问题
已知积分区域 $\Sigma$ $=$ $\Sigma_{1}$ $+$ $\Sigma_{2}$ , 则，根据第二类曲面积分的性质，$\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $?$选项
[A]. $\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $\iint_{\frac{1}{\Sigma_{1}}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $+$ $\iint_{\frac{1}{\Sigma_{2}}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$[B]. $\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $\iint_{\Sigma_{1}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $\times$ $\iint_{\Sigma_{2}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$
[C]. $\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $\iint_{\Sigma_{1}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $-$ $\iint_{\Sigma_{2}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$
[D]. $\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $\iint_{\Sigma_{1}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $+$ $\iint_{\Sigma_{2}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$
$\iint_{\Sigma}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $=$ $\iint_{\Sigma_{1}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$ $+$ $\iint_{\Sigma_{2}}$ $P$ $\mathrm{~d} y$ $\mathrm{~d} z$