Answer
$\pi$
Work Step by Step
The arc length can be computed as: $L=\int_a^b \sqrt {1+(y^{\prime})^2} \ dx $
We are given that $y=\sqrt {4-x^2}$ and $y^{\prime}=-\dfrac{x}{\sqrt {4-x^2}}$
Now, the arc length is: $L=\int_{0}^{2} 1+(-\dfrac{x}{\sqrt {4-x^2}})^2 \ dx \\=\int_0^2 \sqrt {\dfrac{4}{4-x^2}}\ dx \\=2 \lim\limits_{a \to 2^{-}}\int_0^a \dfrac{1}{\sqrt {4-x^2}}\ dx\\=2 \lim\limits_{a \to 2^{-}} [\arcsin (x/2)]_0^a\\=2 \lim\limits_{a \to 2^{-}}[\arcsin (a/2)-\arcsin (0)]\\=2 (\dfrac{\pi}{2}-0)\\=\pi$
