Question

This question is a little out there but has interested me for some time now. If given a continuously differentiable function f(x)= (x,y) over the interval (x1,x2) and this R3 function is rotated about the x axis (surface of revolution) that has a resultant surface area of (Sa), what is the relationship between the original function SOR, Sa and the Sa of the resultant SOR of the first derivative of the original function over the same interval ?

Answer 1

Could you clarify?

Usually if I want to create a surface area SA of a tube-like object, I'm rotating a continuously differentiable function y = f(x) that is defined on the interval (x1,x2)
\[f : (x_1,x_2) \rightarrow \mathbb{R}\]
(which I would call an R1 or C1 function ...). I'm guessing that if this surface area is called SA(f), you're interested in comparing it to SA(g), where
\[g(x) = f'(x)\]
... The two surfaces of revolution are somewhat independent; for example you can make another SA(f) arbitrarily large by adding a big constant to make the tube larger
\[f_2(x) = f(x)+C\]
while SA(g) remains unchanged.