Line Integrals and Green s Theorem Jeremy Orloff

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1 Line Integrals and Green s Theorem Jerem Orloff Vector Fields (or vector valued functions) Vector notation. In 8.4 we will mostl use the notation (v) = (a, b) for vectors. The other common notation (v) = ai + bj runs the risk of i being confused with i = especiall if I forget to make i boldfaced. Definition. A vector field (also called called a vector-valued function) is a function F(, ) from 2 to 2. That is, F(, ) = (M(, ), N(, )), where M and N are regular functions on the plane. In standard phsics notation F(, ) = M(, )i + N(, )j = (M, N). Algebraicall, a vector field is nothing more than two ordinar functions of two variables. Eample GT.. Here are a number of standard eamples of vector fields. (a.) Force: constant gravitational field (a.2) Velocit: V(, ) = ( 2 + 2, F(, ) = (, g). ) ( = r 2, ) r 2. (Here r is our usual polar r.) It is a radial vector field, i.e. it points radiall awa from the origin. It is a shrinking radial field like water pouring from a source at (,). This vector field ehibits another important feature for us: it is not defined at the origin because the denominator becomes zero there. We will sa that V has a singularit at the origin. (a.3) Unit tangential field: F = (, ) /r. Tangential means tangent to circles centered at the origin. We know it is tangential because it is orthogonal to the radial vector field in (a.2). F also has a singularit at the origin. We (a.4) Gradient field: F = f, e.g., f(, ) = 2 f = ( 2, 2 ).. Visualization of vector fields This can be summarized as: draw little arrows in the plane. More specificall, for a field F, at each of a number of points (, ) draw the vector F(, ) Eample GT.2. Sketch the vector fields, (a.), (a.2) and (a.3) from the previous eample.

2 2 DEFINITION AND OMPUTATION OF LINE INTEGALS ALONG A PAAMETIZED UVE2 (a.) onstant vector field (a.2) Shrinking radial field (a.3) Unit tangential field 2 Definition and computation of line integrals along a parametrized curve Line integrals are also called path or contour integrals. We need the following ingredients: A vector field F(, ) = (M, N) A parametrized curve : r(t) = ((t), (t)), with t running from a to b. Note: since r = (, ), we have dr = (d, d). Definition. The line integral of F along is defined as F dr = (M, N) (d, d) = M d + N d. omment: The notation F dr is common in phsics and M d + N d in thermodnamics. (Though everone uses both notations.) We ll see what these notations mean in practice with some eamples. Eample GT.3. Let F(, ) = ( 2, 2 ) and let be the curve r(t) = ( t, t 2), with t running from to. ompute the line integral I = F dr. Do this first using the notation M d + N d. Then repeat the computation using the notation F dr. answer: First we draw the curve, which is the part of the parabola = 2 running from (, ) to (, ).

3 3 WOK DONE BY A FOE ALONG A UVE 3 (i) Using the notation M d + N d. We have r = (, ), so = t, = t 2. In this notation F = (M, N), so M = 2 and N = 2. We put everthing in terms of t: d = dt d = 2t dt M = (t 2 )(t 2 ) = t 4 N = t 2t 2 Now we can put all of these in the integral. Since t runs from to, these are our limits. I = M d + N d = t 4 dt + (t 2t 2 )2t dt = t 4 + 2t 2 4t 3 dt = 2 5. (ii) Using the notation F dr. Again, we have to put everthing in terms of t: F = (M, N) = ( t 4, t 2t 2) dr dt dr = (, 2t), so d r = dt = (, 2t) dt dt Thus, F dr = ( t 4, t 2t 2) (, 2t) dt = t 4 + (t 2t 2 )2t dt. So the integral becomes I = F dr = This is eactl the same integral as in method (i). t 4 + (t 2t 2 )2t dt. 3 Work done b a force along a curve Having seen that line integrals are not unpleasant to compute, we will now tr to motivate our interest in doing so. We will see that the work done b a force moving a bod along a path is naturall computed as a line integral. Similar to integrals we ve seen before, the work integral will be constructed b dividing the path into little pieces. The work on each piece will come from a basic formula and the total work will be the sum over all the pieces, i.e. an integral.

4 3 WOK DONE BY A FOE ALONG A UVE 4 3. Basic formula: work done b a constant force along a small line We ll start with the simplest situation: a constant force F pushes a bod a distance s along a straight line. Our goal is to compute the work done b the force. The figure shows the force F which pushes the bod a distance s along a line in the direction of the unit vector T. The angle between the force F and the direction T is θ. F length= s θ T=unit vector vector= r = s T We know from phsics that the work done b the force on the bod is the component of the force in the direction of motion times the distance moved. That is, work = F cos(θ) s We want to phrase this in terms of vectors. Since T = we know F T = F cos(θ). Using this in the formula for work we have work = F T s. () Equation is important and we will see it again. For now, we want to make one more substitution. We ll call the vector s T = r. This is the displacement of the bod. (Note, it is essentiall the same as our formula ds dt T = dr.) Using this, Equation becomes dt work = F r. (2) This is the basic work formula that we ll use to compute work along an entire curve 3.2 Work done b a variable force along an entire curve Now suppose a variable force F moves a bod along a curve. Our goal is to compute the total work done b the force. The figure shows the curve broken into 5 small pieces, the jth piece has displacement r j. If the pieces are small enough, then the force on the jth piece is approimatel constant. This is shown as F j. r 5 r 4 F 3 F 4 F F 2 r 3 r 2 r F 5

