# Relation between surface area and area under curve (easy)

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?

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. $$S= 4\pi$$.