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!
 A: Let's start with the difficult one; $$x^2+y^2 \leq \pi y \iff x^2+y^2 -\pi y \leq 0$$ which is a circle(with all the points inside the circle, can't remember the english word) with center $C(0, \frac{\pi}{2})$ and radius $r=\frac{\pi}{2}$
You can parameterize this cirle as $$-\frac{\pi}{2}<x<\frac{\pi}{2}, \;  \frac12 (\pi-\sqrt{\pi^2-4 x^2)} \leq y \leq \frac12 (\sqrt{\pi^2-4 x^2}+\pi)$$
Now we have left;
$$0 <y \leq x$$
So $x$ has to be positive. Using the logical AND arguement, we have to combine these the solution from above with this one, thus $$0<x<\frac{\pi}{2}$$ and $$x \geq y$$ thus $$\frac12 (\pi-\sqrt{\pi^2-4 x^2)} \leq y \leq x \land 0<x<\frac{\pi}{2}$$
The greater than instead of greater or equal than is throwing me off but I guess it could go like this(I may be wrong);
$$\int_0^{\frac{\pi}{2}} \int_{\frac12 (\pi-\sqrt{\pi^2-4 x^2)}}^x \frac{x\sin y}{y} \mathrm{d}y \mathrm{d}x $$
Don't take my word for it though. It's my first time doing boundary conditions of this level. This was supposed to be a comment but it turned out rather long.
A: $$
\iint_E {x\sin(y) \over y}\ \rm{dx\ dy},\ \ \ \ 
E=\Big\{(x,y) \in \mathbb{R^2} \mid 0<y\le x\ \ \  \land\ \ \  x^2+y^2 \le \pi y\Big\}\\
\\
\begin{cases}
x^2+y^2\le\pi y \iff x^2\le\pi y -y^2 \iff \vert x\vert \le \sqrt{\pi y -y^2}\\
0<y\le x
\end{cases} \implies \\
\implies 
0<y\le x \le \sqrt{\pi y -y^2} \implies 
\begin{cases} 
\sqrt{\pi y -y^2} \ge y \\
\pi y -y^2\ge 0\\
y >0
\end{cases}
\iff 0 < y \le \pi/2
$$
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. 