5 4 GAD, UL AND DIV 5 Also, if the pieces are small enough, then each segment is approimatel a straight line and the force is approimatel constant. So we can appl our basic formula for work and approimate the work done b the force moving the bod along the jth piece as W j F j r j. The total work is the sum of the work over each piece. total work = W j F j r j. Now, as usual, we let the pieces get infinitesimall small, so the sum becomes an integral and the approimation becomes eact. We get: total work = F dr. The subscript indicates that it is the curve that has been split into pieces. That is, the total work is computed as a line integral of the force over the curve! 4 Grad, curl and div Gradient. For a function f(, ): gradf = f = (f, f ). url. For a vector in the plane F(, ) = (M(, ), N(, )) we define curlf = N M. NOTE. This is a scalar. In general, the curl of a vector field is another vector field. For vectors fields in the plane the curl is alwas in the k direction, so we simpl drop the k and make curl a scalar. Sometimes it is called the bab curl. Divergence. The divergence of the vector field F = (M, N) is divf = M + N. 5 Properties of line integrals In this section we will uncover some properties of line integrals b working some eamples. Eample GT.4. First look back at the value found in Eample GT.3. Now, use the same vector field as in that eample, but, in this case, let be the straight line from (, ) to (, ), i.e. same endpoints, but different path. ompute the line integral F dr. answer: As alwas, start b sketching the curve:

6 5 POPETIES OF LINE INTEGALS 6 We ll use the notation M d + N d. Parametrize the curve: = t, = t, with t from to. Put everthing in terms of t: Now we put this into the integral I = M d + N d = d = dt d = dt M = 2 = t 3 N = 2 = t t 3 dt t dt = t 3 t dt = 4. This is a different value from Eample GT.3, which leads to the important principle: Important principle for line integrals. Line integrals over two different paths with the same endpoints ma be different. Eample GT.5. Again, look back at the value found in Eample GT.3. Now, use the same vector field and curve as Eample GT.3 ecept use the following (different) parametrization of. = sin(t), = sin 2 (t); t π/2. ompute the line integral F dr. answer: We won t sketch the curve it is identical to the one in Eample GT.3. Putting everthing in terms of t we have d = cos(t) dt We put these in the integral I = I = = = π/2 π/2 d = 2 sin(t) cos(t) dt M = 2 = sin 2 (t) sin 2 (t) = sin 4 (t) N = 2 = sin(t) 2 sin 2 (t) M d + N d and compute sin 4 (t) cos(t) dt + (sin(t) 2 sin 2 (t))2 sin(t) cos(t) dt ( sin 4 (t) + 2 sin 2 (t) 4 sin 3 (t) ) cos(t) dt (Let u = sin(t), du = cos(t) dt.) = 2 5. u 4 + 2u 2 4u 3 du

7 5 POPETIES OF LINE INTEGALS 7 This is the same value we got in Eample GT.3! In fact, the u substitution led to eactl the same integral! This leads us to the important principle: Important principle for line integrals. The parametrization of the curve doesn t affect the value of line the integral over the curve. You should note that our work with work make this reasonable, since we developed the line integral abstractl, without an reference to a parametrization. 5. List of properties of line integrals. Independent of parametrization: The value of the line integral F dr is independent of the parametrization of. 2. eversing direction on the curve changes the sign: If is a curve, then we write for the same curve traversed in the opposite direction. In this case F dr = F dr. (See the net eample.) Eample GT.6. Let be the curve from Eample GT.3. Sketch and and give a parametrization of. answer: follows the parabola = 2 from (,) to (,), so the curve covers the same section of the parabola, but goes from (,) to (,), i.e. we reversed the direction of the arrow. goes from (,) to (,) goes from (,) to (,) The curve can be parametrized as r(t) = ( t, t 2), with t running from to. The easiest wa to reverse this is to have t run from to With this parametrization the t limits on the integral are reversed, which, we know from 8., changes the sign of the integral. If ou insist on an increasing parameter, we can parametrize b r(u) = ( u, ( u) 2), with u runnning from to. 3. (Intrinsic formula) We can write the line integral as F dr = F T ds where T = unit tangent vector to and ds = differential of arclength.

