Cylindrical limits of integration for a particular triple integral In cylindrical coordinates, what would be the limits of integration for the triple integral serving to find the volume of the region in $\mathbb R^3$ bounded by:
$x^2 + y^2 = y$ and the sphere of center $O$ and radius $1$, in the first octant?
$\theta$ moves between $0$ and $\pi/2$, and $z$ moves between $0$ and $\sqrt{1 - r^2}$; but what is the behaviour of $r$?
Thanks a lot.
 A: $x^2+y^2 = y$ describes an infinite cylinder $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$
The interior of intersection of two regions is thus given by
$$
   \mathcal{R} = \left\{  x^2 + y^2 + z^2 \leqslant 1, x^2 + \left(y - \frac{1}{2}\right)^2 \leqslant \left(\frac{1}{2}\right)^2, x \geqslant 0, y \geqslant 0, z \geqslant 0 \right\}
$$
which is 
$$
     \mathcal{R} = \left\{  x^2 + y^2 + z^2 \leqslant 1, x^2 + y^2 \leqslant y, x \geqslant 0, y \geqslant 0, z \geqslant 0\right\}
$$
Switching to polar coordinates $x = r \cos(\theta)$, $x = r \sin(\theta)$ we have
$$\begin{eqnarray}
   \mathcal{R} &=& \left\{ r^2 + z^2 \leqslant 1, r \leqslant \sin(\theta), 0 \leqslant \theta \leqslant \frac{\pi}{2}, r \geqslant 0, z \geqslant 0\right\} \\
  &=& \left\{0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r  \leqslant \sin(\theta), 0 \leqslant z \leqslant \sqrt{1-r^2}  \right\}
\end{eqnarray}
$$
Therefore the volume
$$\begin{eqnarray}
   \operatorname{Vol}\left(\mathcal{R}\right) &=& \int_0^{\pi/2} \mathrm{d} \theta \int_0^{\sin \theta} r \mathrm{d} r \int_0^{\sqrt{1-r^2}} \mathrm{d} z \\ 
 &=& \int_0^{\pi/2} \mathrm{d} \theta \int_0^{\sin \theta} r \sqrt{1-r^2} \mathrm{d} r  \\ 
  &=& \int_0^{\pi/2} \frac{1}{3} \left(1-\cos^3(\theta)\right) \mathrm{d} \theta = \frac{\pi}{6} - \frac{2}{9} \approx 0.301377
\end{eqnarray}
$$
Confirming in Mathematica:
In[1]:= NIntegrate[1, 
 Element[xvec, 
  RegionIntersection[Cuboid[{0, 0, 0}, {1, 1, 1}], Ball[{0, 0, 0}], 
   Cylinder[{{0, 1/2, 0}, {0, 1/2, 1}}, 1/2]]]]

Out[1]= 0.301377

