Students who compute \(d\boldsymbol{\vec{r}}_1\) and \(d\boldsymbol{\vec{r}}_2\) directly can easily get confused, since they may try to eliminate \(x\) or \(y\), rather than \(u\) or \(v\).

Along the curve \(v=\hbox{constant}\), one has \(dy=v\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = dx\,\boldsymbol{\hat{x}} + dy\,\boldsymbol{\hat{y}} = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}})\,dx\), which some students will want to write in terms of \(x\) alone. But one needs to express this in terms of \(du\)! This can be done using \(du = x\,dy + y\,dx = x (v\,dx) + y\,dx = 2y\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}}) \,\frac{du}{2y}\). A similar argument leads to \(d\boldsymbol{\vec{r}}_2 = (-\frac{1}{v}\,\boldsymbol{\hat{x}}+\boldsymbol{\hat{y}})\,\frac{x\,dv}{2}\) for \(u=\hbox{constant}\), so that \(d\boldsymbol{\vec{S}} = d\boldsymbol{\vec{r}}_1\times d\boldsymbol{\vec{r}}_2 = \boldsymbol{\hat{z}} \,\frac{x}{2y}\,du\,dv = \boldsymbol{\hat{z}} \,{du\,dv\over2v}\). This calculation can be done without solving for \(x\) and \(y\), provided one recognizes \(v\) in the penultimate expression.

Emphasize that one must choose parameters, both on the region, and on each curve, and that \(u\) and \(v\) are chosen to make the limits easy.