8 6 ETANGULA AND IULA PATHS 8 eason: We know from our work on parametrized curves that dr dt = Tds. So, dr = T ds. dt (A comparison of Equations and 2 above, essentiall shows the same thing.) 4. If is a closed curve we use the notation F dr = M d + N d. The little circle on the integral sign indicates the curve is closed, i.e. starts and ends at the same point. 6 ectangular and circular paths Eample GT.7. Evaluate I = from (,) to (,) shown below. d + ( + 2) d (, ) where is the rectangular path 2 answer: The path is given in two pieces labeled and 2. This means we will have to split the integral into two pieces, i.e. I = d + ( + 2) d = d + ( + 2) d + d + ( + 2) d. 2 We ll do the integration one piece at a time. First, d + ( + 2) d. Parametrize : We ll use as the parameter: Put everthing in terms of : =, =, with running from to =, =, d = d, d =, M =, N (skip, since d = ). Put this in the integral and compute: M d + N d = M d = Net, the integral over 2. d =. Parametrize 2 : Use parameter : =, =, runs from to. Put everthing in terms of : =, =, d =, d = d, M (skip, since d = ), N = + 2 = + 2

9 7 SOME SUPE-DUPE, EALLY SEIOUSLY IMPOTANT EXAMPLES 9 Put this in the integral and compute M d + N d = N d = 2 2 Adding, the pieces we have I = 2 =. + 2 d = 2. Shorthand. Because d = on and d = on 2 we can write M d + N d = + 2 M d + N d. 2 Using the shorthand will save us some writing in the future. Eample GT.8. Evaluate I = d + d where is the unit circle traversed in a counterclockwise (W) direction. answer: Parametrization: = cos(t), = sin(t), t 2π. So, d = cos(t) dt, d = sin(t) dt. We get I = 2π sin t( sin t) dt + cos t(cos t) dt = 2π dt = 2π. 7 Some super-duper, reall seriousl important eamples In these eamples we are going to integrate a tangential field around a closed loop. These will be ke computations as we eplore Green s theorem and gradient fields. In the following r is the usual polar distance r 2 = Eample GT.9. Let F =, r 2 r, and let be the unit circle traversed in a counterclockwise (W) direction. ompute I = F 2 dr answer: Sketch and the vector field F.

10 7 SOME SUPE-DUPE, EALLY SEIOUSLY IMPOTANT EXAMPLES Parametrize : = cos(t), = sin(t), t 2π. Put everthing in terms of t: (Note, on the unit circle r =.) d = cos(t) dt, d = sin(t) dt, M = r 2 = sin(t), N = r 2 = cos(t). Put this in the integral and compute: I = 2π sin(t)( sin(t)) dt + cos(t)(cos(t)) dt = 2π sin 2 (t) + cos 2 (t) dt = 2π dt = 2π. Eample GT.. Let F be the same as the previous eample. Let 2 be the unit circle centered on (2,) traversed counterclockwise. ompute I 2 = F dr. 2 answer: Parametrize 2 : = 2 + cos(t), = sin(t), t from to 2π. Put everthing in terms of t: (Note, r 2 is not constant.) d = sin(t) dt d = cos(t) dt Put this in the integral: I 2 = M d + N d = 2 r 2 = = (2 + cos(t)) 2 + sin 2 (t) = cos(t) M = r 2 = sin(t) cos(t) N = r 2 = 2 + cos(t) cos(t) 2π sin 2 (t) + 2 cos(t) + cos 2 (t) cos(t) O! We put this into Wolfram Alpha and found I 2 =. dt = 2π + 2 cos(t) cos(t) dt Note. We should suspect that the value of is no accident. This is true and we will see it easil once we learn Green s theorem. Avoiding actuall computing an integral like this should be motivation enough for us to learn Green s theorem. 8. challenge. ompute the integral for I 2. Hints: You can use the substitution u = tan(t/2) and partial fractions. It s best to use smmetr and compute 2 times the integral from to π.

11 8 GADIENT AND ONSEVATIVE FIELDS 8 Gradient and conservative fields We will now focus on the important case where F is a gradient field. That is, for some function f(, ), F = f = (f, f ). Note. We will learn to call f a potential function for F. 8. The fundamental theorem for gradient fields Theorem GT.. (fundamental theorem for gradient fields) Suppose that F = f is a gradient field and is an path from point P to point Q. The fundamental theorem for vector fields sas F dr = f(, ) Q P = f(q) f(p ) = f(, ) f(, ). (3) where P = (, ) and Q = (, ). That is, for gradient fields the line integral depends onl on the endpoints of the path and is independent of the path taken. 2 Q = (, ) P = (, ) If F = f then F dr = F dr = f(q) f(p ). 2 The proof of the fundamental theorem is given after the net eample Eample GT.2. Let f(, ) = and let be the curve shown. ompute F = f and compute F dr two was: (i) directl as a line integral, (ii) using the fundamental theorem. (, 2) (, )

