Polar integrals with trigonometric Problem The prompt is to evaluate the double integrals using polar system. The given constraints are: $$ \iint_D (x^2 + y^2) \, dx \, dy $$ 
$$ D = \{(x,y): x \ge 0, y \ge x, x^2 + y^2 \le 2y \}$$
Upon graphing we have a circle placed at the origin of 1 on the y-axis and line cutting through the origin and the circle. According to the constraints the the $\theta$ and $r$ values are $ \{\pi/2 \le \theta \le \pi/4 \}$ and $r$ has been derived as:
$$ x^2+y^2-2y \le 0$$
$$r^2 = 2\sin \theta $$ $$ r = 2\sin\theta$$
$$ 0 \le r \le 2\sin\theta$$
The integral we get is $\int_{\pi/4}^{\pi/2} \int_0^{2\sin \theta} (r)^3 \, dr \, d\theta$
After evaluating a couple of steps I get stuck here, $\displaystyle \left. \int_{\pi/4}^{\pi/2} \frac {r^4}4 \right|_0^{2\sin\theta}$
I'm not really sure of the $r$ as well, any help and suggestions? How do I conclude this? I'm kinda new to this, please help.

The double integral set up should be: $$ \int_{\pi/4}^{\pi/2} \int_0^{2\sin \theta} r^3 \, dr \, d\theta= \int_{\pi/4}^{\pi/2} 4\sin^4\theta d\theta= \frac 1 4 \int_{\pi/4}^{\pi/2} (1-\cos(2\theta))^2 \, d\theta=\cdots.$$ Also use $\cos^2(2\theta) = \dfrac{1+\cos(4\theta)}{2}$ in expanding the integrand above.
