Chapter 12.1, Problem 109E

Surfaces of revolution Let C be the curve x = f(t), y = g(t), for a ≤ t ≤ b, where f′ and g′ are continuous on [a, b] and C does not intersect itself, except possibly at its endpoints. If g is nonnegative on [a, b], then the area of the surface obtained by revolving C about the x-axis is

$S={\displaystyle {\int }_a^{b}2\pi g\left(t\right)\sqrt{f\prime {\left(t\right)}^2+g\prime {\left(t\right)}^2}dt}.$

Likewise, if f is nonnegative on [a, b], then the area of the surface obtained by revolving C about the y-axis is

$S={\displaystyle {\int }_a^{b}2\pi f\left(t\right)}\sqrt{f\prime {\left(t\right)}^2+g\prime {\left(t\right)}^2}dt.$

(These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve y = f(x).)

109. Find the area of the surface obtained by revolving one arch of the cycloid x = t − sin t, y = 1 − cos t, for 0 ≤ t ≤ 2π