## Calculus (3rd Edition)

$4$
First, we can rewrite the integral as follows $$\int\int_R(2+x^2y)dA=\int\int_R2dA+\int\int_Rx^2ydA$$ Now, the integral $\int\int_R(2+x^2y)dA=2(1-0)(1-(-1))=4.$ The integral $\int\int_Rx^2ydA=0$. We know this because $x^2(-y)=-x^2y$, and because of symmetry, the (negative) signed volume of the region below the $xy$-plane where $-1 \leq y \leq 0\pi$ cancels with the (positive) signed volume of the region above the xy-plane where $0 \leq y \leq1$. Thus, result is $4$.