Question

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

Answer 1

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`.