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It's known that the volume of revolution of the function $f(x)$ (assuming it's real, continuous...) is $$V=\pi\int_a^b f(x)^2dx$$

This can be modelized as if we add together all the infinitesimal cylinders of radius $|f(x)|$ integrating over all the domain $x \in [a,b]$

However, what's the geometric interpretation of the surface

$$S=2\pi\int_a^b f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx$$

And this isn't the sum of the perimeter of all the small circumference with radius $|f(x)|$ (at least I can figure out why) ?

It will be $$S=2\pi\int_a^b ydx$$ Can anyone explain the geometric meaning of $\sqrt{1+\bigg(\dfrac{dy}{dx}\bigg)^2}$

Thanks in advance.

player3236
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  • I've found the answer https://math.stackexchange.com/questions/1691147/why-is-surface-area-not-simply-2-pi-int-ab-y-dx-instead-of-2-pi-in?rq=1 – MatMorPau22 Feb 02 '21 at 10:48

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I moved this answer to the more active: Why is surface area not simply $2 \pi \int_{a}^{b} (y) dx$ instead of $2 \pi \int_a^b (y \cdot \sqrt{1 + y'^2}) dx$?

Jap88
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