Hello, Ideasman!

We canIdentify the surface, which is defined by the following parametric equations:

. . (1) x .= .2·cos(u)·sin(v)

. . (2) y .= .2·sin(u)·sin(v)

. . (3) z .= .6·cos(v)tryto eliminate the parameters and get it in terms of x, y, z.

Square (1): . x² .= .4·cos²(y)·sin²(v)

Square (2): . y² .= .4·sin²(u)·sin²(v)

Add: . x² + y² .= .4·sin²(v)[sin²(u) + cos²(u)]

. . . . .x² + y² .= .4·sin²(v) .(4)

From (3), we have: . cos(v) .= .z/6 . → . cos²(v) .= .z²/36

. . Then: .sin²(v) .= .1 - cos²(v) .= .1 - z²/36 .= .(36 - z²)/36

Substitute into (4): . x² + y² .= .4(36 - z²)/36

. . which simplifies to: . 9x² + 9y² + z² .= .36

. . . . . . . . . . . x² . .y² . . z²

And we have: . --- + --- + --- .= .1 . . . . an ellipsoid

. . . . . . . . . . . .4 . . 4 . . 36