Given $z=f(x)$ for $a<x<b$ differentiable and $f(x) > 0$ for all $x\in (a,b)$. By revolution of the graph around the x-axis in $\mathbb{R}^{3}$ there evolves a revolutionary area $S$.
We parametrize: $\phi(s,t)= s, f(s) \sin(t), f(s) \cos(t)$
This gives: $$D\phi = \begin{pmatrix}1 & 0 \\ f' \sin(t) & f \cos(t) \\ f' \cos (t) & -f \sin(t) \end{pmatrix}$$ So : $$\|\partial_{1} \phi \times \partial_2 \phi \| = \sqrt{f^{2}f'+f^2 \sin^2 +f^2 \cos^2} = f\sqrt{1+f'(x)^2} \tag{$\ast$} $$
So now :$$\int_{0}^{2\pi} \int_{a}^{b} f \sqrt{1+f'(x)^2} dxdt = 2\pi \int_{a}^{b} f \sqrt{1+f'(x)^2} dx .$$
Is this last star step correct? Thanks for all efforts.
