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

y=\sqrt(4-x^2)

1 Answer
Jun 2, 2018

pi/3 approx 1.0472

Explanation:

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

s = int_a^b sqrt(1+(dy/dx)^2) dx

In this example y = sqrt(4-x^2)

dy/dx = 1/2(4-x^2)^(-1/2) * (-2x)

= (-x)/sqrt(4-x^2)

:. s = int_0^1 sqrt((1+x^2/(4-x^2))) dx

= int_0^1 sqrt((4-x^2+x^2)/(4-x^2)) dx

= int_0^1 2/sqrt(4-x^2) dx

Let u = x/2 -> dx =2du

s= int_0^(1/2) (2xx2)/sqrt(4-4u^2) du

= 4/2*int_0^(1/2) 1/sqrt(1-u^2) du

Apply standard integral

s = 2* [arcsinu]_0^(1/2)

= 2 [arcsin(1/2) - arcsin (0)]

= 2 * [pi/6 -0] = pi/3

approx 1.0472

Hence the length of y from 0 to 1 is pi/3 units.