Using integration, what is the area enclosed by the circle $x^2 + y^2 = 16$?

Explanation

1. 01

   By symmetry the area is $4\int_0^4 \sqrt{16 - x^2}\,dx$, four quarter-circles.

2. 02

   $\int_0^4 \sqrt{16 - x^2}\,dx = \left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}\right]_0^4 = 4\pi$, so the area is $16\pi$.