Can someone help me out and explain how to find the boundary I need in order to solve the following double integral, I posted below what I did so far but I am stuck here and I don't know if it's even correct what I did until now. Evaluating the integral is not needed, I just want to learn how to find the boundary.
$$ \int \int _D\frac{yln\left(x^2+y^2\right)}{x^2}dxdy$$ $where$ $D:\:1\le x^2+y^2\le e^2;$ $2x\le y\le x\le 0$
I have done the following so far:
$x^2+y^2=1$ circle with the center O(0,0) and R=1
$x^2+y^2=e^2$ circle with the center O(0,0) and R=e
$2x=y$ first bisector
$x=0$ at Oy with $x\le 0$ in $\left[\frac{\pi }{2},\pi \right]$ and $\left[\pi ,\frac{3\pi \:}{2}\right]$
$$T:D'\rightarrow D,T:x=\delta cos\theta ,y=\delta sin\theta, \delta \in \left[1,e\right], \theta \in \left[\frac{5\pi }{4} ,\frac{3\pi }{2}\right]$$
I chose $D'=\left[1,e\right]$x$\left[\frac{5\pi }{4} ,\frac{3\pi }{2}\right]$ but i am not sure if it's the right boundary or not especially on the 2nd bracket, I am almost certain one of them is $\frac{3\pi }{2}$ but the other one I am not sure.
edit: added the desmos
$x = r \cos\theta, y = r\sin\theta$. Bounds of $\theta$ is given by line $y = x$ and $y = 2x$
$r \cos\theta = r\sin\theta \implies \theta = \frac{5 \pi}{4}$ and that is one of the bounds you correctly found.
$2 r \cos\theta = r\sin\theta \implies \theta = (\pi + \arctan 2)$ is the other bound.
So bounds are $1 \leq r \leq e, \frac{5 \pi}{4} \leq \theta \leq (\pi + \arctan 2)$