Combine inverse-sine integration with two rotation radii

$\int\sin^{-1}(3x)\,dx=x\sin^{-1}(3x)-\int3x/\sqrt{1-9x^2}\,dx=x\sin^{-1}(3x)+\tfrac13\sqrt{1-9x^2}+C$. This real formula applies on the inverse-sine domain, with its derivative interpreted in the interior.

Solving $1-(\sqrt3/\pi)x^2=\tfrac13\cos x$ in the first quadrant gives $(\alpha,\beta)\approx(1.2810352,0.09524112)$. At height $y\in[\beta,1]$, the outer radius satisfies $x^2=(\pi/\sqrt3)(1-y)$. For $\beta\le y\le1/3$ there is an inner hole of radius $\cos^{-1}(3y)$; above $1/3$ there is no hole.

The source washer setup is

$\begin{aligned}&V\\ & =\pi\int_\beta^1\frac\pi{\sqrt3}(1-y)\,dy\\ &{}-\pi\int_\beta^{1/3}[\cos^{-1}(3y)]^2\,dy\approx1.75.\end{aligned}$

Units are cubic. As an independent check, cylindrical shells give $V=2\pi\int_0^\alpha x[1-(\sqrt3/\pi)x^2-\tfrac13\cos x]\,dx$, with the same value.