The statement is: Use Green's Theorem to compute the value of the line integral $\int_{\gamma}y\,dx + x^2\,dy$, where $\gamma$ is the circle given by $g(t) = (\cos t, \sin t), 0 \leq t \leq 2\pi$
By Green's theorem,
$$\int_\gamma y\,dx + x^2\,dy = \iint\limits_{\text{D}} 2x-1 \,dA$$
I made a change to polar coordinates, $x= r \cos \theta; y=r \sin \theta $ with $r∈[0,1]$ and $\theta ∈ [0,2 \pi]$
\begin{align}
\int_0^{2\pi} \int_0^1 (2r \cos \theta-1) \, dr \, d\theta = & \int_0^{2\pi} \Big[r^2 \cos \theta-r\Big]_0^1 \, d\theta\\ = & \int_0^{2\pi} \Big[\cos \theta-1\Big]_0^1 \, d\theta\\ = & \Big[\sin \theta- \theta\Big]_0^1=-2 \pi\\
\end{align}
But the textbook says the answer is $- \pi$
Best Answer
After performing change to polar coordinates, we should use $r\,dr\,d\theta$ on $dA$, not just $dr\,d\theta$. Hence the integral is $$I=\int_0^{2\pi} \int_0^1 (2r^2\cos\theta-r) \, dr \, d\theta = \int_0^{2\pi} \bigg(\frac23\cos\theta-\frac12\bigg)\,d\theta=-\pi.$$