<s>[INST] 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!
 [/INST] 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
$$ 
\begin{eqnarray*}
\iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\left( x^{2}+y^{2}\right) ^{2}}
&=&\int_{r=1}^{2}\int_{\theta =0}^{\pi }\frac{1}{r^{4}}r\;\mathrm{d}r\mathrm{
d}\theta  \\
&=&\int_{0}^{\pi }\left( \int_{1}^{2}\frac{1}{r^{3}}\mathrm{d}r\right) \mathrm{d}
\theta  \\
&=&\int_{0}^{\pi }\frac{3}{8}\mathrm{d}\theta  \\
&=&\frac{3}{8}\pi. 
\end{eqnarray*}
$$
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