# Surface area of a solid of revolution: Why does not $\int_{b}^{a} 2\pi \,f(x) \,dx$ work? [duplicate]

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?

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.

• 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? – user1904218 Dec 19 '14 at 22:04
• 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. – Simon S Dec 19 '14 at 22:06
• 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. – Simon S Dec 19 '14 at 22:07

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 $f(x).$