Evaluate integral using mapping Given $D = \{(x, y) \in \mathbb{R}^2 \mid 2 \ge x^2 + y^2 \ge 1 , x \ge y \ge -x\}$ and let $(u, v) = G(x, y) = (x^{2} + y^{2}, 2\cdot x \cdot y)$.
Evaluate

$$\iint\limits_D (x^2 - y^2) \cos\left(\frac{\pi xy}{x^2 + y^2}\right) dA$$ using this given mapping.

I end up getting the integral
$$\frac{1}{4} \cdot \int_{-2}^{2} \int_{1}^{2} \cos(\frac{\pi v}{2 u}) dudv$$
Using change of variables
But how do I evaluate this?
 A: If you add $u$ and $v$ together, you'll get
$$u+v=x^2+y^2+2xy=(x+y)^2.$$
Since $x+y\ge0$ due to the given constraint $y\ge-x$, we can take the square root, so
$$x+y=\sqrt{u+v}.$$
Similarly, if you subtract them, you'll get
$$u-v=x^2+y^2-2xy=(x-y)^2.$$
Again, since $x-y\ge0$ due to the given constraint $x\ge y$, we can take the square root, so
$$x-y=\sqrt{u-v}.$$
Now you can solve the system of equations
$$\left\{\begin{align} x+y&=\sqrt{u+v} \\ x-y&=\sqrt{u-v} \end{align}\right.$$

UPDATE: now that the change of variable issue has been resolved, and we need to finish integration of the new double integral.
Your limits of integration for $v$ are incorrect. If you graph the region, you'll see that in the $(u,v)$-plane it's not going to be a rectangular region, and the limits of integration for $v$ are not from $-2$ to $2$ (even though these actually are its least and greatest values). Here are the correct constraints and how we can find them.


*

*From $y\le x$, we have an "upper" boundary given by
$$y=x \; \Rightarrow \; x-y=0 \; \Rightarrow \; \sqrt{u-v}=0 \; \Rightarrow \; u-v=0 \; \Rightarrow \; v=u.$$

*From $-x\le y$, we have a "lower" boundary given by
$$y=-x \; \Rightarrow \; x+y=0 \; \Rightarrow \; \sqrt{u+v}=0 \; \Rightarrow \; u+v=0 \; \Rightarrow \; v=-u.$$
So $v$ ranges within $-u\le v\le u$ (reaching the values of $-2$ and $2$ only when $u=2$). And the correct double integral is
$$\frac{1}{4} \cdot \int_1^2 \int_{-u}^u \cos\left(\frac{\pi v}{2u}\right)\,dv\,du.$$
I hope you can take it from here. The inside integral is now with respect to $v$, which is actually doable.
