# Find the arc length of the curve along the interval 0\lex\le1?

## $y = \setminus \sqrt{4 - {x}^{2}}$

Jun 2, 2018

$\frac{\pi}{3} \approx 1.0472$

#### Explanation:

The length $\left(s\right)$ along a curve $\left(y\right)$ from $a$ to $b$ is given by:

$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

In this example $y = \sqrt{4 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right)$

$= \frac{- x}{\sqrt{4 - {x}^{2}}}$

$\therefore s = {\int}_{0}^{1} \sqrt{\left(1 + {x}^{2} / \left(4 - {x}^{2}\right)\right)} \mathrm{dx}$

$= {\int}_{0}^{1} \sqrt{\frac{4 - {x}^{2} + {x}^{2}}{4 - {x}^{2}}} \mathrm{dx}$

$= {\int}_{0}^{1} \frac{2}{\sqrt{4 - {x}^{2}}} \mathrm{dx}$

Let $u = \frac{x}{2} \to \mathrm{dx} = 2 \mathrm{du}$

$s = {\int}_{0}^{\frac{1}{2}} \frac{2 \times 2}{\sqrt{4 - 4 {u}^{2}}} \mathrm{du}$

$= \frac{4}{2} \cdot {\int}_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - {u}^{2}}} \mathrm{du}$

Apply standard integral

$s = 2 \cdot {\left[\arcsin u\right]}_{0}^{\frac{1}{2}}$

$= 2 \left[\arcsin \left(\frac{1}{2}\right) - \arcsin \left(0\right)\right]$

$= 2 \cdot \left[\frac{\pi}{6} - 0\right] = \frac{\pi}{3}$

$\approx 1.0472$

Hence the length of $y$ from $0$ to $1$ is $\frac{\pi}{3}$ units.