Using Cylindrical Coordinates to find volume of a solid How do I use Cylindrical Coordinates to find volume of a solid in the first Octant that is bounded by the cylinder $x^2 + y^2 = 2y$, the half cone $z = \sqrt{x^2 + y^2}$, and the $xy$-plane.
I have drawn the region of integration and obtained this:
$\int_0^2 \int_0^\sqrt{2y-y^2}\int_0^\sqrt{x^2 + y^2} dzdxdy$
Is this correct and from here were do I apply the cylindrical coordinates?

The question makes only sense if the solid is the volume outside the half cone.
Otherwise the volume is infinite, because (almost) half of the cylinder which extends
to $+\infty$ in $z$ is in it. So we have to integrate over $x^2+y^2\ge z^2$.
For constant $z$ the section through the solid
is the area inside the circle $x^2+y^2-2y=0$ defined by the cylinder, which is inside the circle $x^2+(y-1)^2=1$,
but outside the circle $z^2=x^2+y^2$ defined by the cone. And since we're supposed to look only at one octant,
$z\ge 0, x\ge 0, y\ge 0$. The intersection of the 2 circles vanishes if the radius of
the circle at (0,0) is too large, which means $z\le 2$.
$$
V=\int_0^2 dz \int\int_{x^2+y^2\ge z^2, x^2+(y-1)^2\le 1} dx dy
$$
[Note that the lower limit in the partial integral $\int_0^{\sqrt{2y-y^2}}dx$ of the OP is wrong.]
In cylindrical coordinates $r^2=x^2+y^2$, $x=r\cos\theta, y=r\sin\theta$
we find the limits of $\theta=[\theta_0,\pi/2]$ where the circle $x^2+y^2=r^2$ intersects
the circle $x^2+(y-1)^2=1$:
$$
r^2\cos^2\theta_0+(r\sin\theta_0-1)^2=1;
$$
$$
r^2-2r\sin\theta_0+1=1;
$$
$$
\to \theta_0=\arcsin (r/2);
$$
So with the usual Jacobi Determinant factor $r$ in cylindrical coordinates
$$
V=\int_0^2 dz \int_z^2 r dr \int_{\arcsin(r/2)}^{\pi/2}d\theta
$$
$$
=\int_0^2 dz \int_z^2 r dr (\pi/2 - \arcsin(r/2))
$$
Here $\int r\arcsin (r/2) dr= (r^2/2-1)\arcsin(r/2)+r\sqrt{1-r^2/4}/2$, so
$$
V=\int_0^2 dz \left[\frac{\pi}{2} \frac{r^2}{2}-\left((\frac{r^2}{2}-1)\arcsin\frac{r}{2}+\frac{r\sqrt{1-r^2/4}}{2}\right)\right]\mid_{r=z}^2
$$
$$
=\int_0^2 dz \left[\frac{\pi}{2} - \frac{\pi}{2} \frac{z^2}{2}+\left((\frac{z^2}{2}-1)\arcsin\frac{z}{2}+\frac{z\sqrt{1-z^2/4}}{2}\right)\right]
$$
and this is evaluated via
$$
\int (\frac{z^2}{2}-1)\arcsin(z/2) dz = \frac{z^3}{6}\arcsin\frac{z}{2}-z\arcsin\frac{z}{2}+\frac{z^2-10}{9}\sqrt{1-z^2/4}
$$
and
$$
\int \frac{z}{2}\sqrt{1-z^2/4} dz= -\frac{1}{12}(4-z^2)^{3/2}.
$$
$$
V=16/9.
$$
