The curl of a vector field in space. The curl of conservative fields. Stokes Theorem in space. Idea of the proof of Stokes Theorem.

Transcription

1 The tokes Theorem. (ect. 16.7) The curl of a vector field in space. The curl of conservative fields. tokes Theorem in space. Idea of the proof of tokes Theorem. The curl of a vector field in space. Definition The curl of a vector field F = F 1, F 2, F 3 in R 3 is the vector field curl F = ( 2 F 3 3 F 2 ), ( 3 F 1 1 F 3 ), ( 1 F 2 2 F 1 ). Remark: ince the following formula holds, i j k curl F = F 1 F 2 F 3 curl F = ( 2 F 3 3 F 2 ) i ( 1 F 3 3 F 1 ) j + ( 1 F 2 2 F 1 ) k, then one also uses the notation curl F = F.

2 The curl of a vector field in space. Find the curl of the vector field F = xz, xyz, y 2. olution: ince curl F = F, we get, i j k F = x y z xz xyz y 2 = ( y ( y 2 ) z (xyz) ) i ( x ( y 2 ) z (xz) ) j + ( x (xyz) y (xz) ) k, We conclude that = ( 2y xy ) i ( x ) j + ( yz ) k, F = y(2 + x), x, yz. The tokes Theorem. (ect. 16.7) The curl of a vector field in space. The curl of conservative fields. tokes Theorem in space. Idea of the proof of tokes Theorem.

3 The curl of conservative fields. Recall: A vector field F : R 3 R 3 is conservative iff there exists a scalar field f : R 3 R such that F = f. Theorem If a vector field F is conservative, then F =. Remark: This Theorem is usually written as ( f ) =. The converse is true only on simple connected sets. That is, if a vector field F satisfies F = on a simple connected domain D, then there exists a scalar field f : D R 3 R such that F = f. Proof of the Theorem: F = ( y z f z y f ), ( x z f z x f ), ( x y f y x f ) The curl of conservative fields. Is the vector field F = xz, xyz, y 2 conservative? olution: We have shown that F = y(2 + x), x, yz. ince F, then F is not conservative. Is the vector field F = y 2 z 3, 2xyz 3, 3xy 2 z 2 conservative in R 3? olution: Notice that F = i j k x y z y 2 z 3 2xyz 3 3xy 2 z 2 = (6xyz 2 6xyz 2 ), (3y 2 z 2 3y 2 z 2 ), (2yz 3 2yz 3 ) =. ince F = and R 3 is simple connected, then F is conservative, that is, there exists f in R 3 such that F = f.

4 The tokes Theorem. (ect. 16.7) The curl of a vector field in space. The curl of conservative fields. tokes Theorem in space. Idea of the proof of tokes Theorem. tokes Theorem in space. Theorem The circulation of a differentiable vector field F : D R 3 R 3 around the boundary of the oriented surface D satisfies the equation F dr = ( F) n dσ, where dr points counterclockwise when the unit vector n normal to points in the direction to the viewer (right-hand rule). n r (t) r (t)

5 tokes Theorem in space. 2 Verify tokes Theorem for the field F = x 2, 2x, z 2 on the ellipse = {(x, y, z) : 4x 2 + y 2 4, z = }. olution: We compute both sides in F dr = ( F) n dσ. x 1 z z n y We start computing the circulation integral on the ellipse x 2 + y 2 = We need to choose a counterclockwise parametrization, hence the normal to points upwards. We choose, for t [, 2π], r(t) = cos(t), 2 sin(t),. 2 x 1 2 y Therefore, the right-hand rule normal n to is n =,, 1. tokes Theorem in space. Verify tokes Theorem for the field F = x 2, 2x, z 2 on the ellipse = {(x, y, z) : 4x 2 + y 2 4, z = }. olution: Recall: F dr = ( F) n dσ, with r(t) = cos(t), 2 sin(t),, t [, 2π] and n =,, 1. The circulation integral is: = 2π F dr = 2π F(t) r (t) dt cos 2 (t), 2 cos(t), sin(t), 2 cos(t), dt. F dr = 2π [ cos 2 (t) sin(t) + 4 cos 2 (t) ] dt.

