Let $y = f(x)$ be a smooth curve.

$A$ = area bounded by the curve, $x$-axis, $x = a$ and $x = b$.

$S$ = area of the surface generated by revolving the curve about $x$-axis between $x = a$ and $x = b$.

Then is $2 \pi A \leq S$ or $S \leq 2 \pi A$.

Answer given: $S \leq 2 \pi A$

$$***$$ My answer: $ 2 \pi A \leq S$

We know that, Area under curve = $\int_{a}^{b}f(x)dx$

Surface area by axis revolution = 2$\pi \int_{a}^{b}f(x)\sqrt{1+(\frac{dy}{dx})^2}dx$

Now, $2\pi A$ = $2\pi \int_{a}^{b}f(x)dx$

Since, $\sqrt{1+(\frac{dy}{dx}})^2$ > 1, since $(\frac{dy}{dx})^2 > 0$, therefore, 2$\pi \int_{a}^{b}f(x)$ is getting multiplied by a factor > $1$ in S. So, $ S \geq 2 \pi A$.


My reasoning must be incorrect. Why is $S \leq 2 \pi A $ correct?


1 Answer 1


Your argument and conclusion is correct.

To verify, just consider a specific example where $f(x)=x$, with $a=0$ and $b=1$. So, it is a cone with a circular base. Then, you have $2\pi A= \pi$ and $S=\pi\sqrt2$, which verifies $S > 2\pi A$.

Another convenient case to check is a half-circle with unit radius, for which you have $2\pi A = \pi^2$ vs. $S= 4\pi$.


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