onservative fields and potential functions. (Sect. 16.3) eview: Line integral of a vector field. onservative fields. The line integral of conservative fields. Finding the potential of a conservative field. omments on eact differential forms. The line integral of a vector field along a curve. Definition The line integral of a vector-valued function F : D n n, with n = 2, 3, along the curve associated with the function r : t, t 1 ] D 3 is given b F dr = F(t) r (t) dt t F r emark: An equivalent epression is: r (t) F dr = F(t) t r (t) r (t) dt, s1 F dr = ˆF û ds, s where û = r (t(s)) r (t(s)), and ˆF = F(t(s)).
Work done b a force on a particle. Definition In the case that the vector field F : D n n, with n = 2, 3, represents a force acting on a particle with position function r : t, t 1 ] D 3, then the line integral W = F dr, is called the work done b the force on the particle. F r A projectile of mass m moving on the surface of Earth. The movement takes place on a plane, and F =, mg. W in the first half of the trajector, and W on the second half. onservative fields and potential functions. (Sect. 16.3) eview: Line integral of a vector field. onservative fields. The line integral of conservative fields. Finding the potential of a conservative field. omments on eact differential forms.
onservative fields. Definition A vector field F : D n n, with n = 2, 3, is called conservative iff there eists a scalar function f : D n, called potential function, such that F = f. F r A projectile of mass m moving on the surface of Earth. The movement takes place on a plane, and F =, mg. F = f, with f = mg. onservative fields. 1 Show that the vector field F = (1 2 + 2 2 + 3 2 1, 2, 3 is )3/2 conservative and find the potential function. Solution: The field F = F 1, F 2, F 3 is conservative iff there eists a potential function f such that F = f, that is, F 1 = 1 f, F 2 = 2 f, F 3 = 3 f. Since i ( (1 2 + 2 2 + 3 2 = 2 )3/2 i 1 + 2 2 + 3 2 ) ] 1/2, i = 1, 2, 3, then we conclude that F = f, with f = 1. 1 2 + 2 2 + 3 2
onservative fields and potential functions. (Sect. 16.3) eview: Line integral of a vector field. onservative fields. The line integral of conservative fields. Finding the potential of a conservative field. omments on eact differential forms. The line integral of conservative fields. Definition A set D n, with n = 2, 3, is called simpl connected iff ever two points in D can be connected b a smooth curve inside D and ever loop in D can be smoothl contracted to a point without leaving D. emark: A set is simpl connected iff it consists of one piece and it contains no holes. D Simpl connected Not simpl connected D
The line integral of conservative fields. Notation: If the path n, with n = 2, 3, has end points r, r 1, then denote the line integral of a field F along as follows r1 F dr = F dr. (This notation emphasizes the end points, not the path.) Theorem A smooth vector field F : D n n, with n = 2, 3, defined on a simpl connected domain D n is conservativewith F = f iff for ever two points r, r 1 D the line integral F dr is independent of the path joining r to r 1 and holds r1 F dr = f (r 1 ) f (r ). r emark: A field F is conservative iff F dr is path independent. r The line integral of conservative fields. Summar: F = f equivalent to Proof: Onl ( ). r1 r F dr = r1 r f dr = r1 r F dr = f (r 1 ) f (r ). where r(t ) = r and r(t 1 ) = r 1. Therefore, ( f ) r (t) dt, t r(t) r1 r F dr = t d ] f (r(t) dt = f (r(t1 )) f (r(t )). dt We conclude that r1 r F dr = f (r 1 ) f (r ). (The statement ( ) is more complicated to prove.)
