2

I happened to notice that the surface $$ x = \sin(u-v), y = \sin(v), z = \sin(-u) $$ or equivalently (if I haven't blundered) $$ x^4 + y^4 + z^4 - 2 x^2 y^2 - 2 x^2 z^2 - 2 y^2 z^2 + 4 x^2 y^2 z^2 = 0 $$ resembles the octahemioctahedron in the same way Steiner's Roman surface resembles the tetrahemihexahedron.

Has it a name?

Anton Sherwood
  • 628
  • The plane $z=0$ (for example) intersects your parametrically given surface at a segment of the line pair $(x=\pm y; z=0)$, but meets the algebraic surface along the whole of the line pair $(x=\pm y; z=0)$. – John Bentin Aug 21 '15 at 21:43
  • The parametric surface is of course bounded. The real algebraic variety is, I think, the union of that and six lines (the ones @JohnBentin described and those obtained from them by permuting $x,y,z$) – Robert Israel Aug 22 '15 at 00:05
  • Thank you. Can the algebraic equation be amended to avoid that flaw? – Anton Sherwood Aug 22 '15 at 04:04

1 Answers1

0

Mathworld calls this the sine surface.

sine surface

Apocheir
  • 232