Answer
$\frac{8}{3}$
Work Step by Step
V=$\int\int f(x,y)dA$
=$\int^{1}_{-1} \int^{1}_{-1}(x^2+y^2)dy dx$
=$\int^{1}_{-1}[x^2y+\frac{1}{3}y^3]^1_{-1}dx$
=$\int^{1}_{-1}(2x^2+\frac{2}{3})dx$
=$[\frac{2}{3}x^3+\frac{2}{3}x]^1_{-1}$
=$\frac{8}{3}$
