I am looking at the proof of the divergence theorem and I have some questions.
The proof of the divergence theorem
$$\iiint_D \nabla \cdot \overrightarrow{F} dV= \iint_S \overrightarrow{F} \cdot \hat{n} ds$$
is the following:
$$z_1=f_1(x,y), \ \ \ \ \ \ \ \ \ \ (x,y) \in \mathbb{R}$$ $$z_2=f_2(x,y), \ \ \ \ \ \ \ \ \ \ (x,y) \in \mathbb{R}$$
$$\overrightarrow{F}=M\hat{\imath}+N \hat{\jmath}+P \hat{k}$$
$$\hat{n}=\hat{\imath} \cos{\alpha}+\hat{\jmath} \cos{\beta}+ \hat{k} \cos{\gamma}$$
$$\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma=\iint_S (M \cos{\alpha}+N \cos{\beta}+P \cos{\gamma}) d \sigma=\iiint_D(\frac{\partial{M}}{\partial{x}}+\frac{\partial{N}}{\partial{y}}+\frac{\partial{P}}{\partial{z}})dxdydz$$
$$\iint_S M \cos{\alpha}d \sigma=\iiint_D \frac{\partial{M}}{\partial{x}}dxdydz$$ $$\iint_S N \cos{\beta}d \sigma=\iiint_D \frac{\partial{N}}{\partial{y}}dxdydz$$ $$\iint_S P \cos{\gamma}d \sigma=\iiint_D \frac{\partial{P}}{\partial{z}}dxdydz$$
$$\iint_S P \cos{\gamma}d \sigma=\iint_{S_1} P \cos{\gamma}d \sigma+\iint_{S_2} P \cos{\gamma}d \sigma$$
$$\hat{k} \cdot \hat{n_2}=\cos{\gamma_2}>0$$ $$\cos{\gamma_2}d \sigma=dxdy$$
$$\cos{\gamma_1}<0$$ $$\cos{\gamma_1} d \sigma=-dxdy$$
$$\\$$
$$\iint_S P \cos{\gamma} d \sigma=\iint_R P(x,y,z_2) dxdy-\iint_R P(x,y,z_1)dxdy=\iint_R[P(x,y,z_2)-P(x,y,z_1)]dxdy=\iint_R[\int_{z_1}^{z_2}\frac{\partial{P}}{\partial{z}}dz]dxdy=\iiint_D \frac{\partial{P}}{\partial{z}}dxdydz$$
So $$\iiint_D \nabla \cdot \overrightarrow{F} d \overrightarrow{V}=\iint_S \overrightarrow{F} \cdot \hat{n}d \sigma$$
$$$$
Are $z_1$ and $z_2$ the surfaces of $S_1$ and $S_2$ respectively??
Why do we take at the beginning: $\hat{n}=\hat{\imath} \cos{\alpha}+\hat{\jmath} \cos{\beta}+ \hat{k} \cos{\gamma}$ ??
Could you explain me the part where we prove that: $\iint_S P \cos{\gamma}d \sigma=\iiint_D \frac{\partial{P}}{\partial{z}}dxdydz$ ??