# [SOLVED] Derivation of formula using calculus?

• Jul 1st 2008, 05:01 AM
fardeen_gen
[SOLVED] Derivation of formula using calculus?
Derive formula of lateral(curved) surface area of cone using integral calculus?

I checked the wikipedia link but it has no diagram for the proof. I couldn't understand the way they have divided the cones into triangles.
Cone (geometry)/Proofs - Wikipedia, the free encyclopedia

Can anybody help me with the proof? (It would be nice if anybody cud make the diagram)
• Jul 1st 2008, 05:50 AM
Soroban
Hello, fardeen_gen!

Quote:

Derive formula of lateral (curved) surface area of cone using integral calculus.
I read that proof, and don't care for it.
I think those "triangles" are very clumsy to handle.

I would approach it like this . . .
Code:

        |       r +              * (h,r)         |          *  :         |      *      :         |  *          :     - - * - - - - - - - + --         |              h
We have a right triangle formed by the x-axis, $\displaystyle y \:=\:\frac{r}{h}x$, and $\displaystyle x = h$

Rotate the triangle about the x-axis to form a cone with radius $\displaystyle r$ and height $\displaystyle h.$

The surface area is given by: .$\displaystyle S \;=\;2\pi\int^a_b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$

So we have: .$\displaystyle S \;=\;2\pi\int^h_0\left(\frac{r}{h}x\right)\sqrt{1 + \left(\frac{r}{h}\right)^2}\,dx \;=\;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\int^h_0 x\,dx$

. . . . . . . $\displaystyle = \;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\cdot\frac{1}{2}x^2\,\bigg]^h_0 \;=\;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\cdot \frac{1}{2}h^2$

Therefore: .$\displaystyle S \;=\;\pi r\sqrt{r^2+h^2}$

• Jul 1st 2008, 05:52 AM
fardeen_gen
It was great help!