What is the arc length of #f(x)=sqrt(4-x^2) # on #x in [-2,2]#?

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Answer 1 · 2015-11-08T02:51:33

arc length #=intsqrt [1+(dy/dx)^2]dx#

#dy/dx = (-x)/[sqrt(4-x^2)]#

#(dy/dx)^2=x^2/(4-x^2)#

arc length #=int sqrt[1+x^2/(4-x^2)]dx=2int1/sqrt(4-x^2)dx#

#=arcsin(x/2)+C#

Finally, evaluate over [-2,2]

arc length #=2pi#

hope that helped