# Proving equivalence of integral and differential form of curl

I am trying to prove the equivalence of integral and differential form of curl in terms of curvilinear coordinates using this image: https://imgur.com/9uhgHAY

in integral form:

$$(\nabla \times \mathbf{a}) \cdot \mathbf{\hat n} = \lim_{A\to 0} \frac{1}{A} \oint_{C} \mathbf{a \cdot dr}$$

based on the image, the solution given by my textbook is:

$$\oint_{PSRQ} \mathbf{a \cdot dr} = a_2h_2\Delta u_2 + \biggl[a_3h_3 + \frac {\partial}{\partial u_2}(a_3h_3)\Delta u_2 \biggl]\Delta u_3 \\ -\biggl[a_2h_2 + \frac{\partial}{\partial u_3}(a_2h_2)\Delta u_3 \biggl]\Delta u_2 - a_3h_3 \Delta u_3 \\ = \Biggl[ \frac {\partial}{\partial u_2}(a_3h_3) - \frac{\partial}{\partial u_3}(a_2h_2) \Biggl] \Delta u_2 \Delta u_3 \tag{1}$$

this can be used to find $$(\nabla \times \mathbf{a})_1$$ (with respect to $$\mathbf{\hat e_1}$$) and hence, $$(\nabla \times \mathbf{a})_2$$ and $$(\nabla \times \mathbf{a})_3.$$

From my understanding of $$(1)$$:

$$\oint_{PSRQ} \mathbf{a \cdot dr} = \int_{PS} \mathbf{a \cdot dr} + \int_{SR} \mathbf{a \cdot dr} + \int_{RQ} \mathbf{a \cdot dr} + \int_{QP} \mathbf{a \cdot dr}$$

I can understand that:

$$\int_{PS} \mathbf{a \cdot dr} = a_2h_2\Delta u_2 \\ \int_{QP} \mathbf{a \cdot dr} = a_3h_3 \Delta u_3$$

because it is along the coordinate axis

However I cannot understand why:

$$\int_{SR} \mathbf{a \cdot dr} = \biggl[a_3h_3 + \frac {\partial}{\partial u_2}(a_3h_3)\Delta u_2 \biggl]\Delta u_3 \\ \int_{RQ} \mathbf{a \cdot dr} = \biggl[a_2h_2 + \frac{\partial}{\partial u_3}(a_2h_2)\Delta u_3 \biggl]\Delta u_2$$

why are there extra partial derivative terms if $$SR$$ is parallel to $$PQ$$ and $$RQ$$ is parallel to $$SP$$? Shouldn't $$SR = PQ$$ and $$RQ = SP$$?

Also in this case of curvilinear coordinates, since $$h_i$$ is the scale factor where $$\mathbf{e_1} = h_i \mathbf{\hat e_1}$$ and hence, $$h_i = ||\mathbf{e_1}||$$, can it be brought out of the partial derivative since it is a constant?