# Theorem of Pappus

Given a surface of revolution $$S$$ which can be parametrized by the map $$\mathbf x(u,v) = (f(v)\cos u,f(v)\sin u,g(v)),$$ over the open set $$U =\{(u,v) \in \mathbb R^2 \mid 0 < u < 2\pi, a < v < b\}$$, I computed the area of $$S$$ to be \begin{align*} \int_a^b\int_0^{2\pi} |\mathbf x_u \times \mathbf x_v| \, du \, dv = 2\pi\int_a^b |f(v)| \sqrt{(f'(v))^2+(g'(v))^2} \, dv. \end{align*} If $$l$$ is the length of the generating curve $$C$$, how does one then get the area of $$S$$ to also be written $$2\pi \int_0^l \rho (s) \, ds,$$ where $$\rho=\rho(s)$$ is the distance to the rotation axis of the point $$C$$ corresponding to $$s$$? I think that the arc length $$s=\int_a^b |\alpha'(t)| \, dt$$, where $$\alpha$$ is the space curve, but I'm not sure in particular how one changes the interval $$[a,b]$$ to $$[0,l]$$ when changing the variable $$v$$ to $$s$$.

You need to reparametrize your curve, replacing $$v$$ by $$s$$, where $$s(v)=\int\limits_a^v\sqrt{f'(v)^2+g'(v)^2}\,dv$$ hence $$ds=\sqrt{f'(v)^2+g'(v)^2}\,dv$$. When $$v$$ takes values in $$[a,b]$$, $$s$$ takes values in $$[0,l]$$ where $$l$$ is the length of the curve. Also, $$|f(v)|$$ is just the distance from the point to the rotation axis, so it is precisely $$\rho(s)$$ (you can do the computation, but I do not think there is a need for it),

Consider $$2\pi \int_0^l \rho (s) \, ds$$ We know $$s(t)=\int_{a}^t||\alpha'(z)||dz$$. For regular curves this is a diffeomorphism from $$(a,b)$$ to $$(0,l)$$ where $$l$$ is the arclength Thus using change of variable formula rewrite the integral as $$2\pi\int_a^b\rho(s(t))|s'(t)|dt=2\pi\int_a^b|f(t)|\sqrt{f'(t)^2+g'(t)^2}dt$$