Sketching in the $(r,\theta)$ plane

Question

Considering the integral

$ \int\int_R x^2log(x^2+y^2)dxdy $ where $R$ is the annulus $1\leq x^2+y^2\leq 2$.

Change the variables to polar coordinates and sketch the region $R$ in the $(x,y)$ plane and also the corresponding region in the $(r,\theta)$ plane.

So after changing the variables I find the integral turns in to:

$\int^{2\pi}_0\int^{2}_1r^3cos\theta log(r^2)drd\theta$.

However when it comes to sketching I am confused, in the $x,y$ plane I can see our region is a donut shape enclosed between two discs. However how can I sketch the corresponding region?

Based on searching online I found that the region turns in to $1\leq r \leq \sqrt2$ and $0\leq \theta \leq 2\pi$.

What does this even mean however, what region should I be sketching/looking for?

Answer 1

You have the bounds right for the integral when you've transformed it, so just look at those bounds. They are numbers, not variables, so that means your variable goes from one constant to another constant. Since that happens for both variables, the region is called rectangular.

Answer 2

Hint. From $$ x=r\cos \theta,\quad y=r\sin \theta, $$ one gets $$ x^2+y^2=r^2(\cos^2 \theta+\sin \theta^2)=r^2 \cdot 1 $$ then $$ 1\le x^2+y^2 \le 2 $$ reads $1\le r^2 \le 2$ that is $$ 1\le r \le \sqrt{2} $$ since $r\ge0$.

One thus obtains $$ \begin{align} \int\int_R x^2\log(x^2+y^2)dxdy&=2\int^{2\pi}_0\int^{\sqrt{2}}_1r^3\cos^2\theta \log(r)dr\:d\theta \\&=2\left(\int^{2\pi}_0\cos^2\theta \:d\theta\right)\left(\int^{\sqrt{2}}_1r^3\log(r)dr\right) \end{align} $$ Can you finish it?