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.