Volume of the region lying inside circular and parabolic cylinders I have the following problem:
Find the volume of the region lying inside the circular cylinder $x^2+y^2=2y$ and inside the parabolic cylinder $z^2 = y$.
To solve the problem, I did the following:
$x^2+y^2=2y \implies x^2 +(y-1)^2=1$. So we have a circle centered at $(0,1)$.
I think writing volume $V=\displaystyle\int_{-1}^{1}\int_{-2}^2(?)dA$ is not useful at this point. We have $z^2 =y$ too so the lower bound for $y$ must be $0$.
So, how can I calculate the volume, i.e., write down a double integral? Thanks.
 A: Here's the region in question

First note that it's bounded above and below by the two halves of the parabolic cylinder, so we can write the $z$ limits as 
$$ -\sqrt{y} \le z \le \sqrt{y} $$
Now we need to put limits on the intersection curve, projected down to the $xy$-plane. This is just a circle centered at $(0,1)$ with radius $1$. You have two choices for the limits
$$ -\sqrt{1-(y-1)^2} \le x \le \sqrt{1-(y-1)^2}, \ 0 \le y \le 2 $$
or
$$ 1 - \sqrt{1-x^2} \le y \le 1 + \sqrt{1-x^2}, \ -1 \le x \le 1 $$
Putting it together, you'll get something like
$$ V = \int dV = \int_{-1}^{1} \int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}} \int_{-\sqrt{y}}^{\sqrt{y}} dz\ dy\ dx $$

If you know how to integrate in cylindrical coordinates, the problem could be potentially easier. Let
$$ x = r\cos \theta, \ y = r\sin \theta, \ z = z $$
Then the intersection circle has the form
$$ r = 2\sin \theta $$
And the integral looks like
$$ V = \int dV = \int_{0}^{\pi} \int_{0}^{2\sin\theta} \int_{-\sqrt{r\sin\theta}}^{\sqrt{r\sin\theta}} r \ dz \ dr \ d\theta $$
A: $z$ is bounded below by $-\sqrt y$ and above by $\sqrt y$.
$x^2 + (y - 1)^2 = 1 \implies y = 1 \pm \sqrt{1 - x^2}$
So $y$ is bounded at the "back" by $1 - \sqrt{1 - x^2}$ and at the "front" by $1 + \sqrt{1 - x^2}$.
$x$ is bounded to the left by $-1$ and to the right by $1$ (these are the minimum and maximum $x$-values that exist on the circle).
So$$V=\int_{-1}^{1} \int_{1 - \sqrt{1 - x^2}}^{1 + \sqrt{1 - x^2}} \int_{-\sqrt y}^{\sqrt y} 1 \, \mathrm dz \, \mathrm dy \, \mathrm dx$$
