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?


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. $S= 4\pi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.