Finding the volume with triple integrals I want to find the volume of a function described by:
$$ G= \{(x,y,z)|\sqrt{x^2+y^2} \le z \le 1, (x-1)^2+y^2 \le 1\}$$
This question can be best solved in cylindrical coordinates. So if I follow that process, I get the following limits:
$$ r \le z \le 1$$
$$ 0 \le r \le 2\cos(\theta)$$
$$ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$
I am fairly certain that these limits are correct. So continuing, to find the volume of G:
$$\begin{align}
\iiint r \ dz\ dr\ d\theta
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \int^{2\cos(\theta)}_{0} \int^{1}_{r} r \ dz \ dr \ d\theta\\
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \int^{2\cos(\theta)}_{0} r-r^2  \ dr \ d\theta\\
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{(2\cos(\theta))^2}{2}-\frac{(2\cos(\theta))^3}{3}   \ d\theta\\
&= \pi - \frac{32}{9}
\end{align}$$
Solving the above integral I get a negative answer which does not make sense considering the physical quantity is volume, which has to be positive. Where am I going wrong?

Your limits are not completely correct. Note that it should be
$$0 \le r \le 2\cos(\theta)\quad\textbf{and}\quad 0 \le r \leq 1$$
that is the upper limit is $1$ for $|\theta| \le \frac{\pi}{3}$ and it is $2\cos(\theta)$ for $\frac{\pi}{3}\leq|\theta| \le \frac{\pi}{2}$.
Therefore, by symmetry,
$$\begin{align}
V&=2\int^{\frac{\pi}{3}}_{0} \int^{1}_{0} (r-r^2) \ dr \ d\theta+2\int^{\frac{\pi}{2}}_{\frac{\pi}{3}} \int^{2\cos(\theta)}_{0} (r-r^2)  \ dr \ d\theta\\
&=\frac{4\pi}{9}+\frac{3\sqrt{3}}{2}-\frac{32}{9}\approx 0.439.
\end{align}$$
