## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{1}_{0}\frac{4ds}{\sqrt{4-s^2}}=\frac{2\pi}{3}$$
$$A=\int^{1}_{0}\frac{4ds}{\sqrt{4-s^2}}=4\int^{1}_{0}\frac{ds}{\sqrt{2^2-s^2}}$$ Since we have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$ Therefore, $$A=4\sin^{-1}\Big(\frac{s}{2}\Big)\Big]^{1}_{0}$$ $$A=4\Big(\sin^{-1}\frac{1}{2}-\sin^{-1}0\Big)$$ $$A=4\Big(\frac{\pi}{6}-0\Big)$$ $$A=\frac{2\pi}{3}$$