I wanted to derive the formula for the multivariable change of basis in an integral on my own (for the 2 by 2 case). What I did was:
$$x=f(u,v)$$
$$y=g(u,v)$$
so $$dx = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv$$ $$dy = \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv$$
Then, $$dx \wedge dy = (\frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv) \wedge (\frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv) = (\frac{\partial f}{\partial u}\frac{\partial g}{\partial v} - \frac{\partial g}{\partial u}\frac{\partial f}{\partial v}) du \wedge dv$$
I recognize that term as the Jacobian. Then: $$dx\wedge dy = J(u,v) du \wedge dv$$
but I don't want to be working with bivectors, I want to work with scalars. I take the absolute value on both sides and since dx is perpendicular to dy:
$$dx dy = J(u,v) \sin(\theta) du dv $$
where $\theta$ is the angle between the two vectors. This is not the formula I learned in my undergraduate studies. How did the original formula work even if dx and dy were scalars?