I am learning about surface integrals. One question that I am not understanding the answer to is: What does it measure? When you have an answer - what does that answer represent?
I can plug the numbers all I want, but without this understanding, I don't know what I'm doing at all!
Thank you for your help!
On
Once the OP question is about the meaning of a surface integral, I will not be much rigorous in what follows. Let us consider the surface integral defined to be: $$\int\int_{S}\vec{F}\cdot \hat{n}dS$$ where $\vec{F}$ is a vector field in $\mathbb{R}^{3}$ and $\hat{n}$ is a unit normal associated to the oriented surface $S$ in which you are performing the integral.
To understand this definition, let us consider the following situation: you have a a plane, say, $y=0$ and a vector field pointing in the direction $\hat{y}$ 'everywhere', i.e. something like $\vec{F}(x,y,z) = F(x,y,z)\hat{y}$. The meaning of this vector field depends on the type of problem you are dealing with: it can be the electric field crossing the surface, a fluid passing through it (in which case each vector $F(x,y,z)\hat{y}$ represents the velocity of the point particle of the fluid in the position $(x,y,z)$) and so on. In addition, let us demand that $\vec{F}$ to be position-independent, i.e. $\vec{F}(x,y,z) = F\hat{y}$ for some constant $F$. Let us calculate the surface integral (or flux) of this vector field through a 'portion' of the plane with area $A$. $$\int \int_{A}\vec{F}\cdot\hat{n}dS = F\int \int_{A}dS = FA$$ which can be interpreted as 'the amount' of the vector field passing through the 'portion' of the plane with area $A$.
This analysis was simplified by two assumptions: first I supposed that the field was always poiting $\hat{y}$ and second that it is position-independent. Now, if the field $\vec{F}(x,y,z) = F\hat{p}$, where $\hat{p}$ is some unitary vector, then we would expect that the intuitive notion of 'the amount of $\vec{F}$ that passes through the portion of the plane' would change, since now we must take the component of $\vec{F}$ which, in fact, passes through the portion of the plane. For instance, if $\hat{p} = \frac{1}{\sqrt{2}}(\hat{y}+\hat{z})$, then $\vec{F}$ has, at each point $(x,y,z)$ on the plane $y=0$, a $\hat{y}$ component which is 'crossing' the plane and a $\hat{z}$ component which is parallel to the plane. Thus, if we want to keep the notion of 'amount of' $\vec{F}$ that passes through the plane, we would consider only the $\hat{y}$ contribution to the surface integral. This is why the define the surface integral with the dot product $\vec{F}\cdot \hat{n}$. In a more general case, where $\vec{F}$ depends on $x,y,z$, we must find ways to compute the integral, but its interpretation is the same.
When you evaluate a surface integral, you are measuring the net movement of a fluid through the surface you are integrating over. Typically, you are taking the surface integral over of a vector field, $F$, over an oriented surface, $S$; this surface integral is called flux in many textbooks. When we say $F$ is a vector field, we can imagine that at each point in space, there is a velocity vector of that fluid. When you calculate this surface integral to be positive, that means that there is a net positive movement of fluid towards the side that is indicated as "positive", and it's a similar story for negative.
In a problem, we would like to measure how much fluid is passing through the surface in question, $S$, which is the flux across $S$. We imagine computing the flux across an infinitesimal section of the surface, with area $dS$, and then "adding up" every infinitesimal area over $S$ with an integral. Let $n$ be a unit normal vector to the surface at a point, then $F\cdot n$ is the projection of $F$ onto the direction of $n$, so it measures how fast the fluid is moving across the surface. In one unit of time the fluid moving across the surface will fill a volume of $F\cdot ndS$, which represents the rate at which the fluid is moving across a small patch of the surface and we arrive at the total flux across $S$ is: $$ \iint_SF\cdot n dS=\iint_SF\cdot d\mathbf{S}$$ where $d\mathbf{S}\triangleq n dS$. Usually, the questions will tell you the orientation of the surface you're concerned with, and some theorems require that you pick a normal vector such that your surface is oriented in a particular way. Some surfaces may not even be orientable (Mobius strip). One example that may help with your physical intuition is Maxwell's equation asserting that magnetic monopoles do not exist. Namely, $$ \nabla \cdot \mathbf{B} = 0\iff \oint_S\mathbf{B}\cdot d\mathbf{A}=0$$ Here, $S$ is any closed surface, and $d\mathbf{A}$ is a vector whose magnitude is, again, an infinitesimal piece of the surface $S$ with direction given by the outward pointing normal of the surface. So the entire integral represents the net flux of the magnetic field out of the surface, and this law states that it is always zero. Thus magnetic monopoles cannot exist, since if they could, then the divergence, and the corresponding integral, would be nonzero, as is the case of Gauss' Law: $$ \nabla \cdot \mathbf{D} = \rho\iff \oint_S\mathbf{D}\cdot d\mathbf{A}=Q_{enc}$$ since electrostatic monopoles do exist (protons and electrons).