问题
已知 $\Sigma$ 为有向曲面，$\Sigma^{-}$ 与 $\Sigma$ 的法向量相反，则，根据第二类曲面积分中积分区域的方向性，以下选项中，正确的是哪个？选项
[A]. $\begin{cases} \iint_{\Sigma^{-}} P \mathrm{~d} y \mathrm{~d} z=-\iint_{\frac{1}{\Sigma}} P \mathrm{~d} y \mathrm{~d} z, \\ \iint_{\Sigma^{-}} Q \mathrm{~d} z \mathrm{~d} x=-\iint_{\frac{1}{\Sigma}} Q \mathrm{~d} z \mathrm{~d} x, \\ \iint_{\Sigma^{-}} R \mathrm{~d} x \mathrm{~d} y=-\iint_{\frac{1}{\Sigma}} R \mathrm{~d} x \mathrm{~d} y. \end{cases}$[B]. $\begin{cases} \iint_{\Sigma^{-}} P \mathrm{~d} y \mathrm{~d} z=\iint_{\Sigma} P \mathrm{~d} y \mathrm{~d} z, \\ \iint_{\Sigma^{-}} Q \mathrm{~d} z \mathrm{~d} x=\iint_{\Sigma} Q \mathrm{~d} z \mathrm{~d} x, \\ \iint_{\Sigma^{-}} R \mathrm{~d} x \mathrm{~d} y=\iint_{\Sigma} R \mathrm{~d} x \mathrm{~d} y. \end{cases}$
[C]. $\begin{cases} \iint_{\Sigma^{-}} P \mathrm{~d} y \mathrm{~d} z=-\iint_{\Sigma} P \mathrm{~d} y \mathrm{~d} z, \\ \iint_{\Sigma^{-}} Q \mathrm{~d} z \mathrm{~d} x=-\iint_{\Sigma} Q \mathrm{~d} z \mathrm{~d} x, \\ \iint_{\Sigma^{-}} R \mathrm{~d} x \mathrm{~d} y=-\iint_{\Sigma} R \mathrm{~d} x \mathrm{~d} y. \end{cases}$
[D]. $\begin{cases} \iint_{\Sigma^{-}} P \mathrm{~d} y \mathrm{~d} z=\iint_{\frac{1}{\Sigma}} P \mathrm{~d} y \mathrm{~d} z, \\ \iint_{\Sigma^{-}} Q \mathrm{~d} z \mathrm{~d} x=\iint_{\frac{1}{\Sigma}} Q \mathrm{~d} z \mathrm{~d} x, \\ \iint_{\Sigma^{-}} R \mathrm{~d} x \mathrm{~d} y=\iint_{\frac{1}{\Sigma}} R \mathrm{~d} x \mathrm{~d} y. \end{cases}$
$\begin{cases} \iint_{\Sigma^{-}} P \mathrm{~d} y \mathrm{~d} z=-\iint_{\Sigma} P \mathrm{~d} y \mathrm{~d} z, \\ \iint_{\Sigma^{-}} Q \mathrm{~d} z \mathrm{~d} x=-\iint_{\Sigma} Q \mathrm{~d} z \mathrm{~d} x, \\ \iint_{\Sigma^{-}} R \mathrm{~d} x \mathrm{~d} y=-\iint_{\Sigma} R \mathrm{~d} x \mathrm{~d} y. \end{cases}$