How to apply Fubini's theorem here? Given $f:\mathbb{R}^2\times \mathbb{R}^2 \to \mathbb{R}$, my classmate and I are discussing how to apply Fubini's theorem in the next integral with $x=(x_1,x_2), y=(y_1,y_2)$
$$\int_{x\in C}\int_{y\in C-x}{f(x,y)}dydx,$$
where ,$$C=\{x=(x_1,x_2)\in \mathbb{R}^2:2<x_1^2+x_2^2<3\}$$ and
$$ C-x=\{y-x:y\in C\}.$$
Our first attempt was
$$\int_{x\in C}\int_{y\in C-x}{f(x,y)}dydx=\int_{y\in C}\int_{x\in C+y}{f(x,y)}dxdy,$$ we discarded this option by means of the simpler 1D example:
$$\int_{x=-1}^{x=1}\int_{y= -1-x}^{y=1-x}f(x,y)dydx\neq\int_{y=-1}^{y=1}\int_{x= -1-y}^{x=1-y}{f(x,y)}dxdy.$$
Does anyone have any hint that can help us?
 A: Let us consider $D$, a subset in $\Bbb R^2\times\Bbb R^2\cong \Bbb R^4$, with:
$$
\begin{aligned}
D 
&=\{\ (x,y)\in \Bbb R^2\times \Bbb R^2\ :\ x\in C\ ,\ y\in C-x\ \}\\
&=\{\ (x,y)\in \Bbb R^2\times \Bbb R^2\ :\ x\in C\ ,\ x+y\in C\ \}\ .
\end{aligned}
$$
Then
$$
\int_{x\in C}dx\int_{y\in C-x}f(x,y)\; dy
=
\iint_D f(x,y)\; dx\;dy
\ .
$$
And now, in order to get the "other Fubini order of integration", we have to get first the projection of $D$ onto the second $\Bbb R^2$, those cartesian component where $y$ lives in. Well, taking $x$ near $(-3,0)$ in $C$, and $y$ near $(6,0)$ in a convenient manner, we still can produce $x+y\in C$ near $(3,0)$. So the "second or $y$-projection" $D_2$ of $D$ is "something different", it does not look like $C$, it is rather $D_2=C-C=C+C$, a subset of the open disk centered in zero with radius six. Well, is it this full open disk. Then all we can write is now:
$$
\iint_D f(x,y)\; dx\;dy
=
\int_{y\in D_2}dy\int_{x\in C\cap(C-y)}f(x,y)\; dx
\ .
$$

In one dimension, use instead $C=[-1,1]$ and $D_2=C-C=C+C=[-2,2]$.
