ENGI Surface Integrals Page 2.01

Transcription

1 ENGI 543. urface Integrals Page.1. urface Integrals This chapter introduces the theorems of Green, Gauss and tokes. Two different methods of integrating a function of two variables over a curved surface are developed. The sections in this chapter are:.1 Line Integrals. Green s Theorem.3 Path Independence.4 urface Integrals - Projection Method.5 urface Integrals - urface Method.6 Theorems of Gauss and tokes; Potential Functions

2 ENGI Line Integrals Page..1 Line Integrals Two applications of line integrals are treated here: the evaluation of work done on a particle as it travels along a curve in the presence of a [vector field] force; and the evaluation of the location of the centre of mass of a wire. Work done: The work done by a force F in moving an elementary distance Δr along a curve is approximately the product of the component of the force in the direction of Δr and the distance Δr travelled: ΔW FiΔr = Fcosθ Δ r Integrating along the curve yields the total work done by the force F in moving along the curve : W = Fdr i dx dy dz = f dx + f dy + f dz = f + f + f dt dt dt dt ( ) t t 1 dr W = Fdr i = Fi dt dt

3 ENGI Line Integrals Page.3 Example.1.1 Find the work done by F = y, x, z in moving around the curve (defined in parametric form by x = cos t, y = sin t, z =, t π ). F = y, x, z = sin t,cos t, dr d = cos t,sin t, = sin t,cos t, dt dt dr F i = sin t + cos t + = 1 dt π dr π W = F i dt = 1dt = π dt dr F i Note that Fv = Fv i ˆ = dt = 1 everywhere on the curve, so that dr dt W = 1 = π (the length of the path around the circle). Also note that F = y, x, z curlf = kˆ 3 everywhere in. The lesser curvature of the circular lines of force further away from the z axis is balanced 3 exactly by the increased transverse force, so that curl F is the same in all of. We shall see later (tokes theorem, page.4) that the work done is also the normal component of the curl integrated over the area enclosed by the closed curve. In this case W = Fn i ˆ A = kk ˆ i ˆ π 1 = π. ( ) ( ) ()

4 ENGI Line Integrals Page.4 Example.1.1 (continued) Example.1. Find the work done by F = x, y, z in moving around the curve (defined in parametric form by x = cos t, y = sin t, z =, t π ). F = x, y, z = cos t,sin t, = r dr dt d = cos t,sin t, = sin t,cos t, dt dr F i = cost sin t + sin t cost + = dt π W = dt = In this case, the force is orthogonal to the direction of motion at all times and no work is done.

5 ENGI Line Integrals Page.5 If the initial and terminal points of a curve are identical and the curve meets itself nowhere else, then the curve is said to be a simple closed curve. Notation: When is a simple closed curve, write Fdr i as Fdr i. F is a conservative vector field if and only if Fdr i = for all simple closed curves in the domain. Be careful of where the endpoints are and of the order in which they appear (the t1 t dr orientation of the curve). The identity F dt F dt t dt dr i i leads to the result t dt 1 Fdr i = Fdr i simple closed curves Another Application of Line Integrals: The Mass of a Wire Let be a segment (t t t 1 ) of wire of line density ρ (x, y, z). Then ( x, y, z) Δm ρ Δs First moments about the coordinate planes: ds ds m = ρ ds = ρ dt = ρ dt dt t1 dt t Δ M = r Δm ρ r Δs M = t1 t ρ r ds dt dt The location r of the centre of mass of the wire is = M r, where m t1 t ds 1 ds ds dr dx dy dz M = ρr dt, m = ρ dt and = = + + t dt t dt dt dt dt dt dt.

