# Confused with the surface integral in the proof of the divergence theorem

I know how to work with the triple integral of the divergence of F part of the theorem, but in many textbooks, they don't explain the surface integral component. I don't understand how they go from here: $$\iint_{\delta W} F\boldsymbol{\cdot}kdS = \iint_{S_1} F\boldsymbol{\cdot}kdS_1 + \iint_{S_2} F\boldsymbol{\cdot}kdS_2$$

to here: "The surface $$S_1$$ is defined by $$g_1(x,y)$$, and $$\begin{equation}dS_1 = \Big( \frac{\delta g_1}{\delta x}i + \frac{\delta g_1}{\delta y}j - k \Big)dxdy\end{equation}$$ Therefore, $$\iint_{S_1} F\boldsymbol{\cdot}kdS_1 = - \iint_D F(x,y,g_1(x,y))dxdy$$

Similarly, for the top face $$S_2$$, $$\begin{equation}dS_2 = \Big( -\frac{\delta g_2}{\delta x}i - \frac{\delta g_2}{\delta y}j + k \Big)dxdy\end{equation}$$ Therefore, $$\iint_{S_2} F\boldsymbol{\cdot}kdS_2 = \iint_D F(x,y,g_2(x,y))dxdy$$"

I understand why one is negative and the other isn't, but I don't understand how they got the derivative of the surface like that.

• What is relation between $\partial W,S_1 ,S_2$ ? – Ajay Mishra Jul 9 at 10:11
• $\delta W$ is the entire closed surface that encloses the 3D region $W$, and $S_1$ is the bottom surface and $S_2$ is the top surface of the region, which means $\delta W= S_1 + S_2$ – ororodislikesphysicalexertion Jul 9 at 10:34

In general \begin{align} \iint_S \mathbf F \cdot \mathbf {dS} &= \iint_S( \mathbf F \cdot \mathbf n )dS \\ &= \iint_S\Bigg(\mathbf F \cdot \cfrac{\frac{\partial \mathbf x}{\partial s} \times \frac{\partial \mathbf x}{\partial t}}{\Big|\Big| \frac{\partial \mathbf x}{\partial s} \times \frac{\partial \mathbf x}{\partial t} \Big|\Big|} \Bigg) \Bigg|\Bigg| \dfrac{\partial \mathbf x}{\partial s} \times \dfrac{\partial \mathbf x}{\partial t} \Bigg|\Bigg| ds dt \\ &= \iint_S \mathbf F \cdot \bigg( \cfrac{\partial \mathbf x}{\partial s} \times \cfrac{\partial \mathbf x}{\partial t} \bigg) dsdt\end{align} Whose implication is $$\mathbf {dS} = \Bigg(\cfrac{\partial \mathbf x}{\partial s} \times \cfrac{\partial \mathbf x}{\partial t} \Bigg) dsdt$$
Where $$\mathbf x$$ is position vector with of point $$x \hat i + y \hat j + f(x,y) \hat k$$, considering $$s = x, t = y$$ $$\implies \cfrac{\partial \mathbf x}{\partial x} = 1 \hat i + 0 \hat j + f_x(x,y) \hat k , \cfrac{\partial \mathbf x}{\partial y} = 0 \hat i + i \hat j + f_y(x,y) \hat k$$