"Consider $f:[a,b]\rightarrow\mathbb{R}$ of class $C^1$, limited, such that $f(x)\neq 0$ for all $a\leq x\leq b$. After this, consider the revolution surface by turning graphic of $f$ around the $x$ axis. Call this surface by $S$.

Show that for any $R>0$ there is a bijection preserving the areas between $S$ and a cylinder which basis is a circle with radius $R$ and height $H$, such that the area of $S$ is equal to $2\pi RH$."

I know that the cylindrical projection(from a sphere to a cylinder) gives me a bijection preserving areas, so I tried to construct a bijection from $S$ to some sphere and a bijection between two cilinders, then composing this bijections I would have the bijection of the problem. But constructing a bijection from $S$ to a sphere is being pretty hard. I'm starting to think this is not good.

I need some help here, thank you!

• You haven't told us what $\operatorname{f}$ is! – Fly by Night Apr 29 '13 at 17:38
• is any $C^1$ function, going from $[a,b]$ to $\mathbb{R}$, probably you must consider $f$ limited. – diff_math Apr 29 '13 at 17:40
• You said "...such that $f(x)$..." in your post. I assumed you'd missed something. What is the condition on $f$? – Fly by Night Apr 29 '13 at 17:41
• true, I forgot this part, its fixed now. – diff_math Apr 29 '13 at 17:42
• I believe your first step would be to find the area of revolution, no? – Mark Ping May 1 '13 at 4:01

For a start, you might think about making a projection between two such cylinders, one with radius $R$ and height $H$ and another with radius $R'$ and height $H'=H\frac R{R'}$ These two cylinders satisfy the hypotheses of your problem. Clearly you need to stretch in one axis by the same amount as you compress in the other.
For the more general case, you do the same thing, but the amount you stretch depends on the local radius, which is $f(x)$