6 ENGI Line Integrals Page.6 Example.1.3 Find the mass and centre of mass of a wire (described in parametric form by x = cos t, y = sin t, z = t, π t π ) of line density ρ = z. Let c = cos t, s = sin t. dr r = cst,, = sc,,1 dt ds ( ) = s + c + 1 = dt [The shape of the wire is one revolution of a helix, aligned along the z axis, centre the origin.] ρ = z = t π π 3 π ds t ρ dt π π 3 π m = ρ ds = dt = t dt = m = 3 π 3 M t1 = ρ r = dt t π ds dt t c t s t 3 dt,, π x component: Integration by parts. ( ) tcdt= t s+ tc π ( ) tcdt= t s+ tc π π ( ) ( ) = π + π = 4π π

7 ENGI Line Integrals Page.7 Example.1.3 (continued) y component: For all integrable functions f (t) and for all constants a note that a a () f t dt = a ( ) if f t is an ODD function () f () t f t dt if is an EVEN function t sin t is an odd function π tsdt= π z component: t 3 is also an odd function π t dt = π 3 Therefore r M = 4π M 3 ˆ 6 = = 4π i = 3 m π π ˆi ˆi 6 The centre of mass is therefore at,, π

8 ENGI 543. Green s Theorem Page.8. Green s Theorem ome definitions: r ˆ i ĵ, a t b) is A curve on ú (defined in parametric form by ( t) = x( t) + y( t) closed iff (x(a), y(a)) = (x(b), y(b)). 1 for all t 1, t such that a < t 1 < t < b ; (that is, the curve neither touches nor intersects itself, except possibly at the end points). The curve is simple iff r( t ) r( t ) Example..1 Two simple curves: open closed Two non-simple curves: open closed

9 ENGI 543. Green s Theorem Page.9 Orientation of closed curves: A closed curve has a positive orientation iff a point r(t) moves around in an anticlockwise sense as the value of the parameter t increases. Example.. Positive orientation Negative orientation Let D be the finite region of ú bounded by. When a particle moves along a curve with positive orientation, D is always to the left of the particle. For a simple closed curve enclosing a finite region D of ú and for any vector function F = f1, f that is differentiable everywhere on and everywhere in D, Green s theorem is valid: f f1 Fdr i = da x D The region D is entirely in the xy-plane, so that the unit normal vector everywhere on D is k. Let the differential vector da = da k, then Green s theorem can also be written as ( ) ˆ Fdr i = F ikda = ( curl F ) i da D D f f Fdr i = f, f i dx, dy f dx+ f dy = da 1 1 and f f = = = T 1 det x det T x f1 f F 1 ( ) x D z component of F Green s theorem is valid if there are no singularities in D.

10 ENGI 543. Green s Theorem Page.1 Example..3 x F =, : r Green s theorem is valid for curve 1 but not for curve. There is a singularity at the origin, which curve encloses. Example..4 For F = x + y, x y and as shown, evaluate Fdr i. F dr = F dr + F i i idr + Fi dr PQ QR RP

11 ENGI 543. Green s Theorem Page.11 Example..4 (continued) F = x + y, x y Everywhere on the line segment from P to Q, y = x (and the parameter t is just x) d = x, x r r = 1, 1 and F =,x dx Fdr i = ( ( x ) ) dx = ( 4 x) dx = 4x x PQ = ( ) (8 4) = 4 Everywhere on the line segment from Q to R, y = + x d = x,+ x r r = 1,1 and F = x+, dx = x+ dx = xdx = x Fdr i (( ) ) QR = 4 = 4 Everywhere on the line segment from R to P, y = dr r = x, = 1, and F = x, x dx x Fdr i = ( x+ ) dx = xdx = RP = = Fdr i = =

12 ENGI 543. Green s Theorem Page.1 Example..4 (continued) OR use Green s theorem! det x y = ( x y) ( x+ y) = 1 1 = x f1 f everywhere on D = f f1 da = da x D By Green s theorem it then follows that D Fdr i =

13 ENGI 543. Green s Theorem Page.13 Example..5 Find the work done by the force F = x y, y in one circuit of the unit square. By Green s theorem, W f f = = 1 Fdr i x D da f f1 = ( y ) ( xy) = x x x The region of integration is the square < x < 1, < y < ( ) W = xda = x dy dx D = x dy dx = x[ y] dx = x( 1 ) dx 1 x 1 1 = = + = Therefore W = 1 The alternative method (using line integration instead of Green s theorem) would involve four line integrals, each with different integrands!

