Volume enclosed by the surface of revolution of a Reuleaux triangle

Shown below is a Reuleaux triangle with the $x$-axis as one of its symmetry axes and the origin as one of its corners.

Let points $A$, $B$, and $C$ be the corners of the Reuleaux triangle with width $w=1$ shown below. These are also the corners of the inscribed equilateral triangle whose side length is $w=1$. It can be seen that $A=(0,0)$, $B=\left(\frac{\sqrt{3}}{2},-\frac12\right)$, and $C=\left(\frac{\sqrt{3}}{2},\frac12\right)$.

A circle centered at $(h,k)$ has the equation $(x-h)^2+(y-k)^2=r^2$. From this, we find the equation of the curve above the $x$-axis

\[f(x)=\left\{\begin{array}{ll} \sqrt{-x^2+\sqrt{3}x+\frac14}-\frac12&,0\le x\le\frac{\sqrt{3}}{2}\\ \sqrt{1-x^2}&,\frac{\sqrt{3}}{2}\le x\le 1\\ \end{array}\right.\]

The solid is bounded by the planes $x=0$ and $x=1$ and the curve $y=f(x)$ revolved around the x-axis. Its volume is

\begin{align*} V &=\int_0^1\pi\left(f(x)\right)^2\mathrm{d}x=\int_0^\frac{\sqrt{3}}{2}\pi\left(f(x)\right)^2\mathrm{d}x+\int_\frac{\sqrt{3}}{2}^1\pi\left(f(x)\right)^2\mathrm{d}x\\ &=\int_0^\frac{\sqrt{3}}{2}\pi\left(\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}-{\textstyle\frac12}\right)^2\!\mathrm{d}x+\int_\frac{\sqrt{3}}{2}^1\pi\left(\sqrt{1-x^2}\right)^2\!\mathrm{d}x\\ &=V_1+V_2 \end{align*}

The second integral is \[V_2=\int_\frac{\sqrt{3}}{2}^1\pi\left(1-x^2\right)\mathrm{d}x=\pi\left[x-{\textstyle\frac13}x^3\right]_\frac{\sqrt{3}}{2}^1=\pi\left({\textstyle\frac23-\frac{3\sqrt{3}}{8}}\right)\]

The first integral is \[V_1=\int_0^\frac{\sqrt{3}}{2}\pi\left(-x^2+\sqrt{3}x+{\textstyle\frac14}-\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}+{\textstyle\frac14}\right)\mathrm{d}x\]

To find $\displaystyle\int\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}\,\mathrm{d}x$, let $u=x-\frac{\sqrt{3}}{2}$. Thus, $\mathrm{d}u=\mathrm{d}x$. \[\int\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}\,\mathrm{d}x=\int\sqrt{1-\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)^2}\,\mathrm{d}x=\int\sqrt{1-u^2}\,\mathrm{d}u\]

To find $\int\sqrt{1-x^2}\,\mathrm{d}x$, use integration by parts.

$\int f(x)\frac{\mathrm{d}g}{\mathrm{d}x}\,\mathrm{d}x=f(x)g(x)-\int g(x)\frac{\mathrm{d}f}{\mathrm{d}x}\,\mathrm{d}x$

Choose $f(x)=\sqrt{1-x^2}$ and $\frac{\mathrm{d}g}{\mathrm{d}x}=1$. Thus, $\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{-x}{\sqrt{1-x^2}}$ and $g(x)=x$.

\begin{align*} \int\sqrt{1-x^2}(1)\,\mathrm{d}x &=\sqrt{1-x^2}(x)-\int(x)\frac{-x}{\sqrt{1-x^2}}\,\mathrm{d}x\\ \int\sqrt{1-x^2}\,\mathrm{d}x &=x\sqrt{1-x^2}-\int\frac{-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}-\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}-\int\sqrt{1-x^2}\mathrm{d}x+\arcsin(x)+C_1 \end{align*}\begin{align*} 2\int\sqrt{1-x^2}\,\mathrm{d}x &=x\sqrt{1-x^2}+\arcsin(x)+C_1\\ \int\sqrt{1-x^2}\,\mathrm{d}x &={\textstyle\frac12}x\sqrt{1-x^2}+{\textstyle\frac12}\arcsin(x)+C \end{align*}

Thus, \[\int\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}\mathrm{d}x=\int\sqrt{1-\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)^2}\mathrm{d}x\] \[=\int\sqrt{1-u^2}\mathrm{d}u={\textstyle\frac12}u\sqrt{1-u^2}+{\textstyle\frac12}\arcsin(u)+C\] \[={\textstyle\frac12}\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)\sqrt{1-\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)^2}+{\textstyle\frac12}\arcsin\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)+C\] Now,

\begin{align*} V_1 &=\int_0^\frac{\sqrt{3}}{2}\pi\left(-x^2+\sqrt{3}x+{\textstyle\frac14}-\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}+{\textstyle\frac14}\right)\mathrm{d}x\\ &=\pi\left[-{\textstyle\frac13}x^3+{\textstyle\frac{\sqrt{3}}{2}}x^2+{\textstyle\frac14}x-\left({\textstyle\frac12}\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)\sqrt{1-\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)^2}\right.\right.\\ &\left.+{\textstyle\frac12}\arcsin\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)\Bigg)+{\textstyle\frac14}x\right]_0^\frac{\sqrt{3}}{2}=\pi\left({\textstyle\frac{\sqrt{3}}{2}}-\left({\textstyle\frac{\sqrt{3}}{8}}+{\textstyle\frac{\pi}{6}}\right)\right) \end{align*}

Finally, $V=V_1+V_2=\pi\left(\frac{3\sqrt{3}}{8}-\frac{\pi}{6}\right)+\pi\left(\frac23-\frac{3\sqrt{3}}{8}\right)=\pi\left(\frac23-\frac{\pi}{6}\right)$.

For a width $w$, the volume is $\color{lightgray}\boxed{\color{black}{\textstyle\pi\left(\frac23-\frac{\pi}{6}\right)w^3}}$.