Transformation to polar coordinates in an integral

Question

Suppose that the domain of integration for a double integral is: $\{(x,y), - \infty < x \le a, -\infty < y \le a \}$. If I want to do a change of variable (to polar coordinates), how do I express the new domain? I first thought it is $0 \le \theta \le 2\pi$, $0 \le r \le \alpha \sqrt 2$, but it seems incorrect

Answer 1

I think the easiest thing you could do is make a change of variables $(x,y)\to(x',y')$, such that $x' = x - a, y' = y = a$, this allows you to easily make the transformation to polar, and just worry about the region $\{(x',y'),\ x' \leq 0, y' \leq 0\}$ which could correspond to $\pi \leq \theta \leq \frac{3\pi}{2},\ 0 \leq r < \infty$.

In the space you originally defined, you constrained yourself to a rectangle (because $x$,$y$ were constrained individually). If you are interested in constraining to a circle (making a direct constraint on radius), you would have the space $R=\{(x,y): \{\pi/2 \leq \theta \leq 2\pi,-\infty< x \leq 0,-\infty< y\leq 0\}\cup\{0\leq \theta\leq \pi/ 2,\ x^2 + y^2 \leq \alpha\}$ Not sure what you want to do with that extra quadrant.

Answer 2

domain/range

$0 \le \theta \le 2\pi$ and $- \infty \le r \le \alpha \sqrt 2$.

Answer 3

Hint:

Use: $$ x=a-r\cos \theta \qquad y=a-r\sin \theta $$

for $\pi<\theta < \frac{3}{2}\pi$ and $r\ge 0$.