Checking a double integral in polar coordinates

Question

I'm trying to evaluate the following double integral:

$\int_F \arctan(\frac{x}{y})\ dxdy, F:=\{(x,y)\in\mathbb{R^2}:1\leq x^2+y^2\leq4, |y|\leq|x|\}$.

By switching to polar coordinates and splitting I get:

$\int_F \arctan(\frac{x}{y})\ dxdy= \int_0^{\pi/4}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta + \int_{3\pi/4}^{5\pi/4}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta + \int_{7\pi/4}^{2\pi}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta = \int_0^{\pi/4}\theta d\theta\int_{r=1}^{r=2}rdr+ \int_{3\pi/4}^{5\pi/4}\theta d\theta\int_{r=1}^{r=2}rdr + \int_{7\pi/4}^{2\pi}\theta d\theta\int_{r=1}^{r=2}rdr = \frac{1}{4}(\frac{\pi^2}{16}\cdot 3)+\frac{1}{4}(\frac{25\pi^2}{16}-\frac{9\pi^2}{16})\cdot 3+ \frac{1}{4}(4\pi^2-\frac{49\pi^2}{16})\cdot 3= \frac{3\pi^2}{64}+\frac{48\pi^2}{64}+\frac{45\pi^2}{64}=\frac{96\pi^2}{64}=\frac{3\pi^2}{2}$.

Can someone double-check this?

Thanks a lot.

Answer 1

Your region $F$ can besplited into three regions, parametrized in polar coordinates as

$$F_1=\{(r,\theta) \in \mathbb{R}^2 / 1\le r \le 2, 0 \le \theta \le {\pi \over 4}\}$$

$$F_2=\{(r,\theta) \in \mathbb{R}^2 / 1\le r \le 2, {3 \pi \over 4} \le \theta \le {5 \pi \over 4}\}$$

$$F_3=\{(r,\theta) \in \mathbb{R}^2 / 1\le r \le 2, {7 \pi \over 4} \le \theta \le {2 \pi }\}$$

Then, your integral $I=I_1+I_2+I_3 $ where

$$I_1=\int_{0}^{\pi \over 4} \int_1^2 r \theta \ dr \ d\theta = {3 \over 2} \int_{0}^{\pi \over 4} \theta d\theta ={3 \over 2} {\pi^2 \over 32}$$

$$I_2=\int_{3 \pi \over 4}^{5\pi \over 4} \int_1^2 r (\theta - \pi) \ dr \ d\theta = {3 \over 2} \int_{3 \pi \over 4}^{5 \pi \over 4} \theta - \pi d\theta = 0$$

And

$$I_3=\int_{7\pi \over 4}^{2\pi} \int_1^2 r (\theta -2\pi) \ dr \ d\theta = {3 \over 2} \int_{7\pi \over 4}^{2\pi} \theta -2\pi d\theta ={3 \over 2} {-\pi^2 \over 32}$$

Then, $I=0$, as you suggested

As another user has commented, we typically take the Principal Branch of the arctan, defined between ${-\pi \over 2}$ and ${\pi \over 2}$. See this link for more information about it. It may be the error you are looking for.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ \underbrace{\arctan\pars{x \over y} \bracks{1 < x^{2} + y^{2} < 4}\bracks{\rule{0pt}{5mm}\verts{x} < \verts{y}}} _{\ds{\mbox{An}\ \color{red}{odd}\ \mbox{function of}\ x}} \dd x\,\dd y}} = \bbx{0} \end{align}

where $\ds{\arctan: \mathbb{R} \to \pars{-\,{\pi \over 2},{\pi \over 2}}}$.