$\theta$ for Triple Integral above paraboloid $z = x^2 + y^2$ and below $z = 2y$ [Stewart P1011 15.8.37]

Solution 1:

1. The projection is a circle as you indicated.
2. The circle passing through the origin.
3. 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 $y\ge 0$.

1. The radial arm ranges from o to 1
2. The radial arm starts at $\theta=0$ on the positive x axis
3. The radial arm covers the disk at $\theta=\pi$ on the negative x axis.

The following figures represent the projection on xy-plane

Solution 2:

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.