## Thomas' Calculus 13th Edition

a)2$\pi$ b)$\pi$
a) $\int_{-2}^2 \sqrt{4-x^2}dx$ =$\frac{1}{2}[x(2)^2]$ =$2\pi$ b) $\int_0^2\sqrt{4-x^2}dx$ =$\frac{1}{4}[x(2)^2]$ =$\pi$