# 7.4 | Arc Length and Surfaces of Revolution

Let $$f$$ be a continuous function defined on $$[a,b],$$ and let $$P=\{x_0,x_1, \ldots, x_n \}$$ be a regular partition of $$[a,b].$$ The arc length of the graph of $$f$$ from $$P(a,f(a))$$ to $$Q(b,f(b))$$ is
$$L = \lim_{n \rightarrow \infty} \sum_{k=1}^ \infty d(P_{k-1},P_k)$$
if the limit exists.
Let $$f$$ be smooth on $$[a,b].$$ Then the arc length of the graph of $$f$$ from $$P(a,f(a))$$ to $$Q(b,f(b))$$ is
$$L = \int_a^b \sqrt{1+ [f'(x)]^2} \; dx.$$
Let $$f$$ be smooth on $$[a,b].$$ The arc length function $$s$$ for the graph of $$f$$ is defined by
$$s(x) = \int_a^x \sqrt{1 + [f'(t)]^2} \; dt$$
with domain $$[a,b].$$
The lateral surface area, $$S$$, of a frustum is given by
$$S = 2 \pi r L,$$
where $$r = \frac{1}{2}(r_1 + r_2)$$ and $$L$$ is the length of the shortest line segment on the surface connecting the bases.
The surface area of the surface of revolution obtained by revolving the graph of $$y = f(x)$$ from $$x =a$$ to $$x = b$$ about the $$x$$-axis is given by
$$s = \int_a^b 2 \pi f(x) \sqrt{1+[f'(x)]^2} \; dx$$