**I am trying to solve this problem. I would appreciate if someone can check if my approach is correct.
Problem:
D is the plane region defined by: $$x^2+y^2-2x \le0\ and\ y\le |x| $$
$a)\ Calculate\ the\ Area\ D$ $(b)\ Evaluate$ $$\iint_D \frac{|y|}{x^2+y^2} \,\mathrm dxdy$$
SOLUTION $$Intersection\ of\ the\ two\ functions\ gives\ Points\ (0,0)\ and\ (1,1)$$ $$x^2+y^2-2x\le 0 => (x=1)^2 +y^2 \le1\ which \ is\ a\ circle\ with\ radius \ and\ centre\ (1,0)$$ $$\begin{align}\text{Area }A&=\iint_D \,\mathrm dA\qquad\qquad \mathrm dA=r\,\mathrm dr\,\mathrm d\theta\\&=\int_0^\frac{3\pi}4\int_0^{2cos\theta}r\,\mathrm dr\,\mathrm d\theta\\&=\int_0^\frac{3\pi}4\left[\frac{r^2}2\right]_0^{2cos\theta}\,\mathrm d\theta\\&=\left[\frac{3\pi -2}4\right]\end{align}$$
(b) using the limits $\frac{-\pi}{2} to \frac{\pi}{4} $ $$\iint_D \frac{|y|}{x^2+y^2} \,\mathrm dxdy =\int_\frac{-\pi}2^\frac{\pi}4\int_0^{2cos\theta}\frac{rsin\theta}{r^2}r \,\mathrm drd\theta\ $$
$$\iint_D \frac{|y|}{x^2+y^2} \,\mathrm dxdy =\int_\frac{-\pi}2^\frac{\pi}4\int_0^{2cos\theta}\sin\theta \,\mathrm drd\theta\ $$
$$\iint_D \frac{|y|}{x^2+y^2} \,\mathrm dxdy =-\frac{1}{2} $$
Working to check the solution:
$$\frac34 \cdot \pi \cdot 1^2 + \frac12 \cdot 1^2 $$
should be the correct answer as that is the area of $\frac34$ of the circle plus the area of a triangle.
Your mistake is the range of the angle:
The angle should be from $-\frac{\pi}2$ to $\frac{\pi}4$,
\begin{align}&\int_{-\frac{\pi}2}^\frac{\pi}4 2 \cos^2 \theta \, d\theta \\ &=\int_{-\frac{\pi}2}^\frac{\pi}4 \cos (2\theta)+1 \, d\theta \\ &= \frac{ \sin (2\theta)}{2}+\theta \mid_{-\frac{\pi}2}^\frac{\pi}4\\ &=\frac{1}{2}+\frac{3\pi}{4}\end{align}
Edit:
\begin{align} \int_{-\frac{\pi}2}^{\frac{\pi}4} \int_0^{2\cos\theta} \frac{r|\sin \theta|}{r^2} \cdot r\, \, dr d\theta &= \int_{-\frac{\pi}2}^{\frac{\pi}4} \int_0^{2\cos\theta} |\sin \theta|\, \, dr d\theta \\ &= \int_{-\frac{\pi}2}^{\frac{\pi}4} 2\cos\theta|\sin \theta|\, d\theta \\ &= \int_{-\frac{\pi}2}^{0} 2\cos\theta|\sin \theta|\, d\theta +\int_{0}^{\frac{\pi}4} 2\cos\theta|\sin \theta|\, d\theta \\ &= -\int_{-\frac{\pi}2}^{0} 2\cos\theta\sin \theta\, d\theta +\int_{0}^{\frac{\pi}4} 2\cos\theta\sin \theta\, d\theta \\ \end{align}