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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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    $\begingroup$ Kind of like considering $dL = dx$ in the formula for arc length is wrong, you have to use $dL = \sqrt{dx^{2}+dy^{2}}$ $\endgroup$ – Joshua Wang Mar 4 at 22:40
  • $\begingroup$ See this answer. $\endgroup$ – Paramanand Singh Mar 7 at 16:24
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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.

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  • $\begingroup$ I see. How does the volume formula account for slope then? It looks like it doesn't; is that because volume can be approximated by the cylinders irrespective of the slope, but the surface area is very much dependent on the slope? $\endgroup$ – A-Level Student Mar 4 at 22:43
  • $\begingroup$ @A-LevelStudent: That is exactly right! $\endgroup$ – Brian Tung Mar 4 at 23:01
  • $\begingroup$ Ok, thank you very much for your help! $\endgroup$ – A-Level Student Mar 4 at 23:02

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