324. $y=\frac{1}{2} x^{3}, x=-2, x=4$ and $y=0$.
Solution. The graph of the function $y=\frac{1}{2} x^{3}$ lies below the $O x$ axis on the interval $[-2,0]$, and above the $O x$ axis on the interval $[0,4]$ (Fig. 171). Therefore,

$S=\left|\int_{-2}^{0} \frac{1}{2} x^{3} d x\right|+\int_{0}^{4} \frac{1}{2} x^{3} d x=\left|\frac{1}{8} x^{4}\right|_{-2}^{0}+\left.\frac{1}{8} x^{4}\right|_{0} ^{4}=2+32=34$ (sq. units).