Suppose $C$ is a closed curve parametrised by $x=\cos(2t)$ and $y=t\sin(t)$.
Using Stokes' Theorem, I calculate the result is $0$. But the answer is $\frac{32}9$. What is going wrong? Here is my caculation: $ \int_C x dy =\int_{0}^{2\pi} cos(2t)(tcos(t)+sin(t)) dt =0$
The problem is that the curve with $\theta=0\rightarrow\theta=\pi$ in clock-wised and with $\theta=\pi\rightarrow\theta=2\pi$ is anti clock-wised. So if you do all the integration anti clock-wised as you should, you get
$$\iint_{\rm D}{\rm d}x{\rm d}y=\int_{\partial\rm D}x{\rm d}y=\\=-\int_{0}^{\pi}\cos\left(2t\right)\left(t\cos t+\sin t\right){\rm d}t+\int_{\pi}^{2\pi}\cos\left(2t\right)\left(t\cos t+\sin t\right){\rm d}t=\frac{32}{9}$$
as needed.