Answer
$\dfrac{8}{3}$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{-1}^{1} \int_{-1}^{1} (x^2+y^2) dy dx=\int_{-1}^{1} (x^2y+\dfrac{y^3}{y}]_{-1}^{1} dx$
This implies that
$\int_{-1}^{1} [2x^2+\dfrac{2}{3}] dx=(2)(\dfrac{x^3}{3}]_{-1}^{1} dy$
Hence, $(\dfrac{2}{3})[1^3-(-1)^3]=\dfrac{8}{3}$
