Chapter 6.4, Problem 25E

To determine

Find the exact length of the given curve.

Answer to Problem 25E

The volume of the curve is $\ln \left(1+\sqrt{2}\right)$

Explanation of Solution

Given information: Given curves $y=\ln \left(\cos x\right),0\le x\le \frac{\pi }{4}$

Calculation:

To the length of the curve, to calculate: $L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(y^{\prime }\right)}^2}}dx.$

As know that ${\left(\ln x\right)}^{\prime }=\frac{1}{x}$ ,to find $y^{\prime }$ :

  $\begin{array}{l}{y}^{\prime }={\left(\ln \left(\cos x\right)\right)}^{\prime }\\ =\frac{1}{\cos x}{\left(\cos x\right)}^{\prime }\\ =\frac{1}{\cos x}\left(-\sin x\right)\\ =-\frac{\sin x}{\cos x}\\ =-\tan x\end{array}$

Finally ,to find the length of the curve;

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(-\tan x\right)}^2}}dx.\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(\tan x\right)}^2}}dx\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{{\left(\sec x\right)}^2}}dx\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\sec x\right)}dx\\ =\ln {\left|\sec x+\tan x\right|}_0^{\frac{\pi }{4}}\\ =\ln \left|\sec \frac{\pi }{4}+\tan \frac{\pi }{4}\right|-\ln \left|\sec 0+\tan 0\right|\\ =\ln \left|\sqrt{2}+1\right|-\ln \left|1+0\right|\\ =\ln \left(1+\sqrt{2}\right)\end{array}$

Thus, the volume of the curve is $\ln \left(1+\sqrt{2}\right)$