Simple Surface of Revolution:

Let the line segment with end points P₁ [1 1 0] and P₂ [6 2 0] lying in the xy-plane. Rotating the line around the x-axis yields a conical surface. Find out the point on this surface at u = 0.5, φ = π/3.

The parametric equation for the line segment from P₁ to P₂ is following

               P(u) = [x(u)   y(u)  z(u)] = P₁ + (P₂ - P₁) u     0 < u  < 1

∴             x(u) = x₁ + (x₂ - x₁) u = 1 + 5u

              y(u) = y₁ + (y₂ - y₁) u = 1 + u

              z(u) = z₁ + (z₂ - z₁) u = 0

∴ The point Q (1/2, π/3) on the surface of revolution is

Q (1/2, π/3) = [1 + 5t  (1 + t) cos φ  (1 + t) sin φ]

= ?[7 /2   3/2  cos ( π/3)   3 sin( π/3) ]

= [(7/2) (3/2) 3√3 /4] = [ 3.5 0.75 1.3]