Answer
$\frac{160}{3}$
Work Step by Step
V=$\int\int f(x,y)dA$
=$\int^{2}_{0} \int^{2}_{0}(16-x^2-y^2)dy dx$
=$\int^{2}_{0}[16-x^2y-\frac{1}{3}y^3]^2_0dx$
=$\int^{2}_{0}(\frac{88}{3}-2x^2)dx$
=$[\frac{88}{3}x-\frac{2}{3}x^3]^2_0$
=$\frac{160}{3}$
