18.02 Multivariable Calculus (Spring 2009): Lecture 27Simply Connected RegionsApril 14Reading Material: From Lecture Notes: V5 and V6.Last time: Green’s Theorem: why it works. Examples.Today: Simply Connected Regions.2 Generalized (or Extended) Green’s TheoremConsider a region R with n holes. Call C0the outside boundary and C1, ....., Cnthe inside boundariesof the holes. Orient the boundaries so that the region R is always on the left while wolking on thesecurves. Also assume that these curves are simple and closed:Assume that a vector filed~F is continuously differentiable on R. Then the Green’s theorem saysthatnXi=0ICi~F · d~r =ZZR∂N∂x−∂M∂ydA.Proof: Let’s consider for simplicity a region with two holes. We cut through them and we split theregion R into two new regions R1and R2that are now without holes. Let’s call their boundariesB1and B2respectively:1For each of the two regions we apply the usual Green’s theoremIB1~F · d~r +IB2~F · d~r =ZZR1∂N∂x−∂M∂ydA +ZZR2∂N∂x−∂M∂ydA =ZZR∂N∂x−∂M∂ydA.On the other hand, in the left hand side, since the line integral on the segment introduced in thecutting appears twice but in opposite directions, we have that the only contribution remaining isrelative only to the outside boundary C0and the boundaries of the two holes C1and C2. thisconcludes the proof.3 Simply connected regionsConsider the two vector fields~G =−yx2+ y2ˆi +xx2+ y2ˆj. (3.1)~F =−yx2ˆi +xx2ˆj. (3.2)Question: Are these two vectors fields conservative in their domains?We first consider the vector field~G that we already discussed. It’s domain is the whole space minus(0, 0). We know thatMy=x2− y2(x2+ y2)2= Nx, (3.3)but in this case having zero curl it is not enough to claim that~G is conservative since we already(easily) computed thatIC~G · d~r = 2π (3.4)where C is any circle centered at (0, 0), oriented counterclockwise. Since C is closed~G cannot beconservative.We can remark though, using the generalized Green’s theorem, that any other simple closed curve˜C containing (0, 0), oriented counterclockwise,is such thatI˜C~G · d~r = 2π.2To see this take a little circle crof radius r centered at (0, 0) and contained inside˜C. Orient crclockwise and let R be the region below:ThenI˜C~G · d~r +Icr~G · d~r =ZZR∂N∂x−∂M∂ydA,but the Right Hand Side is zero by (3.3) and by (3.4) we obtainI˜C~G · d~r = −Icr~G · d~r = 2π.We consider now the vector filed~F . Its domain is the whole plane minus the y-axis x = 0:We also have in this domain thatMy= −1x2= Nx, (3.5)So in each half plane we can actually use a generalization of the Gradient test (see Theorem 1below) to show that each closed simple curve in the domain has zero line integral. One can also finda potential functionf(x, y) =yx+ C.The difference between the domains of the vector field~G and~F is that the first one is connectedbut not simply connected and the second one is made of two disconnected pieces both of themsimply connected, accordingly to the following definition:Definition 1. A two-dimensional region D of the plane consisting of one single piece is calledsimply connected if it has this property: whenever a simple closed curve C lies entirely in D, thenits interior part also lies entirely in D3Examples of non-simply connected regions:Examples of simply connected regions:We have the following theoremTheorem 1 (Generalized Gradient Test). If~F is continuously differentiable on a simply connectedregion, then~F is conservative if and only ifNx− My= 0.Moreover there exists a potential function f for~F .Intuition: Of course in this case one can repeat the proof that was given last time by cutting theregion into small rectangles. The point is that by definition each rectangular curve that sits insidethe simply-conncetd region has all its interior inside it too.On the other hand consider a vector field that is singular at a point (its domain is then not simplyconnnected). We can take a curve around that point where the vector field is well defined. It is truethat zero curl (Nx− My= 0) inside the curve implies nothing going around hence no work alongthe curve, but we don’t know what happens at the missing point, and that could account for workdone on the curve. In a similar language, but using flux, what we just said can be rephrased in thefollowing way: consider a vector filed that is singular at a point and has zero divergence everywhereelse (Mx+ Ny= 0). Again let’s pick a curve around the singular point. Of course z ero divergenceinside the region enclosed by the curve means that there are no sources and no sinks, but we don’tknow what happens at the singular point. That point may have a source or sink and that couldaccount for flux trough the curve.Exercise 1. Consider the vector filed~F =2x(y − x2+ 2)2ˆi −1(y − x2+ 2)2ˆj.Find its domain D and decide if it is conservative or not.4Solution: We observe that D is the whole plane minus the parabola y = x2− 2We observe that D = D1∪ D2and both D1and D2are simply-connected regions. We then look atthe curl of~F and we see thatNx= My= −4x(y − x2+ 2)3hence the generalized gradient test shows that~F is conservative on each region D1and D2andhence on D. One can also easily show thatf(x, y) =1y − x2+ 2is a potential function.4 What we did in 2D and what we need to do in 3D2D 3DDouble integrals Triple integralsPolar coordinates Cylindrical & spherical coordinates2D Applications 3D ApplicationsLine integrals(Line integrals in 3DSurface integralsGreen’s Theorem for flux Gauss (divergence) TheoremGreen’s Theorem for work Stoke’s Theorem5 Introduction to triple integralsThe 2D case: We recall that given a function f(x, y) and a region R5we can writeZZRf(x, y) dA =Zba"Zg (x)h(x)f(x, y) dy#dx.To determine the limits of the other variable we look at the shadow of R on the axis relative to theother variable.The 3D case: Here we usually have a function f(x, y, z) and a 3D region D and we want to defineZZZDf(x, y, z) dVwhere dV denotes the infinitesimal element of volume. The definition of the integral above shouldbe such thatZZZD1 dV = Vol(D).We will start from here next lecture!6Study Guide 1. Think about the following questions:• Can you give another example of a vector field which domain is a simply connected region?• Is it true that if the domain of a vector field is not simply connected, even if the curl is zero,the vector field is never
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