Jacobians change of variable integrals

[Liquidxlax]

Homework Statement

edit* proper problem

change to polar

Integral(0 to infinity)Integral(0 to infinity)e-sqrt(x² + y^2[/SUP)dxdy

the one below isn't the right problem, but i'd still appreciate help on the one below

Make the change of variables u=x-y, v=x+y to evaluate the integral

Integral(0 to 1)dy Integral(0 to (1-y)) e^(x-y)(x+y)dx

Homework Equations

J = J((x,y)/(u,v)) = partial((x,y)/(u,v)) = det(partial((x,y)/(u,v)))

The Attempt at a Solution

PROBLEM 1
Not sure how to do a jacobian for this or even if its needed. It looks as if i would just stick in y=rsint x=rcost.

PROBLEM 2

I did the determinant which came to

{1, 1}, {-1, 1} = 2


integral()integral()f|J|dudv = integral()integral()2e^uvdudv

not sure if this is right, but if it is I'm not sure how to change my limits

thanks

 

[jackmell]

For (2), you need to determine how the area of integration in x and y is transformed to a new area of integration in u and v under the transformations u=x-y and v=x+y. So first draw a simple plot of the area in x and y for:

\int_0^1 \int_0^{1-y} f dxdy

That's just the triangular region between (0,0), (1,0) and (0,1) right? Ok, so just take the line segment along the x-axis from 0 to 1 and see how that line is transformed to a new line in the u-v coordinate system. So, y=0 for that and x goes from 0 to 1 so:

u=x
v=x
v=u

then the line along the real axis from 0 to 1 in the x-y coordinate plane is mapped to the line v=u from u=0 to u=1 in the u-v coordinate plane. Now, do that for the other two lines around this area, and that will give you the new region in the u-v plane to integrate over.

 

[Liquidxlax]

can anyone help with my first question.

 

[HallsofIvy]

You don't "need" the Jacobian if you remember that the "differential of area" in polar coordinates is r drd\theta.

Of course, "r" is the Jacobian:
\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{array}\right|= \left|\begin{array}{cc}cos(\theta) & -r sin(\theta) \\ sin(\theta) & r cos(\theta)\end{array}\right|= r

 

[Liquidxlax]

You don't "need" the Jacobian if you remember that the "differential of area" in polar coordinates is r drd\theta.

Of course, "r" is the Jacobian:
\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{array}\right|= \left|\begin{array}{cc}cos(\theta) & -r sin(\theta) \\ sin(\theta) & r cos(\theta)\end{array}\right|= r

yes i seen that in previous chapter in my textbooks. but I'm not sure how easy this question is supposed to be. When i applied what you put up top


re^{-\sqrt{(rcos\vartheta)^{2}+(rsin\vartheta)^{2}}}drd\vartheta

re^sqrt(...) since i can't get the text to sit right
which would reduce to re^-r