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Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ yield the correct answer when calculating the surface area of a solid of revolution?

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Because the 'infinitesimal' line element which you are rotating about a circle of circumference $2\pi f(x)$ doesn't have length $dx$; rather the length is $\sqrt{1 + f'(x)^2} \ dx$, which is always longer for any $x$ for which $f'(x) \neq 0$.

For a cylinder where $f(x)$ is a constant, then $f'(x) = 0$ and your proposed expression does work.

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  • $\begingroup$ Forgive me if this is a silly question, but why can I use dx when calculating the volume of a solid of revolution? Why is it different in this case? $\endgroup$ – user1904218 Dec 19 '14 at 22:04
  • $\begingroup$ Because in the volume calculation you are adding up 'discs' of cross-sectional area $\pi f(x)^2$ and width $dx$. There no sort of adjustment is required. $\endgroup$ – Simon S Dec 19 '14 at 22:06
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    $\begingroup$ It's worth going back to the Riemann sums in both cases to make sure you understand how the SA and volume integrals are constructed. $\endgroup$ – Simon S Dec 19 '14 at 22:07
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For a thin slice surface area of a cone = $ 2 \pi \,f(x) \,ds $ which you should integrate.

If it is a flat disc of height tending to zero there would be a big error without multiplying by $ \sec \phi $ due to slope.

$ \int_{b}^{a} 2\pi \,f(x) \,dx $ yields correct answer only when calculating the surface area of a cylinder, zero slope, constant radius $ a = f(x) $.

No area is represented by $ \int_{b}^{a} 2\pi \,f(x) \,dx,$ for variable $ f(x). $

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