6 tokes Theorem in space. Verify tokes Theorem for the field F = x 2, 2x, z 2 on the ellipse = {(x, y, z) : 4x 2 + y 2 4, z = }. 2π [ olution: F dr = cos 2 (t) sin(t) + 4 cos 2 (t) ] dt. The substitution on the first term u = cos(t) and du = sin(t) dt, implies ince 2π 2π cos 2 (t) sin(t) dt = F dr = 2π cos 2 (t) dt = u 2 du =. 2π cos(2t) dt =, we conclude that 2 [ 1 + cos(2t) ] dt. F dr = 4π. tokes Theorem in space. 2 Verify tokes Theorem for the field F = x 2, 2x, z 2 on the ellipse = {(x, y, z) : 4x 2 + y 2 4, z = }. olution: F dr = 4π and n =,, 1. We now compute the right-hand side in tokes Theorem. x 1 z n 1 I = ( F) n dσ. 2 i j k F = x y z x 2 2x z 2 y F =,, 2. is the flat surface {x 2 + y 2 1, z = }, so dσ = dx dy. 2 2

7 tokes Theorem in space. Verify tokes Theorem for the field F = x 2, 2x, z 2 on the ellipse = {(x, y, z) : 4x 2 + y 2 4, z = }. olution: F dr = 4π, n =,, 1, F =,, 2, and dσ = dx dy. Then, ( F) n dσ = x x 2,, 2,, 1 dy dx. The right-hand side above is twice the area of the ellipse. ince we know that an ellipse x 2 /a 2 + y 2 /b 2 = 1 has area πab, we obtain ( F) n dσ = 4π. This verifies tokes Theorem. tokes Theorem in space. Remark: tokes Theorem implies that for any smooth field F and any two surfaces 1, 2 having the same boundary curve holds, ( F) n 1 dσ 1 = ( F) n 2 dσ 2. 1 Verify tokes Theorem for the field F = x 2, 2x, z 2 on any half-ellipsoid 2 = {(x, y, z) : x 2 + y z2 = 1, z }. a2 olution: (The previous example was the case a.) n 2 x 1 z a n y 2 We must verify tokes Theorem on 2, F dr = ( F) n 2 dσ 2. 2

8 tokes Theorem in space. Verify tokes Theorem for the field F = x 2, 2x, z 2 on any half-ellipsoid 2 = {(x, y, z) : x 2 + y z2 = 1, z }. a2 olution: F dr = 4π, F =,, 2, I = ( F) n 2 dσ 2. 2 z n n 2 2 a 2 is the level surface F = of 2 x y F(x, y, z) = x 2 + y z2 a 2 1. n 2 = F F, F = 2x, y 2, 2z a 2, ( F) n 2 = 2 2z/a2 F. dσ 2 = F F k = F 2z/a 2 ( F) n 2 dσ 2 = 2. tokes Theorem in space. Verify tokes Theorem for the field F = x 2, 2x, z 2 on any half-ellipsoid 2 = {(x, y, z) : x 2 + y z2 = 1, z }. a2 olution: F dr = 4π and ( F) n 2 dσ 2 = 2. Therefore, ( F) n 2 dσ 2 = 2 dx dy = 2(2π). 2 1 We conclude that ( F) n 2 dσ 2 = 4π, no matter what is the value of a >. 2

9 The tokes Theorem. (ect. 16.7) The curl of a vector field in space. The curl of conservative fields. tokes Theorem in space. Idea of the proof of tokes Theorem. Idea of the proof of tokes Theorem. plit the surface into n surfaces i, for i = 1,, n, as it is done in the figure for n = 9. F dr = = n F dr i i=1 i=1 i n F d r i ( i the border of small rectangles); i n ( F) n i da (Green s Theorem on a plane); i=1 Ri ( F) n dσ.

10 The Divergence Theorem. (ect. 16.8) The divergence of a vector field in space. The Divergence Theorem in space. The meaning of urls and Divergences. Applications in electromagnetism: Gauss law. (Divergence Theorem.) Faraday s law. (tokes Theorem.) The divergence of a vector field in space. Definition The divergence of a vector field F = F x, F y, F z is the scalar field Remarks: div F = x F x + y F y + z F z. It is also used the notation div F = F. The divergence of a vector field measures the expansion (positive divergence) or contraction (negative divergence) of the vector field. A heated gas expands, so the divergence of its velocity field is positive. A cooled gas contracts, so the divergence of its velocity field is negative.

