Answer
${\rm{Area}}\left( T \right) = 4{\pi ^2}ab$
Work Step by Step
Recall that the torus in Exercise 45 is obtained by rotating the circle in the $yz$-plane of radius $a$ centered at $\left( {0,b,0} \right)$ about the $z$-axis (Figure 24). Thus, the center of mass of the circle is located at $\left( {0,b,0} \right)$. As it is rotated about the $z$-axis, the distance traversed by the center of mass is $2\pi b$.
In this case, the generating curve is the circle of radius $a$ whose length $L$ is $2\pi a$.
By the Pappus's Theorem: the area of a surface of revolution $T$ is equal to the length $L$ of the generating curve times the distance traversed by the center of mass. Therefore,
${\rm{Area}}\left( T \right) = \left( {2\pi a} \right)\cdot\left( {2\pi b} \right) = 4{\pi ^2}ab$
Thus, the surface area of the torus is ${\rm{Area}}\left( T \right) = 4{\pi ^2}ab$.