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?

5. Jan 24, 2009

### kidsmoker

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

$$I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{\sqrt2}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta$$ ... ?

6. Jan 24, 2009

### 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.

7. Jan 25, 2009

### 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!

8. Jan 26, 2009

### CompuChip

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

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:

9. Jan 26, 2009

### 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!