The line integral of conservative fields. Evaluate I = (1,2,3) (,,) 2 d + 2 d + 2z dz. Solution: I is a line integral for a field in 3, since I = (1,2,3) (,,) 2, 2, 2z d, d, dz. Introduce F = 2, 2, 2z, r = (,, ) and r 1 = (1, 2, 3), then I = r1 r F dr. The field F is conservative, since F = f with potential f (,, z) = 2 + 2 + z 2. That is f (r) = r 2. Therefore, I = r1 r f dr = f (r 1 ) f (r ) = r 1 2 r 2 = (1 + 4 + 9). We conclude that I = 14. The line integral of conservative fields. (Along a path.) Evaluate I = (1,2,3) (,,) 2 d + 2 d + 2z dz along a straight line. Solution: onsider the path given b r(t) = 1, 2, 3 t. Then r() =,,, and r(1) = 1, 2, 3. We now evaluate F = 2, 2, 2z along r(t), that is, F(t) = 2t, 4t, 6t. Therefore, I = t F(t) r (t) dt = 1 2t, 4t, 6t 1, 2, 3 dt I = 1 (2t + 8t + 18t) dt = 1 ( t 2 28t dt = 28 2 1 ). We conclude that I = 14.
onservative fields and potential functions. (Sect. 16.3) eview: Line integral of a vector field. onservative fields. The line integral of conservative fields. Finding the potential of a conservative field. omments on eact differential forms. Finding the potential of a conservative field. Theorem (haracterization of potential fields) A smooth field F = F 1, F 2, F 3 on a simpl connected domain D 3 is a conservative field iff hold 2 F 3 = 3 F 2, 3 F 1 = 1 F 3, 1 F 2 = 2 F 1. Proof: Onl ( ). Since the vector field F is conservative, there eists a scalar field f such that F = f. Then the equations above are satisfied, since for i, j = 1, 2, 3 hold F i = i f i F j = i j f = j i f = j F i. (The statement ( ) is more complicated to prove.)
Finding the potential of a conservative field. Show that the field F = 2, ( 2 z 2 ), 2z is conservative. Solution: We need to show that the equations in the Theorem above hold, that is 2 F 3 = 3 F 2, 3 F 1 = 1 F 3, 1 F 2 = 2 F 1. with 1 =, 2 =, and 3 = z. This is the case, since 1 F 2 = 2, 2 F 1 = 2, 2 F 3 = 2z, 3 F 2 = 2z, 3 F 1 =, 1 F 3 =. Finding the potential of a conservative field. Find the potential function of the conservative field F = 2, ( 2 z 2 ), 2z. Solution: We know there eists a scalar function f solution of F = f f = 2, f = 2 z 2, z f = 2z. f = 2 d + g(, z) f = 2 + g(, z). f = 2 + g(, z) = 2 z 2 g(, z) = z 2. g(, z) = z 2 d +h(z) = z 2 +h(z) f = 2 z 2 +h(z). z f = 2z + z h(z) = 2z z h(z) = f = ( 2 z 2 ) +c.
onservative fields and potential functions. (Sect. 16.3) eview: Line integral of a vector field. onservative fields. The line integral of conservative fields. Finding the potential of a conservative field. omments on eact differential forms. omments on eact differential forms. Notation: We call a differential form to the integrand in a line integral for a smooth field F, that is, F dr = F, F, F z d, d, dz = F d + F d + F z dz. emark: A differential form is a quantit that can be integrated along a path. Definition A differential form F dr = F d + F d + F z dz is called eact iff there eists a scalar function f such that emarks: F d + F d + F z dz = f d + f d + z f dz. A differential form F dr is eact iff F = f. An eact differential form is nothing else than another name for a conservative field.
omments on eact differential forms. Show that the differential form given below is eact, where F dr = 2 d + ( 2 z 2 ) d 2z dz. Solution: We need to do the same calculation we did above: Writing F dr = F 1 d 1 + F 2 d 2 + F 3 d 3, show that 2 F 3 = 3 F 2, 3 F 1 = 1 F 3, 1 F 2 = 2 F 1. with 1 =, 2 =, and 3 = z. We showed that this is the case, since 1 F 2 = 2, 2 F 1 = 2, 2 F 3 = 2z, 3 F 2 = 2z, 3 F 1 =, 1 F 3 =. So, there eists f such that F dr = f dr. Green s Theorem on a plane. (Sect. 16.4) eview: Line integrals and flu integrals. Green s Theorem on a plane. irculation-tangential form. Flu-normal form. Tangential and normal forms equivalence.
