$\theta$ for Triple Integral above paraboloid $z = x^2 + y^2$ and below $z = 2y$ [Stewart P1011 15.8.37] 
$\bf\sf37.$  Evaluate $\iiint_E z\,dV,$ where $E$ lies above the paraboloid $z=x^2+y^2$ and below the plane $z=2y.$  In cylindrical coordinates the paraboloid is given by $z=r^2$ and the plane by $z=2r\sin\theta$ and they intersect in the circle $r=2\sin\theta$. Then $\iiint_E z\,dV=\int_0^\pi\int_0^{2r\sin\theta}rz\,dz\,dr\,d\theta=\tfrac{5\pi}6.$ [using a CAS]

The solution claims that the domain of integration $= \{ \, 0 \le \theta \le  \color{red}{  \pi }, 0 \le r \le 2\sin\theta, r^2 \le z \le 2r \sin\theta \, \} $. Why isn’t the upper bound on $\theta$: $\color{red}{ 2 \pi } $?
$1.$ I know $ x^2 + y^2 = 2y \iff r = 2\sin\theta$. But how does this become the upper bound on $r$? 
$2.$ Conforming to the solution, the projection (onto $z=0$) of the intersection of paraboloid and plane $= x^2 + y^2 = 2y \iff x^2 + (y - 1)^2 = 1$, a circle of radius 1 centered at $(0, 1)$.
Aren’t we concerned with this entire circle?  
I tried to plot this: https://mathematica.stackexchange.com/q/45059/9983
 A: *

*The projection is a circle as you indicated.  

*The circle passing through the origin.  

*The circle has center at $\color{red}{(0,1)}$ not $(0,0.5)$ and it has radius $\color{red}{1}$


So, the circle lies in the upper half of the xy-plane where $y\ge 0$.   


*

*The radial arm ranges from o to 1  

*The radial arm starts at $\theta=0$ on the positive x axis  

*The radial arm covers the disk at $\theta=\pi$ on the negative x axis.  


The following figures represent the projection on xy-plane 
 
A: Okay, so the first thing to take into account, is that the intersection is a circle, but it is not parallel with the $xy$ plane, it is instead at an angle, which is what leads to what I would call an unorthodox parametrisation.
If we consider $\{0\le\theta\le\pi,0\le r\le 2\sin\theta\}$, this parametrises a circle, (it is actually the circle presented in Semsem's answer), I have entered if here in wolfram: https://www.wolframalpha.com/input/?i=plot+r%3D2sin%28x%29+for+x+from+0+to+pi
The reason that $2\sin\theta$ is the upperbound, is that it represents the length of the line from the origin $(0,0)$ to the point $(\theta,r)$, (when you consider the image in my link), (in my link just think of $x$ as $\theta$).
So all in all this answers $(1)$ and $(2)$, its just not the standard way of parameterising a circle.
the maximum value for $\theta$ is $\pi$ because $0\le\theta\le\pi$, is all that is needed to produce the circle, if you had $0\le\theta\le 2\pi$, this would go round the circle twice.
