Algebraic manipulation of the product of multiple differentials

Question

I want to correct some of my misunderstandings in multivariable differentiation from a purely algebraic and non-geometrical viewpoint (assuming one does not know anything about geometry and no intuition relating double integrals to surfaces, and just literally understanding double integrals as limits of Riemann sums). To this end, I have boiled down the problem to a simple discrete analog. Assume that one wants to calculate $\sum\limits_{i = 0}^0 {\sum\limits_{j = 0}^0 {\Delta i\Delta j} } $. Obviously since $i$ and $j$ are integer counters, ${\Delta i}={\Delta j}=1$ and so $\sum\limits_{i = 0}^0 {\sum\limits_{j = 0}^0 {\Delta i\Delta j} } = \sum\limits_{i = 0}^0 {\sum\limits_{j = 0}^0 1 } = 1$. Now assume a transformation from the $ij$ space to the $pq$ space defined as $\left. \begin{array}{l} i = p + q\\ j = - 2p - q \end{array} \right\} \leftrightarrow \left\{ \begin{array}{l} p = - i - j\\ q = 2i + j \end{array} \right.$. Obviously, since the delta is a linear operator, one has $\left\{ \begin{array}{l} \Delta p = - \Delta i - \Delta j\\ \Delta q = 2\Delta i + \Delta j \end{array} \right.$ so $\left\{ \begin{array}{l} \Delta p = - 2\\ \Delta q = 3 \end{array} \right.$. Also, according to the transformation, for the only term in the series which is $i=j=0$, we have $p=q=0$, hence:$$\begin{array}{l} \sum\limits_{i = 0}^0 {\sum\limits_{j = 0}^0 {\Delta i\Delta j} } = - \sum\limits_{p = 0}^0 {\sum\limits_{q = 0}^0 {\left( {\Delta p + \Delta q} \right)\left( {2\Delta p + \Delta q} \right)} } = \\ - 2\sum\limits_{p = 0}^0 {\sum\limits_{q = 0}^0 {{{\left( {\Delta p} \right)}^2}} } - 3\sum\limits_{p = 0}^0 {\sum\limits_{q = 0}^0 {\Delta p\Delta q} } - \sum\limits_{p = 0}^0 {\sum\limits_{q = 0}^0 {{{\left( {\Delta q} \right)}^2}} } = \\ - 2 \times 4 + 3 \times 6 - 9 = 1. \end{array}$$ As easily seen, the answer is all correct here. The problem is while one could calculate a term like $\sum\limits_p {\sum\limits_q {{{\left( {\Delta p} \right)}^2}} } $ in the discrete case without any problem, this term does not translate into a well-defined two-dimensional integral, contrary to a term like $\sum\limits_p {\sum\limits_q {\Delta p\Delta q} } $ which obviously corresponds to $\int\limits_u {\int\limits_v {dudv} }$. Note that for this problem, the Jacobian would also not help much since it could be seen that it would equate to unity for this simple example.