Chapter 6.4, Problem 9E

To determine

Find the length of the curve $x=y^{\frac{3}{2}},0\le y\le 1$

Answer to Problem 9E

The length of the curve is $\frac{13\sqrt{13}-8}{27}$

Explanation of Solution

Given information: Given curve $x=y^{\frac{3}{2}},0\le y\le 1$

Calculation:

To find the length of the curve $x=y^{\frac{3}{2}},0\le y\le 1$ ,

Use the formula:

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+{\left(f^{\prime }\left(y\right)\right)}^2}}dy\\ L={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+{\left({\left(y^{\frac{3}{2}}\right)}^{\prime }\right)}^2}}dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+{\left(\frac{3}{2}{y}^{\frac{3}{2}-1}\right)}^2}}dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+{\left(\frac{3}{2}{y}^{\frac{1}{2}}\right)}^2}}dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+\frac{9}{4}}y}dy\end{array}$

To solve the integral

  ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+\frac{9}{4}}y}dy.$

Put that $1+\frac{9}{4}y=u,$ then it is $\frac{9}{4}dy=du\to dy=\frac{4}{9}du.$

  $\begin{array}{l}\text{If }y=o,\text{then }1+\frac{9}{4}\cdot 0=1=u.\\ \text{If }y=1,\text{then }1+\frac{9}{4}\cdot 1+\frac{9}{4}=\frac{13}{4}=u\end{array}$

Now, solve the integral

  $\begin{array}{l}L={\displaystyle \underset{1}{\overset{\frac{13}{4}}{\int }}\frac{9}{4}\sqrt{u}}du\\ =\frac{9}{4}{\displaystyle \underset{1}{\overset{\frac{13}{4}}{\int }}{u}^{\frac{1}{2}}}du\\ =\frac{9}{4}{\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_1^{\frac{13}{4}}\\ =\frac{9}{4}\frac{2}{3}{\left[u^{\frac{3}{2}}\right]}_1^{\frac{13}{4}}\\ =\frac{8}{27}\left({\left(\frac{13}{4}\right)}^{\frac{3}{2}}-1^{\frac{3}{2}}\right)\\ =\frac{8}{27}\left(\frac{13\sqrt{13}}{4\sqrt{4}}-1\right)\\ =\frac{8}{27}\left(\frac{13\sqrt{13}}{8}-1\right)\\ =\frac{8}{27}\frac{13\sqrt{13}}{8}-1\\ =\frac{8}{27}\frac{13\sqrt{13}-8}{8}\\ =\frac{13\sqrt{13}-8}{27}\end{array}$

Thus the length of the curve is $\frac{13\sqrt{13}-8}{27}$