Proof that this surface is of revolution

Question

I have a surface with parametric equation
$$\mathbf{x}(u,v)=(u\cos(v),u\sin(v),u^2),$$ $u$ is any real number, $v$ is between $0$ and $2\pi$. I don't know how to show that this is surface of revolution. Please help. Thanks.

Answer 1

If you fix $u$ (say $u=c$), then $$ x(c,v) = (c\cos v, c\sin v, c^2) $$ So $v \mapsto x(c,v)$ is the equation of a circle of radius $c$ lying in the plane $z=c^2$ (you can check that $x^2 + y^2 = c^2)$. We have shown that the curves of the form $u = \text{constant}$ are circles lying in parallel planes, so the surface is a surface of revolution.

Answer 2

If you call $(x,y,z) = (u\cos(v),u\sin(v),u^{2})$, we have $z = x^{2}+y^{2}$, which is the equation of a paraboloid of revolution. You rotate the regular parabola $y=x^{2}$ around the central axis to get the surface.

Answer 3

You need to remember some trigonometry: As $v$ changes, the pair $(\cos v,\sin v)$ follows a circle.