Trouble setting up double integrals

Question

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!

Answer 1

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 $v=x-y$.)

Answer 2

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 $x,y$: $$ \begin{eqnarray*} I &=&\int_{0}^{1/2}\left( \int_{-x}^{x}e^{x+y}\sin (x-y)\mathrm{d}y\right) \mathrm{d}x \\ &&+\int_{1/2}^{\pi /2}\left( \int_{-x}^{1-x}e^{x+y}\sin (x-y)\mathrm{d} y\right) \mathrm{d}x \\ &&+\int_{\pi /2}^{(1+\pi )/2}\left( \int_{x-\pi }^{1-x}e^{x+y}\sin (x-y) \mathrm{d}y\right) \mathrm{d}x. \end{eqnarray*} $$

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