Evaluating $\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz$, wrong book solution? 
Evaluate $$\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz,$$ where
  $$\Omega = \{(x, y, z) \in \mathbb R^3 \mid y \geq 0,\ z \geq 0,\ x^2 + y^2 \leq 4,\ (y - 1)^2 + z^2 \leq 1\}.$$

From the second and last conditions we get a semi-cylinder parallel to the $x$ axis, tangent to it and with radius $1$. The other two conditions give another cylinder, this time parallel to the $z$ axis. Therefore I choose to integrate as follows:
$$\iint_K\int_0^{\sqrt{1 - (y - 1)^2}} z\,\mathrm dz\,\mathrm dx\mathrm dy,$$
where $K$ is the projection on the $xy$ plane of the intersection of the two cylinders. That is, the red region in this graph:

The integral becomes
$$\begin{align}
\iint_K\left[\frac12 z^2\right]_0^{\sqrt{2y - y^2}}\,\mathrm dx\mathrm dy &= \frac12\int_0^1\int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}}(2y - y^2)\mathrm dx\mathrm dy =\\
&= \int_0^1(2y - y^2)\sqrt{4 - y^2}\mathrm dy =\\
&= \frac{16}3 - 2\sqrt3 + \frac{\sqrt3}4 - \frac\pi3 =\\
&= \frac{16}3 - \frac{7\sqrt3}4 - \frac\pi3
\end{align}$$
However, the book's solution is just
$$\frac{16}3 - \pi$$
Is there an error in my solution or in the book?

I don't think you interpreted the integral correctly. These are the limits of integration I got.
\begin{align}
-2\leq &x\leq 2\\
0\leq &y\leq \sqrt{ 4-x^2}\\
0\leq &z\leq \sqrt{2y-y^2}
\end{align}
Then the integral is
\begin{gather}
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\int_0^{\sqrt{2y-y^2}}z\,dz\,dy\,dx\\
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\left.\frac{z^2}{2}\right|_0^{\sqrt{2y-y^2}}\,dy\,dx\\
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\frac{2y-y^2}{2}\,dy\,dx\\
\int_{-2}^2 \left.\frac{y^2}{2}-\frac{y^3}{6}\right|_0^{\sqrt{ 4-x^2}}\,dx\\
\int_{-2}^2 \frac{4-x^2}{2}-\frac{\left(4-x^2\right)^\frac{3}{2}}{6}\,dx\\
\int_{-2}^2 \left(4-x^2\right)\frac{3-\sqrt{4-x^2}}{6}\,dx\\
\frac{16}{3}-\pi
\end{gather}
I evaluated the last integral using wolfram alpha. Nevertheless, I believe your error stems from your interpretation of the volume. The cylinders intersect so that you can find the boundaries of $z$ and $y$ directly from the inequalities already given to you. $K$ in your solution should be the entire semicircle above the $x$ axis. If you change your $y$ integration limits to be from $0$ to $2$ you get the correct answer.
