Line Integrals and Green’s Theorem Jeremy Orlo

1 Line Integrals and Green s TheoremJeremy Orloff1 Vector Fields (or vector valued functions)Vector we will mostly use the notation (v) = (a,b) for vectors. Theother common notation (v) =ai+bjruns the risk ofibeing confused withi= 1 especially if I forget to vector field (also called called a vector-valued function) is a functionF(x,y)fromR2toR2. That is,F(x,y) = (M(x,y), N(x,y)),whereMandNare regular functions on the plane. In standard physics notationF(x,y) =M(x,y)i+N(x,y)j= (M,N).Algebraically, a vector field is nothing more than two ordinary functions of two are a number of standard examples of vector fields.( ) Force: constant gravitational fieldF(x,y) = (0, g).( ) Velocity:V(x,y) =(xx2+y2,yx2+y2)=(xr2,yr2).

2 (Hereris our usual polarr.) It is a radial vector field, it points radially away from theorigin. It is a shrinking radial field like water pouring from a source at (0,0).This vector field exhibits another important feature for us: it is not defined at the originbecause the denominator becomes zero there. We will say thatVhas a singularity at theorigin.( ) Unit tangential field:F= ( y,x)/r. Tangential means tangent to circles centeredat the origin. We know it is tangential because it is orthogonal to the radial vector field in( ).Falso has a singularity at the origin. We( ) Gradient field:F= f, ,f(x,y) =xy2 f=(y2,2xy). Visualization of vector fieldsThis can be summarized as: draw little arrows in the plane.

3 More specifically, for a fieldF,at each of a number of points (x,y) draw the vectorF(x,y)Example the vector fields, ( ), ( ) and ( ) from the previous DEFINITION AND COMPUTATION OF LINE Integrals ALONG A PARAMETRIZED CURVE2xy( ) Constant vector fieldxy( ) Shrinking radial fieldxy( ) Unit tangential field2 Definition and computation of line Integrals along a parametrizedcurveLine Integrals are also called path or contour need the following ingredients:A vector fieldF(x,y) = (M,N)A parametrized curveC:r(t) = (x(t), y(t)), withtrunning : sincer= (x,y), we havedr= (dx, dy). line integral ofFalongCis defined as CF dr= C(M,N) (dx,dy) = CM dx+N : The notationF dris common in physics andM dx+N dyin thermodynamics.

4 (Though everyone uses both notations.)We ll see what these notations mean in practice with some (x,y) =(x2y, x 2y)and letCbe the curver(t) =(t, t2), withtrunning from 0 to 1. Compute the line integralI= CF this first using the notation CM dx+N dy. Then repeat the computation using thenotation CF :First we draw the curve, which is the part of the parabolay=x2running from(0,0) to (1,1).3 WORK DONE BY A FORCE ALONG A CURVE3xyC11(i) Using the notation CM dx+N haver= (x,y), sox=t,y=t2. In this notationF= (M, N), soM=x2yandN=x put everything in terms oft:dx=dtdy= 2tdtM= (t2)(t2) =t4N=t 2t2 Now we can put all of these in the integral. Sincetruns from 0 to 1, these are our CM dx+N dy= 10t4dt+ (t 2t2)2tdt= 10t4+ 2t2 4t3dt= 215.

5 (ii) Using the notation CF , we have to put everything in terms oft:F= (M, N) =(t4, t 2t2)drdt= (1,2t),sodr=drdtdt= (1,2t)dtThus,F dr=(t4, t 2t2) (1,2t)dt=t4+ (t 2t2)2tdt. So the integral becomesI= CF dr= 10t4+ (t 2t2) is exactly the same integral as in method (i).3 Work done by a force along a curveHaving seen that line Integrals are not unpleasant to compute, we will now try to motivateour interest in doing so. We will see that the work done by a force moving a body along apath is naturally computed as a line to Integrals we ve seen before, the work integral will be constructed by dividing thepath into little pieces. The work on each piece will come from a basic formula and the totalwork will be the sum over all the pieces, an WORK DONE BY A FORCE ALONG A Basic formula: work done by a constant force along a small lineWe ll start with the simplest situation: a constant forceFpushes a body a distance salong a straight line.

