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

1 Answer
Mar 23, 2015

The answer is: #A=2piint_a^bf(x)sqrt(1+[f'(x)]^2)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(x)#, 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^dsqrt(1+[f'(x)]^2)dx#.

Since the lateral area of a cylinder is: #A=2pirh#, than

#A=2piint_a^bf(x)sqrt(1+[f'(x)]^2)dx#.