# 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.

I see several troubles in your solution.

1. 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. $[-\pi/2,\pi/2]$.
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}

• Thanks! Just before your message I actually realized that the of course the middle integral limits should be $[0,2cos(\theta)]$, but the outermost integral limits were still wrong and I didn't get the right answer. The outermost limits make me feel especially stupid as that one was even given in the text for the exercise I was doing... – juahan Oct 5 '17 at 12:26

## 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 $$V=V_2-V_1=3\pi-\frac58\pi=\frac{19}{8}\pi$$

• Great! This helps me to understand the other approach as well, thank you. – juahan Oct 5 '17 at 13:03
• @juahan I've added also direct integration. – alexjo Oct 5 '17 at 14:38