I have seen the fact that in certain instances, the Surface Integral over a Vector Field gives the quantity of fluid flowing through the surface in unit time (as in here, or in any standard Vector Calculus textbook).
I am having trouble exactly seeing this from the formula. Usually a rate is defined by a derivative, I am unclear on how it emerges from this integral. I hope I am being clear, I would simply like some motivation in how this interpretation comes out of the surface integral. Thanks.
On
In the case of which you speak, the surface integral is
$$\int_S \mathbf{v} \cdot \mathbf{n} dS$$
where $\mathbf{v}$ is a velocity of a fluid flowing through $S$.
At each surface element $dS$, the normal component of the velocity provides the time rate you seek. If $\mathbf{r}$ is the position of an element of fluid that passes through the surface element $dS$, then, at the instant that particle is at that surface element, $\mathbf{v} = d\mathbf{r}/dt$; in this case, across the surface, the surface integral is
$$\frac{d}{dt} \int_S \mathbf{r} \cdot \mathbf{n} dS$$
This integral whose time derivative is taken is the volume of the fluid across the surface; thus, the surface integral in question is the volume flow rate of fluid across the surface $S$.
(I of course assumed that the surface itself is not moving. This is a pretty big assumption, as moving surfaces appear frequently in physical applications.)
On
let me try to answer you in a very simple way using different kinds of interpretation for the operation which is used in the equation you are asking about
let's start with dot product :- dot product can be understood from the fact that it get the amount of projected vector field of another vector that's because you are multiplying the magnitude of vector A as example times the cosine of the angle so you are projecting magnitude of A on B
form this face if you take a look at the equation you are asking about you will find that dS is translated into ndS where n is the normal component on the surface you are integrating , so if you thought for a little you will find that the meaning of the surface integral as following :-
at every single point on the surface you take your vector field and project it using dot product on the normal component to the surface and then you are integrating with respect to this surface " and integration is summation of all the magnitudes you already got" that's why you get amount of fluid going out Note 1 : you have to keep in mind that this happen for any surface not suppose to be closed one Note 2 : if we are talking about vector field it will actually be found all over the volume of the surface whether outside it of inside the volume " if it is closed" so you can simply imagine that there is a dotted of imaginary version of your surface and you are projecting your vector field on it
If you're asking about $\int \vec{v} \cdot d\vec{S}$, it might be best to think about a flat surface first. We assign to an infinitesimal "patch" of the surface a vector $d\vec{S}$ which points perpendicular to that "patch" (some people call these "areal (normal) vectors); for a flat surface, all of these point in the same direction.
The vector field described by $\vec {v}$ (or, say, $\vec {v}(x,y,z)$ to make it a function) might represent the "flow" of something. At each patch on the surface, these "flow" vectors are going to make some angle to each areal vector $d\vec{S}$. If $\vec{v}$ and $d\vec{S}$ are parallel, the greatest possible amount of flow "outward" through the surface occurs; if they are anti-parallel, the greatest possible "inward" flow occurs there; if they are perpendicular (the flow is "along the surface"), then the flow through the surface at that point is zero; intermediate values are possible for other relative angles.
So the dot product $\vec{v} \cdot d\vec{S}$ gives the amount of flow at each little "patch" of the surface, and can be positive, zero, or negative. The integral $\int \vec{v} \cdot d\vec{S}$ carried out over the entire surface will give the net flow through the surface; if that sum is positive (negative), the net flow is "outward" ("inward"). An integral value of zero would mean that over the entire surface, there is as much inward as outward flow, so that the net flow is zero.
There is no reason that the full surface has to be flat, since we are integrating all of these infinitesimal patches. We can as easily make a surface which is a closed spherical shell (like a bubble). A physical example which uses this is the electric field of a charge. A single positive electrically-charged particle produces a field $\vec{E}$ which points radially away from the particle. On a sphere, all of the areal vectors $d\vec{S}$ point perpendicularly outward from the sphere's center in their individual patches. So the field $\vec{E}$ is aligned everywhere on this "bubble" with every areal vector $d\vec{S}$, and we have $\vec{E} \cdot d\vec{S} = E dS \cos(0) = E dS$ (here, E is the strength of the electric field at the surface of this sphere and has the same value everywhere on it). The "electric flux" through this spherical shell is then the result of summing E times the total area of all the patches covering the "bubble" of radius R: $\int \vec{E} \cdot d\vec{S} = \int E dS = E \int dS = E \cdot 4\pi R^2$, since $4 \pi R^2$ is the surface area of the spherical "bubble" we've chosen. If the electric charge were negative, the electric field would point toward the charge, giving us an "inward" flux $\int \vec{E} \cdot d\vec{S} = \int -E dS = -E \cdot 4\pi R^2$. If we place equal amounts of positive and negative charge inside the spherical shell, the net electric flux would add up to zero.
This is a very simple and symmetrical example of a surface integral, but the same principle applies to more complicated vector fields and more complicated (open or closed) surfaces.