Proving the surface area and volume of revolution formula

Question

Title Is there any way to prove the formula or surface area and volume of revolution of a function? The derivation of the formula I have found online does not really seem like a proof, but more like using intuition to explain it. Thanks for your time and sorry if it is a stupid question lol

Answer 1

I am trying to give a derivation of the two formula. Please let me know if you find any mistakes:

The proof of surface area is to divide the volumn into frustrums on $[x_{i-1},x_i]$, where the surface area is given by $$A=2\pi rl$$, where $l$ is length of slant of the frustrum, and $r=\frac{r_1+r_2}{2}$, where $r_1$ and $r_2$ are radius of left and right radius. Then we parametrize $r_1=f(x_i)$, $r_2=f(x_{i-1})$, and $$l=\sqrt{1+f'(x_i^*)^2}\Delta{x}$$, where $x_i^*\in[x_{i-1},x_i]$. Let $\Delta{x}\to 0$, $$A_i=2\pi\frac{f(x_i)+f(x_{i-1})}{2}\sqrt{1+f'(x_i^*)^2}\Delta{x}\to 2\pi f(x_i^*)\sqrt{1+f'(x_i^*)^2}\Delta{x}$$ The whole surface area $S$ is given by $$S=\lim_{n\to\infty}\sum_{i=1}^{n}A_i=\int_{a}^{b}2\pi f(x)\sqrt{1+f'(x)^2}dx$$

The proof of volumes under revolution starts from the volumns of cylinders, i.e. $$\Delta{V}=\pi y^2\Delta{x}$$ (replacing $r^2$ with $y^2$). Then the volumn is just $$V=\lim_{\Delta{x}\to 0}\sum_{x=a}^{b}\Delta{V}=\lim_{\Delta{x}\to 0}\sum_{x=a}^{b}\pi y^2\Delta{x}=\int_{a}^{b}\pi y^2dx$$

If you want to rotate around different axis, just switch the order of the functions, like replacing $y=f(x)$ to $x=h(y)$. Hope this helps.