$$\oint_{\partial D} P\;dx + Q\;dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\;dA$$

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.

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    $\begingroup$ Using single variable calculus does not relieve you from the obligation to get the orientation right. You are just so used to integrating over an interval from left to right that you may not notice that this (LeftToRight) is a choice of orientation. This shows in 2D-problems, when you use a parametric presentation of the boundary curve. As an example consider the upper semicircle with parametrization $x=\cos t$, $y=\sin t$, $0\le t\le \pi$. This parametrization traverses the curve from right to left, and you get the wrong sign from $\int y\,dx=\int_{t=0}^\pi y(t)x'(t)\,dt$. $\endgroup$ – Jyrki Lahtonen Nov 16 '11 at 5:10
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    $\begingroup$ Same thing happens when you're integrating a function of one variable. Remember the result that $\int_a^b{f(t)}dt\;=-\int_b^a{f(t)}dt$? $\endgroup$ – Patch Nov 16 '11 at 5:24
  • $\begingroup$ The area formula coming from Green's theorem happens to be the same as the surveyor's formula (or the shoelace formula) extended to convex regions with smooth boundaries. And the shoelace formula in its original form is oriented. $\endgroup$ – Divide1918 Jun 29 '20 at 15:21

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.

  • $\begingroup$ I had my $D$ and $\partial D$ flipped. Fixed. $\endgroup$ – Patch Nov 16 '11 at 5:31
  • $\begingroup$ Thank you for your answer! I'm actually brushing up on on my vector-calc before i make my attempt at calculus from a differential forms perspective. This time im trying to gain an intuitive understand of all this vector stuff rather than when i learned it mechanically the first time (just plug and chug). Is there any sites or books you would recommend for learning differential forms? $\endgroup$ – skyfire Nov 16 '11 at 15:53
  • $\begingroup$ A decent read is Spivak's "Calculus on Manifolds". It's a quick read and covers differentiation in multiple variables, tensor products of vector spaces, integration of differential manifolds on chains (essentially formal sums of hyper-surfaces embedded in some $\mathbb{R}^n$), and even a little about differential forms on manifolds. Sometimes his notation is a bit unconventional, but it's still good to start with, in my opinion. As far as full-out Riemannian geometry goes, I still haven't found a text I'm in love with yet. $\endgroup$ – Patch Nov 16 '11 at 20:07

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 $B.$

  • $\begingroup$ very helpful, I'm studying as an EE so I appreciate the visualization with amperes law. $\endgroup$ – skyfire Nov 16 '11 at 15:55

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