Doubt about a double integral $\int_1^2\rho d\rho\int_\frac{-\pi}{4}^{\frac{\pi}{4}}\theta d\theta$

Question

I'm trying to integrate $$\iint_\Omega \arctan\left(\frac{y}{x}\right) dxdy$$

Where $\Omega=\{(x,y)\in\mathbb{R}^2 : 1<x^2+y^2<4,|y|<|x|\}$

Using polar coordinates we have that

${\Omega'=\{(\rho\cos\theta,\rho\sin\theta)\in\mathbb{R}^2 : 1<\rho<2,-\frac{\pi}{4}<\theta<\frac{\pi}{4}}\}$

So the integral becomes

$$\int_1^2\rho d\rho\int_\frac{-\pi}{4}^{\frac{\pi}{4}}\theta d\theta$$

Now, I know that if a function is odd and it is integrated on a symmetric integral the integral is zero. And so I would say that the integral is zero.

But on my textbook it is $\frac{5}{4}$.

Am I wrong (if yes, where?) or is there a typo on my textbook?

Answer 1

Your set $ \Omega'$ is not correct !

Correct is

$\Omega'=\{(\rho\cos\theta,\rho\sin\theta)\in\mathbb{R}^2 : 1<\rho<2,-\frac{\pi}{4}<\theta<\frac{\pi}{4}\} \cup \{(\rho\cos\theta,\rho\sin\theta)\in\mathbb{R}^2 : 1<\rho<2,-\frac{-3 \pi}{4}<\theta<\frac{3 \pi}{4}\} $.