Calculate
$$\iint_D {1 \over {(x^2+y^2)^2}} dxdy$$
when: $$D= \{\space (x,y) \in \mathbb R ^2 \space |\space {1 \over 2} \le x \space,\space x \le y \le \sqrt 3 x\space,\space x^2+y^2 \le 1\space \}$$
I change to polars $\space x=r \cos \theta$, $\space y=r \sin \theta \space$, and I can see that ${\pi \over 4}\le \theta \le {\pi \over 3}$, but what are the limits of $r$?
Even when I draw this I can't figure what they are.
Note: I just want to understand how to find the new limits, I'll calculate the rest by myself.
For each $\theta$, you need to integrate from the vertical line $x = \frac{1}{2}$ to the circle $r=1$. The upper limit is obviously $1$.
As for the lower limit, the equation $x=\frac{1}{2}$ translated to polar coordinates becomes $$ r \cos\theta = \frac{1}{2} \implies r = \frac{1}{2\cos\theta} = \frac{1}{2}\sec\theta $$ This means that $$ D = \left\{(r,\theta)\mid \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3},\ \frac{1}{2} \sec\theta \leq r \leq 1\right\} $$