Help with tricky double integral 
Consider the region $R$ bounded by the circles
  $$x^{2}+y^{2}=Ax$$$$x^{2}+y^{2}=Bx$$$$x^{2}+y^{2}=Cy$$$$x^{2}+y^{2}=Dy$$
  where $B>A$ and $D>C$.
Use the change of variables
  \begin{cases} u=\frac{x}{x^2+y^2}\\
 v=\frac{y}{x^2+y^2} \end{cases} 
to evaluate the integral $$\int\int_R  \frac{dxdy}{(x^2+y^2)^3}\quad
 (1)$$

My work so far: the Jacobian is
$$\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ 
 \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}$$
With $\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}=\frac{-2xy}{(x^2+y^2)^2} $ and $\frac{\partial u}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2}$
and $\frac{\partial v}{\partial y}=\frac{x^2-y^2}{(x^2+y^2)^2}$
: $$J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}=\frac{-1}{(x^2+y^2)^2} \quad(2)$$ 
My first approach to get the limits is rearranging $x^{2}+y^{2}=Ax$ and hence $\frac{1}{A}=\frac{x}{x^2+y^2}$ hence $1/A$ is a limit and the rest is $1/B$, $1/C$ and $1/D$. 
My questions:


*

*How do I find the limits of integration?

*How do I express the integral is terms of $u$ and $v$ ?


Any tips will be appreciated thanks!!

*

*You pretty much got it:
$$
R(u,v)=\left\{(u,v)\;|\; \frac{1}{B}\le u\le\frac{1}{A} , \frac{1}{D}\le u\le \frac{1}{C}\right\}
$$


The Figures below show the case where $A=C=1$, $B=D=2$:

So the region we are interested in is



*We are given \begin{cases} u=\frac{x}{x^2+y^2}\\
 v=\frac{y}{x^2+y^2} \end{cases}
Solving for $x$ and $y$ yields:
\begin{cases} x=\frac{u}{u^2+v^2}\\
 y=\frac{v}{u^2+v^2} \end{cases}
The Jacobian of such a transformation equals $\frac{1}{(u^2+v^2)^2}$, and $\frac{1}{(x^2+y^2)^3}=(u^2+v^2)^3$ therefore the integral equals
$$
\int_{1/B}^{1/A}\int_{1/D}^{1/C}(u^2+v^2)^3\frac{dvdu}{(u^2+v^2)^2}=\int_{1/B}^{1/A}\int_{1/D}^{1/C}u^2+v^2\;dvdu 
$$


This is very easy to integrate, although the final result does not simplify too well. The case with $A=C=1$, $B=D=2$ gives $\frac{7}{24}$.
