invariance of formula for surface integration with respect to area under change of variables

First, we can use the chain rule for Jacobians to see how one of the terms in parentheses transforms:

$$\frac{\partial (x,y)}{\partial (u,v)}=\frac{\partial (x,y)}{\partial (u^{\prime },v^{\prime })}\frac{\partial (u^{\prime },v^{\prime })}{\partial (u,v)}$$

A similar story holds for the other two factors. Combining them, we conclude that

$$\sqrt{{\left(\frac{\partial (x,y)}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (y,z)}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (z,x)}{\partial (u,v)}\right)}^2}=$$

$$\sqrt{{\left(\frac{\partial (x,y)}{\partial (u^{\prime },v^{\prime })}\frac{\partial (u^{\prime },v^{\prime })}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (y,z)}{\partial (u^{\prime },v^{\prime })}\frac{\partial (u^{\prime },v^{\prime })}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (z,x)}{\partial (u^{\prime },v^{\prime })}\frac{\partial (u^{\prime },v^{\prime })}{\partial (u,v)}\right)}^2}=$$

$$\frac{\partial (u^{\prime },v^{\prime })}{\partial (u,v)}\sqrt{{\left(\frac{\partial (x,y)}{\partial (u^{\prime },v^{\prime })}\right)}^2+{\left(\frac{\partial (y,z)}{\partial (u^{\prime },v^{\prime })}\right)}^2+{\left(\frac{\partial (z,x)}{\partial (u^{\prime },v^{\prime })}\right)}^2}$$

Since the factor in parentheses in front of the square root is the Jacobi determinant, we can apply the rule change of variables in multidimensional integrals to conclude that

$$\int f(u,v)\sqrt{{\left(\frac{\partial (x,y)}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (y,z)}{\partial (u,v)}\right)}^2+{\left(\frac{\partial (z,x)}{\partial (u,v)}\right)}^2}𝑑u𝑑v=$$

$$\int f(u^{\prime },v^{\prime })\sqrt{{\left(\frac{\partial (x,y)}{\partial (u^{\prime },v^{\prime })}\right)}^2+{\left(\frac{\partial (y,z)}{\partial (u^{\prime },v^{\prime })}\right)}^2+{\left(\frac{\partial (z,x)}{\partial (u^{\prime },v^{\prime })}\right)}^2}𝑑u^{\prime }𝑑v^{\prime },$$

which shows that our formula gives the same answer for ${\int }_{S}f(u,v)d^{2}A$, no matter how we choose to parameterize $S$.