What is the length of the path along the graph of y=√(4-x^2) between x =1 and x = 2?

Question

a) 2π/3
b) π/3
c) π/6
d) π/4

Answer 1

The calculus way:
y' = -x/√(4-x^2)
∫[1,2] √(1+x^2/(4-x^2)) dx
= ∫[1,2] √(2/(4-x^2)) dx = 2arcsin(x/2) [1,2] = 2(π/2 - π/6) = 2π/3

The easy way:
The angle formed by the line from (0,0) to (1,√3) is π/3
s = rθ = 2π/3