11 The divergence of a vector field in space. Find the divergence and the curl of F = 2xyz, xy, z 2. olution: Recall: div F = x F x + y F y + z F z. x F x = 2yz, y F y = x, z F z = 2z. Therefore F = 2yz x 2z, that is F = 2z(y 1) x. Recall: curl F = F. i j k F = x y z 2xyz xy z 2 = ( ), ( 2xy), ( y 2xz) We conclude: F =, 2xy, (2xz + y). The divergence of a vector field in space. Find the divergence of F = r, where r = x, y, z, and ρ3 ρ = r = x 2 + y 2 + z 2. (Notice: F = 1/ρ 2.) olution: The field components are F x = x ρ 3, F y = y ρ 3, F z = z ρ 3. x F x = x [ x ( x 2 + y 2 + z 2) 3/2] x F x = ( x 2 + y 2 + z 2) 3/2 3 2 x( x 2 + y 2 + z 2) 5/2 (2x) x F x = 1 ρ 3 3x 2 ρ 5 y F y = 1 ρ 3 3y2 ρ 5, zf z = 1 ρ 3 3z2 ρ 5. F = 3 ρ 3 3(x 2 + y 2 + z 2 ) ρ 5 We conclude: F =. = 3 ρ 3 3ρ2 ρ 5 = 3 ρ 3 3 ρ 3.

12 The Divergence Theorem. (ect. 16.8) The divergence of a vector field in space. The Divergence Theorem in space. The meaning of urls and Divergences. Applications in electromagnetism: Gauss law. (Divergence Theorem.) Faraday s law. (tokes Theorem.) The Divergence Theorem in space. Theorem The flux of a differentiable vector field F : R 3 R 3 across a closed oriented surface R 3 in the direction of the surface outward unit normal vector n satisfies the equation F n dσ = ( F) dv, where V R 3 is the region enclosed by the surface. Remarks: The volume integral of the divergence of a field F in a volume V in space equals the outward flux (normal flow) of F across the boundary of V. The expansion part of the field F in V minus the contraction part of the field F in V equals the net normal flow of F across out of the region V. V

13 The Divergence Theorem in space. Verify the Divergence Theorem for the field F = x, y, z over the sphere x 2 + y 2 + z 2 = R 2. olution: Recall: F n dσ = ( F) dv. We start with the flux integral across. The surface is the level surface f = of the function f (x, y, z) = x 2 + y 2 + z 2 R 2. Its outward unit normal vector n is n = f f, f = 2x, 2y, 2z, f = 2 x 2 + y 2 + z 2 = 2R, We conclude that n = 1 x, y, z, where z = z(x, y). R ince dσ = f f k dx dy, then dσ = R dx dy, with z = z(x, y). z V The Divergence Theorem in space. Verify the Divergence Theorem for the field F = x, y, z over the sphere x 2 + y 2 + z 2 = R 2. olution: Recall: n = 1 R x, y, z, dσ = R dx dy, with z = z(x, y). ( z F n dσ = x, y, z 1 ) R x, y, z dσ. F n dσ = 1 ( x 2 + y 2 + z 2) dσ = R dσ. R The integral on the sphere can be written as the sum of the integral on the upper half plus the integral on the lower half, both integrated on the disk R = {x 2 + y 2 R 2, z = }, that is, F n dσ = 2R R R z dx dy.

14 The Divergence Theorem in space. Verify the Divergence Theorem for the field F = x, y, z over the sphere x 2 + y 2 + z 2 = R 2. R olution: F n dσ = 2R dx dy. R z Using polar coordinates on {z = }, we get 2π R F n dσ = 2 R 2 r dr dθ. R 2 r 2 The substitution u = R 2 r 2 implies du = 2r dr, so, F n dσ = 4πR 2 R 2 u F n dσ = 2πR 2( 2u 1/2 R 2 ) 1/2 ( du) 2 R 2 = 2πR 2 u 1/2 du F n dσ = 4πR 3. The Divergence Theorem in space. Verify the Divergence Theorem for the field F = x, y, z over the sphere x 2 + y 2 + z 2 = R 2. olution: F n dσ = 4πR 3. We now compute the volume integral F dv. The divergence of F is F = , that is, F = 3. Therefore ( 4 F dv = 3 dv = 3 3 πr3) We obtain V V F dv = 4πR 3. We have verified the Divergence Theorem in this case. V V

15 The Divergence Theorem in space. Find the flux of the field F = r across the boundary of the region ρ3 between the spheres of radius R 1 > R >, where r = x, y, z, and ρ = r = x 2 + y 2 + z 2. olution: We use the Divergence Theorem F n dσ = ( F) dv. ince F =, then V V ( F) dv =. Therefore F n dσ =. The flux along any surface vanishes as long as is not included in the region surrounded by. (F is not differentiable at.) The Divergence Theorem. (ect. 16.8) The divergence of a vector field in space. The Divergence Theorem in space. The meaning of urls and Divergences. Applications in electromagnetism: Gauss law. (Divergence Theorem.) Faraday s law. (tokes Theorem.)

