Chapter 8.3, Problem 68E

To determine

To choose : the correct option.

Answer to Problem 68E

The correct option is (d) $\frac{\pi }{2}\left[e^2+e^{-2}-2\right]$ .

Explanation of Solution

Given information : Let $R$ be the region in the first quadrant bounded by the graph of $y=e^{-x}$ and $x=1$ .

  $\begin{array}{l}V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left[R^2\left(x\right)-r^2\left(x\right)\right]dx}\\ V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left[{\left(e^x\right)}^2-{\left(e^{-x}\right)}^2\right]dx}\\ V=\frac{\pi }{2}\left[e^2+e^{-2}-2\right]\end{array}$