Volume bounded by the three cylinders and the two tangent planes

Question

Three identical cylinders of radius $R$ meters, with $0 < R < 1$, are placed so that their axes form an equilateral triangle of side $2$ meters. Calculate the volume bounded by the three cylinders and the two tangent planes to the three cylinders.

Answer 1

If we consider $A$ as the origin of coordinates, the cylinder $C_1$, which has its axis $\overline{AB}$ on the $x$-axis, is defined as:

$C_1 = \displaystyle \left\{ (x, y, z) \in R^3 \ \ \left( y^2+z^2=R^2 \right) \wedge \left( 0 \leq x \leq 2\sqrt{3} \right) \right\}$

The parametric equations of $C_1$ are:

$C_1 \equiv \begin{cases} x &= t \\ y &= R\cos(\alpha) \\ z &= R\sin(\alpha) \\ \end{cases} \quad \left( 0 \leq t \leq 2\sqrt{3} \right) \wedge (0 \leq \alpha < 2\pi)$...(1)

The axis $\overline{AC}$ of the cylinder $C_2$ is contained in the line with equation $y= \displaystyle \tan \left( \frac{\pi}{3} \right) x= \sqrt{3}x$.

$C_2$ cylinder is the result of the rotation of $\displaystyle \frac{\pi}{3}$ radians around the $Z$ axis of the $C_1$ cylinder:

$C_2 \equiv \begin{cases} x &= \frac{t}{2}-\frac{\sqrt{3}}{2} R \cos(\alpha) \\ y &= \frac{\sqrt{3}t}{2}+\frac{R}{2}\cos(\alpha) \\ z &= R\sin(\alpha) \\ \end{cases} \quad \left( 0 \leq t \leq 2\sqrt{3} \right) \wedge (0 \leq \alpha < 2\pi)$...(2)

Clearing $\alpha$ from (2), we obtain that the implicit equation of $C_2$ is: $3x^2+y^2+4z^2-2\sqrt{3}xy-4R^2=0$.

The slice produced at $C_1$ by the plane $z=z$ turns out to be:

$y^2=R^2-z^2 \implies \left( y = \sqrt{R^2-z^2} \right) \vee \left( y = -\sqrt{R^2-z^2} \right)$

The slice produced at $C_2$ by the plane $z=z$ turns out to be:

$\begin{array}{ll} & 3x^2+y^2-2\sqrt{3}xy-4(R-z)^2=0 \implies \\ & \left( y = \sqrt{3}x+2(R-z) \right) \vee \left( y = \sqrt{3}x+2(z-R) \right) \end{array}$

Let $A_t(z)$ be the area of trapezoid $A^{\prime}B^{\prime}EF$ :

$\begin{array}{lll} A_t(z) &=& \displaystyle \frac{\lvert \overline{A^{\prime}B^{\prime}} \rvert + \lvert \overline{EF} \rvert}{2} \cdot y \\ &=& \displaystyle \frac{2}{2} \left( \sqrt{3}-\frac{\sqrt{R^2-z^2}+2(R-z)}{\sqrt{3}}+\sqrt{3}-\frac{2(R-z)}{\sqrt{3}} \right) \sqrt{R^2-z^2} \\ &=& \displaystyle \left( 2\sqrt{3} - \frac{\sqrt{R^2-z^2}+4R}{\sqrt{3}} \right) \sqrt{R^2-z^2} \end{array}$

Let $H$ be the triangular prism defined by the two tangent planes to the $3$ cylinders and with a triangular base $\triangle ABC$. The volume $V_{1t}$ of the part of cylinder $C_1$ between the intersections with $C_2$ y $C_3$ within the triangular prism is:

$ \begin{array}{lll} V_{1t} &=& \displaystyle \int_{-R}^R A_t(z) \mathrm{d}z \\ &=& \displaystyle \int_{-R}^R \left( 2\sqrt{3} - \frac{4R}{\sqrt{3}} \right) \sqrt{R^2-z^2} \mathrm{d}z - \int_{-R}^R \frac{\sqrt{R^2-z^2}}{\sqrt{3}} \sqrt{R^2-z^2} \mathrm{d}z \\ &=& \displaystyle \sqrt{3} \pi R^2 -\frac{2\sqrt{3}\pi R^3}{3} - \frac{4R^3}{3} = \sqrt{3} \pi R^2 - \frac{(2\sqrt{3}\pi+4)R^3}{3} \end{array}$

The area $A_p(z)$ of the parallelogram $ADA'E$ is:

$\begin{array}{lll} A_p(z) &=& \displaystyle \lvert \overline{AE} \rvert \cdot y = \left( \frac{\sqrt{R^2-z^2}+2(R-z)}{\sqrt{3}} \right) \sqrt{R^2-z^2} \\ &=& \displaystyle \frac{2\sqrt{3}}{3}(R-z) \sqrt{R^2-z^2} + \frac{\sqrt{3}}{3}(R^2-z^2) \end{array}$

The volume $V_{12}$ of the intersections of the cylinders $C_1$ and $C_2$ within the triangular prism $H$ is:

$\begin{array}{lll} V_{12} &=& \displaystyle \int_{-R}^R A_p(z) \mathrm{d}z \\ &=& \displaystyle \int_{-R}^R \left( \frac{2\sqrt{3}}{3}(R-z) \sqrt{R^2-z^2} + \frac{\sqrt{3}}{3}(R^2-z^2) \right) \mathrm{d}z \\ &=& \displaystyle \frac{\sqrt{3}R^3 \pi }{3} + \frac{4\sqrt{3}}{9} R^3 = \frac{\sqrt{3}(3\pi+4)}{9}R^3 \end{array}$

The volume $V_p$ of the triangular prism defined between the two tangent planes to the 3 cylinders with triangular base $\triangle ABC$ is:

$V_p = \displaystyle [ABC] \cdot 2R = \frac{\sqrt{3}}{4} \left( 2\sqrt{3} \right)^2 \cdot 2R = \displaystyle \frac{4 \cdot 3 \cdot 2 \cdot \sqrt{3}}{4}R = 6 \sqrt{3}R$

By symetry, the volume $V(R)$ we are looking for is:

$\begin{array}{lll} V(R) &=& Vp-3 \left( V_{12} + V_{1t} \right) \\ &=& \displaystyle 6 \sqrt{3}R - 3 \left( \frac{\sqrt{3}(3\pi+4)}{9}R^3 + \sqrt{3} \pi R^2 - \frac{(2\sqrt{3}\pi+4)R^3}{3} \right) \\ &=& 6 \sqrt{3}R +\sqrt{3}\pi(R-3)R^2-\frac{4}{3} \left( \sqrt{3} - 3 \right)R^3 \end{array}$