Integrated volume of sliced cylinders (polar coordinates) There is again something that I just can't figure out.
I'm trying to integrate a volume for two cylinders one of which is inside the other, but not with same center point.
The $r=cos(\theta)$ has central point in $(x,y)=(\frac{1}{2},0)$ and radius for it is $r=\frac{1}{2}$. The volume I'm after is bounded by following:
$$r = cos(\theta)$$
$$r = 2 cos(\theta)$$
$$z=x+2$$
$$z=0$$
Which means that there is one larger sliced cylinder with a centerpoint at $(x,y)=(1,0)$ with $r=1$ and smaller at $(x,y)=(\frac{1}{2},0)$ with $r=\frac{1}{2}$.
Therefore (just to make it easier to understand/graph to me) in Cartesian coordinates the bottom circles are:
$$A: (x-(\frac{1}{2})^2)+y^2=(\frac{1}{2})^2$$
$$B: (x-1^2)+y^2=1^2$$
And the height is limited by (surfaces):
$$0\leq z \leq x+2$$
Now I see that the smaller sliced cylinder is within the other and therefore I proceed to calculate first the volume for the larger one, and then just substract the smaller volume from it.
$$V_A=\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{r*cos(\theta)+2} r \,dz\,dr\,d\theta = 2\pi$$
$$V_B=\int_{0}^{2\pi} \int_{0}^{\frac{1}{2}} \int_{0}^{r*cos(\theta)+2} r \,dz\,dr\,d\theta = \frac{\pi}{2}$$
$$V_A-V_B = \frac{3\pi}{2}=\frac{12\pi}{8}$$
So far everything is quite straightforward, but then I try to check my calculation against the sliced cylinder volume equation $V=\pi r^2h=\pi r^2\frac{h_1+h_2}{2}$
Now the h for the larger one is $\frac{2+4}{2}=3$ and for smaller one $\frac{2+3}{2}=\frac{5}{2}$, which makes the volumes to be:
$$V_A= \pi r^2\frac{h_1+h_2}{2}=\pi 1^2\frac{2+4}{2} = 3\pi$$
$$V_B= \pi r^2\frac{h_1+h_2}{2}=\pi (\frac{1}{2})^2\frac{2+3}{2} = \frac{5\pi}{8}$$
$$V_A-V_B = \frac{24\pi}{8}-\frac{5\pi}{8}=\frac{19\pi}{8}$$
And as you can see, the volumes I get, do not match. I've been now going through the reasoning and all the calculations for a few hours and I just can't find any mistakes. Could someone help me a bit.
 A: Direct integration
The volume is described by $$\{(r,\theta,z)\in\Bbb R^3: \cos\theta\le r\le 2\cos\theta, -\frac\pi 2\le\theta\le\frac\pi 2, 0\le z\le r\cos\theta+2\}$$
and then 
$$
\begin{align}
V&=\int_{-\pi/2}^{\pi/2}\int_{\cos\theta}^{2\cos\theta}\int_0^{2+r\cos\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta\\
&=\int_{-\pi/2}^{\pi/2}\int_{\cos\theta}^{2\cos\theta}(2r+r^2\cos\theta)\,\mathrm dr\,\mathrm d\theta\\
&=\int_{-\pi/2}^{\pi/2}\left(3\cos^2\theta+\frac73\cos^4\theta\right)\,\mathrm d\theta\\
&=\frac{19}{8}\pi
\end{align}
$$
Another approach to confirm the result
The circle $r=\cos\theta$ is 
$$
\gamma_1=\{(x,y)\in\Bbb R^2:(x-1/2)^2+y^2=1/4\}
$$
and putting $x-\frac12=\rho\cos\phi$ and $y=\rho\sin\phi$ for $\phi\in[0,2\pi]$ and $\rho\in\left[0,\frac12\right]$
so the volume of the Cylinder $\Gamma_1$ sliced by $z=x+2$ and basis $\gamma_1$ is
$$
\begin{align}
V_1&=\int_0^{2\pi}\int_0^\frac12\int_0^{\rho\cos\phi+\frac52}\rho\,\mathrm dz\,\mathrm d\rho\,\mathrm d\phi\\
&=\int_0^{2\pi}\int_0^\frac12 \left(\rho^2\cos\phi+\frac52\rho\right)\,\mathrm d\rho\,\mathrm d\phi\\
&=\int_0^{2\pi} \left(\frac{1}{24}\cos\phi+\frac{5}{16}\right)\,\mathrm d\phi=\frac58\pi
\end{align}
$$
The circle $r=2\cos\theta$ is 
$$
\gamma_2=\{(x,y)\in\Bbb R^2:(x-1)^2+y^2=1\}
$$
and putting $x-1=\varrho\cos\varphi$ and $y=\varrho\sin\varphi$ for $\varphi\in[0,2\pi]$ and $\varrho\in\left[0,1\right]$
so the volume of the Cylinder $\Gamma_2$ sliced by $z=x+2$ and basis $\gamma_2$ is
$$
\begin{align}
V_2&=\int_0^{2\pi}\int_0^1\int_0^{\varrho\cos\varphi+3}\varrho\,\mathrm dz\,\mathrm d\varrho\,\mathrm d\varphi\\
&=\int_0^{2\pi}\int_0^1(\varrho^2\cos\varphi+3\varrho)\,\mathrm d\varrho\,\mathrm d\varphi\\
&=\int_0^{2\pi} \left(\frac{1}{3}\cos\varphi+\frac32\right)\,\mathrm d\varphi=3\pi
\end{align}
$$
Finally the volume required is
$$
V=V_2-V_1=3\pi-\frac58\pi=\frac{19}{8}\pi
$$
A: I see several troubles in your solution.


*

*As I understand $r=\cos\theta$ and $r=2\cos\theta$ are supposed to describe bases of the cylinders, but in the integral you take the bounds for $\theta$ as $[0,2\pi]$. It is wrong. To begin with the variable $r$ cannot be negative, however, $\cos\theta$ is negative on a part of $[0,2\pi]$. The correct bounds are e.g. $[-\pi/2,\pi/2]$.

*The bounds for $r$ are wrong as well. For example, the larger cylinder has bounds $0\le r\le2\cos\theta$, not $[0,1]$, because it is not centered at the origin.


With this in mind, we get
$$
V_A=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}\int_0^{r\cos\theta+2}r\,dzdrd\theta=[...]=3\pi.
$$
Similarly for the smaller cylinder.
P.S. After you have understood the cartesian coordinates for the cylinders, it seems easier to calculate the volumes in $x,y,z$ as
$$
V_A=\iint_A\int_0^{x+2}1\,dz\,dxdy
$$
introducing the shifted polar coordinates for $A$
\begin{align}
x&=1+r\cos\phi,\qquad && 0\le r\le 1,\\
y&=r\sin\phi, && 0\le\phi\le 2\pi.
\end{align}
