$$ D= \left\{ (x,y)\in\mathbb{R}^{2} \Bigg| 1\leq x^2+y^2\leq 2^2,~~0\leq y\leq x \right\} $$
$$ I:=\iint_{D} x\log \left(x^2+y^2 \right) \mathrm{d}x \mathrm{d}y $$
I want to know the exact domain of $~D~$
I think that at least the green-painted domain is a subdomain of $~D~$
I've come up with the dought that black domain may also be a subdomain of $~D~$
Can it be?
Moreover, I've been studying math completely with self-taught one so I can't have a confidence that whether I can replace $~\log~$ to $~\ln~$ without point(s) subtracted of an exam.
Owing to @Anna Bauval, I think I got the solution of the double integral.
Since it is too natural for me using $~\ln~$ than $~\log~$,I use natural logarithm here for now without asking the person who made this math problem.
$$ D= \left\{ (x,y)\in\mathbb{R}^{2} \Bigg| 1\leq x^2+y^2\leq 2^2,~~0\leq y\leq x \right\} $$
$$ I=\iint_{D} x\log \left(x^2+y^2 \right) \mathrm{d}x \mathrm{d}y $$
$$ E:= \left\{ (r,\theta)~~ \Bigg| 1\leq r\leq 2,~~ 0\leq\theta\leq {\pi \over 4 } \right\} $$
$$ \begin{cases} x=r\cos(\theta)\\ y=r\sin(\theta) \end{cases} $$
$$\begin{align} I&=\iint_{E} r^2\cos(\theta)\ln \left(r^2 \right) \mathrm{d}r \mathrm{d}\theta\\ &=\cdots\\ &= \left(\int_{1}^{2}r^2 \ln \left(r^2 \right) \mathrm{d}r\right) \left(\int_{0}^{{\pi \over 4 } } \cos(\theta) \mathrm{d}\theta \right)\\ &=\cdots\\ &= \sqrt{2} \left\{ {4 \over 3 }\ln(4)- {7 \over 9 } \right\} \end{align}$$