Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 431: 79

$2\pi$

We use the washer method (see graph below) $V=\displaystyle \int_{a}^{b}\pi([R(x)]^{2}-[r(x)]^{2})dx$ $V=\displaystyle \pi\int_{0}^{2}(\cosh^{2}x-\sinh^{2}x)dx$ Apply the identity: $\qquad \cosh^{2}x-\sinh^{2}x=1$ $=\displaystyle \pi\int_{0}^{2}1dx$ $=\pi[x]_{2}^{0}$ $=2\pi$