12 8 GADIENT AND ONSEVATIVE FIELDS 2 answer: F(, ) = f (f, f ) = ( 3 + 2, 3 2) (i) Parametrize : = t, = 2t, t runs from to. Write everthing in terms of t: d = dt, d = 2 dt, M = = 8t 3 + 2t, M = 3 2 = 2t 3. Put all this into the integral and compute: F dr = ( 3 + 2) d d = (ii) B the fundamental theorem F dr = You decide which method is easier! (8t 3 + 2t) dt + 2t 3 2 dt = f dr = f(, 2) f(, ) = 9. 32t 3 + 2t dt = Proof of the fundamental theorem Proof. The proof uses the definition of line integral together with the chain rule and the usual fundamental theorem of calculus. We assume the following: (i) F = f (ii) The curve is parametrized b r(t) = ((t), (t)), with t running from t to t and r(t ) = P, r(t ) = Q. First recall the for a parametrized curve r(t) the chain rule sas So, F dr = df(r(t)) dt f dr = d = f dt + f d dt t t = f dr dt. f dr t dt dt = df(r(t)) dt t dt = f(r(t)) t t = f(r(t )) f(r(t )) = f(q) f(p ) QED 8.3 Path independence Definition. For a vector field F we sa that the line integrals F dr are path independent if for an two points P and Q the integral ields the same value for ever path connecting P to Q. From thefundamental theorem we can conclude: if F = f is a gradient field, then the integrals F dr are path independent.

13 8 GADIENT AND ONSEVATIVE FIELDS 3 The following theorem offers an alternative wa to epress path independence. Theorem GT.3. For a given vector field F, the line integrals F dr independent is equivalent to F dr = for an closed path c. c Proof. To show equivalence we need to show two things: (i) Path independence implies the line integral around an closed path is. (ii) The line integral around all closed paths is implies path independence. are path Proof (i). To start, note that the constant path where r(t) = P, with t running from to has line integral F dr = F dt =. Assume path independence and consider the closed path c shown in Figure (i) below. Since both c and have the same start and end points, path independence sas the line integrals are the same, i.e. F dr = c F dr =. This proves (i). (ii) Assume F dr = c for an closed curve. onsider two paths between P and Q as shown in Figure (ii). The curve c = 2 is a closed curve starting and ending at P. Therefore, b assumption F dr =. So c = F dr = c F dr = 2 F dr F dr 2 This implies F dr = F dr. That is, the integral is the same for all paths from P 2 to Q, i.e. the line integrals are path independent. P = Q c Q 2 P Figure (i) Figure (ii) 8.4 onservative vector fields Given a vector field F, Theorem GT.3 in the previous section said that the line integrals of F were path independent is equivalent to the line integral of F around an closed path

14 8 GADIENT AND ONSEVATIVE FIELDS 4 is. Following phsics terminolog, we call such a vector field a conservative vector field. The fundamental theorem sas the if F is a gradient field: F = f, then the line integral F dr is path independent. That is, a gradient field is conservative. The following theorem sas that the converse is also true on connected regions. B a connected region we mean that an two points in the region can be connected b a continuous path that lies entirel in the region. Theorem GT.4. A conservative field on a connected region is a gradient field. Proof. all the region D. We have to show that if we have a conservative field F = (M, N) on D the there is a potential function f with F = f. The eas part will be defining f. The trickier part will be showing that its gradient is F. So, first we define f: Fi a point (, ) in D. Then for an point (, ) in D we define f(, ) = F dr, where γ is an path from (, ) to (, ). γ Path independence guarantees that f(, ) is well defined, i.e. it doesn t depend on the choice of path. Now we need to show that F = f, i.e if F = (M, N), then we need to show that f = M and f = N. We ll show the first case, the case f = N is essentiall the same. First, note that b definition df( + t, ) f (, ) = dt. t= The function f( + t, ) is defined in terms of a line integral. So, we need to write down this integral and differentiate it. In the figure below, f(, ) is the integral along the path γ from (, ) to (, ) and f( + t, ) is the integral along γ + γ 2, where γ 2 is the horizontal line segment from (, ) to ( + t, ). This means that f( + t, ) = F dr + F dr = f(, ) + F dr. γ γ 2 γ 2 D (, ) γ 2 ( + t, ) γ (, ) We need to parameterize the horizontal segment γ 2. Notationwise, the letters, and t are alread taken, so we let γ 2 (s) = (u(s), v(s), where u(s) = + s, (s) = ; with s t

15 9 POTENTIAL FUNTIONS 5 On this segment du = ds and d =. So, f( + t, ) = f(, ) + t M( + s, ) ds The piece f(, ) is constant as t varies, so the fundamental theorem of calculus sas that Setting t = we have f (, ) = df( + t, ) dt df( + t, ) dt = M( + t, ). = M(, ). t= This is eactl what we needed to show! We summarize in a bo: On a connected region a conservative field is a gradient field. Eample GT.5. If F is the electric field of an electric charge it is conservative. Eample GT.6. A gravitational field of a mass is conservative. 9 Potential functions Definition. If F = f is a gradient vector field then we call f a potential function for F. Note. The usual phsics terminolog would be to call f the potential function for F. This section is devoted to answering two questions.. How do we know if a vector field F is a gradient vector field, i.e. if F = f for some potential function f? 2. If it eists, how do we find the potential function f? 9. First answers to our questions Theorem GT.7. Suppose F = (M, N). We have the following answer to our two questions. (a) If F = f, then M = N, i.e. M = N. (b) If M = N in the whole plane then F is a gradient vector field, i.e F = f for some potential function f. (c) If F is conservative on a connected region, then F is a gradient field. Notes. The restriction that F is defined on the whole plane is too stringent for our needs. Below, we will give what we call the Potential theorem, which onl requires that F be defined and differentiable on what is called a simpl connected region. Proof of (a). If F = (M, N) = f then M = f and N = f. This implies, M = f and N = f, i.e M = N. QED. (Provided f has continuous second partials.)