16 The meaning of urls and Divergences. Remarks: The meaning of the url and the Divergence of a vector field F is best given through the tokes and Divergence Theorems. 1 F = lim F dr, {P} A() where is a surface containing the point P with boundary given by the loop and A() is the area of that surface. 1 F = lim F ndσ, R {P} V (R) where R is a region in space containing the point P with boundary given by the closed orientable surface and V (R) is the volume of that region. The Divergence Theorem. (ect. 16.8) The divergence of a vector field in space. The Divergence Theorem in space. The meaning of urls and Divergences. Applications in electromagnetism: Gauss law. (Divergence Theorem.) Faraday s law. (tokes Theorem.)

17 Applications in electromagnetism: Gauss Law. Gauss law: Let q : R 3 R be the charge density in space, and E : R 3 R 3 be the electric field generated by that charge. Then q dv = k E n dσ, R that is, the total charge in a region R in space with closed orientable surface is proportional to the integral of the electric field E on this surface. The Divergence Theorem relates surface integrals with volume integrals, that is, E n dσ = ( E) dv. Using the Divergence Theorem we obtain the differential form of Gauss law, R E = 1 k q. Applications in electromagnetism: Faraday s Law. Faraday s law: Let B : R 3 R 3 be the magnetic field across an orientable surface with boundary given by the loop, and let E : R 3 R 3 measured on that loop. Then d dt B n dσ = E dr, that is, the time variation of the magnetic flux across is the negative of the electromotive force on the loop. The tokes Theorem relates line integrals with surface integrals, that is, E dr = ( E) n dσ. Using the tokes Theorem we obtain the differential form of Faraday s law, t B = E.

18 Review for Exam 4. ections , 16.7, minutes. 5 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed. Review for Exam 4. (16.1) Line integrals. (16.2) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area, surface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem.

19 Line integrals (16.1). Integrate the function f (x, y) = x 3 /y along the plane curve given by y = x 2 /2 for x [, 2], from the point (, ) to (2, 2). olution: We have to compute I = f ds, by that we mean f ds = t1 t f ( x(t), y(t) ) r (t) dt, where r(t) = x(t), y(t) for t [t, t 1 ] is a parametrization of the path. In this case the path is given by the parabola y = x 2 /2, so a simple parametrization is to use x = t, that is, r(t) = t, t2, t [, 2] r (t) = 1, t. 2 Line integrals (16.1). Integrate the function f (x, y) = x 3 /y along the plane curve given by y = x 2 /2 for x [, 2], from the point (, ) to (2, 2). olution: r(t) = f ds = f ds = t1 t, t2 2 for t [, 2], and r (t) = 1, t. t f ( x(t), y(t) ) r (t) dt = 2 f ds = We conclude that 2 t t t 2 /2 2 dt, 2t 1 + t 2 dt, u = 1 + t 2, du = 2t dt. 5 1 u 1/2 du = 2 3 u3/2 5 1 = 2 3 ( 5 3/2 1 ). f ds = 2 3( ).

20 Review for Exam 4. (16.1) Line integrals. (16.2) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area, surface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. Vector fields, work, circulation, flux (plane) (16.2). Find the work done by the force F = yz, zx, xy in a moving particle along the curve r(t) = t 3, t 2, t for t [, 2]. olution: The formula for the work done by a force on a particle moving along given by r(t) for t [t, t 1 ] is W = F dr = t1 t F(t) r (t) dt. In this case r (t) = 3t 2, 2t, 1 for t [, 2]. We now need to evaluate F along the curve, that is, F(t) = F ( x(t), y(t), z(t) ) = (t 2 )t, t(t 3 ), (t 3 )t 2 We obtain F(t) = t 3, t 4, t 5.

21 Vector fields, work, circulation, flux (plane) (16.2). Find the work done by the force F = yz, zx, xy in a moving particle along the curve r(t) = t 3, t 2, t for t [, 2]. olution: F(t) = t 3, t 4, t 5 and r (t) = 3t 2, 2t, 1 for t [, 2]. The Work done by the force on the particle is W = t1 t F(t) r (t) dt = 2 t 3, t 4, t 5 3t 2, 2t, 1 dt W = 2 ( 3t 5 + 2t 5 t 5) 2 dt = 4t 5 dt = t6 We conclude that W = 2 7 /3. = Vector fields, work, circulation, flux (plane) (16.2). Find the flow of the velocity field F = xy, y 2, yz from the point (,, ) to the point (1, 1, 1) along the curve of intersection of the cylinder y = x 2 with the plane z = x. olution: The flow (also called circulation) of the field F along a curve parametrized by r(t) for t [t, t 1 ] is given by F dr = t1 t F(t) r (t) dt. We use t = x as the parameter of the curve r, so we obtain r(t) = t, t 2, t, t [, 1] r (t) = 1, 2t, 1. F(t) = t(t 2 ), (t 2 ) 2, t 2 (t) F(t) = t 3, t 4, t 3.

