Answer
14
Work Step by Step
$\int^0_{-2}$$\int^2_{-1}$ $( x^{2} + y^{2} )dx dy$
Evaluating Inner Integral First :
$\int^2_{-1}$$( x^{2} + y^{2} )dx$
$[\frac{x^3}{3}+xy^2]^2_{-1}$
$[\frac{2^3}{3}+2y^2]$ - $[\frac{(-1)}{3}+(-1)y^2]$
$[\frac{8}{3}+2y^2]$ - $[\frac{-1}{3}-y^2]$
$\frac{8}{3}+2y^2 +\frac{1}{3}+y^2$ = $\frac{9}{3} +3y^2=3+3y^2$
Evaluating Outer Integral :
$\int^0_{-2}$$( 3 + 3y^{2} )dy$
$[{3y}+y^3]^0_{-2}$
$[{3(0)}+0^3]$ - $[{3(-2)}+(-2)^3]$
0-[-6-8]=-[-14] = 14
