Double integral on domain $x^2+y^2 \leq 1$ and $0 \leq x \leq 1/2$ I need to compute the following double integral. I am stuck. 


  Let $$\mathbb D = \left\{ (x,y)\in\mathbb R^2 : 0 \leq x \leq \frac12, x^2 + y^2 \leq 1 \right\}$$ Find the following integral $$I = \iint_{\mathbb D}\sqrt{x^2+y^2} \, \mathrm d x \, \mathrm d y$$
By symmetry, the integral equals $2$ times the integral of $r$ over the sector $S=\{(r,\theta),r\in[0,1], \theta\in[\pi/3,\pi/2]\}$ and the triangle $R=\{(r,\theta),r\in[0,1/(2\cos\theta)], \theta\in[0,\pi/3]\}$.
$$I=2\int_{\pi/3}^{\pi/2}\int_0^1r\cdot rdrd\theta + 2\int_{0}^{\pi/3}\int_0^{1/(2\cos\theta)}r\cdot rdrd\theta$$
By symmetry, the integral equals $2$ times the integral of $r$ over the sector $S=\{(r,\theta),r\in[0,1], \theta\in[\pi/3,\pi/2]\}$ and the triangle $R=\{(r,\theta),r\in[0,1/(2\cos\theta)], \theta\in[0,\pi/3]\}$.
$$I=2\int_{\pi/3}^{\pi/2}\int_0^1r\cdot rdrd\theta + 2\int_{0}^{\pi/3}\int_0^{1/(2\cos\theta)}r\cdot rdrd\theta$$