In an exercise, I tried to integrate a function over the area for the fourth quadrant of the unit circle $(3\pi/2 \rightarrow 2\pi)$. I am pretty confused about the answer.
I tried translating the original integral to polar coordinates as follows $$ \iint_D xy~dxdy = \int_{\frac{3\pi}{2}}^{2\pi}\int_0^1 r\cos\theta\sin\theta~drd\theta $$ but I get twice the area that I'm supposed to get ($-1/4$ instead of $-1/8$). I can't find anything that looks off with my integral. Am I doing this wrong, and what?
Also... wouldn't you rather have a positive area?
You have \begin{align*}\int\limits_{\theta=1.5\pi}^{2\pi}\ \int\limits_{r=0}^{1}r\cos\theta\sin\theta\text{d}r\text{d}\theta&=\frac{1}{2}\int\limits_{1.5\pi}^{2\pi}\sin(2\theta)\text{d}\theta\int\limits_{0}^{1}r\text{d}r \\&=-\frac{1}{8}\Big.\cos(2\theta)\Big\vert_{1.5\pi}^{2\pi}\cdot \Big.r^2\Big\vert_{0}^{1}=-\frac{1}{4}\end{align*}