How do you find the arc length of the curve #sqrt(4-x^2)# from [-2,2]?

Question

Cameron Alexander · Accepted Answer

The answer turned out to be #2pi# units. The arc length is essentially the usage of the distance formula with a small, independent change in x and a small change in y, where y changes according to #f(x)#. Thus: #D = sqrt((Deltax)^2 + (Deltay)^2)# #= sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)# #= sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax# Make it a sum, and you've got the arc length over an interval #[a,b]#. #s = sum_a^b sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax# Turn it into an integral expression to get:
#s = int_a^b sqrt(1 + ((dy)/(dx))^2)dx# So, we could take the derivative first to get: #(dy)/(dx) = 1/(2sqrt(4-x^2))*-2x = (-x)/(sqrt(4-x^2))# Then, let's square it and plug it in. #((dy)/(dx))^2 = x^2/(4-x^2)# #-> s =  int_a^b sqrt(1 + x^2/(4-x^2))dx# Cross-multiply:
#=  int_a^b sqrt((4-x^2 + x^2)/(4-x^2))dx# #=  int_a^b sqrt((4)/(4-x^2))dx# #=  2int_a^b 1/sqrt(4-x^2)dx# And we've got ourselves a u-substitution, if you know that #d/(dx)[arcsin(x)] = 1/sqrt(1-x^2)#. #=  2int_a^b 1/sqrt4*1/sqrt(1-x^2/4)dx# #= int_a^b 1/sqrt(1-(x/2)^2)dx# Let #u = x/2# and you get: #dx = 2du# #= 2int_a^b 1/sqrt(1-u^2)du# #= [2 arcsin(x/2)]|_(-2)^(2)# #= 2 arcsin((2)/2) - 2 arcsin((-2)/2)# #= 2 arcsin(1) - 2 arcsin(-1)# Let:
#y = arcsin(1) -> siny = 1 -> y = pi/2# #y = arcsin(-1) -> siny = -1 -> y = (3pi)/2# #-> 2(pi/2) - 2((3pi)/2) -> 2(pi/2) - 2((-pi)/2)#
(so that our answer is positive) #= pi + pi = color(blue)(2pi " u")#