Jacobian transformation and 2D curl

1 reply · 3K views · last reply
Feodalherren
Messages
604
Reaction score
6
Handwritten solution: the Jacobian ∂(x,y)/∂(u,v) computed as x²/(x+y), the mass integral of 1/x² over u and v, and the result ln(27) grams.

Umm what just happened?
I understand as far as u=x+y and v = y/x and when he does the 2d curl. What I don't get is the step thereafter when he flips it. How does he know to flip it? Further, when he flips it wouldn't that make the dvdu inside the integral cancel and hence leave him with dxdy?
 
The Jacobian has the property
[itex]J=\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{ \partial (x,y)}}[/itex]
so the change of variables can be written as:
[itex]\int_{S}f(x,y)dxdy=\int_{S'}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{ \partial(u,v)}\right|dudv=\int_{S'}f(x(u,v),y(u,v))\frac{1}{\left|\frac{\partial(u,v)}{ \partial (x,y)}\right|}dudv[/itex]

This is why you can first write [itex]u=u(x,y); v=v(x,y)[/itex] and then find the inverse of the corresponding Jacobian. Of course, you can easily verify that the result comes out the same if you use their inverses directly, that is, write [itex]x=x(u,v);y=y(u,v)[/itex] and find the corresponding Jacobian, it's just that in this particular example the first method works faster. As you can see, the [itex]dudv[/itex] part remains unaffected either way.