My question is if I am doing the following change of coordinates correctly, since I am not getting the right answer.
The exercise is to find the line integral of $\int_{\gamma}{y^{3}dx+x^{3}dy}$, where $\gamma$ is the curve of the ellips segment $x^{2}+y^{2}/4=1$ in the second, third and fourth quadrant, from (0,2) to (1,0).
$ I=\int_{\partial D}{y^{3}dx+x^{3}dy}$, where D is the area of the quarter ellips bounded by $x^{2}+y^{2}/4=1$ and the x- and y-axis. Now, Green's theorem gives us $I=\iint_{D}{(3x^{2}-3y^{2})dxdy}\implies\{x=r\cos{\theta},y=2r\sin{\theta}\}\implies \\3\iint_{D}{((r\cos{\theta})^{2}-(2r\sin{}\theta)^{2})2rdrd\theta}=\\6\iint_{D}{(\cos^{2}{\theta}-4\sin^{2}{\theta})r^{3}drd\theta}=6\int_{\pi/2}^{\pi}{d\theta}\int_{0}^{1}{(\cos^{2}{\theta}-4\sin^{2}{\theta})r^{3}dr}=\\6/4\int_{\pi/2}^{\pi}(1-5/2(1-\cos{2\theta))}{d\theta}=-9\pi/8.$
The book arrives at the answer $-27\pi/8$, a factor of 3 of my answer.
I checked the Jacob determinent of the variable change. I won't write out the calculation here, but I got the answer $J=2r$, so when we go from $(x,y)\rightarrow(r,\theta)$ we get $dxdy=2rdrd\theta$.