## Thomas' Calculus 13th Edition

$\frac{16}{3}$
V=$\int \int f(x,y)dA$ =$\int^{1}_{0} \int^{2}_{0}(4-y^2)dy dx$ =$\int^{1}_{0}[4y-\frac{1}{3}y^3]^2_0dx$ =$\int^{1}_{0}(\frac{16}{3})dx$ =$[\frac{16}{3}dx]$ =$\frac{16}{3}$