Let f be a scalar field, and recall the definition of a path integral. If the curve represents a wire, and f is a function that gives the density of the wire at every point, the path integral computes the total mass of the wire. The integrand is the product of density times speed, so as you move along, tracing the path of the wire, the integrand measures the amount of mass whizzing by per second. Integrate with respect to time and get the total mass of the wire. An analogous formula holds for surface area.
The surface integral is the double integral of surface area times f, where f is the density function. You can use any of the formulas for surface area, including the formulas for surfaces of revolution. Just multiply by f and integrate.
Let the density of a sphere be proportional to its z coordinate. In other words, the equator is light as a feather and the top is heavy. What is the mass of the top half of this sphere?
This is a surface of revolution, so surface area is 2π times the integral of r/sqrt(1-r2). The denominator is the z coordinate. Multiply by z and we are left with r. Integrate from 0 to 1, and the mass is π. This is half the surface area. The mass would equal the surface area if the density were 1 everywhere.
Now for the surface normal integral in a vector field. The line integral measures the "agreement" between the path and the vector field. How much of the path flows with the force field and how much pushes against it. Well a surface normal integral measures the amount of surface that is perpendicular to the field. How much field flows through the surface. What is the flux?
Let's consider a real world example. Why is the equator hot, while the poles are cold? The earth is 93 million miles from the sun, so the extra 4,000 miles from equator to pole shouldn't make that much difference. Something else is at work.
consider a patch of ground, one meter square, at the equator. It faces the sun dead-on, and receives a square meter of sun light. Now move to the north pole and consider a square meter of ground at that location. Stand back and look at this patch of earth. From the sun's point of view, it is practically edge-on. It looks like a slit. A small amount of sun light passes through this slit and spreads across the square. That's not much heat per square meter by day, and at night heat is radiated into space with the same efficiency, whether you're at the equator or the poles. So - with far less sunlight per square meter, the poles have a much lower equilibrium temperature. And the differences would be more extreme if we didn't have an atmosphere and oceans to move the heat around.
The surface normal integral measures the total amount of heat, or light, received by a surface. The area of each tiny patch on the surface is multiplied by the intensity of the field at that point, times the sine of the angle between the field and the surface. If the light is streaming directly into the surface, the angle is 90°, the sine is 1, and the sunshine is working at full capacity. Near the north pole, the surface and the sun's rays are almost parallel. The angle is near 0, the sine is near 0, and there is very little heat flowing into the surface at that point.
Now for the magic. Recall our formula for surface area. The area of a small patch on the surface is the length of the cross product of two vectors that define a tiny parallelogram at that point. If the vector field is approximately equal to v throughout this patch, then we are interested in the length of v times the area of the parallelogram times the sine of the angle between them. But aha, this is the triple scalar product.
If ge maps uv into xyz, and defines a surface in 3 space, and f is a vector field, the surface normal integral is the integral, over the region in the uv plane, of (g∂u×g∂v).f. In other words, we are integrating the triple scalar product.
Let h(x,y) define the umbrella, a surface over a region in the xy plane. Let the region have area a. A small square on the ground maps to a parallelogram on the surface, having vectors 1,0,h∂x and 0,1,h∂y. Take the cross product of these vectors, and dot this with the vector 0,0,1, which represents the sunlight streaming down from above. The triple scalar product evaluates to 1. Integrate 1 over an area of a, and the amount of sunlight blocked by the umbrella is indeed a.
If the sun streams down unevenly, e.g. there are some branches over head, let f(x,y) describe the intensity of the sunlight at each xy location on the ground. Open your h(x,y) umbrella as before, and compute the surface normal integral. This time we are dotting with 0,0,f, rather than 0,0,1. The sunlight blocked by your umbrella is now the double integral of f over a. This does not depend on the shape of the umbrella, and if you pull the umbrella away, the formula gives the amount of solar energy deposited on the flat patch of ground beneath. Everything works out.
The concept of a surface integral, in a scalar field or a vector field, and even the umbrella theorem, generalizes to higher dimensions.