# Math Help - Jacobian help

1. ## Jacobian help

My double integral is (x+y)
where R is the region on the xy plane bounded by the lines
x+y=3, x-y=3, and the hyperbola x^2-y^2=3
(u=x^2 - y^2....v=x+y)

I calculated the jacobian d(u,v)/d(x,y) determinant to be 2(x+y)=2v
and the reciprical = 1/2v

I dont understand what the new integral is and its limits?
my book gives like one example then goes to the next section

2. so x=v/y implies v goes from -3 to 3 right

3. You have to find the functions $x=g(u,v)$ and $y=h(u,v)$ by
solving the equations for u and v and then plug those back into the function you're integrating, which is $f(x,y)=x+y$

$u=x^2-y^2=(x+y)(x-y)=v(x-y)=vx-vy$

$u=vx-vy$
$v=x+y$

This gives $x=\frac{u+v^2}{2v}$ and $y=\frac{v^2-u}{2v}$

So now the integral will take the form

$\int_S\int f(\frac{u+v^2}{2v},\frac{v^2-u}{2v})\mid \frac{\partial (x,y)}{\partial (u,v)}\mid dudv$

4. I calculated the jacobian d(u,v)/d(x,y) determinant to be 2(x+y)=2v
and the reciprical = 1/2v
I don't see what you did here. The defintion of the Jacobian is:

$\mid \frac{\partial (x,y)}{\partial (u,v)}\mid =\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}$

5. partial derivatives of u wrt x,y
partial derivaitves of v wrt x,y

6. Originally Posted by begin
partial derivatives of u wrt x,y
partial derivaitves of v wrt x,y
I think you're confused by the notation. The defintion in terms of a determinant is this:

$\frac{\partial (x,y)}{\partial (u,v)}=\left( \begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right)$

7. yes and the reciprical =d(u,v)/d(x,y) which is easier to compute

8. I calculated it to $\frac{1}{2v}$, which is what you obtained, but I don't know what your method was. I'm not familiar with it...

9. yea thats what i got..thanks for the help so far...i got the limits for v=-3,3 and u=0-sqrt(3)
is that right?

I calculated it to $\frac{1}{2v}$, which is what you obtained, but I don't know what your method was. I'm not familiar with it...