# How do you find the arc length of x=2/3(y-1)^(3/2) between 1<=y<=4?

Aug 26, 2014

It can be found by $L = {\int}_{0}^{4} \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$.

Let us evaluate the above definite integral.

By differentiating with respect to y,
$\frac{\mathrm{dx}}{\mathrm{dy}} = {\left(y - 1\right)}^{\frac{1}{2}}$

So, the integrand can be simplified as
$\sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} = \sqrt{1 + {\left[{\left(y - 1\right)}^{\frac{1}{2}}\right]}^{2}} = \sqrt{y} = {y}^{\frac{1}{2}}$

Finally, we have
$L = \setminus {\int}_{0}^{4} {y}^{\frac{1}{2}} \mathrm{dy} = {\left[\frac{2}{3} {y}^{\frac{3}{2}}\right]}_{0}^{4} = \frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(0\right)}^{\frac{3}{2}} = \frac{16}{3}$

Hence, the arc length is $\frac{16}{3}$.

I hope that this helps.