Substitution problem

[kyle_can]

In making a substitution in a double integral from rectangular coordinates (x,y) to new coordinates (u,v), John always replaces the differentials dx dy by du dv. Briefly explain why this is wrong when u = 3x+y, v = x + 3y. Then show a corrected version by showing the appropriate Jacobian.

my answer: because you need the Jacobian as the scaling factor?

 

[HallsofIvy]

Basically, yes. You could also note that with u= 3x+ y and v= x+ 3y, \(\displaystyle x= \frac{3}{8}u- \frac{1}{8}v\) and \(\displaystyle y= -\frac{1}{8}du+ \frac{3}{8}v\) so that \(\displaystyle dx= \frac{3}{8}du- \frac{1}{8}dv\) and \(\displaystyle dy= -\frac{1}{8}du+ \frac{3}{8}dv\).

Then \(\displaystyle dxdy= (\frac{3}{8}du- \frac{1}{8}dv)(-\frac{1}{8}du+ \frac{3}{8}dv)= -\frac{3}{64}dudu+ \frac{9}{64}dudv+\frac{1}{64}dvdu= \frac{3}{64}dvdv\)

But, of course, the product of differentials is anti-commutative, dudu= dvdv=0 and dvdu=-dudv, so \(\displaystyle dxdy= (\frac{9}{64}- \frac{1}{64})dudv= \frac{1}{8} dudv\). (Which is basically the reason for the "Jacobian".)