14 ENGI Path Independence Page.14.3 Path Independence Gradient Vector Fields: If F = φ, then F φ φ f =, f 1 = φyx φxy x x (provided that the second partial derivatives are all continuous). It therefore follows, for any closed curve and twice differentiable potential function φ that φidr Path Independence If F = φ (or F = φ ), then φ is a potential function for F. Let the path travel from point P o to point P 1 : φ φ φ Fdr i = φidr = dx + dy + dz = dφ x z [chain rule] P1 [ φ] φ( ) φ( ) = = P P 1 P which is independent of the path between the two points. Therefore work done difference in φ by = φ between endpoints of φ = φ( P) φ( P) = idr [work done = potential difference]

15 ENGI Path Independence Page.15 Domain A region Ω of is a domain if and only if 1) For all points P o in Ω, there exists a circle, centre P o, all of whose interior points are inside Ω; and ) For all points P o and P 1 in Ω, there exists a piecewise smooth curve, entirely in Ω, from P o to P 1. Example.3.1 Are these domains? { (x, y) y > } { (x, y) x } YE (but not simply connected) NO If a domain is not specified, then, by default, it is assumed to be all of ú.

16 ENGI Path Independence Page.16 When a vector field F is defined on a simply connected domain Ω, these statements are all equivalent (that is, all of them are true or all of them are false): F = φ for some scalar field φ that is differentiable everywhere in Ω; F is conservative; Fdr i is path-independent (has the same value no matter which path within Ω is chosen between the two endpoints, for any two endpoints in Ω); Fdr i = φend φstart (for any two endpoints in Ω); Fdr i = for all closed curves lying entirely in Ω; f f1 = everywhere in Ω; and x F = everywhere in Ω (so that the vector field F is irrotational). There must be no singularities anywhere in the domain Ω in order for the above set of equivalencies to be valid. Example.3. ( x y dx x y dy) where is any piecewise-smooth curve from Evaluate ( + ) + ( + 3 ) (, ) to (1, ). F = x + y, x+ 3y is continuous everywhere in Ω= f x f = 1 = 1 F is conservative and F = φ φ φ = x + y and = x+ 3y x 3 A potential function that has the correct first partial derivatives is φ = x + xy + y ( 1,) Fdr i = φ = ( ) ( + + ) (,) Therefore (( x y) dx ( x 3y ) dy) = 11

17 ENGI Path Independence Page.17 Example.3.3 (A ounterexample) Evaluate Fdr y x i, where F =, x + y x + y origin. and is the unit circle, centre at the F is continuous everywhere except (, ) Ω is not simply connected. [ Ω is all of except (, ).] f x y f = 1 x ( x + y ) = everywhere in Ω We cannot use Green s theorem, because F is not continuous everywhere inside (there is a singularity at the origin). Let c = cos t and s = sin t then r = cs, t< π r = sc, ( ) s c F =, = s, c c + s c + s ( s c ) dt 1dt t Fdr i = = = Therefore Note: Fdr i π π π, but Fdr i = π everywhere on,, where Arctan x Ω F = φ φ = + k y The problem is that the arbitrary constant k is ill-defined.

18 ENGI Path Independence Page.18 Example.3.3 (continued) Let us explore the case when k =. ontour map of φ = x Arctan y + We encounter a conflict in the value of the potential function φ. olution: hange the domain Ω to the simply connected domain except the Ω = non-negative x axis then the potential function φ can be well-defined, but no curve in Ω can enclose the origin.

19 ENGI urface Integrals - Projection Method Page.19.4 urface Integrals - Projection Method urfaces in ú 3 In ú 3 a surface can be represented by a vector parametric equation r = x uv, ˆi + y uv, ˆj + z uv, kˆ where u, v are parameters. Example.4.1 ( ) ( ) ( ) The unit sphere, centre O, can be represented by r θ, φ = sinθ cos φ, sinθ sin φ, cosθ ( ) θ π and φ < π declination azimuth If every vertical line (parallel to the z-axis) in ú 3 meets the surface no more than once, then the surface can also be parameterized as r x, y = x, y, f x, y or as z = f x, y Example.4. ( ) ( ) ( ) {( ) } z x y x y x y = 4,, + 4 is a hemisphere, centre O. A simple surface does not cross itself. If the following condition is true: { r( u1, v1) = r( u, v) ( u1, v1) = ( u, v) for all pairs of points in the domain} then the surface is simple. The converse of this statement is not true. This condition is sufficient, but it is not necessary for a surface to be simple. The condition may fail on a simple surface at coordinate singularities. For example, one of the angular parameters of the polar coordinate systems is undefined everywhere on the z-axis, so that spherical polar (,, ) and (,, π) both represent the same artesian point (,, ). Yet a sphere remains simple at its z-intercepts.

