The length (s)(s) along a curve (y)(y) from aa to bb is given by:
s = int_a^b sqrt(1+(dy/dx)^2) dxs=∫ba√1+(dydx)2dx
In this example y = sqrt(4-x^2)y=√4−x2
dy/dx = 1/2(4-x^2)^(-1/2) * (-2x)dydx=12(4−x2)−12⋅(−2x)
= (-x)/sqrt(4-x^2)=−x√4−x2
:. 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.