Identify the surface, which is defined by the following parametric equations:
x=2cos(u)sin(v), y=2sin(u)sin(v), z=6cos(v)
How can you tell?
Hello, Ideasman!
We can try to eliminate the parameters and get it in terms of x, y, z.Identify 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)
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