The area of a surface of revolution $y(x)$ is express by the integral $$A=\int\limits_a^b 2\pi y(x)\underbrace{\sqrt{1+y'(x)^2}}_\text{arc length of y(x) at x}dx$$
But this seems counter-intuitive to me. Consider the volume integral $$V=\int\limits_a^b \pi y(x)^2dx$$ Here we don't use the arc length, but why? In the integral for $A$, it seems that we just need to take infinitesimal circle lengths at $x$ multiplied by infinitesimal width $dx$: $$A=\int\limits_a^b 2\pi y(x)dx$$
I would appreciate if someone could please clarify this for me strictly conceptually / geometrically.