Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The region $D$ can be represented as a difference of two simpler regions, $D = D_1 \backslash D_2$, where $$ D_1 = \{ (x,y) \in \mathbb{R}^2: x \geqslant 0, y \geqslant 0, x+y \leqslant 4 \} \quad D_2 = \{ (x,y) \in \mathbb{R}^2: x \geqslant 0, y \geqslant 0, x^2+y^2 < 4 \} $$ enter image description here

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: $$ \int_{D_2} \frac{y}{x^2+1} \mathrm{d} x \mathrm{d} y = \int_0^2 r \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \phi \frac{r \sin(\phi)}{r^2 \cos^2(\phi)+1} = \frac{5}{2} \arctan(2) - 1 $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.