# Chap12 Green's Theorem

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<p>Greens theorem 1Chapter 12 Greens theoremWe are now going to begin at last to connect dierentiation and integration in multivariablecalculus. In addition to all our standard integration techniques, such as Fubinis theorem andthe Jacobian formula for changing variables, we now add the fundamental theorem of calculusto the scene. In fact, Greens theorem may very well be regarded as a direct application ofthis fundamental theorem.A. The basic theorem of GreenConsider the following type of region R contained in R2, which we regard as the xy plane.We use the standard orientation, so that a 90 counterclockwise rotation moves the positivex-axis to the positive y-axis. Then we assume the existence of two continuous functions a(y)and b(y), dened for c y d, where a(y) < b(y) for c < y < d andR = {(x, y) | a(y) x b(y)}.xydcGiven a function R FR of class C1, we then compute by means of Fubinis theorem__RFx dxdy =_ dc_ b(y)a(y)Fx dxdyFTC=_ dcF(x, y)x=b(y)x=a(y)dy=_ dcF(b(y), y)dy _ dcF(a(y), y)dy.We now write the right side of this equation as_bdRFdy,2 Chapter 12where we mean by this notation the counterclockwise integration of F, restricted to bdR, withrespect to y. (The actual denition is given in the formula.)xyNotice that on a horizontal portion of bdR, y is constant and we thus interpret dy = 0 there.B. Line integralsWe have now met an entirely new kind of integral, the integral along the counterclockwisebdR seen above. Before proceeding further, we need to discuss this sort of oriented integral.It will prove useful to do this in more generality, so we consider a curve in Rnwhich isof class C1. Thus, [a, b] Rnis of class C1. We typically denote the independent variable(the parameter) as t. If f is a continuous real-valued-function dened on Rn, we then denethe line integral_fdxi =_ baf((t))</p>
<p>i(t)dt.Here of course 1 i n. The notation is intended to be very suggestive and to lead us tothe appropriate substitutions x = (t), xi = i(t), and dxi = </p>
<p>i(t)dt.It is an easy matter to imagine some useful properties of this sort of integral, and eveneasier to prove them.INDEPENDENCE OF PARAMETER CHANGESuppose rst that (t), a t b, is represented instead as (h(s)), c s d, where h isa C1function such that h(c) = a and h(d) = b. Then using the parametrization (h(s)) leadsto the line integral_ dcf((h(s))) ddsi(h(s))ds =_ dcf((h(s)))</p>
<p>i(h(s))h</p>
<p>(s)dst=h(s)=_ baf((t))</p>
<p>i(t)dt,Greens theorem 3which is the original line integral.stbac dThis proves the desired independence.On the other hand, if instead h(c) = b and h(d) = a, then we obtain_ dcf((h(s))) ddsi(h(s))ds = _ baf((t))</p>
<p>i(t)dt,so we get the anticipated change of sign.stbac dLINEARITYThis is virtually obvious from the denition:_afdxi = a_fdxi if a is a constant;_(f + g)dxi =_fdxi +_gdxi.COMBINING CURVESSuppose we are given curves and such that the nal point of equals the initial pointof . Then we can think of a new curve that covers and then . This curve will be continuousbut not necessarily C1. We denote this curve loosely as + , and then the clear result is_+fdxi =_fdxi +_fdxi.GENERALIZATIONIf f1, . . . , fn are given functions, then we dene_f1dx1 + + fndxn =_f1dx1 + +_fndxn.4 Chapter 12Another way to think of this is to arrange the functions into a vector F = (f1, . . . , fn) andformally writex = (x1, . . . , xn),dx = (dx1, . . . , dxn),F dx = f1dx1 + + fndxn.Thus we have the abbreviation_F dx =_f1dx1 + + fndxn.FUNDAMENTAL THEOREM OF CALCULUSSuppose that f is a real-valued function of class C1and thatF = f = (f/x1, . . . , f/xn).Then we calculate directly_f dx =_ baf((t)) </p>
<p>(t)dtchain rule=_ baddtf((t))dtFTC= f((t))ba= f((b)) f((a)).Thus we have proved_f dx = f(nal point of ) f(initial point of ).We shall say much more about this equation in Section E.C. A general Greens theoremWe now return to the formula of Section A,__RFx dxdy =_bdRFdy. ()Greens theorem 5The right side is now completely understood as a line integral taken along the curve bdR withits counterclockwise orientation. We presently have severe restrictions on what the region Rcan be, and we now show how we may easily remove many of these restrictions.Rather than give precise denitions to delineate the allowable regions, we prefer to rely onpictures. Suppose then we have a region in R2that looks like