Green s Theorem Jerem Orloff Line integrals and Green s theorem. Vector Fields 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). Eamples: (a) Force, constant gravitational field F(, ) = (, g). (b) Velocit V(, ) = ( 2 + 2, ) ( 2 + 2 = r 2, ) r 2. (This is a shrinking radial field like water pouring from a source at (,).) (c) Unit tangential field F = (, ) /r. (d) Gradient field F = f, e.g., f(, ) = 2 f = ( 2, 2 )... Visualization of vector fields Draw little arrows in the plane. (a) onstant vector field (b) Shrinking radial field (c) Unit tangential field
LINE INTEGALS AND GEEN S THEOEM 2..2 Work done b a force over a curve Sa F is a constant force, applied over a displacement r Work = F cos θ r = F r. (Note F cos θ = force in direction of motion = onl part of force that does work.) For a variable force over a curve the total work is found b summing over pieces, i.e., integrating: F dr. F θ r The force vector is at an angle to the curve. This angle varies over the curve. F dr F dr dr F..3 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. Div. The divergence of the vector field F = (M, N) is divf = M + N. This is just for completeness. We probabl won t see this in 8.4..2 Line integrals (Also called path or contour integrals) Ingredients Vector field F(, ) = (M, N) urve : r(t) = ((t), (t)) Definition. The Line integral of F over is computed b F dr = (M, N) (d, d) = M d + N d. (F dr used in phsics, M d + N d used in thermodnamics)
LINE INTEGALS AND GEEN S THEOEM 3.2. omputing line integrals Eample GT.. Evaluate I = = 2 from (, ) to (, ). 2 d + ( 2) d over the part of the parabola answer: Parametrize curve: = t, = t 2, t. d = dt, d = 2t dt I = t 2 (t 2 ) dt + (t 2t 2 )2t dt = eason for substitution: M d = t 4 + 2t 2 4t 3 dt = 2 5. M d dt dt etc. Eample GT.2. Same integral as previous eample ecept is the straight line from (, ) to (, ). answer: Parametrize curve: = t, = t, t. d = dt, d = dt I = t 2 t dt + (t 2t) dt = (Different value from Eample GT..) t 3 t dt = 4. Eample GT.3. Same as Eample GT. ecept we use a different parametrization of. = sin t, = sin 2 t, t π/2 d = cos t dt, d = 2 sin t cos t dt. I = = π/2 π/2 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 = sin5 t + 2 5 3 sin3 t sin 4 t π/2 = 2 5. This is the same value as in Eample GT.. That is, the parametrization doesn t affect the value of the line integral.
LINE INTEGALS AND GEEN S THEOEM 4.2.2 Properties of line integrals. Independent of parametrization. 2. everse direction on curve change sign. That is, F dr = F dr. (Here, means the same curve traversed in the opposite direction.) 3. If is closed we use the notation F dr = M d + N d. Eample GT.4. Evaluate I = shown. d + ( + 2) d where = + 2 is the curve (, ) 2 answer: Have 2 pieces so I = d + ( + 2) d + d + ( + 2) d. 2 Note, we don t alwas need to introduce the variable t. In this case we can easil use or as parameters. : =, use as parameter. d = d, d =. So, d + (2 + 2) d = d =. 2 : =, use as parameter. goes from to. So, d + ( + 2) d = 2 ( + 2) d = + 2 d = 2. answer: I = 2 =. Eample GT.5. Evaluate I = counterclockwise (W) direction. d + d where is the unit circle traversed in a 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.
