I am having difficulty understanding how this practice question is supposed to be solved.
Here it is:
Use polar coordinates to find the exact value of the double integral, $\iint_D x$ $dA$ where $D$ is the region bounded above by the line $y = x$ and below by the circle $x^2 + (y-1)^2 = 1$. The polar equation for the circle is given by $r = 2\sin\theta$. The intersection points of the line and the circle are $(0,0)$ and $(1,1)$.
The provided solution uses this step:
let $x=r\cos\theta, y = r\sin\theta$
$\iint_D x dA = \int_0^{\pi/4}\int_0^{2\sin\theta} (r \cos\theta) r\,dr\,d\theta$
From here on it's pretty straightforward. I understand the outer limits (since $y=x$ forms an angle of $\pi/4$ with the $x$-axis) and why $x$ is replaced with $(r \cos\theta) r\,dr\,d\theta$. But why is $2\sin\theta$ the upper limit for this inner integral, and why is $0$ the lower limit, considering the diagram should look like this:
(Enclosed area $D$ shaded in diagram)
The inner integral has a constant $\theta$, say $\theta_0$. It is integrating along the ray $\theta = \theta_0$ (in polar coordinates), which starts at the origin and moves outwards at angle $\theta_0$ to the $x$-axis. As is also true in cartesian coordinates, it has to intergrate over all parts of the ray that are inside your area $D$.
The part of the ray inside $D$ starts already at the origin, so integration starts at $r=0$, as the solution says. The ray leaves $D$ when it hits he circle. As you were given in the problem statement, the polar equation of that circle is $r=2 \sin(\theta)$, so for this ray it ends at $r=2 \sin(\theta_0)$, which is the upper end of the ingetral. The only difference is that it uses simply $\theta$ instead of $\theta_0$.