Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 960: 47

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 ${\rm{Area}}\left( T \right) = 4{\pi ^2}ab$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.