Hello dear mathematics community.
I was wondering if you could help me to understand the following issue please.
In my engineering calculus course, I was taught that there are only two possible versions of surface integrals:
First type surface integrals (surface integrals of scalar fields):
If a scalar function $U = U(x, y, z)$ is defined on a smooth or piecewise smooth surface $\sigma$, there can be defined the surface integral of the first type as $J = \int \int U(x, y, z) d \sigma $
Second type surface integrals (surface integrals of vector fields):
If a vector field $\vec a$ is defined on a smooth or piecewise smooth oriented surface $\vec \sigma$, there can be defined the surface integral of the second type as $J = \int \int ( \vec a \cdot \vec n ) d \sigma $ (where $\vec n$ is, obviously, the outward unit normal for $\sigma$)
My issue is that, obviously, the second type surface integral gives me flux through the surface. But I want to integrate the full vector over the surface - not only the normal component of the vector. This is where my question formulation comes from: what if I want to integrate the component of the vector field that is tangent to the surface. There are only two types of surface integrals and none of them allows me to do that ...
Therefore my specific questions are as follows.
Can I do a surface integral like $J = \int \int \vec a ~ d \sigma$ ? I.e. when I have vector field (as in the second type surface integral) and not-oriented surface (as in the first type surface integral).
If I can't do that, then why? What is the mathematical principle behind it?
If I can, then I have another follow-up question.
I, obviously, tried doing that. I wanted to integrate the $x$ component of my vector field. Therefore, I got this: $$ J = \int \int \vec a d \sigma = \int \int (\vec a_x + \vec a_y + \vec a_z) d \sigma = \int \int \vec a_x d \sigma + \int \int \vec a_y d \sigma + \int \int \vec a_z d \sigma = J_x + J_y + J_z $$ $$ J_x = \int \int \vec a_x d \sigma = \int \int a_x \vec i d \sigma $$ I don't know what to do with $\vec i$. Can I take it out of the integral? If so, then why, what mathematical principle allows me to do that? I do want to take it out, in this case I will have the first type surface integral which I can evaluate. But I don't know how to justify whether I can or cannot do that.
Thank you very much in advance.
Ivan
P.S. I did have a look at the Wikipedia surface integrals article. I didn't find the answer there. The author says a relevant thing in the first sentence of the second paragraph in the part called "Surface integrals of vector fields". Quote:
The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector.
But I can't understand this statement: if I follow the "definition of the surface integral of a scalar field" then how can I get that "the result is a vector"? The result of a surface integral of a scalar field is a scalar ...
