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 $\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0$ ?
