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When I learnt about the derivation for the formula $$V=\pi\int_{x_1}^{x_2} y^2~dx$$ where $V$ is volume of the solid generated when $y=f(x)$ is rotated about the $x$ axis by $2\pi$ radians between $x_1$ and $x_2$, my teacher showed that this works by approximating the volume by splitting up the solid generated into very thin cylinders. Each cylinder would have width $\delta x$ and radius $y$, so we'd have $$V=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}\pi y^2\delta x=\pi\int_{x_1}^{x_2} y^2~dx$$


I thought of doing the same thing to find the surface area of thus solid that we have produced. Once more, consider the very thin cylinders that we used before. The surface area of the outer strip of each cylinder is $2\pi y\delta x $. Hence, I thought that the surface area should be equal to $$SA=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}2\pi y\delta x=2\pi\int_{x_1}^{x_2}y~dx$$ But this is wrong! Why is that? What is wrong my reasoning?

Thank you for your help.

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Your formula accounts for the fact that the surface gets larger (in terms of its circumference) as the curve gets further from the axis of revolution, but it does not account for the slope of that curve. Therefore, your integrand has to be multiplied by the infinitesimal arclength, which accounts for this slope.

Brian Tung
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