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?
 A: 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,
$$\begin{align}
V&=2\int^{\frac{\pi}{3}}_{0} \int^{1}_{0} (r-r^2) \ dr \ d\theta+2\int^{\frac{\pi}{2}}_{\frac{\pi}{3}} \int^{2\cos(\theta)}_{0} (r-r^2)  \ dr \ d\theta\\
&=\frac{4\pi}{9}+\frac{3\sqrt{3}}{2}-\frac{32}{9}\approx 0.439.
\end{align}$$
A: Your limits of integration aren't quite right. Since $r \leq z \leq 1$, it follows that $r \leq 1$. But your limits for $r$ exceed $1$. Here's a picture of the domain in the $(x, y)$-plane. It's the intersection of two overlapping circles, each of which is centered on the the other.

If you're going with cylindrical coordinates, you probably want to exploit symmetry and double the integral for $0 \leq \theta \leq \frac{\pi}{2}$. However, you'll have to break it into the sum of two integrals:
$$
0 \leq \theta \leq \frac{\pi}{3} 
\quad\text{and}\quad
\frac{\pi}{3} \leq \theta \leq \frac{\pi}{2} 
$$
A: Here is your domain:

You should be inside the red cylinder, above the cone, but below the teal plane. So start with $z$, then calculate $r$, then calculate the angle range:

*

*$z$ varies from $0$ to $1$

*$r$ varies between $0$ and $z$

*For $\theta$ we need to calculate intersections of the circle of radius $r$ with the off axis circle:
$$x^2+y^2=r^2\\(x-1)^2+y^2=1$$
Subtracting the second from the first and expanding the square, you get $$x=\frac{2^2}2$$
If you write $x=r\cos\theta$, you get $$\cos\theta=\frac r2$$Therefore the limits of $\theta$ are $\pm\arccos(r/2)$
