I have been doing one example, and I do not understand something about it:
- In the description of the problem they say that $\hat N$ is an outward normal vector, and in the first sentence of the solution they say that it is a normal vector field?
- If we assume that $\hat N=(A,B)$ is pointed outward, as specified in the solution, then $\hat T= (-B,A)$. Why is that?
- Consequently, they say, $(P,Q) \cdot (A\cdot B)=(-Q, P)\cdot(-B,A)$. Why is that?
Here is the exercise problem:
The normal vector is indeed a vector normal to $\partial D$ which lies parallel to the $x\text{-}y$ plane since $\hat{\mathbf{N}}$ has only $x\text{ and }y$ components, but since we are operating in $\mathbb{R}^3$, then $\hat{\mathbf{N}}$ forms a vector field in $\mathbb{R}^3$ that doesn't change with respect to its $z$-coordinate, but does change based on which segment of $\partial D$ is being considered.
$\hat{\mathbf{T}}$ is the tangent vector, so by definition
$$\hat{\mathbf{N}}\cdot\hat{\mathbf{T}}=0$$
Which implies that $\hat{\mathbf{T}}=\left<-B,A\right>$
$\left<B,-A\right>$ also satisfies, the only difference is direction. Counterclockwise direction is taken to be positive orientation, which $\left<-B,A\right>$ describes for positive $A$ and $B$
The integral changes form from a flux integral (using $\hat{\mathbf{N}}$) to a circulation integral (using $\hat{\mathbf{T}}$), but it's value must be conserved. Changing to a circulation integral allows for the application of Green's theorem
$$\int_{\partial D}\left<-Q,P\right>\cdot\hat{\mathbf{T}}ds=\int_{\partial D}\left<-Q,P\right>\cdot d\mathbf{r}$$