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I have equations derived from Stokes's/Gauss's Theorems that I want to know if the following can be computed given the integrand used: $$\int_S (\hat{n} \cdot\nabla\chi) \cdot ds = \int_D \alpha d\textbf{x}$$. I don't think I can do this because the integrand yields a scalar value due to the dot product which should give $$\hat{n} \cdot\nabla\chi = |n||\frac{\partial \chi}{\partial x}+\frac{\partial \chi}{\partial y}|\cos(0)$$ (where $\cos(0)$ comes from the gradient and normal vector being parallel). The problem is that I can't evaluate a dot product with the differential $ds$ if I have a scalar. Does anyone have any ideas?


ADDITIONAL CONTEXT BELOW


This comes from the following provided relationships: $$(1a).\nabla^2 \chi = \alpha \\(1b). \nabla^2 \psi= \zeta \\(2a). \partial_n \chi - \partial_s \psi = V_n \\ (2b). \partial_s \chi + \partial_n \psi = V_s$$ where $\partial_n = \hat{n}\cdot\nabla$ where $\hat{n}$ is normal to the boundary of the surface, and $\partial_s = \hat{s}\cdot\nabla$ where $\hat{s}$ is parallel to the boundary of the surface.

According to the paper I am reading from, Gauss's and Stokes's Theorem were both applied to (1) and (2) to obtain the integrals:

$$(3a)\int_S V_n \cdot ds = \int_D \alpha d\textbf{x} \\ (3b) \int_S V_s \cdot ds = \int_D \zeta d\textbf{x}$$ where the n subscript denotes a vector pointed normal to the surface, and ds is differentiated along (parallel) to the surface. (2a) was plugged into (3a) to get the integral I presented in the question.

Side note: The paper I am reading from also states that $\int_S \partial_s\chi ds = 0$ but, I'm pretty sure $\partial_s \chi = 0$ inherently since $\hat{s}$ is normal to the gradient so the dot product would yield 0.


EDIT: I made a mistake reading the paper

It should be $$\int_S V_n ds = \int_S \partial_n \chi ds= \int_S n \cdot \nabla \chi ds$$. which we can work on. I also noticed another point of confusion where I think the paper meant to use $dS$ i.e. a capital S representing the surface. It had just defined lowercase s as a vector so I thought it was integrating with respect to a vector. Now viewing the theorems themselves again thanks to Kurt, I can now see that it might have been a mistake. Also there was no dot product on $\cdot ds$, I misread it.

I think we should get: $$\int_S (\frac{\partial \chi}{\partial x} + \frac{\partial \chi}{\partial y}) dS$$

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Too long for a comment:

  • First of all the names of those giants were Stokes and Gauss. So their theorems should be called Stokes$\color{red}{\,'}$ and Gauss$\color{red}{\,'}$ theorem.

  • Strictly speaking, the integrals they relate with each other are integrals of scalars: $$\tag{Gauss} \oint_{\partial V}\underbrace{\boldsymbol{F}\cdot\boldsymbol{n}}_{scalar}\,dS=\int_V\underbrace{\nabla\cdot\boldsymbol{F}}_{scalar}\,dV\,, $$ $$\tag{Stokes} \int_{\partial S}\underbrace{\boldsymbol{F}\cdot\,ds}_{scalar}=\int_V\underbrace{(\nabla\times\boldsymbol{F})\cdot\boldsymbol{n}}_{scalar}\,dS\,. $$

  • The notation I am using is

  • $\boldsymbol{n}$ unit normal vector to the surface $S$

  • $dS$ surface area element, $ds$ line element, $dV$ volume element

  • If your $\alpha$ is the divergence of $\nabla\chi$ and if you do me the favour to tell me what your $ds,S,D$ are then your first formula looks like Gauss$\color{red}{\,'}$ theorem in its full glory.

  • The question arises why you think you cannot do this because the integrands are scalars. Very confusing.

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  • $\begingroup$ Thank you! Examining this I can see it was a bit of my fault for not rereading the theorems, and the paper's usage of $ds$ instead of $dS$ throwing me off. Exactly as it appears in the paper, the integral was $\int_S ds V_n = \int_D dx \alpha$ where it oddly put the variable of integration $ds$ before the integrand $V_n$. There were some rather odd conventions I had not seen before used throughout like that. I couldn't tell if it was integrating wrt the vector s or what was going on. Also, I thought I saw an extra dot product where there wasn't one. $\endgroup$ Commented Sep 26, 2023 at 8:15
  • $\begingroup$ I am consistently terrible with names regardless of context. I even forget my own friend's names and people I met yesterday sadly. I apologize for the mistake! $\endgroup$ Commented Sep 26, 2023 at 8:17

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