LINE INTEGALS AND GEEN S THEOEM 5.3 Gradient and conservative fields.3. The fundamental theorem for gradient fields Theorem GT.6. (Fundamental theorem for gradient fields) Suppose that F = f is a gradient field. Then, F dr = f(, ) P P = f(p ) f(p ) = f(, ) f(, ) where P = (, ) and P = (, ) are the endpoints of the curve. That is, for gradient fields the line integral is independent of the path taken. P = (, ) P = (, ) The proof of the fundamental theorem is given after the net eample Eample GT.7. Let f(, ) = 3 + 2. So, F = f = ( 3 + 2, 3 2). Let be the curve shown and compute I = F dr. Do this both directl and using the fundamental theorem. (, 2) (, )
LINE INTEGALS AND GEEN S THEOEM 6 answer: Method : parametrize : =, = 2,. Then, I = ( 3 + 2) d + 3 2 d = (8 3 + 2) d + 2 3 2 d. = 32 3 + 2 d = 9. Method 2: f dr = f(, 2) f(, ) = 9..3.2 Proof of the fundamental theorem This just requires some algebra and the chain rule: t [ f dr = f d + f d = f ((t), (t)) d dt + f ((t), (t)) d ] dt dt = t t t d dt f((t), (t)) dt = f((t), (t)) t t = f(p ) f(p ).3.3 Path independence Definition. For a vector field F, the line integral F dr is called path independent if, for an two points P and Q, the line integral has the same value for ever path between P and Q. Theorem GT.8. F dr is path independent is equivalent to F dr = for an closed path. Note. These statements are technicall a little slopp. The should all include the qualifing phrase in a region A for F, the points and the paths. Proof. We have to show two things: (i) Path independence implies the line integral around an closed path is. (ii) If the line integral around all closed paths is then we have path independence. To see (i), assume path independence and consider the closed path shown in figure (i) below. Since the starting point P is the same as the endpoint P the line integral F dr must have the same value as the line integral over the curve consisting of the single point P. Since that is clearl we must have the integral over is. To see (ii), assume F dr = for an closed curve. onsider the two curves and 2 shown in figure (ii). Both start at P and end at P. B the assumption that integrals around closed paths are, we have F dr =. So, 2 F dr = F dr. 2 That is, an two paths from P to P have the same line integral. We have shown that the line integrals are path independent.
LINE INTEGALS AND GEEN S THEOEM 7 P = P P 2 P Figure (i) Figure (ii).3.4 onservative vector fields Given a vector field F, Theorem GT.8 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 is. 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.9. 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= This means we need to write down the line integral for f( + t, ). In the figure below, f(, ) is the integral along the path γ from (, ) to (, ) and f( + t, ) is the integral along γ followed b the horizontal line γ 2 to ( + t, ). This means that f( + t, ) = F dr + F dr = f(, ) + F dr. γ γ 2 γ 2
LINE INTEGALS AND GEEN S THEOEM 8 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 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.. If F is the electric field of an electric charge it is conservative. Eample GT.. A gravitational field of a mass is conservative..4 Potential functions Definition. If F is a gradient field with F = f, then we call f a potential function for F. Note. The usual phsics terminolog would be to call f the potential function for F. We have just seen that if F = f (a gradient field) then F is conservative. That is, P P F dr, is independent of the path from P to P and, Questions:. How can we tell if a vector field F is a gradient field? 2. If F is a gradient field, how do we find a potential function f. F dr =, where is an closed path.
LINE INTEGALS AND GEEN S THEOEM 9 We start to answer these questions in the net theorem. Theorem GT.2. Take F = (M, N) (a) If F = f then curlf = N M =. (b) If F is differentiable and curlf = is conservative. in the whole plane then F = f for some f, i.e. F 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 s called a simpl connected region. Proof of (a): If F = (M, N) = f then we have M = f and N = f. Thus, M = f, and N = f. This proves (a) (provided f has continuous second partials). The proof of (b) will be postponed until after we have proved Green s theorem and we can state the Potential Theorem. The eamples below will show how to find f Eample GT.3. For which values of the constants a and b will F = ( a, 2 + b ) be a gradient field? answer: M = a, N = 2 a = 2 and b is arbitrar. Eample GT.4. Is F = ( (3 2 + ), e ) conservative? answer: M =, N = e. Since M N, F is not conservative. (, ) Eample GT.5. Is 2 + 2 conservative? answer: NO! The reasoning is a little trickier than in the previous eample. First, M = 2 2 ( 2 + 2 ) 2 = N. BUT, since the field is not defined for all (, ) Theorem GT.2b does not appl. The answer turns out to be no, F is not conservative. We can see that as follows. onsider = unit circle parametrized b = cos t, = sin t. omputing the line integral directl we have F dr = 2 + 2 d + 2π 2 + 2 d = dt = 2π. Since this is not the field is not conservative. Eample GT.6. Is F = (, ) 2 + 2 conservative? answer: Again we find N = M, BUT since F is not defined at (, ) Theorem GT.2b does not appl. HOWEVE, it turns out that F = ln( 2 + 2 ) = ln r Since we found a potential function we have shown directl that F is conservative on the region consisting of the plane minus the origin.
