Derive Equation of Parabola Rotated in R3

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So for this question:

"Derive the equation of the surface obtained when the parabola $x=y^2$ is rotated about the x-axis. Identify this surface."

I'm not sure why the answer is $x=y^2+z^2$ (a circular paraboloid). Is it just because the shape that the function $x=y^2$ sweeps out is a circle in the xz-plane?

However, what happens if you rotate a function that's not a parabola about the x-axis (like a hyperbola)? Would you still get 2 circular paraboloids? Or would that be a hyperboloid of 2 sheets? For functions that aren't parabolas, how would you derive the equation of it being rotated about the x-axis?

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Revolving a curve about the $x$-axis amounts to making the substitution $y \mapsto \sqrt{y^2 + z^2}$. More precisely, suppose that a curve in 2 dimensions is defined by $f(x,y) = 0$, and that a surface is generate by rotating this curve about the $x$-axis. Then the equation of the resulting surface is given by $$ g(x,y,z) = f\left(x,\sqrt{y^2 + z^2}\right) = 0. $$ For the case of your parabola, the curve corresponding to $f(x,y) = x - y^2 = 0$ became the surface $g(x,y,z) = x - y^2 + z^2 = 0$.

Rotating a hyperbola about the $x$-axis could result either in a hyperboloid of two sheets or a hyperboloid of one sheet. For example, if we begin with the curve $$ f(x,y) = x^2 - y^2 - 1 = 0, $$ then the resulting surface satisfies $x^2 - y^2 - z^2 - 1 = 0$ and can be seen to be a hyperboloid of two sheets. On the other hand, if we begin with the curve $$ f(x,y) = x^2 - y^2 + 1 = 0, $$ then the resulting surface satisfies $x^2 - y^2 - z^2 + 1 = 0$ and can be seen to be a hyperboloid of one sheet.

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Any 2d function

$$y=f(x)$$

when rotated about x-axis can be changed to a surface of revolution of 3d equation

$$ r=\sqrt {y^2+z^2}=f(x)$$

The simplest example of a straight line

$$ y=x$$

on rotation becomes a cone of varying radius $r$,

$$ r=\sqrt {y^2+z^2}=x $$

For a particular paraboloid,

$$ r=\sqrt {y^2+z^2}=\sqrt {x} $$

For a particular two sheeted hyperboloid,

$$ r=\sqrt {y^2+z^2}=\sqrt{x^2-1}. $$