Chapter 6.4, Problem 26E

To determine

Find the lengthof the given curve.

Answer to Problem 26E

The volume of the curve is $2-\sqrt{2}-\ln \sqrt{3}+\ln \left(1+\sqrt{2}\right)$

Explanation of Solution

Given information: Given curves $y=\ln x,,1\le x\le \sqrt{3};$

Calculation:

To the length of the curve, to calculate: $L={\displaystyle \underset{1}{\overset{\sqrt{3}}{\int }}\sqrt{1+{\left(y^{\prime }\right)}^2}}dx.$

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

Finally, to find the length of the curve;

  $\begin{array}{l}L={\displaystyle \underset{1}{\overset{\sqrt{3}}{\int }}\sqrt{1+{\left(\frac{1}{x}\right)}^2}}dx.\\ ={\displaystyle \underset{1}{\overset{\sqrt{3}}{\int }}\sqrt{\frac{x^2+1}{x^2}}}dx\\ ={\displaystyle \underset{1}{\overset{\sqrt{3}}{\int }}\frac{\sqrt{x^2+1}}{x}}dx\\ ={\left|\sqrt{x^2+1}-\ln \left|\frac{1+\sqrt{x^2+1}}{x}\right|\right|}_0^{\sqrt{3}}\\ =\left(\sqrt{1+3}-\ln \left|\frac{1+\sqrt{1+3}}{\sqrt{3}}\right|\right)-\left(\sqrt{1+1}-\ln \left|\frac{1+\sqrt{1+1}}{1}\right|\right)\\ =2-\ln \sqrt{3}-\sqrt{2}+\ln \left(1+\sqrt{2}\right)\\ =2-\sqrt{2}-\ln \sqrt{3}+\ln \left(1+\sqrt{2}\right)\end{array}$

Thus, the volume of the curve is $2-\sqrt{2}-\ln \sqrt{3}+\ln \left(1+\sqrt{2}\right)$