Trouble setting up double integrals I am studying for an entrance exam and have troubles to set up double integrals appropriately. Actually my biggest problem is that I don't get the notation.
$$ A = \{(x,y) | 0 \leq x + y \leq 1,  0 \leq x-y \leq \pi\} $$
$$ \iint_A e^{x+y} \sin(x-y)\mathrm dx\mathrm dy $$ 
and for another example:
$$ R = \{(x,y) | 1 \leq x^2 + y^2 \leq 4,  y \geq 0\} $$
$$ \iint_R \frac{\mathrm dx\mathrm dy}{(x^2+y^2)^2} $$
Normally I would go draw a picture of something like a line or circle.
When I have an equation like $x+y=z$ for example.
I tried interpreting the inequality as an area so for the first one I came up with the between 0 and the line $y=1-x$ which seemed reasonable for I failed to connect this to the other inequality.
I hope somebody can point out how to set this up or give some hints how to interpret this notation.
Any help is greatly appreciated!

For the first example. The double inequality $0\leq x+y\leq 1$ means that
$$
\left\{ 
\begin{array}{c}
0\leq x+y \\ 
x+y\leq 1
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
y\geq -x \\ 
y\leq 1-x
\end{array}
\right. 
$$
and $0\leq x-y\leq \pi $ means that
$$
\left\{ 
\begin{array}{c}
0\leq x-y \\ 
x-y\leq \pi 
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
y\leq x \\ 
y\geq x-\pi. 
\end{array}
\right. 
$$
So the conditions $0\leq x+y\leq 1$ and $0\leq x-y\leq \pi $ are equivalent
to the system of four inequalities
$$
\left\{ 
\begin{array}{c}
y\geq -x \\ 
y\leq 1-x \\ 
y\leq x \\ 
y\geq x-\pi. 
\end{array}\tag{1}
\right. 
$$
The region $A$  is a rectangle limited by the four lines $y=-x$, $y=1-x$, $y=x$, $y=x-\pi $  (see figure). 

To evaluate 
$$
I:=\iint_{A}e^{x+y}\sin (x-y)\;\mathrm{d}x\mathrm{d}\tag{2}y
$$
we may consider the rotated system of coordinates $x',y'$ with respect to the $x,y$ system, the rotation angle being $\theta =-\pi /4$, as shown in the figure. This corresponds to the following transformation of coordinates
$$
\begin{eqnarray*}
x^{\prime } &=&x\cos \left( -\frac{\pi }{4}\right) +y\sin \left( -\frac{\pi 
}{4}\right) =\frac{1}{2}\sqrt{2}x-\frac{1}{2}\sqrt{2}y \\
y^{\prime } &=&-x\sin \left( -\frac{\pi }{4}\right) +y\cos \left( -\frac{\pi 
}{4}\right) =\frac{1}{2}\sqrt{2}x+\frac{1}{2}\sqrt{2}y,
\end{eqnarray*}
$$
whose inverse is
$$
\begin{eqnarray*}
x &=&x^{\prime }\cos \left( -\frac{\pi }{4}\right) -y^{\prime }\sin \left( -
\frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}%
y^{\prime } \\
y &=&x^{\prime }\sin \left( -\frac{\pi }{4}\right) +y^{\prime }\cos \left( -
\frac{\pi }{4}\right) =-\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}%
y^{\prime }.
\end{eqnarray*}
$$
Since $\frac{\partial (x,y)}{\partial (x^{\prime },y^{\prime })}=1$, the
integral $I$ is transformed into
$$
\begin{eqnarray*}
I &=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\left( \int_{x^{\prime }=0}^{\pi \sqrt{
2}/2}e^{\sqrt{2}y^{\prime }}\sin (\sqrt{2}x^{\prime })\mathrm{d}x^{\prime
}\right) \mathrm{d}y^{\prime } \\
&=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\sqrt{2}e^{y^{\prime }\sqrt{2}}\mathrm{d}
y^{\prime } \\
&=&e-1,\tag{3}
\end{eqnarray*}
$$
because 
$$
\begin{eqnarray*}
x-y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\left( -
\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }\right) =\sqrt{2
}x^{\prime } \\
x+y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\frac{1
}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }=\sqrt{2}y^{\prime }.
\end{eqnarray*}
$$
Alternatively we could split $A$ into three regions, a triangle ($0\le x\le 1/2$), a quadrilateral ($1/2\le x\le π/2$) and a triangle ($\pi/2\le x\le (1+\pi)/2$), and evaluate $I$ in the original variables $x,y$:
$$
\begin{eqnarray*}
I &=&\int_{0}^{1/2}\left( \int_{-x}^{x}e^{x+y}\sin (x-y)\mathrm{d}y\right) 
\mathrm{d}x \\
&&+\int_{1/2}^{\pi /2}\left( \int_{-x}^{1-x}e^{x+y}\sin (x-y)\mathrm{d}
y\right) \mathrm{d}x \\
&&+\int_{\pi /2}^{(1+\pi )/2}\left( \int_{x-\pi }^{1-x}e^{x+y}\sin (x-y)
\mathrm{d}y\right) \mathrm{d}x.
\end{eqnarray*}
$$

As for the second example $R$ is the semi-annulus  centered at $(0,0)$ with outer radius equal to 2, inner radius 1 and $y\ge 0$.  The Jacobian of the transformation of Cartesian to polar coordinates is $\frac{\partial \left( x,y\right) }{\partial \left(
r,\theta \right) }=\sqrt{x^{2}+y^{2}}=r$.  Hence
$$ 
\begin{eqnarray*}
\iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\left( x^{2}+y^{2}\right) ^{2}}
&=&\int_{r=1}^{2}\int_{\theta =0}^{\pi }\frac{1}{r^{4}}r\;\mathrm{d}r\mathrm{
d}\theta  \\
&=&\int_{0}^{\pi }\left( \int_{1}^{2}\frac{1}{r^{3}}\mathrm{d}r\right) \mathrm{d}
\theta  \\
&=&\int_{0}^{\pi }\frac{3}{8}\mathrm{d}\theta  \\
&=&\frac{3}{8}\pi. 
\end{eqnarray*}
$$
