Jacobian transformation and 2D curl

Feodalherren
Messages
604
Reaction score
6
jacobian.png

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?
 
Physics news on Phys.org
The Jacobian has the property
J=\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{ \partial (x,y)}}
so the change of variables can be written as:
\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

This is why you can first write u=u(x,y); v=v(x,y) 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 x=x(u,v);y=y(u,v) and find the corresponding Jacobian, it's just that in this particular example the first method works faster. As you can see, the dudv part remains unaffected either way.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
2
Views
3K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
Back
Top