LINE INTEGALS AND GEEN S THEOEM.4. Finding the potential function Eample GT.7. Show F = ( 3 2 + 6, 3 2 + 6 ) is conservative and find the potential function f such that F = f. answer: First, M = 6 = N. Since F is defined for all (, ) Theorem GT.2 implies F is conservative. Method (for finding f): Since F dr = f(p ) f(p ) we have f(p ) = f(p )+ F dr for an path from P to P. We let P = (, ) be an arbitrar point. We can fi P and an wa we want. For this problem take P = (, ) and = + 2 as the path shown. 2 (, ) : =, =, so d =, d = d, 2 : =, =, so d = d, d = f(, ) f(, ) = Thus, = = F dr = M(, ) d + 6 d + = 3 2 + 3 + 3 2 M d + N d N(, ) d 3 2 + 6 d f(, ) f(, ) = 3 2 + 3 + 3 2 = 3 2 + 3 + 3 2. Switching from (, ) to (, ) and letting f(, ) =, we have f(, ) = 3 2 + 3 + 3 2 +. Method 2: We start with f = 3 2 + 6. Integrating with respect to gives f(, ) = 3 + 3 2 + g(). We need to add g() as the constant of integration since is a constant when differentiating with respect to.
LINE INTEGALS AND GEEN S THEOEM Now integrating f = 3 2 + 6 with respect to gives f(, ) = 3 2 + 3 2 + h() omparing the two epresions for f we see g() = 3 2 + and h() = 3 +. So, (The same as method.) In general I prefer method. f(, ) = 3 + 3 2 + 3 2 +. Eample GT.8. Let F = ( ( + 2 ), (2 + 3 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.2, F is conservative. Method : Use the path shown. f(p ) f(, ) = F dr = M d + N d. : =, =, d = d, d = M(, ) =. 2 : =, =, d =, d = d N(, ) = 2 + 3 2. F dr = d + 2 + 3 2 d = 2 /2 + 2 +. 3 Thus, f(, ) f(, ) = 2 /2 + 2 + 3. Equivalentl f(, ) = 2 /2 + 2 + 3 +. (, ) 2 Path for Eample 6 Method 2: f = + 2, so f = 2 /2 + 2 + g(). f = 2 + 3 2, so f = 2 + 3 + h(). omparing the two epressions for f we get f(, ) = 2 /2 + 2 + 3 +..5 Green s Theorem Ingredients: a simple closed curve (i.e. no self-intersection), and the interior of.
LINE INTEGALS AND GEEN S THEOEM 2 must be positivel oriented (traversed so interior region is on the left) and piecewise smooth (a few corners are oka). Theorem GT.9. Green s Theorem: If the vector field F = (M, N) is defined and differentiable on then M d + N d = N M da. () In vector form this is written F dr = where the curl is defined as curlf = (N M ). curlf da. Eample GT.2. Use the right hand side of Equation to find the left hand side. Use Green s Theorem to compute I = 3 2 2 d + 2 2 ( + ) d where is the circle shown. a answer: B Green s Theorem I = 6 2 + 4 6 2 da = 4 da. We could compute this directl, but here s a trick. We know the center of mass is cm = da = a. A Thus da = πa 3, so I = 4πa 3.
LINE INTEGALS AND GEEN S THEOEM 3 Eample GT.2. (Use the left hand side of Equation to find the right hand side.) Use Green s Theorem to find the area under one arch of the ccloid. answer: The ccloid is = a(θ sin(θ)), = a( cos(θ)). The picture shows the curve = + 2 surrounding the area under one arch. 2a 2 πa 2πa B Green s Theorem, d = da = area. So, area = d = + 2 d + d 2 area = Other was to compute area: da = d = d = 2 2π a 2 ( cos(θ)) 2 dθ = 3πa 2. d + d..5. 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. 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. B direct calculation the right hand side of Green s Theorem (Equation ) is M b d da = M d d. Inner integral: a M(, ) d c = M(, d) + M(, c) c
LINE INTEGALS AND GEEN S THEOEM 4 Putting this into the outer integral we have shown that M da = b For the left hand side of Equation we have M d = M d + M d = bottom b a M(, c) d + top a b a M(, c) M(, d) d. (2) M(, d) d = (since d = along the sides) b a M(, c) M(, d) d. (3) omparing Equations 2 and 3 we find that we have proved Green s Theorem. 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 of the two rectangles equals the line integral over the outside boundar. = 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 close as we want b a sum of rectangles, Green s Theorem must hold on arbitrar regions.
LINE INTEGALS AND GEEN S THEOEM 5.6 Etensions and applications of Green s theorem.6. Simpl connected regions Definition. A region D in the plane is simpl connected if it has no holes. Said differentl, it is simpl connected for ever simple closed curve in D, the interior of is full contained in D. Eamples: D D 2 D 3 D 4 D 5 = whole plane D-D5 are simpl connected. For an simple closed curve inside an of these regions the interior of is entirel inside the region. Note. Sometimes we sa the region is simpl connected if an curve can be shrunk to a point without leaving the region. The regions below are not simpl connected. For each, the interior of the curve is not entirel in the region. Annulus Puntured plane.6.2 Potential Theorem Here is a application of Green s theorem which tells us how to spot a conservative field on a simpl connected region. 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 of Theorem GT.2 above. Theorem GT.22. (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. 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 =. Proof of (a): This was proved in Theorem GT.2.
