I would like to solve the following integral \begin{equation} \int\int_{D} \frac{y}{x^{2}+1} dx dy, \ \ D=\{(x,y)\in R^{2}: x\geq0, y\geq0, x^{2}+y^{2}\geq4 , x+y\leq4 \} \end{equation} using a change of variables, but I don't want to use polar coordinates (they are too complicated for this exercise).
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The region $D$ can be represented as a difference of two simpler regions, $D = D_1 \backslash D_2$, where
$$
D_1 = \{ (x,y) \in \mathbb{R}^2: x \geqslant 0, y \geqslant 0, x+y \leqslant 4 \}
\quad
D_2 = \{ (x,y) \in \mathbb{R}^2: x \geqslant 0, y \geqslant 0, x^2+y^2 < 4 \}
$$
Thus $$ \int_D \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y = \int_{D_1} \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y - \int_{D_2} \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y $$ The integral over $D_1$ is best done directly: $$ \int_{D_1} \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y = \int_0^4 \mathrm{d} x \int_0^{4-x} \mathrm{d} y \frac{y}{x^2+1} = \int_0^4 \mathrm{d} x \frac{1}{2} \frac{(4-x)^2}{x^2+1} = \left.\frac{1}{2}\left( x + 15 \arctan(x) -4 \log(1+x^2) \right)\right|_{x=0}^{x=4} = 2+ \frac{15}{2} \arctan(4) - 2 \log(17) $$ Integral over $D_2$ is best done in polar coordinates: $$ \int_{D_2} \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y = \int_0^2 r \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \phi \frac{r \sin(\phi)}{r^2 \cos^2(\phi)+1} = \frac{5}{2} \arctan(2) - 1 $$
