Let \(f(x)\) be a nonnegative smooth function over the interval \([a,b]\). We wish to find the surface area of the surface of revolution created by revolving the graph of \(y=f(x)\) around the x-axis:
We are going to partition the interval \([a,b]\) and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve. For \(i=1,2,…,n\), construct a line segment from the point \((x_{i-1},f(x_{i-1}))\) to the point \((x_i,f(x_i))\). Now, revolve these line segments around the x-axis to generate an approximation of the surface of revolution:
The shape of each band is a frustrum of a right circular cone. The lateral surface area of a right circular conical frustrum is:
Where \(r_1\) is the radius of the top circle, \(r_2\) is the radius of the bottom circle and \(l\) is the slant height. The proof of this can be found here. The lateral surface area of one of these bands is:
According to the Mean Value theorem, there exists \(x_i^* \in [x_{i-1}, x_i]\) such that \(f'(x_i^*) = \frac{\Delta y_i}{\Delta x}\):
According to the Intermediate Value theorem, there exists \(x_i^{**} \in [x_{i-1}, x_i]\) such that \(f(x_i^{**}) = \frac{f(x_{i-1}) + f(x_i)}{2}\):
The approximate surface area of the whole surface of revolution is:
Taking the limit as \(n→∞\):