What is the relationship between the differential and integral definitions of vector operators?

Say we have the integral definition of divergence:

$$\nabla \cdot \vec F=\lim_{V\to 0} \frac{1}{|V|} \int_{S}\vec{F}\cdot\vec{dS}$$

How does it relate to the, all things considered, more useful definition making use of partial derivatives (and in orthogonal curvilinear coordinates):

$$\nabla \cdot \vec F=\frac{1}{h_{\alpha}h_{\beta}h_{\gamma}}\Bigg\lbrace\frac{\partial(F_{\alpha}h_{\beta}h_{\gamma})}{\partial \alpha}+...\Bigg\rbrace$$

Solution 1:

Take as your volume the cube with a corner at $\mathbf{x}$ with side-length $h$. Then the first definition of divergence becomes $$\nabla\cdot \vec{F} (\mathbf{x})= \lim_{h\to0}\frac{1}{h^3}\int_{B_{\mathbf{x}}(h\vec{e}_1, h\vec{e_2}, h\vec{e}_3)}\vec{F}\cdot d\vec{S}$$. Where $$B_{\mathbf{x}}(\vec{v}_1, \vec{v}_2,\dots)$$ Denotes the cube/square spanned by the vectors starting at $\mathbf{x}$.

This splits into an integral over each of the $6$ sides of the cube, with signs to account for the unit normal pointing opposite ways on opposite sides. $$= \lim_{h\to 0} \frac{1}{h^3}\left[\left(\int_{B_{\mathbf{x} + h\mathbf{e}_1}(h\vec{e}_2, h\vec{e}_3)}\vec{F}\cdot d\vec{S} - \int_{B_{\mathbf{x}}(h\vec{e}_1,he_2)}\vec{F}\cdot d\vec{S} \right )-\left(\int_{B_{\mathbf{x}+h\mathbf{e}_2}(h\vec{e}_1, h\vec{e}_3)} \vec{F}\cdot d\vec{S} - \int_{B_{\mathbf{x}}(h\vec{e}_1, h\vec{e}_3)}\vec{F}\cdot d\vec{S}\right) + \left(\int_{B_{\mathbf{x} + h\mathbf{e}_3}(h\vec{e}_1, h\vec{e}_2)}\vec{F}\cdot d\vec{S}-\int_{B_{\mathbf{x}}(h\vec{e}_1, h\vec{e}_2)}\vec{F}\cdot d\vec{S}\right)\right]$$ Because this is a cube, the normals are just the unit basis vectors, so you get $$= \lim_{h\to 0} \frac{1}{h^3}\left[\left(\int_{B_{\mathbf{x} + h\mathbf{e}_1}(h\vec{e}_2, h\vec{e}_3)}F_1dS - \int_{B_{\mathbf{x}}(h\vec{e}_1,he_2)}F_1 dS \right )-\left(\int_{B_{\mathbf{x}+h\mathbf{e}_2}(h\vec{e}_1, h\vec{e}_3)} F_2 dS - \int_{B_{\mathbf{x}}(h\vec{e}_1, h\vec{e}_3)}F_2 dS\right) + \left(\int_{B_{\mathbf{x} + h\mathbf{e}_3}(h\vec{e}_1, h\vec{e}_2)}F_3 dS-\int_{B_{\mathbf{x}}(h\vec{e}_1, h\vec{e}_2)}F_3 dS\right)\right]$$ Finally, we can compute the integrals. Parametrize every square $B_{\mathbf{x}}(e_i, e_j)$ with $\gamma(t_1, t_2) = \mathbf{x} + t_1\vec{e}_i + t_2\vec{e}_j$. The first pair become: $$\lim_{h\to 0}\frac{1}{h^3}\left(\int_{0}^1\int_0^1 F_1(\mathbf{x} + h\mathbf{e}_1 +t_1 h\mathbf{e}_2 + t_2h\mathbf{e}_3)h^2 dt_1 t_2 - \int_0^1 \int_0^1F_1(\mathbf{x} + t_1h\mathbf{e}_2+t_2 h\mathbf{e}_3)h^2 dt_1 dt_2\right)$$ Now here we can use Taylor's theorem. $$F_1(\mathbf{x} + h\mathbf{v}) = F_1(\mathbf{x}) + h\nabla F_1(\mathbf{x})\cdot \mathbf{v} + R(h)$$ Where $\lim_{h\to 0}\frac{R(h)}{|h|} = 0. $ We can drop the $0$th order term because we're taking the difference. We can also drop the remainder term thanks to the limit. What remains is the linear term: $$= \lim_{h\to 0}\frac{1}{h}\int_0^1 \int_0^1 h\nabla F_1(\mathbf{x} + t_1h\vec{e}_2+ t_2h\vec{e}_3)dt_1 dt_2 = \lim_{h\to 0}\int_0^1 \int_0^1\partial_x F_1(\mathbf{x} + th\vec{e_2}+th\vec{e}_3) = \partial_x F_1(\mathbf{x})$$ The reasoning for the other two terms is identical.