>>12956548The wedges are meant to properly account for jacobian factors.
If you double integrate with respect to dxdy then substitute x=f(u,v), y=g(u,v), you get:
dx = (df/du)*du + (df/dv)*dv
dy = (dg/du)*du + (dg/dv)*dv
Naive multiplication gives:
dxdy= (df/du)*(dg/du)*du*du + [(df/du)*(dg/dv)+(df/dv)*(dg/du)]*du*dv + (df/dv)*(dg/dv)*dv*dv
You ignore the du*du and dv*dv terms since they remain infinitesimal after being integrated once with respect to their corresponding integral.
You are left with [(df/du)*(dg/dv)+(df/dv)*(dg/du)]*du*dv which is wrong (there should be a minus sign.
I did a sleight of hand by switching the order of du and dv for the second term.
If you were to use a wedge product, dx^dy, then replace dx and dy in exactly the same way as above, the du^du and dv^dv terms would automatically die from the rules of the wedge.
The mixed term would be (df/du)*(dg/dv)*du^dv + (df/dv)*(dg/du)*dv^du
= [(df/du)*(dg/dv) - (df/dv)*(dg/du)]*du^dv from rules of the wedge.
The sleight of hand switch mattered and gives a minus sign.
All of this matters because integrals care about orientation (they are directional).
If you swap the bounds of integration on just one integral, you pick up a minus sign even though the interval doesn't change.
This is the effect of reversing 1 axis.
A single axis reversal is the same as swapping two axes (in 2d, reverse the x axis then rotate 90 degrees clockwise to achieve an x-y swap).
This reversal or swapping operation flips the orientation of the space you are working in.
Doing another reversal or swap takes you out of bizzaro world and back to normal orientation which is accounted for by a second minus sign.
