# How do you determine the surface area of a solid revolved about the x-axis?

Mar 23, 2015

The answer is: $A = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$.

If we cut the solid with two parallel planes ($x = c$ and $x = d$), and perpendicular to the x-axis, we obtain a cylinder, whose radius is $f \left(x\right)$, and his height is the lenght of the curve from $x = c$ and $x = d$.

The lenght of the curve from $c$ and $d$ is:

${\int}_{c}^{\mathrm{ds}} q r t \left(1 + {\left[f ' \left(x\right)\right]}^{2}\right) \mathrm{dx}$.

Since the lateral area of a cylinder is: $A = 2 \pi r h$, than

$A = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$.