# Sign of integral of a 2 form

Given a 1-form $\omega$ define on $\mathbb{R}^2$ such that $$d\omega = f \; dx\wedge dy$$ is a 2-form with $f>0$ and consider $D\subset \mathbb{R}^2$

How do I know the orientation of $D$ (or $\partial D$) such that $$\int_D d\omega >0$$

Is there a geometric reason for such orientation (or like rule of the right hand) or there is no convention ?

## 1 Answer

Writing $[a, b]$ to denote the oriented interval with endpoints $a$ and $b$ (so, for example, $[1, 0]$ denotes the unit interval oriented from $1$ to $0$), the oriented boundary of the unit square is the sum of the oriented $1$-chains $$[0, 1] \times \{0\},\qquad \{1\} \times [0, 1],\qquad [1, 0] \times \{1\},\qquad \{0\} \times [1, 0].$$ This notion of orientation is purely algebraic, determined by a fixed ordering of coordinates, and compatible with Green's theorem (i.e., the generalized Stokes theorem in the plane).

Customarily, Cartesian coordinates are drawn so the positive $x$-axis points to the right and the positive $y$-axis points upward. With this additional convention, the orientation of the boundary of the unit square is counterclockwise.

Orientation and the right-hand rule are separate issues: They coincide so long as one's Cartesian coordinate system (specifically, the standard basis in the natural ordering) is "right-handed" (i.e., drawn to a particular geometric convention).