20 ENGI urface Integrals - Projection Method Page. Tangent and Normal Vectors to urfaces A surface is represented by r(u, v). Examine the neighbourhood of a point P o at r(u o, v o ). Hold parameter v constant at v o (its value at P o ) and allow the other parameter u to vary. This generates a slice through the two-dimensional surface, namely a onedimensional curve u containing P o and represented by a vector parametric equation r = r uv, with only one freely-varying parameter (u). ( o ) u v ( u v ) : r, : r, o ( u v) o If, instead, u is held constant at u o and v is allowed to vary, we obtain a different slice containing P o, the curve : r u, v. v ( ) On each curve a unique tangent vector can be defined. o ( o ) At all points along u, a tangent vector is defined by T r( uv, ) u = u [Note that this is not necessarily a unit tangent vector.] At P o the tangent vector becomes Tu P = ( r( uo, vo) ). o u imilarly, along the other curve v, the tangent vector at P o is Tv P = ( r( uo, vo) ). o v If the two tangent vectors are not parallel and neither of these tangent vectors is the zero vector, then they define the orientation of tangent plane to the surface at P o..

21 ENGI urface Integrals - Projection Method Page.1 A normal vector to the tangent plane is ˆi ˆj kˆ x z = det u u u x z v v v ( uo, vo) r r N = Tu Tv = u v ( uo, vo) = ( ) ( ) ( ) ( ) ( ) ( ) yz, zx, xy,,, uv, uv, uv,, ( uo, vo) x ( xy, ) u u where is the Jacobian det. ( uv, ) x v v artesian parameters With u = x, v = y, N = N ˆi + N ˆj + N kˆ 1 3 z = f (x, y), the components of the normal vector are: N 1 ( yz, ) ( xy, ) = = 1 f x f f = x N ( zx, ) ( xy, ) = = f x f 1 f = N 3 ( xy, ) ( xy, ) = = 1 1 = 1 a normal vector to the surface z = f (x, y) at (x o, y o, z o ) is f f N =,, + 1 x ( xo, yo)

22 ENGI urface Integrals - Projection Method Page. If the normal vector N is continuous and non-zero over all of the surface, then the surface is said to be smooth. Example.4.3 A sphere is smooth. A cube is piecewise smooth (six smooth faces) A cone is not smooth ( N is undefined at the apex) urface Integrals (Projection Method) This method is suitable mostly for surfaces which can be expressed easily in the artesian form z = f (x, y). The plane region D is the projection of the surface : f (r) = c onto a plane (usually the xy-plane) in a 1:1 manner. The plane containing D has a constant unit normal ˆn. N is any non-zero normal vector to the surface. Δ A = Δ cosα but cosα = Nn i ˆ N

23 ENGI urface Integrals - Projection Method Page.3 d = N D Nn i ˆ da and N g( r) d = g( r) da Nn i ˆ D For z = f (x, y) and D = a region of the xy-plane, z z N =,, 1 and nˆ = k ˆ x Nn i ˆ = 1 and z z g( r) d = g( r) da x D which is the projection method of integration of g(x, y, z) over the surface z = f (x, y). Advantage: Region D can be geometrically simple (often a rectangle in ). Disadvantage: Finding a suitable D (and/or a suitable 1:1 projection) can be difficult. You may have to split the surface into pieces (such as splitting a sphere into two hemispheres) in order to obtain separate 1:1 projections. The projection fails if part of the surface is vertical (such as a vertical cylinder onto the xy plane).

24 ENGI urface Integrals - Projection Method Page.4 Example.4.4 Evaluate zd, where the surface is the section of the cone z = x + y in the first octant, between z = and z = 4. z = x + y z z = x+ x z x x = = x z x + y By symmetry, z y y = = x + y y z z z x y z d = da = da = + 1 da = da x z z z Use the polar form for da : da = rdrdθ, r 4, θ π r = x + y = z π / 4 zd = r rdrdθ

25 ENGI urface Integrals - Projection Method Page.5 Example.4.4 (continued) π / π / 4 3 r zd = 1dθ r dr = θ 3 4 π 64 8 π 56 = = zd = 8π 3

