Integrating double integral with spherical coordinates problem with interpret a domain Hello i have the following problem i am solving integral with spherical coordinates but i am getting wrong answer - i think i am integrating correct so i think the problem is coming from the constraints. So i have $$\iint_Dxdx$$ and i have $D = (x^2 + y^2 \le 2y,  x \ge 0,  y \ge \frac{1}{2})$. So i begin solving - first i complete the circle and came with $(x-0)^2 + (y-1)^2 = 1$ that is circle centered at $y = 1$ with $radius$ = 1 so i substitute $x^2 + y^2 = 2y$ with $r^2 = 2rsin\theta$ i divide both sides by r i get $r = 2sin\theta$ so i have limits from $0$ to $2sin\theta$ for r also i need to rotate $\phi$ from $0$ to $\pi$ because that is how the circle is placed. And i got $$\int_0^\pi \int_0^{2sin\theta} (rcos\theta)r \,d\varphi\,dr$$ and so after integrating $r$ i am left with $$\frac{1}{3}\int_o^\pi sin^3\theta cos\theta$$ i am using u substitution and say $u = sin\theta$ and $du = cos\theta$ so i have $$\frac{8}{3}\int_0^\pi u^3du$$ and the new limits are from $0$ to $0$ because $sin\theta$ from 0 is 0 and $sin\theta$ from $\pi$ is also $0$ that means the final answer is zero in and it should be $\frac {9}{16}$. I am not good at math so i cannot see where my error is. Thank you for any help in advance.
 A: The limits are $x^2+y^2\le 2y \ ,\ x\ge0 \ , y\ge\frac{1}{2} $
You've found the upper limit i.e, $r = 2\sin\theta$ (upper limit).
For $y\ge\frac{1}{2} \ , \ r\sin\theta \ge\frac{1}{2} $
So, the lower limit of $r$ is $\frac{1}{2\sin\theta}$
Also, $x\ge0$ , $r\cos\theta\ge0$ , $\cos\theta\ge0$ So, $\theta\le\pi/2$
At $y=\frac{1}{2}$, $\sin\theta = \frac{\pi}{6}$ [as, $1.\sin(\pi/6) =1/2$]
$$I = \int^{\pi/2}_{\pi/6}\int^{2\sin\theta}_{\frac{1}{2sin\theta}}r^2\cos\theta \ dr d\theta = \frac{1}{3}\int^{\pi/2}_{\pi/6}\bigg\{8\sin^3\theta - \frac{1}{8sin^3\theta}\bigg\}\cos\theta  d\theta$$
Now let $u = sin\theta$ , $du = \cos\theta d\theta$
At $\theta = \pi/6$, $u=1/2$ and at $\theta = \pi/2$, $u=1$
So, $$I = \frac{1}{3}\int^{1}_{1/2}\bigg\{8u^3- \frac{1}{8u^3}\bigg\}du = \frac{1}{3}\bigg[8\cdot\frac{1}{4}(1-\frac{1}{16}) - \frac{1}{8}\cdot\frac{-1}{2}\bigg(1 - \frac{1}{(1/2)^2}\bigg)\bigg] = \frac{1}{3}$$
$$I = \frac{1}{3}\big[ \frac{15}{8} - \frac{3}{16}\big] = \frac{1}{3}[\frac{27}{16}] = \frac{9}{16}$$

