I need to find a parametrization for the following two surfaces:
(i) The surface obtained by revolving the curve $z = f(x),\ (a \lt x \lt b)$ in the $xz$ plane, around the $z$ axis, where $a \gt 0$.
The solution states that $f(u,v) = (v \cos u,v \sin u,f(v))^{T}$ with $u \in [0,2\pi)$ and $v \in (a,b)$. The part I'm having difficulty with is why it is $v\cos u$ etc and not $f(v)\cos u$?
The second question is the similar except the curve is revolved around the $x$ axis, $f(x) \gt 0$ (and there are no conditions on $a$ and $b$). But in this case, $f(u,v) = (v,f(v)\sin u,f(v)\cos u)^{T}$ - why is it $f(v)$ now?
I've tried doing a few sketches but it still hasn't clicked.