6 Our goal is to compute the work done by the figure shows the forceFwhich pushes the body a distance salong a line in thedirection of the unit vector T. The angle between the forceFand the direction Tis . T=unit vectorlength= svector= r= s TF We know from physics that the work done by the force on the body is the component ofthe force in the direction of motion times the distance moved. That is,work =|F|cos( ) sWe want to phrase this in terms of vectors. Since| T|= 1 we knowF T=|F|cos( ). Usingthis in the formula for work we havework =F T s.(1)Equation 1 is important and we will see it again. For now, we want to make one moresubstitution. We ll call the vector s T= r.

7 This is the displacement of the body. (Note,it is essentially the same as our formuladsdt T=drdt.) Using this, Equation 1 becomeswork =F r.(2)This is the basic work formula that we ll use to compute work along an entire Work done by a variable force along an entire curveNow suppose a variable forceFmoves a body along a curveC. Our goal is to compute thetotal work done by the figure shows the curve broken into 5 small pieces, thejth piece has displacement the pieces are small enough, then the force on thejth piece is approximately is shown asFj. r1 r2 r3 r4 r5F1F2F3F4F54 GRAD, CURL AND DIV5 Also, if the pieces are small enough, then each segment is approximately a straight lineand the force is approximately constant.

8 So we can apply our basic formula for work andapproximate the work done by the force moving the body along thejth piece as Wj Fj total work is the sum of the work over each work = Wj Fj , as usual, we let the pieces get infinitesimally small, so the sum becomes an integraland the approximation becomes exact. We get:total work = CF subscriptCindicates that it is the curve that has been split into pieces. That is, thetotal work is computed as a line integral of the force over the curveC!4 Grad, curl and a functionf(x,y): gradf= f= (fx,fy). a vector in the planeF(x,y) = (M(x,y),N(x,y)) we definecurlF=Nx is a scalar. In general, the curl of a vector field is another vector field. Forvectors fields in the plane the curl is always in the kdirection, so we simply drop the kandmake curl a scalar.

9 Sometimes it is called the baby curl . divergence of the vector fieldF= (M,N) isdivF=Mx+ Properties of line integralsIn this section we will uncover some properties of line Integrals by working some look back at the value found in Example Now, use the samevector field as in that example, but, in this case, letCbe the straight line from (0,0) to(1,1), same endpoints, but different path. Compute the line integral CF :As always, start by sketching the curve:xyC115 PROPERTIES OF LINE INTEGRALS6We ll use the notation CM dx+N the curve:x=t,y=t, withtfrom 0 to everything in terms oft:dx=dtdy=dtM=x2y=t3N=x 2y= tNow we put this into the integralI= CM dx+N dy= 10t3dt tdt= 10t3 tdt= is a different value from Example , which leads to the important principle:Important principle for line Integrals over two different paths with thesame endpoints may be , look back at the value found in Example Now, use the samevector field and curve as Example except use the following (different) sin(t), y= sin2(t).

10 0 t the line integral CF :We won t sketch the curve it is identical to the one in Example Puttingeverything in terms oftwe havedx= cos(t)dtdy= 2 sin(t) cos(t)dtM=x2y= sin2(t) sin2(t) = sin4(t)N=x 2y= sin(t) 2 sin2(t)We put these in the integralI= CM dx+N dyand computeI= /20sin4(t) cos(t)dt+ (sin(t) 2 sin2(t))2 sin(t) cos(t)dt= /20(sin4(t) + 2 sin2(t) 4 sin3(t))cos(t)dt(Letu= sin(t),du= cos(t)dt.)= 10u4+ 2u2 4u3du= PROPERTIES OF LINE INTEGRALS7 This is the same value we got in Example ! In fact, theusubstitution led to exactlythe same integral! This leads us to the important principle:Important principle for line parametrization of the curve doesn t affectthe value of line the integral over the should note that our work with work make this reasonable, since we developed the lineintegral abstractly, without any reference to a List of properties of line integrals1.