eview: The line integral of a vector field along a curve. Definition The line integral of a vector-valued function F : D n n, with n = 2, 3, along the curve r : t, t 1 ] D 3, with arc length function s, is given b s1 F u ds = F(t) r (t) dt, s t where u = r r, and s = s(t ), s 1 = s(t 1 ). z u { z = } emark: Since F = F, F and r(t) = (t), (t), in components, = t t F(t) r (t) dt F (t) (t) + F (t) (t) ] dt. eview: The line integral of a vector field along a curve. Evaluate the line integral of F =, along the loop r(t) = cos(t), sin(t) for t, 2π]. Solution: Evaluate F along the curve: F(t) = sin(t), cos(t). Now compute the derivative vector r (t) = sin(t), cos(t). Then evaluate the line integral in components, F u ds = F (t) (t) + F (t) (t) ] dt, F u ds = F u ds = 2π 2π t ( sin(t))( sin(t)) + cos(t) cos(t) ] dt, sin 2 (t) + cos 2 (t) ] dt F u ds = 2π.
eview: The flu across a plane loop. Definition The flu of a vector field F : {z = } 3 {z = } 3 along a closed plane loop r : t, t 1 ] {z = } 3 is given b F = F n ds, where n is the unit outer normal vector to the curve inside the plane {z = }. z n { z = } emark: Since F = F, F,, r(t) = (t), (t),, ds = r (t) dt, and n = 1 r (t), (t),, in components, F n ds = F (t) (t) F (t) (t) ] dt. t eview: The flu across a plane loop. Evaluate the flu of F =,, along the loop r(t) = cos(t), sin(t), for t, 2π]. Solution: Evaluate F along the curve: F(t) = sin(t), cos(t),. Now compute the derivative vector r (t) = sin(t), cos(t),. Now compute the normal vector n(t) = (t), (t),, that is, n(t) = cos(t), sin(t),. Evaluate the flu integral in components, F n ds = F n ds = 2π F u ds = t F (t) (t) F (t) (t) ] dt, sin(t) cos(t) cos(t)( sin(t)) ] dt, 2π dt F u ds =.
Green s Theorem on a plane. (Sect. 16.4) eview: Line integrals and flu integrals. Green s Theorem on a plane. irculation-tangential form. Flu-normal form. Tangential and normal forms equivalence. Green s Theorem on a plane. Theorem (irculation-tangential form) The counterclockwise line integral F u ds of the field F = F, F along a loop enclosing a region 2 and given b the function r(t) = (t), (t) for t t, t 1 ] and with unit tangent vector u, satisfies that F (t) (t) + F (t) (t) ] dt = F F d d. t z u { z = } Equivalentl, F u ds = ( F F ) d d.
Green s Theorem on a plane. Verif Green s Theorem tangential form for the field F =, and the loop r(t) = cos(t), sin(t) for t, 2π]. Solution: ecall: We found that F u ds = 2π. Now we compute the double integral I = F F d d and we verif that we get the same result, 2π. I = 1 ( 1) ] d d = 2 2π 1 d d = 2 r dr dθ ( r 2 I = 2(2π) 1 ) I = 2π. 2 We verified that F u ds = F F d d = 2π. Green s Theorem on a plane. (Sect. 16.4) eview: Line integrals and flu integrals. Green s Theorem on a plane. irculation-tangential form. Flu-normal form. Tangential and normal forms equivalence.
Green s Theorem on a plane. Theorem (Flu-normal form) The counterclockwise flu integral F n ds of the field F = F, F along a loop enclosing a region 2 and given b the function r(t) = (t), (t) for t t, t 1 ] and with unit normal vector n, satisfies that F (t) (t) F (t) (t) ] dt = F + F d d. t z n { z = } Equivalentl, F n ds = ( F + F ) d d. Green s Theorem on a plane. Verif Green s Theorem normal form for the field F =, and the loop r(t) = cos(t), sin(t) for t, 2π]. Solution: ecall: We found that F n ds =. Now we compute the double integral I = F + F d d and we verif that we get the same result,. I = ( ) + () ] d d = We verified that d d =. F n ds = F + F d d =.
