Now that we’ve covered line integrals, we will begin to look at surface integrals. The integrals we look at today involve vector fields over a flat area. In this case, we’ll look only at flat surfaces that are simply closed (meaning there is no overlap of the line encompassing the area).

Green’s Theorem exists because of the property that integrals over a simply closed surface are equal that that of the integral of the line that surrounds them. This is a pretty handy property because trying to get pretty integrals out of equations like the following don’t look so easy:

∫_{c} xy dx + x^{2} dy over the area between lines y=x and y=x^{2}

In that integral, it means the over the curve C. In that case C would be a line from the origin along y=x^{2} until x=2, then back along the line y=2x until it hit the origin. To do that integral, I’m not even sure where you’d start. Do you start with integrating one part of the equation? What then do you do with the other? I don’t even know. The best answer, though, is to use Green’s Theorem. Let’s take a look at what area we’re talking about here.

No, it wasn’t me who drew those arrows in with MS Paint (seriously, it came from a UCLA homework page). As you can see, the green arrow is the line y=x^{2} and the red line is y=x. Now for the fun part of the theorem… Green’s Theorem specifically states, that an integral of the form ∫_{c} M(x,y) dx + N(x,y) dy can be simplified to a double integral over the area. In this case, M and N are functions of variables x and y. His formula:

∫

_{c}M(x,y) dx + N(x,y) = ∫ ∫_{D}N_{x}– M_{y}dA

In the formula, N is the derivative of the function N_{x} with respect to x (I’ll let you figure out what M_{y} must be then). This combines the two awkward integrals of the closed line integral into a simple iterated integral with respect to dx and dy.

Let’s look back at our example from earlier, ∫_{c} xy dx + x^{2} dy. This would mean that M(x,y)=xy and N(x,y)=x^{2}. Using Green’s Theorem, we first find N_{x} = 2x. Next we find M_{y} = x. Placing them into the intergral we get:

∫ ∫_{D} 2x – x dA = ∫ ∫_{D} x dA

This is a very easy iterated integral. All we have left is to find the limits of integration, which we find by looking back at the graph above. I’m going to integrate in respect to y first, but you could integrate in terms of x first if you really wanted to. I’m choosing to integrate with respect to y first just because we already have the y-limits of integration from the two lines creating curve C.

∫_{0}^{1} ∫_{x2} ^{x} x dy dx

Apologies again for not using a formula editor, but the limits are from 0 to 1 for dx and x^{2} to x for dx. Working it out, we have the following happen:

∫_{0}^{1} ( xy |_{x2} ^{x} ) dx = ∫_{0}^{1} ( x*x)-(x*x^{2}) dx = ∫_{0}^{1} x^{2}– x^{3} dx

Then it becomes a simple single integral, eventually yielding ½ – ⅓ = ⅙. That’s much easier than being stuck at just the equation. What’s good about Green’s Theorem is that a variation of it (Stokes Theorem) also works with 3D objects, which we will discuss in a later post (hopefully).

One last note before I go has to do with conservative vector fields. You’ll notice that if the vector field is conservative, M and N will equal each other leaving you with the integral of ZERO. That should make sense because the line around the area you are integrating is a closed loop (ending where it begins). Therefore, if the field is conservative, your integral MUST equal zero.

Next up, if I get to it is creating parametric surfaces and – perhaps – taking integrals of those surfaces.