Why $\int_0^h 2 \pi \frac{rx}{h} \, dx \neq \pi rl$

Question

I'm new to calculus.
I saw a proof for volume of cone using integral. They taken the cone's vertex at $(0,0,0)$, it's base centre at $(h,0,0)$ and it's radius is $r$
$$V=\int_0^h \pi \left(\frac{rx}{h}\right)^2 \, dx =\frac{\pi r^2h}{3}$$
$R(x)=\frac{rx}{h}$. $R(x)$ gives the radius of the circle on cone at $x.$

Using the same idea I tried to workout the C.S.A of the cone,
Using integral I sum up all the circumference of those circle which must give the C.S.A of cone but, $$S=\int_0^h 2\pi \left(\frac{rx}{h}\right) \, dx =\pi rh$$ where I am doing mistake?

Answer 1

When you integrate for volume the $dx$ part is the infinitesimally small vertical height. However, when you want the curved surface area, you will want a small component $dl$ to be along the length of the slope instead.

This has an easy fix. $x$ and $l$ are related as the the following due to the similar triangles: $$\frac xh=\frac lL$$ Hence, we can substitute as the following-

$$S=\int_{0}^{L} 2\pi \biggr(\frac{rl}{L}\biggr) dl$$ $$S=2\pi\frac rL\frac {L^2}2$$ $$S=\pi RL$$