Chapter A6, Problem 62E

To determine

To find: the volume of the solid.

Answer to Problem 62E

The volume of the solid is $\pi$

Explanation of Solution

Given:

  

Calculation:

As it is clear that to calculate the area or volume between two curves, we find the difference between the values of curves at each point and then integrate it.

The integral hence calculated for volume is,

  ${\displaystyle \underset{-1n\sqrt{3}}{\overset{1n\sqrt{3}}{\int }}\pi {\mathrm{sech}}^2}x\cdot dx$

Therefore the integral can now be evaluated as,

  $\begin{array}{l}{\displaystyle \underset{-1n\sqrt{3}}{\overset{1n\sqrt{3}}{\int }}\pi \sec h^{2}x}\cdot dx=\pi {\left[\tan x\right]}_{-1n\sqrt{3}}^{1n\sqrt{3}}\\ =\pi \left(\frac{e^{1n\sqrt{3}}-e^{-1n\sqrt{3}}}{e^{1n\sqrt{3}}+e^{-1n\sqrt{3}}}-\frac{e^{-1n\sqrt{3}}-e^{1n\sqrt{3}}}{e^{-1n\sqrt{3}}+e^{1n\sqrt{3}}}\right)\\ =2\pi \left(\frac{e^{1n\sqrt{3}}-e^{-1n\sqrt{3}}}{e^{1n\sqrt{3}}+e^{-1n\sqrt{3}}}\right)\\ =2\pi \left(\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\right)\\ \boxed{=\pi }\end{array}$

Conclusion:

Therefore the volume of the solid is $\pi$