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
$$
s = \int_a^b 2 \pi f(x) \sqrt{1+[f'(x)]^2} \; dx
$$

E 7.4 Exercises