this:RxySuch a region fails to satisfy the conditions of Section A. However, the desired equation ()still holds for it. An easy way to see this is to divide R into two regions R1 and R2 by meansof a cut, in such a way that () is valid for each piece separately.xyR1R2We thus have__R1Fx dxdy =_bdR1Fdy,__R2Fx dxdy =_bdR2Fdy.Now add these two equations. The left sides produce the double integral over R itself, and theright sides combine in an interesting way. Namely, the part of each line integral corresponding6 Chapter 12to the cut appears in each, but with opposite orientations. Thus they cancel to produce just_bdRFdy.We therefore nd that () remains valid for this R.Of course, this procedure may be extended to several cuts, so that () remains valid for aregion such asOne more generalization allows holes to appear in R, as for exampleRWe again make cuts to get a couple of regionsR1 R2Greens theorem 7Then we apply () to R1 and R2 and add the results, noting the cancellation of the integrationstaken along the cuts. The result still is (), but with an interesting distinction: the line integralalong the inner portion of bdR actually goes in the clockwise direction.A convenient way of expressing this result is to say that () holds, where the orientationof bdR is such that the interior of R is located to the left of the path of integration.Finally, we close this section with the corresponding result__RFy dxdy = _bdRFdx. ()The minus sign appears naturally, as we see from rewriting the basic case of Section A. Anotherway to think about it is to realize that the coordinate system with y rst, x second, has theopposite orientation to the one we have been working with.Most texts combine these two formulas into a single one by using dierent letters for thetwo cases, and adding. Thus,__R_Qx Py_dxdy =_bdRPdx + Qdy.D. Areas by means of GreenAn astonishing use of Greens theorem is to calculate some rather interesting areas. Thesecome for example by using F = x in () in Section C to obtainarea(R) =_bdRxdy.Likewise, using F = y in () producesarea(R) = _bdRydx.The latter equation resembles the standard beginning calculus formula for area under a graph:xya by f x = ( )R8 Chapter 12The integral_ baf(x)dxis exactly the line integral_ ydxtaken around bdR.It is interesting that the sum of these two formulas is often more easily exploited. Or useQ = x, P = y in the formula at the end of Section C. The result isarea(R) = 12_bdRxdy ydx.This is often useful in situations where x and y appear rather symmetrically.ELLIPSE. We of course know from several investigations that the area inside the ellipsex2/a2+ y2/b2= 1 is equal to ab. This example gives a nice illustration of our new formula.Namely, parametrize the ellipse by x = a cos t, y = b sin t, where 0 t 2. Thenxdy ydx = a cos t b cos tdt b sin t (a sin tdt)= ab(cos2t + sin2t)dt= abdt.Thus the area equals12_ 20abdt = ab.Dont miss the fact that if we had just integrated xdy or ydx to nd this area, the nalintegration would have been signicantly more dicult.FOLIUM OF DESCARTES. Here is a truly signicant example, one for which our formulaprovides what is perhaps the easiest method for calculating a certain area. We rst discussedthis curve in Section 6E, where we used the implicit equation x3+ y3= 3xy. In terms of theratio t = y/x, we have the parametric presentationxyx = 3t1 + t3,y = 3t21 + t3.Greens theorem 9Now we nd the area of the leaf, that region enclosed by the part of the curve in the rstquadrant. Here the parameter t runs between 0 and . We calculatexdy ydx = xd(tx) txdx= x(tdx + xdt) txdx= x2dt= 9t2dt(1 + t3)2= d_ 31 + t3_.Thusarea = 12_ 0d_ 31 + t3_= 3211 + t30= 32.PROBLEM 121. Obtain the area of the leaf of Descartes by using just the formulaarea(R) = _bdRydx.PROBLEM 122. Show that the equationsarea(R) = _bdRydx=_bdRxdycan be regarded as equivalent by means of an integration by parts.10 Chapter 12PROBLEM 123*. Find the area between the folium of Descartes and its asymptotex + y = 1, as shown in the gure.xyPROBLEM 124. Find the area enclosed by the curvex4+ y4= 4xyin the rst quadrant.PROBLEM 125. Consider the curve dened by the equation x4= xy2+ y3.a. Use the parameter t dened by y = tx to express the curve in parametric form.b. Show that this curve has a leaf in the fourth quadrant.c. Calculate the area of the leaf. (ANSWER: 1/210).E. Conservative vector eldsNow we return to our discussion