# Prove that the surface revolution surface is given by $\int_0^1 2\pi f(x)\sqrt{1+f'(x)^2}dx$.

Let $$f:\mathbb R\to \mathbb R$$ a smooth function. We suppose $$f> 0$$. Let $$\gamma =\{(0,x,f(x))\mid x\in [0,1]\}$$ and let $$\Sigma$$ the revolution surface obtained by the rotation of $$\gamma$$ around the axis $$Oy$$. Prove that $$Surface (\Sigma)=\int_0^1 2\pi f(x)\sqrt{1+f'(x)^2}dx,$$ using Riemann sum only. The idea is the following one : an infinitesimal element of area is the red "annulus around a cone". The area is given by $$\mathcal A(x+h)-\mathcal A(x)=2\pi f(x)\sqrt{1+f'(x)^2}h+o(h).$$

Let $$\{x_i^n\}_{i=0}^{m_n-1}$$ subdivision of $$[0,1]$$ s.t. $$\Delta x_i^n:= x_{i+1}^n-x_i^n\to 0$$ when $$n\to \infty$$. We have that

$$Surface (\Sigma)=\lim_{n\to \infty }\sum_{i=0}^{m_n-1}(\mathcal A(x_{i+1}^n)-\mathcal A(x_i^n)$$ $$=\lim_{n\to \infty }\sum_{i=0}^{m_n-1}2\pi f(x_{i}^n)\sqrt{1+f'(x_i^n)}\Delta x_i^n+\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)$$ $$=\int_0^1 2\pi f(x)\sqrt{1+f'(x)^2}dx+\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n).$$ How can I prove now that $$\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0$$ ?

• For such a proof we need the chosen setup for the "area" of curved surfaces. Commented Aug 14, 2019 at 15:36
• @ChristianBlatter: What do you mean ? Here it's normally not a complicated problem, is it ? It's juste the revolution surface obtained by rotating the curve around $Oy$. It's not true that $\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0$ ? If not, could you provide a counter example ? In all example I have in mind it works... Commented Aug 14, 2019 at 16:00