1. ## Finding the Arclength

Use the arc length formula to find the length L of the curve below, 0 ≤ x2. Check your answer by noting that the curve is part of a circle.
y=sqrt(8-x^2)

Here is my work:
ds = sqrt(1+(dy/dx)^2)

dy/dx = x/sqrt(8-x^2)

s=int(sqrt(1-(x^2/(x^2-8)))) from 0 to 2

where ds is a smal portion of the arc and s is the total arclength

My problem comes from where s is equal to the integral, I don't know how to integrate it. Thanks for any help!

2. Originally Posted by dillonmhudson
Use the arc length formula to find the length L of the curve below, 0 ≤ x2. Check your answer by noting that the curve is part of a circle.
y=sqrt(8-x^2)

Here is my work:
ds = sqrt(1+(dy/dx)^2)

dy/dx = x/sqrt(8-x^2)

s=int(sqrt(1-(x^2/(x^2-8)))) from 0 to 2

where ds is a smal portion of the arc and s is the total arclength

My problem comes from where s is equal to the integral, I don't know how to integrate it. Thanks for any help!
let $r^2 = 8$

$y = \sqrt{r^2-x^2}$

$y' = \frac{-x}{\sqrt{r^2-x^2}}$

$(y')^2 = \frac{x^2}{r^2-x^2}$

$1+(y')^2 = \frac{r^2-x^2}{r^2-x^2} + \frac{x^2}{r^2-x^2} = \frac{r^2}{r^2-x^2} = \frac{1}{1 - \left(\frac{x}{r}\right)^2}
$

$\sqrt{\frac{1}{1 - \left(\frac{x}{r}\right)^2}} = \frac{1}{\sqrt{1 - \left(\frac{x}{r}\right)^2}}$

$\int_0^2 \frac{1}{\sqrt{1 - \left(\frac{x}{r}\right)^2}} \, dx$

$r \int_0^2 \frac{\frac{1}{r}}{\sqrt{1 - \left(\frac{x}{r}\right)^2}} \, dx$

$r \left[ \arcsin\left(\frac{x}{r}\right) \right]_0^2$

$r \arcsin\left(\frac{2}{r}\right)$

$2\sqrt{2} \arcsin\left(\frac{1}{\sqrt{2}}\right)$

$2\sqrt{2} \cdot \frac{\pi}{4} = \frac{\pi \sqrt{2}}{2}$