Practice: Green's Theorem

(a) The boundary integral. The integrand is $xdy-ydx$, and the counterclockwise order of vertices is $(0,0)\to (4,0)\to (0,3)\to (0,0)$. Side 1, $(0,0)\to (4,0)$. Here $y=0$ and $dy=0$, so both terms vanish: $xdy=0$ and $ydx=0$. Contribution $0$. Side 2, $(4,0)\to (0,3)$. Parametrise $\mathbf{r}(t)=(4-4t,3t)$ for $0\le t\le 1$, giving $dx=-4dt$ and $dy=3dt$. Then

so the contribution is ${\int }_0^{1}12dt=12$. The $t$ terms cancel exactly. The integrand is constant along the hypotenuse. Side 3, $(0,3)\to (0,0)$. Here $x=0$ and $dx=0$, so again both terms vanish. Contribution $0$.

(b) The check. The legs along the axes have lengths $4$ and $3$, so $\frac{1}{2}\times 4\times 3=6$ ✓. Note that two of the three sides contributed nothing, because each passes through the origin. The whole area came from a single integral along one side. (c) When the boundary integral wins. The elementary formula applies only because this region is a triangle with two sides on the axes. The boundary integral needs no such structure. It needs only a parametrised boundary. So it is the better choice when the boundary is known and the interior is not conveniently described. Concretely: - A polygon with many vertices, where the formula collapses to the shoelace rule and is a short sum, while decomposing the interior into triangles is fiddly.
- A region bounded by a curve given parametrically. An ellipse $(a\cos t,b\sin t)$ gives the area $\pi ab$ immediately, whereas setting up a double integral requires solving for the limits.
- A region whose outline is traced physically or sampled as coordinates, such as a land parcel in a geographic information system, where no formula for the interior exists at all. The deciding property is which description of the region is available. A double integral needs the interior described by limits; the boundary integral needs only the edge. When a region is given by its outline, as measured, digitised or parametrised regions usually are, the boundary integral uses the data as it comes, and the alternative requires converting it first. That is the general shape of the trade, and it is why the formula underlies both the shoelace rule and the mechanical planimeter.

(a) Claim 1 is not established. Equal cross-partials are necessary for a conservative field and sufficient only on a simply connected region. The site contains a point $w$ where the model is undefined, so the domain is the site minus that point, which is not simply connected, because a loop around the well cannot be shrunk to a point without leaving the domain. So the verification they performed, however thorough, does not support the conclusion. This is exactly the punctured-plane situation: the field $(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$ has equal cross-partials at every point of its domain and integrates to $2\pi$ around the unit circle, so it is not conservative there despite passing the test everywhere. Their own report contains the decisive evidence, in claim 3. (b) The area comes from the outline directly. Green's theorem gives

taken counterclockwise, which needs only the boundary, precisely what the survey recorded. For a polygon this collapses to the shoelace formula: summing $x_i{y}_{i+1}-x_{i+1}{y}_i$ over consecutive vertices and halving. The worked triangle in this unit illustrates the mechanics: with vertices $(0,0)$, $(4,0)$, $(0,3)$, the two sides through the origin contribute nothing and the hypotenuse parametrised as $(4-4t,3t)$ gives a constant integrand $12$, so the area is $\frac{1}{2}(0+12+0)=6$, matching $\frac{1}{2}\times 4\times 3$. The reason this suits their situation is the one that decides such choices generally: a double integral needs the interior described by limits, and this needs only the edge. Their data is an outline, so the boundary method uses it as recorded while the alternative would require converting it first. They should confirm the polygon is traversed counterclockwise, since the reverse order returns the negative of the area. (c) The nonzero circulation is very probably not an error. A conservative field gives zero around every closed path, so a nonzero value around the well loop is consistent, and their own cross-partial verification does not contradict it, because that verification never established conservativeness on this domain. Physically, a nonzero circulation around a loop enclosing an extraction well is what an extraction well produces. The local circulation is zero everywhere the model is defined, and the loop still registers a nonzero total: the circulation is concentrated at the point that has been removed from the domain. That is the same structure as the punctured-plane example, where every local measurement reports nothing and every loop around the origin reports $2\pi$. Note also that Green's theorem does not apply to this loop. Its hypothesis is continuous partials on the enclosed region, and the well sits inside, so the theorem offers no route from the zero integrand to a zero loop integral, and there is no inconsistency to resolve. Re-surveying would waste effort and, if the crew were to "correct" the measurement toward zero, would destroy a real signal. (d) The tabulation cannot work as proposed. A single-valued potential tabulated per location requires path independence. A nonzero loop value rules it out directly: going around the loop returns to the starting point having accumulated a nonzero amount, so no function of position can account for the transport. Concretely, transport computed between two points would depend on which side of the well the route passes, and a table indexed by location alone cannot express that. Any such table would silently give one of the two answers and be wrong for routes on the other side. The kind of error that produces plausible numbers indefinitely. The analogy is the polar angle $\arctan (y/x)$, which increases by $2\pi$ per circuit of the origin: it is a perfectly good potential locally and cannot be made single-valued globally. (e) Advice. Keep the area calculation, claim 2 is sound, and the shoelace formula on their recorded outline is the right method. Check the traversal orientation. Retain the circulation measurement as a finding, not an error. Its value quantifies the well's extraction, and the fact that local measurements show zero while the loop does not is the expected signature rather than an inconsistency. Restrict the potential to a region where it exists. The plan works on any simply connected subregion of the site excluding the well, for instance the site cut along a line from the well to the boundary, or any portion lying to one side of the well. On such a region the equal cross-partials do imply a potential, and their tabulation is valid there. This mirrors the punctured-plane field being conservative on the right half-plane with potential $\arctan (y/x)$ while failing on the full punctured plane. What they must not do is tabulate across the whole site and use it for routes that pass on either side of the well. Record the condition in the deliverable. The table should state the region it applies to and that transport across the cut is not covered, because a table without that caveat will eventually be used for a route it cannot describe. Conservativeness is a property of a vector field together with a region, never of a formula alone. Their verification was correct, their conclusion did not follow from it, and the measurement they proposed to discard is the one that shows why.