Change of variables jacobian limits

[ahmed markhoos]

$\int_{0}^{∞}\int_{0}^{∞} \frac{x^2+y^2}{1+(x^2-y^2)^2} e^{-2xy} dxdy$

$u= x^2-y^2$
$v=2xy$I tried to find the jacobian and the area elements,

I found it to be $dA = \frac{1}{v} du dv$

I'm having problem finding the limits of u & v and getting rid of $x^{2}+y^{2}$.

 

[ahmed markhoos]

------------- by mistake

 

[STEMucator]

I could be wrong about this, but I think you should try a polar transformation. All the $x^2$ and $y^2$ terms seem to hint at this.

$$\iint_D \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} dA = \iint_{D'} \frac{r^2}{1 + (r^2 \text{cos}(2 \theta))^2} e^{-r^2 sin(2 \theta)} r dr d\theta$$

Where $D'$ is the part of $D$ inside the circle $x^2 + y^2 \leq c^2$. So $0 \leq r \leq c$ and $0 \leq \theta \leq \frac{\pi}{2}$ define the limits of $D'$.

Then you want to consider when $c \to \infty$, so $D'$ becomes infinitely large.

 

[ahmed markhoos]

I could be wrong about this, but I think you should try a polar transformation. All the $x^2$ and $y^2$ terms seem to hint at this.

$$\iint_D \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} dA = \iint_{D'} \frac{r^2}{1 + (r^2 \text{cos}(2 \theta))^2} e^{-r^2 sin(2 \theta)} r dr d\theta$$

Where $D'$ is the part of $D$ inside the circle $x^2 + y^2 \leq c^2$. So $0 \leq r \leq c$ and $0 \leq \theta \leq \frac{\pi}{2}$ define the limits of $D'$.

Then you want to consider when $c \to \infty$, so $D'$ becomes infinitely large.

Hmmm, seems good. But the question require using Jacobian determinant.

 

[STEMucator]

Hmmm, seems good. But the question require using Jacobian determinant.

When changing to polar co-ordinates:

$$dA = \left| J \right| dA' = \left| J \right| dr d\theta = r \space dr d\theta$$

The Jacobian is just $r$.