I have hard time finding the correct answer to this double integral.
$ \iint_{T}\ ydxdy$
where T is defined as $ T= \{(\rho,\theta)\in \mathbb R^2 : 0\le \theta \le 2\pi , \theta \le \rho \le \theta + 2\pi \}$
I pass to polar coordinates, which gives me $\sin\theta$ of which the integral is $\cos\theta$ that from $0$ to $2\pi$ gives the result equal to $0$.
Not sure if the result is correct.
Sorry if it's like a homework question, hope you understand. Thanks.
Hint: Using change of variables to polar coordinates: $$\begin{cases} x=\rho \cos (\theta) \\ y=\rho \sin (\theta)\end{cases}$$ the integral to calculate is: $$\iint_{T}\ ydxdy =\int_{0}^{2\pi}\int_{\theta}^{\theta+2\pi}\rho^2\sin(\theta)d\rho d\theta= \frac {1} {3}\int_{0}^{2\pi} ((\theta+2\pi)^3-\theta^3)\sin(\theta) d\theta$$
See if you can see why and try to solve it. I hope this helps!