Greens theorem: why does path orientation matter? $$\oint_{\partial D} P\;dx + Q\;dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\;dA$$
Is it correct to interpret this equation as relating the surface area of a three dimensional object (a subset of R3) to the closed path around the object (or the perimeter)? 
Assuming this is the correct interpretation, why does one need to concern themselves with the clockwise/counter-clockwise convention when:
1) The start and endpoint of the path are the same, regardless of how one travels the boundary of the object.
2) What meaning could a negative surface area have?
update: I know wikipedia tells you this a two-dimensional formula, yet from the left side of the equation the integral would look like it resolves into some function F(x2,y2) - F(x1,y1) and furthermore I don't see the use of orientation or vector calculus in 2d when one can use non-vector calculus to find areas in 2d.
 A: Green's theorem is really a special case of the generalized fundamental theorem of calculus. In this theorem, the "differentiation" operation is generalized to allow for the derivative of things known as differential forms. The theorem is incredibly elegant and can be written simply as $$\int_{\partial D}\omega\;=\;\int_{D}{\mathrm{d}\omega},$$ which says that integrating a differential form $\omega$ over the oriented boundary of some region of space $D$ is the same as integrating the exterior derivative of the form (denoted $\mathrm{d}\omega$) over the region $D$. Here the differential form is $P\;dx+Q\;dy$ and its exterior derivative turns out to be $(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dx\wedge dy$ (the "wedge" $\wedge$ is basically multiplication that is anti-commutative, i.e. $dx\wedge dy = -dy\wedge dx$).
The signs become important because orientation on surfaces, curves, etc. is well-defined. You can think of the positives and negatives arising as a by-product of the theorem. If it wasn't for the changes in sign, the theorem would not be true.
1) The start and end of a parametrized curve may be the same, but reversing the parametrization (and hence the orientation) will change the sign of a line integral when you actually compute out the integral by hand.
2)"Negative" area is kind of a tricky. Think about when you are taking a regular integral of a function of one variable. Then $\int_a^b{f(t)}dt$ represents the total area under the graph over the line segment $[a,b]$ if you consider the area under the $x$-axis as "negative". Without going into too much detail, the idea of negative surface area (or negative volume, etc.) comes from the idea that Euclidean space has a positive orientation. If you try to put the $x$-, $y$-, and $z$-axes together in some other arrangement, then we either end up with the same orientation we started with before (where $\mathbf{i}\times\mathbf{j}=\mathbf{k}$) or we get it's negative (where $\mathbf{i}\times\mathbf{j}=-\mathbf{k}$). It again comes down to keeping track of signs so that things remain well defined and so we can properly assign boundaries to objects.
A: No, it is not strictly related to surface area. You can force it to relate to that if, for example, you take $P = - \frac{y}{2}, \; Q = \frac{x}{2} .$ 
I like to show students how this works on a square or rectangle with edges parallel to the axes. So, take the unit square with vertices $ A = (0,0), \; B = (1,0), \; C = (1,1), \; D = (0,1).$ Take the example $P(x,y) = x^2 y^3$ and $Q(x,y) = x^2 + 1.$ Now,
$$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 x y - 3 x^2 y^2  $$ and the double integral becomes 
$$\int_0^1  \left. \; \; \; 2 x y - x^2 y^3 \right|_{y=0}^{y=1} \; \;     dx = \int_0^1  2 x - x^2  dx = \; \; \left. x^2 - \frac{1}{3} x^3 \right|_{x=0}^{x=1} = \frac{2}{3}$$ 
In comparison, along $AB$ we integrate $0 dx$ for 0. Along $BC$ we integrate $2 dy$ giving 2. On $CD$ we integrate $x^2 dx$ to get $1/3$ in the wrong direction. Along $DA$ we integrate $1 dy$ in the wrong direction. All together we get
$$ 0 + 2 - \frac{1}{3} - 1 = \frac{2}{3}.$$
Historically this is most closely related to Ampere's Law in physics, see AMPERE. One could say that the effect of  reversing direction around the boundary is quite simply that of sending current $J$ in the opposite direction, thereby negating the induced magnetic field $B.$