16 9 POTENTIAL FUNTIONS 6 The proof of (b) will be postponed until after we have proved Green s theorem and we can state the Potential theorem. Part (c) is just a restatement of Theorem GT.4. The eamples below will show how to find f Eample GT.8. For which values of the constants a and b will F = ( a, 2 + b ) be a gradient field? answer: M = a, N = 2. To appl the theorem we need M = N in the entire plane. So, a = 2 and b is arbitrar. Eample GT.9. Is F = ( (3 2 + ), e ) conservative? answer: First we check if it is a gradient field: We write F = (M, N) = ( 3 2 +, e ). Then, M =, N = e. Since M N, F is not a gradient field. Now, Theorem GT.7(c) sas it can t be conservative. (, ) Eample GT.2. Is conservative? answer: NO! The reasoning is a little trickier than in the previous eample. First, it is not hard to compute that M = 2 2 ( ) 2 = N. BUT, since the field is not defined for all (, ), Theorem GT.7(b) does not appl. So, all we can sa at this point is that we haven t ruled out its being conservative. To show that it s not conservative we will find a closed path where the line integral is not. In fact, we will use our super-duper important eample from above. Let = unit circle parametrized b = cos(t), = sin(t). Writing everthing in terms of t: d = sin(t) dt, d = cos(t) dt, M = = sin(t), N = = cos(t). Putting this in the line integral F dr = M d + N d = 2π Since this is not, the field is not conservative. Eample GT.2. Is F = (, ) conservative? sin 2 (t) + cos 2 (t) dt = 2π dt = 2π answer: Again it is eas to check that N = M, BUT since F is not defined at (, ) Theorem GT.7(b) does not appl. HOWEVE, it turns out that F = ln( ) = ln r Since F is a gradient field, it is conservative. (Officiall, we should sa, F is conservative on the region consisting of the plane minus the origin.)

17 9 POTENTIAL FUNTIONS Finding the potential function We will show two methods for finding the potential functions. In general, for 8.4 the method of integrating along a rectangular path is more relevant. Eample GT.22. Show that F = ( , ) is conservative and find the potential function f such that F = f. answer: We have F = (M, N), where M = , N = First, M = 6 = N. Since F is defined for all (, ) Theorem GT.7 implies F is a gradient field, hence conservative. Method for finding f. Since F is a gradient field we know F dr = f(q) f(p ) for an path from P to Q. We make use of this b letting be a rectangular path from the origin to an arbitrar point Q = (, ) (see figure). 2 Q = (, ) ectangular path from the origin to Q. We know F dr = f(, ) f(, ). So + 2 f(, ) = F dr + f(, ). + 2 On the rectangular path shown d = on and d = on 2. Therefore, F dr = + 2 N d + M d. 2 These integrals are straightforward to compute. We do each one separatel. Integral over : Parametrize : parameter = : =, =, runs from to. Put everthing we need in terms of the parameter : Put this in the integral and compute d = d, N = = 6. N d = Integral over 2 : Parameter = : =, =, runs from to. 6 d = 3 2.

18 9 POTENTIAL FUNTIONS 8 Put everthing we need in terms of the parameter : d = d, M = = Put this in the integral and compute (remember is a constant): M d = d = So, f(, ) f(, ) = F dr = Now, we are free to choose an value for f(, ), i.e. it is an arbitrar constant of integration c. So, dropping the subscripts on and, we have Method 2 for finding f. f(, ) = c. We know M = f and N = f. We start b integrating M with respect to. f = f(, ) = g(). The function g() is the constant of integration with respect to. Now f = N, so differentiating our epression for f we get f = g () = N = Thus, g () = 6, which implies g() = c. Using this in our epression for f, we have (Same as method.) f(, ) = g() = c. Eample GT.23. Let F = ( ( + 2 ), ( ) ). Show that F is a gradient field and find the potential function using both methods. answer: Testing the partials we have: M = 2 = N, F defined on all (, ). Thus, b Theorem GT.7, F is conservative. Method : Use the path shown. 2 Q = (, ) We know f(, ) f(, ) = + 2 F dr = M d + 2 N d, Parametrize : =, =, runs from to. In terms of the parameter along : d = d, M = + 2 =.

