Post #3179062

@tao@mathstodon.xyz Not really my field at all but I tried to take a crack at this one: If we drop the convexity requirement but keep genus-0 and bounded curvatures ≤ 1, can we achieve volume smaller than the unit ball (4π/3) starting from a small volume genus 2 bowl, then cap the two handles with identical “corks” that are solids of revolution. If smoothing preserves |\kappa_i|\le 1, the result is a genus 0 surface with total volume below 4\pi/3. Cork seed in the plane. Take the equilateral triangle with vertices (0,0),(2,0),(1,\sqrt3). Remove three unit radius sectors of angle \theta=\pi/3. Call the remainder \mathcal H. Rotate \mathcal H about the x axis to obtain the cork \mathcal C. Volume by first moment. V(\mathcal C)=2\pi M_x(\mathcal H) with M_x=\iint y\,dA. Triangle: \operatorname{Area}=\sqrt3, centroid =\sqrt3/3, so M_x(T)=1. Two bottom sectors contribute 2\cdot(\pi/6)\cdot(1/\pi)=1/3. Top sector contributes (\pi/6)(\sqrt3-2/\pi)=\pi\sqrt3/6-1/3. Hence M_x(\mathcal H)=1-\pi\sqrt3/6 and V_{\text{cork}}=2\pi\Bigl(1-\frac{\pi\sqrt3}{6}\Bigr)=\frac{\pi}{3}\,(6-\sqrt3\,\pi)\approx 0.585. Patchwise curvature before smoothing. For a surface of revolution with meridian r(x), \kappa_{\mathrm{mer}}=\frac{|r’’|}{(1+r’^2)^{3/2}},\qquad \kappa_{\mathrm{par}}=\frac{1}{r\sqrt{1+r’^2}}. Spherical patches give \kappa_1=\kappa_2=1. Conical patches with slope \tan 60^\circ=\sqrt3 give \kappa_{\mathrm{mer}}=0 and \kappa_{\mathrm{par}}=1/(2r)\le 1/\sqrt3 since truncation enforces r\ge \sqrt3/2. Reparametrize the meridian by arclength s with angle \alpha(s): x_s=\cos\alpha, r_s=\sin\alpha, \kappa_{\mathrm{mer}}=|\alpha’|, \kappa_{\mathrm{par}}=\cos\alpha/r. Define F=r-\cos\alpha. Then F’=\sin\alpha\,(1+\alpha’).