Suface area calculus definition confusion?

Question

I will be starting my question with an argument on surface area of a frustum.

Let center of smaller face (with radius $r_2$) be placed on the origin of $x$-$y$ coordinate system.

Then, $y = f(x) = mx +$ $r_2$, will be equal the radius function with respect to $x$. For approximation of surface area we can Partition the interval $[0,h]$ into n sub-intervals.

$P = \{x_0, x_1, ...., x_n\}$ where, $x_0$= $0$ and $x_n$ = $h$.

Now, Lower approximation ($L$) for surface area can be done as:

$$L = \sum_{k=1}^n 2\pi f(x_{k-1})\Delta x_k $$

and for Upper approximation ($U$):

$$U = \sum_{k=1}^n 2\pi f(x_{k})\Delta x_k $$

But, $ \lim \limits_{||P|| \to 0} $ (U) = $ \lim \limits_{||P|| \to 0} $ (L) = $\int_0^h 2 \pi f(x) \,dx$

Solving this integral will result in:

Surface Area $= \pi h (r_1 + r_2)$, which is not correct.

Can anyone tell me, what am I doing wrong?? Then I will be posting my question further for the generalized definition of surface area (based on frustums).

Answer 1

This is a classic mistake. You're replacing the area of a frustum with the area of a cylinder, even in the tiny pieces. The height is not the same as the slant height. (There's a reason that the correct calculus formula uses the arclength of the curve, not the vertical height.)

Answer 2

You should consider the slant length $dl$ which is given as $$dl=\sqrt{1+\left(\frac{dy}{dx}\right)^2} dx=\sqrt{1+\left(\frac{d}{dx}(mx+r_2)\right)^2} dx=\sqrt{1+m^2}\ dx$$

The surface area of differential element of frustum of thickness $dx$ is $2\pi y\ dl$. Integrating ths area with proper limits, the surface area $A$ of frustum of cone is by obtained as follows $$A=\int_0^{h}2\pi y\ dl=\int_0^{h}2\pi (mx+r_2)\sqrt{1+m^2}dx$$ Where $m=\frac{r_1-r_2}{h}$