Chapter 8.3, Problem 19E

To determine

To find : the volume of the solid.

Answer to Problem 19E

The volume of the solid is ${\pi }^2-2\pi$ .

Explanation of Solution

Given information : The region bounded by the lines and curves about the x -axis

  $y=\sec x,y=\sqrt{2},-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\le x\le \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ .

Formula used: Volume $=\pi r^2$

Calculation :

The curve $y=\sec x$ and $y=\sqrt{2}$ .

Volume of the solid of revolution:

  $\begin{array}{l}V=\pi {\displaystyle \underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\overset{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\left({\sqrt{2}}^2\right)-{\sec }^{2}xdx}\\ V=\pi {\displaystyle \underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\overset{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\left(2-{\sec }^2\right)xdx}\\ V=\pi {\left[2x-\tan x\right]}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ V=\pi \left[\frac{\pi }{2}-1\right]-\left[-\frac{\pi }{2}-\left(-1\right)\right]\\ V=\pi \left[\pi -2\right]\\ V={\pi }^2-2\pi \end{array}$