help with strange Double Integral: $\iint_E {x\sin(y) \over y}\ \rm{dx\ dy}$ i'm having trouble with this double integral:
$$
\iint_E {x\sin(y) \over y}\ \rm{dx\ dy},\ \ \ \ 
E=\Big\{(x,y) \in \mathbb{R^2} \mid 0<y\le x\ \ \  \land\ \ \  x^2+y^2 \le \pi y\Big\}
$$
i've tried using polar coordinates, but after i made the domain normal i realized that the integrand got a lot more complicated..
then i've tried another transform: $u=y/x, v=y$; with even worse results.
i'm looking mainly for a tip on how to tackle this,
also i'd like to know the reasoning behind an eventual tip... thanks in advance!

$$
\iint_E {x\sin(y) \over y}\ \rm{dx\ dy},\ \ \ \ 
E=\Big\{(x,y) \in \mathbb{R^2} \mid 0<y\le x\ \ \  \land\ \ \  x^2+y^2 \le \pi y\Big\}\\
\\
\begin{cases}
x^2+y^2\le\pi y \iff x^2\le\pi y -y^2 \iff \vert x\vert \le \sqrt{\pi y -y^2}\\
0<y\le x
\end{cases} \implies \\
\implies 
0<y\le x \le \sqrt{\pi y -y^2} \implies 
\begin{cases} 
\sqrt{\pi y -y^2} \ge y \\
\pi y -y^2\ge 0\\
y >0
\end{cases}
\iff 0 < y \le \pi/2
$$
so now we have:
$$
0 < y \le \pi/2 \ \ \ \land \ \ \ y\le x \le \sqrt{\pi y -y^2}
$$
which gives us:
$$
\iint_E {x\sin(y) \over y}\ \rm{dx\ dy} = 
\int_{0}^{\pi/2} {\sin{y} \over y} \ \Big[ \int_{y}^{\sqrt{\pi y -y^2}} x\ \rm{dx} \Big] \rm{dy} = \cdots = {\pi-2 \over 2}
$$
you should be able to solve the single variable integrals yourself, also keep in mind that the integral is improper when y approaches 0. 
