$$I=\iint_{\Sigma} 2 x^{3} \mathrm{~d} y \mathrm{~d} z+2 y^{3} \mathrm{~d} z \mathrm{~d} x+3\left(z^{2}-1\right) \mathrm{d} x \mathrm{~d} y,$$

$$I=\iint_{\Sigma+\Sigma_{1}} 2 x^{3} d y d z+2 y^{3} d z d x+3\left(z^{2}-1\right) d x d y$$
$$-\iint_{\Sigma_{1}} 2 x^{3} d y d z+2 y^{3} d z d x+3\left(z^{2}-1\right) d x d y=I_{1}-I_{2}$$

$$\iint_{\Sigma} P d y d z+Q d z d x+R d x d y=\iiint_{\Omega}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right) d v$$

$$I_{1}=\iiint_{\Omega} 6\left(x^{2}+y^{2}+z\right) d x d y d z=6 \int_{0}^{2 \pi} d \theta \int_{0}^{1} d r \int_{0}^{1-r^{2}}\left(z+r^{2}\right) r d z$$
$=12 \pi \int_{0}^{1} r\left(\frac{z^{2}}{2}+r^{2} z\right)_{0}^{1-r^{2}} d r=12 \pi \int_{0}^{1} r \frac{\left(1-r^{2}\right)^{2}}{2}+r^{3}\left(1-r^{2}\right) d r$
$$=12 \pi\left(-\frac{1}{4} \cdot \frac{\left(1-r^{2}\right)^{3}}{3}+\frac{r^{4}}{4}-\frac{r^{6}}{6}\right)_{0}^{1}=12 \pi \cdot \frac{1}{6}=2 \pi$$

$$I_{2}=\iint_{\Sigma_{1}} 2 x^{3} d y d z+2 y^{3} d z d x+3\left(z^{2}-1\right) d x d y=-\iint_{D} 3(0-1) d x d y=3 \iint_{D} d x d y=3 \pi$$
(其中 $\iint_{D} d x d y$ 为半径为 1 圆的面积, 所以 $\iint_{D} d x d y=\pi \cdot 1^{1}=\pi$ )