Green s Theorem on a plane. Verif Green s Theorem normal form for the field F = 2, 3 and the loop r(t) = a cos(t), a sin(t) for t, 2π], a >. Solution: We start with the line integral F n ds = F (t) (t) F (t) (t) ] dt. t It is simple to see that F(t) = 2a cos(t), 3a sin(t), and also that r (t) = a sin(t), a cos(t). 2π Therefore, F n ds = 2a 2 cos 2 (t) 3a 2 sin 2 (t) ] dt, 2π F n ds = 2a 2 1 1 + cos(2t) 3a 2 1 ] 1 cos(2t) dt. 2 2 2π Since cos(2t) dt =, we conclude F n ds = πa 2. Green s Theorem on a plane. Verif Green s Theorem normal form for the field F = 2, 3 and the loop r(t) = a cos(t), a sin(t) for t, 2π], a >. Solution: ecall: F n ds = πa 2. Now we compute the double integral I = F + F d d. I = I = Hence, (2) + ( 3) ] d d = 2π d d = a (2 3) d d. ( r 2 r dr dθ = 2π a ) = πa 2. 2 F n ds = F + F d d = πa 2.
Green s Theorem on a plane. (Sect. 16.4) eview: Line integrals and flu integrals. Green s Theorem on a plane. irculation-tangential form. Flu-normal form. Tangential and normal forms equivalence. Tangential and normal forms equivalence. Lemma The Green Theorem in tangential form is equivalent to the Green Theorem in normal form. Proof: Green s Theorem in tangential form for F = F, F sas F (t) (t) + F (t) (t) ] dt = F F d d. t Appl this Theorem for ˆF = F, F, that is, ˆF = F and ˆF = F. We obtain F (t) (t) + F (t) (t) ] ( dt = F ( F ) ) d d, t so, F (t) (t) F (t) (t) ] dt = F + F d d, t which is Green s Theorem in normal form. The converse implication is proved in the same wa.
Using Green s Theorem Use Green s Theorem to find the counterclockwise circulation of the field F = ( 2 2 ), ( 2 + 2 ) along the curve that is the triangle bounded b =, = 3 and =. Solution: ecall: F dr = F F d d. F dr = F dr = 3 F dr = (2 2) d d = ( 2 3 3 ) ( 2 )] d = 2 d = 3 3 3 (2 2) d d, 3 ( 2 2 2) d, F dr = 9. Green s Theorem on a plane. (Sect. 16.4) eview of Green s Theorem on a plane. Sketch of the proof of Green s Theorem. Divergence and curl of a function on a plane. Area computed with a line integral.
eview: Green s Theorem on a plane. Theorem Given a field F = F, F and a loop enclosing a region 2 described b the function r(t) = (t), (t) for t t, t 1 ], with unit tangent vector u and eterior normal vector n, then holds: The counterclockwise line integral F u ds satisfies: F (t) (t) + F (t) (t) ] dt = t The counterclockwise line integral F (t) (t) F (t) (t) ] dt = t ( F F ) d d. F n ds satisfies: ( F + F ) d d. eview: Green s Theorem on a plane. z z u n { z = } { z = } irculation-tangential form: F u ds = F F d d. Flu-normal form: F n ds = F + F d d. Lemma The Green Theorem in tangential form is equivalent to the Green Theorem in normal form.