22 Vector fields, work, circulation, flux (plane) (16.2). Find the flow of the velocity field F = xy, y 2, yz from the point (,, ) to the point (1, 1, 1) along the curve of intersection of the cylinder y = x 2 with the plane z = x. olution: r (t) = 1, 2t, 1 for t [, 1] and F(t) = t 3, t 4, t 3. t1 1 F dr = F(t) r (t) dt = t 3, t 4, t 3 1, 2t, 1 dt, t F dr = We conclude that 1 ( t 3 + 2t 5 t 3) dt = 1 F dr = t 5 dt = 2 6 t6 1. Vector fields, work, circulation, flux (plane) (16.2). Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t [, 2π]. olution: The flux (also normal flow) of the field F = F x, F y across a loop parametrized by r(t) = x(t), y(t) for t [t, t 1 ] is given by t1 [ F n ds = Fx y (t) F y x (t) ] dt. t Recall that n = 1 r (t) y (y), x (t) and ds = r (t) dt, therefore F n ds = ( F x, F y 1 ) r (t) y (y), x (t) r (t) dt, so we obtain F n ds = [ F x y (t) F y x (t) ] dt.

23 Vector fields, work, circulation, flux (plane) (16.2). Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t [, 2π]. t1 [ olution: F n ds = Fx y (t) F y x (t) ] dt. We evaluate F along the loop, t F(t) = a cos(t), a [ cos(t) sin(t) ], and compute r (t) = a sin(t), a cos(t). Therefore, F n ds = F n ds = 2π 2π [ a cos(t)a cos(t) a ( cos(t) sin(t) ) ( a) sin(t) ] dt [ a 2 cos 2 (t) + a 2 sin(t) cos(t) a 2 sin 2 (t) ] dt Vector fields, work, circulation, flux (plane) (16.2). Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t [, 2π]. olution: F n ds = ince 2π 2π [ a 2 cos 2 (t) + a 2 sin(t) cos(t) a 2 sin 2 (t) ] dt. 2π F n ds = a 2 [ ] 1 + sin(t) cos(t) dt, F n ds = a 2 2π sin(2t) dt =, we obtain [ sin(2t) ] dt. F n ds = 2πa 2.

24 Review for Exam 4. (16.1) Line integrals. (16.2) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area, surface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. onservative fields, potential functions (16.3). Is the field F = y sin(z), x sin(z), xy cos(z) conservative? If yes, then find the potential function. olution: We need to check the equations y F z = z F y, x F z = z F x, x F y = y F x. y F z = x cos(z) = z F y, x F z = y cos(z) = z F x, x F y = sin(z) = y F x. Therefore, F is a conservative field, that means there exists a scalar field f such that F = f. The equations for f are x f = y sin(z), y f = x sin(z), z f = xy cos(z).

25 onservative fields, potential functions (16.3). Is the field F = y sin(z), x sin(z), xy cos(z) conservative? If yes, then find the potential function. olution: x f = y sin(z), y f = x sin(z), z f = xy cos(z). Integrating in x the first equation we get f (x, y, z) = xy sin(z) + g(y, z). Introduce this expression in the second equation above, y f = x sin(z) + y g = x sin(z) y g(y, z) =, so g(y, z) = h(z). That is, f (x, y, z) = xy sin(z) + h(z). Introduce this expression into the last equation above, z f = xy cos(z) + h (z) = xy cos(z) h (z) = h(z) = c. We conclude that f (x, y, z) = xy sin(z) + c. onservative fields, potential functions (16.3). ompute I = y sin(z) dx + x sin(z) dy + xy cos(z) dz, where given by r(t) = cos(2πt), 1 + t 5, cos 2 (2πt)π/2 for t [, 1]. olution: We know that the field F = y sin(z), x sin(z), xy cos(z) conservative, so there exists f such that F = f, or equivalently df = y sin(z) dx + x sin(z) dy + xy cos(z) dz. We have computed f already, f = xy sin(z) + c. ince F is conservative, the integral I is path independent, and I = (1,2,π/2) (1,1,π/2) [ y sin(z) dx + x sin(z) dy + xy cos(z) dz ] I = f (1, 2, π/2) f (1, 1, π/2) = 2 sin(π/2) sin(π/2) I = 1.