Xrfgtjtk

Is there any intuition behind surface integrals and their applications? I get that there's some stuff with fluid flow but it doesn't really stick that well, especially when it comes to surface orientation.

Is there any surefire way to figure out the orientation of a surface? And is there any sort of intuition behind that or surface integrals?

The surface integral of a vector field, $F$ over a surface $S$ (or flux as it is called) is, as stated above, the net flow of fluid through the surface. The purpose of orientation is to simply indicate which side of the surface is the "positive" side. For instance, if I calculate flux to be +10, this means there is a net movement of 10 units of fluid towards whichever side of the surface is indicated as the positive side. The answer would be -10 if the orientation was reversed (since the "positive" side is now losing 10 units of fluid).

If you are asked to find the flux of $F$ over $S$, they have to tell you the orientation of $S$ in the question. For instance

"Find the flux of $F(x,y,z) = langle 2xz,z-y,y^3 rangle$ DOWNWARD through the section of the plane $x+y+z=6$ in the first octant."

In this question, the region under the plane is the positive side, and a positive flux would indicate a net movement of fluid downward through the plane. A negative answer indicates a net upward movement of fluid. So, how do you figure out orientation for a surface? You don't. It's usual stated in the question. In some cases they don't state orientation because they will accept either the positive or negative answer.

Some theorem's such as Stokes' and Gauss' Divergence Theorem require that the surface in question has a particular orientation. Stokes' theorem for instance, allows us to calculate a line integral along a closed curve by first choosing any surface enclosed by this curve. If the curve is traversed counterclockwise as viewed from above, the orientation of the surface must be "Upward" to use this theorem. So, whatever surface you choose, it's normal vector must point upward. In this case, you are the one choosing the surface so you so have to figure out the orientation yourself.

I should finally point out that I am assuming you are only working with surfaces in $R^3$. Orientation of a surface in $R^n$ (or hypersurfaces in $R^n$) is a very difficult concept since these are not real world objects that you can look at and decide the orientation of. Also, some 2d surfaces such as a Möbius strip, have no orientation since they only have one side.

A surface integral measures the "average outflow" of the vector field from the surface.

One illuminating application is Gauss's Law for electric fields, i.e.,

The electric flux through any closed surface is proportional to the enclosed electric charge.

More precisely,

$$oint_S mathbf{E} cdot mathrm{d}mathbf{A} = frac{Q}{varepsilon_0},$$

where the left-hand side of the equation is a surface integral denoting the electric flux through a closed surface $S$, and the right-hand side of the equation is the total charge $Q$ enclosed by $S$, divided by the electric constant $varepsilon_0$.

Gauss's Law itself has very interesting physical consequences. See the wikipedia page for examples. For instance, the renowned Coulomb's law is equivalent to it. For another: all the charges on a conductor migrate to its surface.

There is also a Gauss's Law for magnetism. It states that the net flux of magnetic fields through any surface is always zero. The difference with electric fields is that there are no magnetic "monopoles"; you always find dipoles. The equation is:

$$ oint_S mathbf{B} cdot mathrm{d}mathbf{A} = 0 $$

$S$ is any closed surface and $B$ is the magnetic field.

Both these Gauss's laws are part of the important Maxwell's equations. Many applications are through integration.

Maybe some trivial examples could show you the importance of orientation. Try to compute the fluid flow through the unit square

$$
S = left{(x,y,z) | 0leq x,y leq 1 , z=0 right}
$$

of some fluid moving upwards at constant speed of $1$m/s.

$$
F(x,y,z) = (0,0,1) .
$$

Then you parametrize your surface, for instance with

$$
varphi (x,y) = (x,y,0) , quad (x,y) in D = left{ (x,y) | 0leq x,y leq 1 right}
$$

The normal field to this parametrization is $varphi_x times varphi_y = (0,0,1)$ and we compute the fluid flow:

$$
int_varphi langle F, dSrangle = int_0^1int_0^1 langle (0,0,1) , (0,0,1) rangle dxdy = 1 .
$$

But why using that particular parametrization? Why not trying with this one?

$$
psi (u,v) = (v,u,0) , quad (u,v) in D = left{ (u,v) | 0leq x,y leq 1 right} .
$$

We compute the normal field again: $psi_u times psi_v = (0,1,0)times (1,0,0) = (0,0,-1)$. And the fluid flow:

$$
int_psi langle F, dS rangle = int_0^1int_0^1 langle (0,0,1) , (0,0,-1) rangle dudv = -1 .
$$

Astonishing, isn't it? -The fluid flow seems to depend on something secondary: the coordinates you use on the surface $S$. This could be even worse, since some surface integrals are used to compute the area of the surface: if the area of a surface depended on the coordinates you put in it, this would be a nonsense.

Fortunately, this isn't true: you can prove that surface integrals do NOT depend on the coordinates you put on the surface EXCEPT (in the case of fluid flows) on the orientation of the coordinates -parametrizations. There are two kinds of orientations and before computing a fluid flow you must choose one of them. For instance, in our examples, if we have said that the fluid is going upwards we have to choose $(0,0,1)$ as the orientation of our surface (the $varphi$ parametrization). Otherwise, the result we get -with the $psi$ parametrization- makes no sense: in one second $-1$ litre has gone trhough the surface!?