Answer
$18$
Work Step by Step
Apply formula (2),
$\displaystyle \int_{-3}^{3}\int_{0}^{\pi/2} (y +y^{2}\cos x)dxdy =\displaystyle \int_{-3}^{3}[\int_{0}^{\pi/2} (y +y^{2}\cos x)dx]dy$
...treat $y$ as a constant in the inner integral.
$=\displaystyle \int_{-3}^{3}[xy+y^{2}\sin x]_{x=0}^{x=\pi/2}dy$
$=\displaystyle \int_{-3}^{3}[((\frac{\pi}{2})y+y^{2}\sin\frac{\pi}{2})-(0+0)]dy$
$=\displaystyle \int_{-3}^{3}(\frac{\pi}{2}y+y^{2})dy$
$=[\displaystyle \frac{\pi}{2}\cdot\frac{y^{2}}{2}+\frac{y^{3}}{3}]_{-3}^{3}$
$=[\displaystyle \frac{9\pi}{4}+9-(\frac{9\pi}{4}-9)]$
$=18$
