Fubini's Theorem, but $dxdy = -dydx$

Question

As differential $2$-forms, clearly $dxdy=-dydx$ by alternation. Yet just as clearly, $\int_0^1 \int_0^1 dxdy \neq -\int_0^1 \int_0^1 dydx$. Where's the abuse of notation here?

Answer 1

$(x,y)\mapsto(y,x)$ is an orientation-reversing map of the plane to itself. For orientation-reversing maps one has to introduce a minus sign factor in the change of variables: $$\int_0^1\int_0^1\,dx\wedge\,dy=-\int_0^1\int_0^1-\,dy\wedge\,dx =\int_0^1\int_0^1\,dy\wedge\,dx.$$

Answer 2

Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is

$$ J(x,y) = \begin{bmatrix} \frac{\partial y}{\partial x} & \frac{\partial x}{\partial x} \\ \frac{\partial y}{\partial y} & \frac{\partial x}{\partial y} \end{bmatrix} =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \, , $$ whose determinant is $-1$. Therefore, $$ \int_0^1 \int_0^1 dx \wedge dy = \int_0^1 \int_0^1 \det(J(x,y)) \, (-dy \wedge dx) = \int_0^1 \int_0^1 dy \wedge dx $$

As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.