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Suppose that $f(x)>0$ is a continuously differentiable function defined on $x\ge 1$. Let $S$ be the surface of revolution of the graph $y=f(x)$ about $x-axis$. Let $E \subset R^3$ be the solid enclosed by $S$ and $x=1$. More precisely:$$E=\left \{ (x,y,z)\in R^3;x\ge 1 \space and \space \sqrt[]{y^2+z^2}\le f(x) \space for \space each \space x\ge 1 \right \}. $$ If the surface area of $S$ is finite, prove that the volume of $E$ must also be finite.

I have already known that the surface area of $S$ is... $$A=2\pi\int_{1}^{\infty }f(x) \sqrt[]{1+\left ( f'{(x)} ^2\right ) } dx =\lim_{b \to \infty}2\pi\int_{1}^{b} f(x) \sqrt[]{1+\left ( f'{(x)} ^2\right ) } dx$$ and the volume of $E$ is...$$V=\pi\int_{1}^{\infty } \left ( f(x) \right ) ^2dx=\lim_{b \to \infty } \pi\int_{1}^{b} \left ( f(x) \right ) ^2dx$$ Can I solve this question by the following?

$$\because \space f'(x) \space is \space continuous \space ,\space \therefore \exists M, \space \space such \space that \space f'(x)\ge M, \space \forall x\ge1$$ then $$A=\lim_{b \to \infty}2\pi\int_{1}^{b} f(x) \sqrt[]{1+\left ( f'{(x)} ^2\right ) } dx\ge \lim_{b \to \infty}2\pi\int_{1}^{b} f(x) \sqrt[]{1+M ^2 } dx = \lim_{b \to \infty}2\pi \sqrt[]{1+M ^2} \int_{1}^{b} f(x) dx$$ Since $A$ is finite, so $\lim_{b \to \infty}\int_{1}^{b} f(x) dx$ is also finite.......$(1)$

Also:$$\because \space f(x) \space is \space continuous \space ,\space \therefore \exists L\ge 0, \space \space such \space that \space f(x)\le L , \space \forall x\ge1$$

So...$$V=\lim_{b \to \infty } \pi\int_{1}^{b} \left ( f(x) \right ) ^2dx \le \lim_{b \to \infty } \pi\int_{1}^{b} L \times f(x) dx = \pi L \lim_{b \to \infty } \int_{1}^{b} f(x) dx ......(2)$$ Finally, because (1) tell me that $\lim_{b \to \infty}\int_{1}^{b} f(x) dx$ is finite, then I have (2) also finite?

mlrofcloud
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    Your two $\therefore$ (for the existence of such $M$ and $L$) are wrong. – Anne Bauval May 23 '23 at 08:00
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    Does this answer your question? Is there a shape with infinite volume but finite surface area? – Anne Bauval May 23 '23 at 08:02
  • Madam, I've seen the link you suggested, and I've also read the proof in https://en.wikipedia.org/wiki/Gabriel%27s_horn, but there is something I haven't figure it out yet. It says $\lim_{t \to \infty} \sup_{x\\ge t} f(x)^2-f(1)^2=\lim_{t \to \infty} \sup \int_{1}^{t}\left ( f(x)^2 \right )'dx\le \int_{1}^{\infty } \left | (f(x)^2)' \right |dx =\int_{1}^{\infty } 2f(x)\left | f'(x) \right | dx \le \int_{1}^{\infty } 2f(x)\sqrt[]{1+f'(x)^2}dx=\frac{A}{\pi } < \infty $, and according to this, there exists a $t _0$ such that the supremum $\sup\left {f(x);x\ge t_0\right } $ is finite. – mlrofcloud May 23 '23 at 10:40
  • hence, it claims, $M=\sup \left { f(x);x\ge 1 \right } $ must be finite, since f is a continuous function, which implies that $f$ is bounded on the interval [1, ∞) – mlrofcloud May 23 '23 at 10:46
  • I know $\int_{1}^{\infty } \left | (f(x)^2)' \right | dx\ge \int_{1}^{\infty } (f(x)^2)'dx$, but WHY $\int_{1}^{\infty } \left | (f(x)^2)' \right | dx\ge \lim_{t \to \infty} \sup \int_{1}^{t } (f(x)^2)'dx$, why we take $limsup$ here? – mlrofcloud May 23 '23 at 10:53
  • And it says $\sup\left {f(x);x\ge t_0\right }$ is finite, $M=\sup \left { f(x);x\ge 1 \right }$ also finite, but is there any possible that $f(x)$ goes to $\infty$ somewhere between $\left ( 1,t_0 \right ) $? – mlrofcloud May 23 '23 at 11:05
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  • $\int_1^t (f(x)^2)'dx\le\int_1^t\left | (f(x)^2)' \right | dx\le\int_{1}^{\infty }\left | (f(x)^2)' \right | dx$ for all $t$ and the final upper bound does not depend on $t$ hence it is also an upper bound for $\limsup_{t\to\infty}\int_1^t (f(x)^2)'dx.$ 2) On $[1,t_0],$ $f$ is continuous hence bounded.
    • – Anne Bauval May 23 '23 at 11:44
  • Thank you so much madam, though I'm not sure if I can understand the idea of limsup or liminf easilyQQ – mlrofcloud May 23 '23 at 15:18
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    $\limsup_{t\to\infty}g(t):=\lim_{t\to\infty}h(t)$ for $h(t):=\sup_{x\ge t}g(x).$ The limit exists because $h$ is non-increasing. – Anne Bauval May 23 '23 at 15:22