The surface integral works much like its younger brother, the line integral. As the line integral came from a closer examination of arc length, the surface integral comes from a closer examination of surface area. These surface integrals play an important role in mathematical physics, especially in the fields of fluid dynamics, electricity, magnetic flux, and elasticity.
The first surface integral-like thing was performed by Archimedes using his "exhaustion" method to find the surface area of a sphere. He found its surface area by inscribing more and more closely approximating polyhedra inside and outside the sphere. In this way he obtained upper and lower bounds for the area of the sphere, from which he inferred the formula: 4πr2.
Tracing the discovery of the surface integral of traditional undergraduate calculus is somewhat more difficult. Surely it was not known to Newton or Leibniz, but it is clear that Gauss knew of them — his theory of Gaussian surfaces depends on them. This places the discovery of the traditional surface area sometime in the early 19th century.
Defining a surface
A surface is a very abstract mathematical entity. For our purposes, a surface in R3 is a subset of R3 which possesses two different coordinate representations - one being the usual (x, y, z); the other being unique to the surface and having only two parameters (u, v). On this surface, u and v serve much the same purpose as "north" and "south" do for us standing on Earth's surface. Part of the surface requirement is a function that transfers us from xyz-space to this uv-"plane". Typically a restriction is placed on the values of u and v that are permitted.
Finding the surface area of a scalar field
Just as we needed to know how to find the arc length of a path before we could find the line integral over it, it is necessary to know something about the surface area of an arbitrary surface before taking surface integrals over it. The simplest surface area is a scalar field over R2, z = f(x, y). In this case the uv-plane is simply the xy-plane. This, then, satisfies our definition for a surface, since (x, y) is mapped to a point in R3 called (x, y, f(x,y)).
Recall that the plane tangent to a point on a scalar field can be formed by using only two things: a point on the scalar field and the vector perpendicular to that field. Getting a point on a scalar field is simple; finding its normal vector is not as simple. The gradient of our scalar field, ∇z, yields a vector (∂xz, ∂yz), from which we can create two directional derivatives; one in the direction of the x-axis (∇z ⋅ i) and one in the direction of the y-axis (∇z ⋅ j). This gives us two vectors which lie on the tangent plane: (1, 0, ∂xz) and (0, 1, ∂yz). To get the vector perpendicular to both of these, we use the cross product and obtain n = (∂xz, ∂yz, -1). We can tell by looking at this that this vector points down; to make it point up we multiply it by -1 and get a more reasonable value for n, (-∂xz, -∂yz, 1).
Now that we know everything we need to know about the tangent plane, we can relate the area of a region of that plane to the corresponding region beneath it on the xy-plane. From projective geometry we know that the area of the projected region on the xy-plane is equal to the real plane's area times the cosine of the angle between them. Thankfully, we know both the normal of the tangent plane and the normal of the xy-plane (k). The angle between them is given by:
cos α = (n ⋅ k)/(||n|| ⋅ ||k||) = 1/√(∂xz2 + ∂yz2 + 1)
This means that, given a small enough region on the xy-plane, we can approximate the area of the scalar field by finding the corresponding area of the tangent plane like so:
dT = dA/cos α = √(∂xz2 + ∂yz2 + 1) dA
Now, depending on the projected region in the xy-plane, you can double integrate dT in whatever manner you'd like, using dA = dx dy or dA = r dr dφ or even dA = ∂(s, t)/∂(x, y) ds dt, just as if it were a regular area on the xy-plane.
If the surface given by φ: (x, y, z) ⇔ (u, v) cannot be represented as a scalar field, we are in a pickle. Look back at the two vectors we used to find the normal vector of a scalar field, (1, 0, ∂xz) and (0, 1, ∂yz). If we thought of a scalar field as a surface X from (x, y) to (x, y, z(x, y)), these two vectors would be ∂xX and ∂yX. With φ there isn't any area on the xy-plane to compare it with anymore, but we can use ∂uφ and ∂vφ to define a really small parallelogram that approximates surface area. That parallelogram's area will be ||∂uφ × ∂vφ||. This is the fundamental vector product, and, if we were to double integrate over it with respect to du dv, it would be equivalent to using the method above for scalar fields.
The second kind of surface integral
Now we are able to talk about the more interesting sort of surface integral which deals with flux. Flux is defined as flow out of an area over a period of time; negative flux is flow into the area. Flux happens over a surface surrounded by another vector field. Let F be a vector field from (x, y, z) to (x, y, z), and let φ be our surface from (x, y, z) to (u, v). Over a really small area, we can approximate this with F ⋅ (∂uφ × ∂vφ). This is like splitting the area and the vector field into its components, finding the flux over each component, and then summing up the component fluxes into one big flux. Then, as above, you can double integrate over du dv any way you'd like.
Orientation preserving parametrizations
As if all that wasn't enough, there's still one more thing you need to worry about when dealing with surfaces. It's possible that the parametrization you find for the surface be negatively oriented, and this will cause your final result to flip signs. It's a nasty thing to happen on a final. Check out the node on orientation for more information about things like that.
This writeup forged from notes from my fourth quarter of calculus. im in yr g3l, n0din' my h0m3workz.