I am fairly new to the generalised Stokes theorem and probably overcomplicating what I am trying to do by using it. As I understand it, the generalised Stokes theorem is given by:
$$\int_{\partial \Omega} \omega=\int_{\Omega} d \omega$$
I have a closed set $X\subseteq \mathbb{R}^N$ and a function $f: \mathbb{R}^N\rightarrow \mathbb{R}$. I want to write something like:
$$\int_{\partial X} f(x) dx=\int_{X} K dx$$
I am a little confused about what $K$ would be here. How can I calculate the value of $K$?
Thank you for any help... I'm very confused and really would appreciate clarity on this.
Perhaps you have something like this in mind. Consider an $(n-1)$-form on $\Bbb{R}^n$, let's say \begin{align} \omega=\sum_{i=1}^nf_i\,dx^1\wedge\cdots\wedge \widehat{dx^i}\wedge\cdots \wedge dx^n. \end{align} Here, $f_1,\dots, f_n:\Bbb{R}^n\to\Bbb{R}$ are smooth functions, and the $\widehat{dx^i}$ means $dx^i$ is omitted from the wedge. Then, its exterior derivative is \begin{align} d\omega&=\sum_{i=1}^ndf_i\wedge dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge\cdots \wedge dx^n\\ &=\sum_{i=1}^n\left(\sum_{j=1}^n\frac{\partial f_i}{\partial x^j}\,dx^j\right)\wedge dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge\cdots \wedge dx^n \\ &=\sum_{i=1}^n\frac{\partial f_i}{\partial x^i}\,dx^i\wedge dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge\cdots \wedge dx^n\\ &=\left(\sum_{i=1}^n(-1)^{i-1}\frac{\partial f_i}{\partial x^i}\right)\,dx^1\wedge \cdots\wedge dx^n. \end{align} (in the third equality, only the $j=i$ term contributes, because for all the others, $dx^j$ appears twice in the wedge, so it is zero). So, Stokes theorem tells you that for nice $X$, you have $\int_{\partial X}\omega=\int_Xd\omega$, i.e \begin{align} \int_{\partial X} \sum_{i=1}^nf_i\,dx^1\wedge\cdots\wedge \widehat{dx^i}\wedge\cdots \wedge dx^n&= \int_X\left(\sum_{i=1}^n(-1)^{i-1}\frac{\partial f_i}{\partial x^i}\right)\,dx^1\wedge \cdots\wedge dx^n. \end{align} Hopefully you can already see the seeds of the various special cases (Green's theorem in the plane, Divergence theorem, which are obtained by choosing the $\omega$ carefully, i.e by choosing the $f_1,\dots, f_n$ carefully. The classical Stokes theorem is also a special case, though we have to take $n=3$ and $\omega=F_1\,dx+F_2\,dy+F_3\,dz$ to be a $1$-form, not an $(n-1)=(3-1)=2$ form as I've described the calculation for above).