# surface Areas using cylindrical shells

If I try to find the surface area of any solid by using cylindrical slices, I'm getting wrong answer. I'm taking this as the formula,

$$S = \int_a^b2\pi y dx$$

where $y$ = height ($2\pi y$ = circumference of the cylinder) $dx$ = width. We can approximate the surface area using cylindrical shells right? If we can approximate volume, we can also approximate surface area right? This is my argument... Please help.

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## 1 Answer

Actually, approximating surface area by cylindrical shells doesn't work, for the same reason that $\pi \neq 4$ in this thread http://www.physicsforums.com/showthread.php?t=452917

Cylindrical shells do not give the correct "small" surface element because they are all "almost" parallel to the axis of revolution.

The correct formula for $y=f(x)$, $a \leq x \leq b$ to find the surface area of the surface formed by revolving $f$ around the $x$-axis is $$S=2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}dx.$$

More information on this topic can be found at http://en.wikipedia.org/wiki/Surface_of_revolution or by googling "surface area by revolution".

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$\pi \ne 4$ is here at math.stackexchange.com/questions/12906/is-value-of-pi-4/… –  Ross Millikan Oct 2 '12 at 15:33