Area of 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 derivative is

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

The area of the surface of revolution is $S=S_1+S_2$, where

\begin{align*} S_1 &=\int_0^{\frac{\sqrt{3}}{2}}2\pi f(x)\sqrt{1+\left({\textstyle\frac{\mathrm{d}}{\mathrm{d}x}f(x)}\right)^2}\mathrm{d}x\\ S_2 &=\int_{\frac{\sqrt{3}}{2}}^12\pi f(x)\sqrt{1+\left({\textstyle\frac{\mathrm{d}}{\mathrm{d}x}f(x)}\right)^2}\mathrm{d}x \end{align*}

The second integral is

\[S_2=\int_{\frac{\sqrt{3}}{2}}^12\pi\sqrt{1-x^2}\sqrt{1+\textstyle\left(\frac{-x}{\sqrt{1-x^2}}\right)^2}\mathrm{d}x\]

where

\[\sqrt{1+\textstyle\left(\frac{-x}{\sqrt{1-x^2}}\right)^2}=\sqrt{1+\textstyle\frac{x^2}{1-x^2}}=\sqrt{\textstyle\frac{1-x^2+x^2}{1-x^2}}=\textstyle\frac{1}{\sqrt{1-x^2}}\]

Thus,

\[S_2=\int_{\frac{\sqrt{3}}{2}}^12\pi\sqrt{1-x^2}{\textstyle\frac{1}{\sqrt{1-x^2}}}\mathrm{d}x=\int_{\frac{\sqrt{3}}{2}}^12\pi\mathrm{d}x=2\pi\Big[x\Big]_{\frac{\sqrt{3}}{2}}^1=\pi(2-\sqrt{3})\]

The first integral is \[S_1=\int_0^{\frac{\sqrt{3}}{2}}2\pi\left(\sqrt{-x^2+\sqrt{3}x+{\textstyle\frac14}}-{\textstyle\frac12}\right)\sqrt{1+\textstyle\left(\frac{-x+\frac{\sqrt{3}}{2}}{\sqrt{-x^2+\sqrt{3}x+\frac14}}\right)^2}\mathrm{d}x\]

Note that $-x^2+\sqrt{3}x+\frac14=1-\left(x-\frac{\sqrt{3}}{2}\right)^2$.

Let $u=x-\frac{\sqrt{3}}{2}$. Thus, $\mathrm{d}u=\mathrm{d}x$. If $x=0$, then $u=-\frac{\sqrt{3}}{2}$. If $x=\frac{\sqrt{3}}{2}$, then $u=0$.

\begin{align*} S_1 &=\int_0^{\frac{\sqrt{3}}{2}}2\pi\left(\sqrt{1-\left(x-{\textstyle\frac{\sqrt{3}}{2}}\right)^2}-{\textstyle\frac12}\right)\sqrt{1+\textstyle\left(\frac{-\left(x-\frac{\sqrt{3}}{2}\right)}{\sqrt{1-\left(x-\frac{\sqrt{3}}{2}\right)^2}}\right)^2}\mathrm{d}x\\ &=\int_{-\frac{\sqrt{3}}{2}}^02\pi\left(\sqrt{1-u^2}-{\textstyle\frac12}\right)\sqrt{1+\textstyle\left(\frac{-u}{\sqrt{1-u^2}}\right)^2}\mathrm{d}u \end{align*}\begin{align*} S_1 &=\int_{-\frac{\sqrt{3}}{2}}^0\textstyle 2\pi\left(\sqrt{1-u^2}-{\frac12}\right)\frac{1}{\sqrt{1-u^2}}\mathrm{d}u\\ &={\textstyle 2\pi}\int_{-\frac{\sqrt{3}}{2}}^0\sqrt{1-u^2}{\textstyle\frac{1}{\sqrt{1-u^2}}}\mathrm{d}u-\pi\int_{-\frac{\sqrt{3}}{2}}^0\frac{1}{\sqrt{1-u^2}}\mathrm{d}u\\ &=2\pi\Big[u\Big]_{-\frac{\sqrt{3}}{2}}^0-\pi\Big[\arcsin(u)\Big]_{-\frac{\sqrt{3}}{2}}^0=\pi(\sqrt{3}-{\textstyle\frac{\pi}{3}}) \end{align*}

Finally, $S=S_1+S_2=\pi(\sqrt{3}-{\textstyle\frac{\pi}{3}})+\pi(2-\sqrt{3})=\pi\left(2-\frac{\pi}{3}\right)$.

For a width $w$, the area is $\color{lightgray}\boxed{\color{black}{\pi\left(2-{\textstyle\frac{\pi}{3}}\right)w^2}}$.