19 GEEN S THEOEM 9 Integrating: M d = d = 2 2. Parametrize 2 : =, =, runs from to. In terms of the parameter along 2 : d = d, N = Integrating: N d = d = So, f(, ) = F dr + f(, ) = f(, ). Letting f(, ) = c and dropping the subscripts on, we have f(, ) = c. Method 2. Since, in 8.4, we are less interested in method 2, we ll leave this to the reader. Green s theorem Green s theorem is the one of the big theorems of multivariable calculus. It relates line integrals and area integrals. Using this relation we can often compute a seemingl difficult integral without integration or reduce it to an eas integral. At the this section we will describe how it is analogous to the fundamental theorem of calculus.. Simple closed curves Definition. A simple closed curve is a closed curve with no self-intersection. A simple closed curve has a well-defined interior. all the interior. We sa that is positivel oriented if is alwas on the left as ou traverse the curve. We call the boundar of. Three positivel oriented simple closed curves bounding a region losed but not simple Note. For smooth curves like the ones shown above the interior is eas to define. For an arbitrar simple closed curves, showing that it has a well-defined interior is more subtle. The theorem that proves this is called the Jordan curve theorem.

20 GEEN S THEOEM 2.2 Green s theorem Theorem GT.24. Green s theorem. Let be a positivel oriented simple closed curve with interior region. We assume is piecewise smooth (a few corners are oka). If the vector field F = (M, N) is defined and differentiable on then M d + N d = N M da. (4) In two dimensions we define curlf = N M. So, in vector form, Green s theorem is written F dr = curlf da. Eample GT.25. (Using the right hand side of Equation 4 to find the left hand side.) Use Green s theorem to compute I = d ( + ) d where is the circle shown. a answer: The line integral is of the form on the left hand side of Green s theorem. So, b Green s theorem we can convert the line integral to an area integral. In the line integral M = 3 2 2, N = 2 2 ( + ), so N M = = 4. Therefore, I = d ( + ) d = N M d d = 4 d d. We could compute this directl, but here s a trick. We know the center of mass is cm = d d = a. A Since area A = πa 2, we have d d = πa 3, so I = 4πa 3. Eample GT.26. (Using the left hand side of Equation 4 to find the right hand side.) Use Green s theorem to find the area under one arch of the ccloid. answer: Our strateg is to use Green s theorem to replace the area integral with a line integral. The figure shows one arch of the ccloid. The region is under the arch and above the -ais. The boundar of the region is = + 2.

21 GEEN S THEOEM 2 2a 2 πa 2πa The trick is to use the vector field F = (, ), so curlf = N M =. With this F, Green s theorem sas F dr = d = N M da = da = area That is, the area is equal to the line integral usual: Parametrize : =, =, runs from to 2πa. So, d. We compute the line integral as d = d, d =, M = =, N (skip N because d = ). Thus F dr =. Parametrize 2 : = a(θ sin(θ)), = a( cos(θ)), θ runs from 2π to. direction θ runs.) So, (Note the d = a( cos(θ)) dθ, d (skip d because N = ), M = = a( cos(θ)), N =. omputing the integral over 2 : 2π F dr = d = a 2 ( cos(θ)) 2 dθ = a 2 ( cos(θ)) 2 dθ = 3πa π (The integral is easil computed b epanding the square and using the half-angle formula.) Adding the two integrals: the area under one arch of the ccloid is 3πa 2..3 Other was to compute area using line integrals The ke to the previous eample was that for F = (, ) we had curlf = N M =. There are other vector fields with the propert curlf =. For eample, F = (, ) or F = ( /2, /2). Using Green s theorem we have Area of = d d = d + d. 2 Here is the positivel oriented curve that bounds..4 Proof of Green s theorem (i) First we ll work on the rectangle shown. Later we ll use a lot of rectangles to approimate an arbitrar region.

22 GEEN S THEOEM 22 d c a b (ii) We ll simplif a bit and assume N =. The proof when N is essentiall the same with a bit more writing. First we consider the right hand side of Green s theorem (Equation 4). B direct calculation (assuming N = ), the right hand side is M da = b d a c M (, ) d d. The inner integral is the integral with respect to of a derivative with respect to. That is, we can compute it using the fundamental theorem of calculus. d c M (, ) d = M(, ) d c = M(, d) + M(, c) Putting this into the outer integral we have shown that M b da = M(, c) M(, d) d. (5) a Net we consider the left hand side of Equation 4. We have (remember N = ) to compute M d. has four sides we parametrized each one: So, M d = bottom: =, = c, runs from a to b, d = d = top: =, = d, runs from b to a, d = d sides: skip because d =. bottom b a M d + M(, c) d + top a b M d M(, d) d = (since d = along the sides) b a M(, c) M(, d) d. (6) omparing Equations 5 and 6 we find that we have proved Green s theorem for the rectangle. Net we ll use rectangles to build up an arbitrar region. We start b stacking two rectangles on top of each other. For line integrals when adding two rectangles with a common edge the common edges are traversed in opposite directions. So, the sum of the line integrals over the two rectangles equals the line integral over the outside boundar.

