Evaluate the integral:

\(\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy\)

where \(\displaystyle D\) is the region \(\displaystyle D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}\)

Tried drawing the region \(\displaystyle D\) and it looks like I have to use polar coordinates.

Using polar coordinates, \(\displaystyle \frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}\).

I'm not sure how to find the limits if integration though, or what to do with the \(\displaystyle dx\;dy\).

Thanks in advance, any pointers would be greatly appreciated.

In general to convert, you do the following:

\(\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)

\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta \)

Where \(\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)} \) is called the jacobian determinant and it's given by:

\(\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)} =

\begin{vmatrix}

\cos \theta & - r \sin \theta \\

\sin \theta & r \cos \theta

\end{vmatrix} = r\)

Hence:

\(\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)

r \, dr \, d\theta \)

Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

\(\displaystyle D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}\)

Now, the 2nd constraint on the region says that \(\displaystyle x^2 + y^2 \leq 3 \). You should be able to see that this equation describes a circle whose radius is less than \(\displaystyle \sqrt{3}\), in other words, the radius is limited by \(\displaystyle 0 \leq r \leq \sqrt{3} \).

Now, the 1st constraint on the region says that \(\displaystyle y \geq x \). Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by \(\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \).

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.

2) Draw a circle centred at the origin with a radius of \(\displaystyle \sqrt{3}\).

3) Draw the line y = x.

4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

\(\displaystyle D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}\)

Hence, in this particular situation:

\(\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta \)