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!

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Picture is almost always necessary. Second integral is natural for polar coordinates, replace the (missing!) $dx\,dy$ by $r\,dr\,d\theta$, $r$ $1$ to $2$, $\theta$ $-\pi/2$ to $\pi/2$. For first integral, draw $x+y=0$, $x+y=1$, $x-y=0$, $x-y=1$. Make natural change of variable $u=x+y$, $v=x-y$. But integral is not terrible to evaluate "as is." – André Nicolas Sep 5 '11 at 9:50
I (almost) take back picture part of comment for first integral, if you are stressed for time. Even without picture $u$ $0$ to $1$, $v$ $0$ to $\pi$ is clear. Don't forget the Jacobian. – André Nicolas Sep 5 '11 at 10:07
Don't take it back; if anything, I would omit the "almost". Always draw a picture. – Robert Israel Sep 5 '11 at 18:33
Thank you somehow it clicked after reading your comments! Sometimes I just need to read the right words and it all comes together. – entrance_exam Sep 6 '11 at 10:20

The first inequality, $0 \le x+y \le 1$, is equivalent to $-x \le y \le 1-x$ (just subtract $x$ everywhere). Hence it represents all points $(x,y)$ lying on or between the two straight lines $y=-x$ and $y=1-x$.

Can you do something similar with the second inequality? In the end you should find that $A$ is a region bounded by a quadrilateral.

(Another hint: Try changing variables to $u=x+y$ and $v=x-y$.)

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Thank you for teaching me about those inequalities! Appreciated! – entrance_exam Sep 6 '11 at 10:19

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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I just glanced at the sketch, and thought "Wow, this person must be Américo" :-) – Srivatsan Sep 6 '11 at 23:28
@Srivatsan: I tried another one before but was not so informative. Thanks! – Américo Tavares Sep 6 '11 at 23:42