23 ANALOGY TO THE FUNDAMENTAL THEOEM OF ALULUS 23 = Similarl when adding a lot of rectangles: everthing cancels ecept the outside boundar. This etends Green s theorem on a rectangle to Green s theorem on a sum of rectangles. Since an region can be approimated as closel as we want b a sum of rectangles, Green s theorem must hold on arbitrar regions. (See figure below.) An region and boundar can be approimated as a sum of rectangles. Analog to the fundamental theorem of calculus M We saw in the proof of Green s theorem that at one ke step we had to integrate d. To do this we literall used the fundamental theorem. There is another wa to view this connection. We will be rather informal in describing it, but it can be made formal and has deep and wide-ranging applications in math and science. To set up the analog we recall the fundamental theorem of calculus b a F () d = F (b) F (a) Here s a picture of the domain of integration: a b Notice that the left hand side of the fundamental theorem involves the integral of the derivative of F over a region (interval), and the right hand side is a sum of F itself over the boundar (endpoints) of the region.

24 2 SIMPLY ONNETED EGIONS 24 Likewise, Green s theorem sas curlf da = F dr. The left hand side involves the integral of a derivative of F (i.e. curlf = N M ) over a region, and the right hand side is an integral (i.e. a sum ) of F itself over the boundar of the region. This is eactl the same language we used to describe the fundamental theorem of calculus. 2 Simpl connected regions We will need the topological notion of a simpl connected region. We will stick with an informal definition of simpl connected that will be sufficient for our purposes. (For those who are interested: We will assume that a simple closed curve has an inside and an outside. This is intuitive and is eas to show if is a smooth curve, but turns out to surprisingl hard if we allow to be strange, e.g. a Koch snowflake.) Definition: A region D in the plane is called simpl connected if, for ever simple closed curve that lies entirel in D the interior of also lies entirel in D. Eamples: D D 2 D 3 D 4 D 5 = whole plane D-D5 are simpl connected, since for an simple closed curve inside them its interior is entirel inside the region. This is sometimes phrased as each region has no holes. Note. An alternative definition, which works in higher dimensions is that the region is simpl connected if an curve in the region can be continuousl shrunk to a point without leaving the region. The regions at below are not simpl connected. That is, the interior of the curve is not entirel in the region. Annulus Puntured plane

25 3 POTENTIAL THEOEM AND ONSEVATIVE FIELDS 25 3 Potential theorem and conservative fields As an application of Green s theorem we can now give a more complete answer to our question of how to tell if a field is conservative. The theorem does not have a standard name, so we choose to call it the Potential theorem. You should check that it is largel a restatement for simpl connected regions of Theorem GT.7 above. Theorem GT.27. (Potential theorem) Take F = (M, N) defined and differentiable on a region D. (a) If F = f then curlf = N M =. (b) If D is simpl connected and curlf = on D, then F = f for some f. Notes.. We know that on a connected region, being a gradient field is equivalent to being conservative. So we can restate the Potential theorem as: On a simpl connected region, F is conservative is equivalent to curlf =. 2. ecall that once we know work integral is path independent, we can compute the potential function f b picking a base point P in D and letting f(q) = F dr, where is an path in D from P to Q. Proof of (a): This was proved in Theorem GT.7. Proof of (b): Suppose is a simple closed curve in D. Since D is simpl connected the interior of is also in D. Therefore, using Green s theorem we have, F dr = curlf da =. D This shows that F is conservative in D. Therefore b Theorem GT.4, F is a gradient field. Summar: Suppose the vector field F = (M, N) is defined on a simpl connected region D. Then, the following statements are equivalent. () (2) Q P F dr is path independent. F dr = for an closed path.

26 4 EXTENDED GEEN S THEOEM 26 (3) F = f for some f in D (4) F is conservative in D. If F is continuousl differentiable then, 2, 3, 4 all impl 5: (5) curlf = N M = in D 3. Wh we need simpl connected in the Potential theorem The basic idea is that if there is a hole in D, then F might not be defined on the interior of. This is illustrated in the net eample. Eample GT.28.(What can go wrong if D is not simpl connected.) Here we will repeat the super-duper reall important Eample GT.9. (, ) Let F = r 2 ( tangential field ). F is defined on D = plane - (,) = punctured plane Puntured plane Several times now we have shown that curlf =. (If ou ve forgotten this, ou should recompute it now.) We also know that on an circle of radius a centered at the origin F dr = 2π. So, the conclusion of the above theorem that curlf = implies F is conservative does not hold. The problem is that D is not simpl connected and, in fact, F is not defined on the entire region inside, so we are not able to appl Green s theorem to conclude that the line integral is. 4 Etended Green s theorem We can etend Green s theorem to a region which has multiple boundar curves. The figures below show regions bounded b 2 or more curves. You will see that this gives us awa to work around singularities in the field F. 4. egions with multiple boundar curves onsider the following three regions.

