Calculate: $$I=\iint\limits_{D}\frac{\text{d}x\text{d}y}{xy(\ln^2x+\ln^2y)}\\ \quad\text{$D$ is an area enclosed by the two functions} \color{violet}{\text{ in the first quadrant}}:\\ D:(\text{enclosed by})\begin{cases} x^2+y^2=1\\ x+y=1 \end{cases} $$
Here's my approach: Integration by substitution: $$ \begin{cases} x=e^{r\cos\theta} \\ y=e^{r\sin\theta} \end{cases} \Rightarrow\text{det}\mathbf{J}=\frac{\partial(x,y)}{\partial(r,\theta)}=re^{r(\cos\theta+\sin\theta)} $$
The boundary $D$ after substitution: $$ \Delta:(\text{enclosed by})\begin{cases} e^{2r\cos\theta}+e^{2r\sin\theta}=1\\ e^{r\cos\theta}+e^{r\sin\theta}=1 \end{cases} $$ Now: $$I=\iint\limits_{\Delta}\frac{\text{d}r\text{d}\theta}{r}=?$$ And I don't know how to continue because of the implicit equations above.
There is no need to find the exact interval after the complicated substitution.
The boundary $D$ after substitution: $$ \Delta:(\text{enclosed by})\begin{cases} (1):\quad e^{r\cos\theta}+e^{r\sin\theta}=1 \label{eq1} \\ (2):\quad e^{2r\cos\theta}+e^{2r\sin\theta}=1 \end{cases}$$
It is easy to find that $\theta \in [\pi,3\pi/2]$.
Now let the function determined by the equation (1) be $r=r(\theta)$.
Hence the equation (2) determines:$r = r(\theta)/2$ and:
$$ I=\iint\limits_{\Delta}\frac{\text{d}r\text{d}\theta}{r}=\int_{\pi}^{3\pi/2}\text{d}\theta\int_{r(\theta)/2}^{r(\theta)}\frac{\text{d}r}{r}=\frac{\pi\ln2}{2} $$