LINE INTEGALS AND GEEN S THEOEM 6 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.9 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. (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.6.3 Wh we need simpl connected in the Potential Theorem If there is a hole then F might not be defined on the interior of. (See the eample on the tangential field below.) D.6.4 Etended Green s Theorem We can etend Green s theorem to a region which has multiple boundar curves. Suppose is the region between the two simple closed curves and 2. 2 2 3 4
LINE INTEGALS AND GEEN S THEOEM 7 (Note is alwas to the left as ou traverse either curve in the direction indicated.) Then we can etend Green s theorem to this setting b F dr + F dr = curlf da. 2 Likewise for more than two curves: F dr + F dr + 2 F dr + 3 F dr = 4 curlf da. Proof. The proof is based on the following figure. We cut both and 2 and connect them b two copies of 3, one in each direction. (In the figure we have drawn the two copies of 3 as separate curves, in realit the are the same curve traversed in opposite directions.) Now the curve = + 3 + 2 3 is a simple closed curve and Green s theorem holds on it. But the region inside is eactl and the contributions of the two copies of 3 cancel. That is, we have shown that curlf da = F dr = + 3 + 2 3 F dr. + 2 This is eactl Green s theorem, which we wanted to prove. 3 3 2 (, ) Eample GT.23. Let F = r 2 ( tangential field ) F is defined on D = plane - (,) = the punctured plane. (Shown below.) It s eas to compute (we ve done it before) that curlf = in D. Question: For the tangential field F what values can F dr take for a simple closed curve (positivel oriented)? answer: We have two cases (i) not around (ii) 2 around
LINE INTEGALS AND GEEN S THEOEM 8 2 In case (i) Green s theorem applies because the interior does not contain the problem point at the origin. Thus, F dr = curlf da =. For case (ii) we will show that F dr = 2π. 2 Let 3 be a small circle of radius a, entirel inside 2. B the etended Green s theorem we have F dr F dr = curlf da =. 2 3 Thus, F dr = F dr. 2 3 Using the usual parametrization of a circe we can easil compute that the line integral is F dr = 3 2π dt = 2π. QED. 3 2 Answer to the question: The onl possible values are and 2π. We can etend this answer in the following wa: If is not simple, then the possible values of F dr are 2πn, times goes (counterclockwise) around (,). where n is the number of Not for class: n is called the winding number of around. n also equals the number of times crosses the positive -ais, counting + from below and from above.
LINE INTEGALS AND GEEN S THEOEM 9 Eample GT.24. Let F = r n (, ). Use etended Green s theorem to show that F is conservative for all integers n. Find a potential function. answer: First note, M = r n, N = r n. Thus, M = nr n 2 = N. Said differentl: curlf =. We can t et claim that F is conservative because the origin is a possible problem point for F. So, we ll show F is conservative b showing F dr = for all simple closed curves. Suppose is a simple closed curve not around then Green s theorem applies and we have F dr =. Net, suppose 3 is a circle centered on (,) then, since F is radial and dr is tangential to the curve we know F dr =. So, F dr =. (If ou don t like this argument, 3 parametrize the circle in the usual wa and show that the resulting integrand is.) Now take the circle 3 so large that it completel surrounds 2. In this case, etended Green s theorem implies (since curlf = between the curves) that F dr = F dr =. 2 3 Thus F dr = for all closed loops. As we know, this implies that F is conservative. 3 2 To find the potential function we use method over the curve shown. The calculation works for n 2. For n = 2 everthing is the same ecept we get natural logs instead of powers. (We also ignore the fact that if (, ) is on the negative -ais we shoud use a different path that doesn t go through the origin. This isn t reall an issue since we alread know a potential function eists, so continuit would handle these points without using an integral.)
LINE INTEGALS AND GEEN S THEOEM 2 2 (, ) (, ) Here s the calculation of the potential: f(, ) = r n d + r n d urve for computing the potential function. Thus, our potential function is = ( + 2 ) n/2 d + = ( + 2 ) (n+2)/2 n + 2 = ( + 2 )(n+2)/2 2 (n+2)/2 n + 2 = (2 + 2 )(n+2)/2 2 (n+)/2 n + 2 ( 2 + ) 2 n/2 d + (2 + 2 )(n+2)/2 n + 2 f(, ) = rn+2 n + 2 +. In the special case of n = 2 we get f(, ) = ln r +. + (2 + 2 )(n+2)/2 ( + 2 ) (n + 2)/2