27 4 EXTENDED GEEN S THEOEM 27 B 3 A The region on the left, A is bounded b and 2. We sa that the boundar is + 2. Note that the wa it is drawn, the region is alwas to the left as ou traverse either boundar curve. The region on the right, is bounded b 7 and 8. We sa that the boundar is 7 8. The reason for the minus sign is that the boundar curves should be oriented so that the region is to our left as ou traverse the curve. As shown, the region is to the right of 8, but to the left of 8. Likewise, in the middle figure, B has boundar You should check that our signs are consistent with the orientation of the curves. 4.2 Etended Green s theorem Theorem GT.29. Etended Green s theorem. Suppose A is the region in the left-hand figure above then, for an vector field F differentiable in all of A we have F dr = curlf d d. + 2 A Likewise for more than two curves: If B has boundar and F is differentiable on all of B then F dr = curlf d d B Proof. We will prove the formula for A. The case of more than two curves is essentiall the same. The ke is to make the cut shown in the figure below, so that the resulting curve is simple. 3 3 A 2 In the figure the curve surrounds the region A. (You have to imagine that the cut is infinitesimall wide so 3 and 3 are right on top of each other.) Now the original Green s theorem applies: F dr = curlf d d

28 4 EXTENDED GEEN S THEOEM 28 Since the contributions of 3 and 3 will cancel, we have proved the etended form of Green s theorem. F dr = curlf d d. QED + 2 (, ) Eample GT.3. Again, let F be the tangential field F = r 2. What values can F dr take for a simple closed positivel oriented curve that doesn t go through the origin? answer: We have two cases (i) does not go around ; (ii) 2 goes around 2 2 ase (i) We know F is defined and curlf = in the entire region inside, so Green s theorem implies F dr = curlf d d =. ase (ii) We can t appl Green s theorem directl because F is not defined everwhere inside 2. Instead, we use the following trick. Let 3 be a circle centered on the origin and small enough that is entirel inside The region 2 has boundar 2 3 and F is defined and differentiable in 2. We know that curlf = in 2, so etended Green s theorem implies F dr = 2 3 curlf d d =. 2 So F dr = 2 F dr. 3 Since 3 is a circle centered on the origin we can compute the line integral directl we ve

29 5 ONE MOE EXAMPLE 29 done this man times alread. F dr = 2π. 3 Therefore, for a simple closed curve and F as given, the line integral 2π or, depending on whether surrounds the origin or not. Answer to the question: The onl possible values are and 2π. F dr is either Eample GT.3. Use the same F as in the previous eample. What values can the line integral take if is not simple. answer: If is not simple we can break it into a sum of simple curves. 2 In the figure, we can think of the entire curve as + 2. Since each of these curves surrounds the origin we have F dr = + 2 F dr + F dr = 2π + 2π = 4π. 2 In general, F dr = 2πn, where n is the number of times goes around (,) in a counterclockwise direction. Aside for those who are interested: The integer n is called the winding number of around. The number n also equals the number of times crosses the positive -ais, counting + if it crosses from below to above and if it crosses from above to below. 5 One more eample Eample GT.32. Let F = r n (, ). For n, F is defined on the entire plane. For n <, F is defined on the -plane minus the origin (the punctured plane). Use etended Green s theorem to show that F is conservative on the punctured plane for all integers n. Then, find a potential function. answer: We start b computing the curl: M = r n M = nr n 2 N = r n N = nr n 2

30 5 ONE MOE EXAMPLE 3 So, curlf = N M =. To show that F is conservative in the punctured plane, we will show that all simple closed curves that don t go through the origin. F dr = for If is a simple closed curve not around then F is differentiable on the entire region inside and Green s theorem implies F dr = curlf d d =. If is a simple closed curve that surrounds, then we can use the etended form of Green s theorem as in Eample GT.3 2 We put a small circle 2 centered at the origin and inside. Since F is radial, it is orthogonal to 2. So, on 2, F dr =, which implies F dr =. 2 Now, on the region with boundar 2 we can appl the etended Green s theorem F dr = curlf d d =. 2 Thus, F dr = F dr, which, as we saw, equals. 2 Thus F dr = for all closed loops, which implies F is conservative. QED To find the potential function we use method over the curve = + 2 shown. 2 (, ) (, ) The following calculation works for n 2. For n = 2 everthing is the same ecept we get natural logs instead of powers. Parametrize using : =, = ; from to. So,

31 5 ONE MOE EXAMPLE 3 d =, d = d, skip M, since d =, N = r n = ( + 2 ) n/2. So, F dr = M d + N d = ( + 2 ) n/2 d = ( + 2 ) (n+2)/2 (u-substitution: u = + 2 ) n + 2 Parametrize 2 using : =, = ( + 2 )(n+2)/2 n + 2 2(n+2)/2 n + 2. = ; from to. So, d = d, d =, M = r n = ( 2 + ) 2 n/2 skip, N since d =. So, F dr = M d + N d = ( 2 + ) 2 n/2 d 2 2 = (2 + 2 )(n+2)/2 (u-substitution: u = n + 2 ) = (2 + 2 )(n+2)/2 n + 2 ( + 2 )(n+2)/2. n + 2 Adding these we get f(, ) f(, ) = (2 + 2 )(n+2)/2 2 (n+2)/2. So, n + 2 f(, ) = rn+2 n c. (If n = 2 we get f(, ) = ln r +.) (Note, we ignored the fact that if (, ) is on the negative -ais we shoud have used a different path that doesn t go through the origin. This isn t reall an issue because we know there is a potential function. Because our function f is known to be a potential function everwhere ecept the negative -ais, b continuit it also works on the negative -ais.)