Green s Theorem on a plane. (Sect. 16.4) eview of Green s Theorem on a plane. Sketch of the proof of Green s Theorem. Divergence and curl of a function on a plane. Area computed with a line integral. Sketch of the proof of Green s Theorem. We want to prove that for ever differentiable vector field F = F, F the Green Theorem in tangential form holds, F (t) (t) + F (t) (t) ] dt = F F d d. We onl consider a simple domain like the one in the pictures. = g () 1 1 = h () = g () 1 Using the picture on the left we show that F (t) (t) dt = F d d; and using the picture on the right we show that F (t) (t) dt = F d d. = h () 1
Sketch of the proof of Green s Theorem. = g () 1 = g () 1 Show that for F (t) = F ((t), (t)) holds F (t) (t) dt = F d d; The path can be described b the curves r and r 1 given b Therefore, r (t) = t, g (t), t, 1 ] r 1 (t) = ( 1 + t), g 1 ( 1 + t) t, 1 ]. r (t) = 1, g (t), t, 1 ] r 1(t) = 1, g 1( 1 + t) t, 1 ]. ecall: F (t) = F (t, g (t)) on r, and F (t) = F (( 1 + t), g 1 ( 1 + t)) on r 1. Sketch of the proof of Green s Theorem. 1 F (t) (t) dt = 1 F (t, g (t)) dt F (( 1 + t), g 1 ( 1 + t)) dt Substitution in the second term: τ = 1 + t, so dτ = dt. 1 1 F (( 1 + t), g 1 ( 1 + t)) dt = F (τ, g 1 (τ)) ( dτ) = Therefore, F (t) (t) dt = We obtain: F (t) (t) dt = 1 1 1 F (τ, g 1 (τ)) dτ. F (t, g (t)) F (t, g 1 (t)) ] dt. g1 (t) g (t) F (t, ) ] d dt.
Sketch of the proof of Green s Theorem. ecall: F (t) (t) dt = 1 g1 (t) g (t) F (t, ) ] d dt. This result is precisel what we wanted to prove: F (t) (t) dt = F d d. We just mention that the result F (t) (t) dt = F d d. 1 = h () is proven in a similar wa using the parametrization of the given in the picture. = h () 1 Green s Theorem on a plane. (Sect. 16.4) eview of Green s Theorem on a plane. Sketch of the proof of Green s Theorem. Divergence and curl of a function on a plane. Area computed with a line integral.
Divergence and curl of a function on a plane. Definition The curl of a vector field F = F, F in 2 is the scalar ( curl F ) z = F F. The divergence of a vector field F = F, F in 2 is the scalar div F = F + F. emark: Both forms of Green s Theorem can be written as: F u ds = curl F d d. F n ds = z div F d d. Divergence and curl of a function on a plane. emark: What tpe of information about F is given in ( curl F ) z? : Suppose F is the velocit field of a viscous fluid and F =, ( curl F ) z = F F = 2. If we place a small ball at (, ), the ball will spin around the z-ais with speed proportional to ( curl F ) z. If we place a small ball at everwhere in the plane, the ball will spin around the z-ais with speed proportional to ( curl F ) z. emark: The curl of a field measures its rotation.
Divergence and curl of a function on a plane. emark: What tpe of information about F is given in div F? : Suppose F is the velocit field of a gas and F =, div F = F + F = 2. The field F represents the gas as is heated with a heat source at (, ). The heated gas epands in all directions, radiall out form (, ). The div F measures that epansion. emark: The divergence of a field measures its epansion. emarks: Notice that for F =, we have ( curl F ) z =. Notice that for F =, we have div F =. Green s Theorem on a plane. (Sect. 16.4) eview of Green s Theorem on a plane. Sketch of the proof of Green s Theorem. Divergence and curl of a function on a plane. Area computed with a line integral.
Area computed with a line integral. emark: An of the two versions of Green s Theorem can be used to compute areas using a line integral. For eample: ( F + F d d = F d F d) If F is such that the left-hand side above has integrand 1, then that integral is the area A() of the region. Indeed: F =, d d = A() = d. F =, d d = A() = d. F = 1 2, d d = A() = 1 2 ( d d ). Area computed with a line integral. Use Green s Theorem to find the area of the region enclosed b the ellipse r(t) = a cos(t), b sin(t), with t, 2π] and a, b positive. Solution: We use: A() = d. We need to compute r (t) = a sin(t), b cos(t). Then, A() = 2π (t) (t) dt = 2π a cos(t) b cos(t) dt. Since A() = ab 2π 2π cos 2 (t) dt = ab 2π cos(2t) dt =, we obtain A() = ab 2 A() = πab. 1 ] 1 + cos(2t) dt. 2 2π, that is,