of line integrals in general, as they were introduced inSection B.DEFINITION. Let D be an open subset of Rn. A vector eld on D is a continuous functionD FRn.Thus F assigns to each point x D a point F(x) Rn. We always think of F(x) as a vectorattached to the point x. Thus we have a picture of the following sort:Greens theorem 11x1x2ODF x ( )F x ( )Using the components of F = (F1, F2, . . . , Fn), we have dened line integrals of the form_F1dx1 + + Fndxn.In Section B we introduced the notation_F dxfor these line integrals.Now we state an important basic theorem about these vector elds.THEOREM. Let D be an open subset of Rn, and let F be a vector eld dened on D. Thenthe following three conditions are equivalent.1. The line integral_ F dx is independent of path, in the sense that its value dependsonly on the initial point of and the nal point of .2. The line integral_ F dx = 0 for every loop in D (a loop is a curve whose initialpoint equals its nal point).3. There exists a function D fR of class C1such thatF = f.12 Chapter 12PROOF. The proof that 1 2 is just based on the simple observation that a loop whichstarts at x0 and ends at x0 and the constant path which just stays at x0 are two paths with thesame initial and nal points. As the line integral over the constant path is zero, condition 1implies that the line integral around the loop is zero as well.The proof that 2 1 is based on another simple observation: if 1 and 2 are two pathswith the same initial and nal points, then the path that goes along 1 and then backwardsalong 2 is a loop. Thus condition 2 implies_1F dx +_2reversedF dx = 0.This proves condition 1.The proof that 3 1 is an immediate consequence of the last formula in Section B.The really signicant part of this theorem is the proof that 1 3.Before we introduce the crucial construction of the function f we notice that we may aswell assume that our open set D is connected. In other words, that any two points in D canbe joined by some curve lying in D. For we may work on each connected component of D andproduce the required function f on each of them in turn.In this context it is important to observe that any two points in the connected set D canin fact be joined by a nice curve, say one of class C1or one which is polygonal. For we needto be able to do line integrals along such curves.DNow our denition of f is actually forced upon us. For given the vector eld F, if there isto be a function such that F = f, then we know from Section B thatf(nal point of ) f(initial point of ) =_F dx.Now choose an arbitrary but xed point b D. It will serve as a base point. Then for anyx D we produce a curve in D whose initial point is b and whose nal point is x. Thenour potential function f has to satisfyf(x) = f(b) +_F dx.Greens theorem 13Since b is xed, f(b) is a constant; we may as well call it 0, as adding a constant to f doesnot disturb our desired equation f = F. Thus we now dene D fR by the equationf(x) =_F dx,where is any nice curve in D starting at b and ending at x.We have now in fact used the hypothesis 1, since the value given for f(x) does not dependon the choice of .Finally we must verify that f = F. Well present two proofs. These proofs are indeedrelated, but they have distinct emphases.FIRST PROOF. Consider any xed point y D. Let 0 be a nice curve in D startingat b and ending at y. Also let B(y, r) be an open ball contained in D. For any unit vectorh Rnand any 0 s < r we use the curve 0 followed by the curve which is the line segmentfrom y to y + sh. Thenf(y + sh) =_0F dx +_[y,y+sh]F dx= constant +_ s0F(y + th) hdt.bg0B y r ( , )y sh +yWe immediately obtain from the FTC the formula for the directional derivativedds(f(y + sh)) = F(y + sh) h.Set s = 0 to achieveDf(y;h) = F(y) h.In particular, when h = unit coordinate vector ei,fxi(y) = Fi(y).Thus the partial derivatives f/xi are equal to Fi and are thus continuous; we conclude thatf is of class C1and f = F.14 Chapter 12SECOND PROOF. This time we go straight for the denition of dierentiability as foundin Section 2E. We have of course a candidate for f(y), namely F(y). So we are led toconsider the quantityf(y + h) f(y) F(y) hfor small h Rn. Then the denition of f givesf(y + h) f(y) F(y) h =_[y,y+h]F dx F(y) h=_ 10F(y + th) hdt F(y) h=_ 10[F(y + th) F(y)] hdt.Given > 0, the continuity of F at y implies that there exists > 0 such that for z y < we have F(z) F(y)...</p>