26 ENGI urface Integrals - urface Method Page.6.5 urface Integrals - urface Method When a surface is defined in a vector parametric form r = r(u, v), one can lay a coordinate grid (u, v) down on the surface. r r A normal vector everywhere on is N =. u v d = d = N du dv Advantage: only one integral to evaluate r r g ( r) d = g ( r) du dv u v Disadvantage: it is often difficult to find optimal parameters (u, v). The total flux of a vector field F through a surface is ˆ r r Φ = Fd i = FN i d = Fi du dv u v r r (which involves the scalar triple product F i ). u v

27 ENGI urface Integrals - urface Method Page.7 Example.5.1: (same as Example.4.4, but using the surface method). Evaluate zd, where the surface is the section of the cone z = x + y in the first octant, between z = and z = 4. hoose a convenient parametric net: u = r = x + y = z and v = θ then r = r cos θ, r sin θ, r ( r 4, θ π ) r = cos θ, sin θ,1 r r and = r sin θ, r cos θ, θ ˆi ˆj kˆ N = ± cosθ sinθ 1 = ± rcos θ, rsin θ, r rsinθ rcosθ N = N = r cos θ + sin θ + 1 = r π / 4 zd zndrdθ r rdrdθ (as before) = = zd = 8π 3

28 ENGI urface Integrals - urface Method Page.8 Just as we used line integrals to find the mass and centre of mass of [one dimensional] wires, so we can use surface integrals to find the mass and centre of mass of [two dimensional] sheets. Example.5. Find the centre of mass of the part of the unit sphere (of constant surface density) that lies in the first octant. artesian equation of the sphere: x + y + z = 1; x>, y >, z > The radius of the sphere is r = 1. For the parametric net, use the two angular coordinates of the spherical polar coordinate r, θ, φ. system ( ) x y z = = = sinθ cosφ sinθ sinφ cosθ < θ < < φ < π π r = cosθ cos φ, cosθ sin φ, sinθ θ r and = sinθ sin φ, sinθ cos φ, φ ˆi ˆj ˆ N k = ± cosθ cosφ cosθ sinφ sinθ sinθ sinφ sinθ cosφ ( ) sin cos, sin sin, sin cos cos sin = ± θ φ θ φ θ θ φ + φ = ± sinθ sinθ cos φ, sinθ sin φ, cosθ = ± sinθ r The outward normal is clearly N = + sinθ r N = N = sinθ r = sinθ

29 ENGI urface Integrals - urface Method Page.9 Example.5. (continued) Mass: N φ m = ρ d = ρ dθ d π/ π/ ρ sinθ dθ dφ = π / π / π / π / sin d d [ cos ] [ ] = ρ θ θ φ = ρ θ φ ( 1 π )( ) = ρ + OR m = ρπ Note that the mass of a complete spherical shell of radius r and constant density ρ is 4π r ρ. Therefore the mass of one eighth of a shell of radius 1 is 4 ρπ ρπ =. 8 By symmetry, the three artesian coordinates of the centre of mass are all equal: x = y = z. Taking moments about the xy plane: π/ π/ ρ ρ ( cosθ) sinθ θ φ M = z d = d d π/ / / 1 π π cosθ ρ sin θ dθ dφ ρ 4 [ φ] = = 1 1 π 1 πρ 1 = ρ + = = 4 4 m M 1 z = = m Therefore the centre of mass is at ( x, y, z ) = ( 1, 1, 1 ) π /

30 ENGI urface Integrals - urface Method Page.3 Example.5.3 Find the flux of the field F = x, y, z the first octant. across that part of x + y + z = 8 that lies in The artesian coordinates x, y will serve as parameters for the surface: r = x, y,8 x y r = 1,, 1 x r and =, 1, ˆi ˆj kˆ N = ± 1 1 = ± 1,,1 1 N = 1,, 1 hoose N to point outwards. Range of parameter values: In the xy plane: x 8 and y 4 But the area is a triangle, not a rectangle, so these inequalities do not provide the correct limits for the inner integral.

31 ENGI urface Integrals - urface Method Page.31 Example.5.3 (continued) Fd F d FN N da FN i da Net flux = Φ = ( ˆ i = N = i )( ) = (where da = dx dy) FN i = x, y, 8 x y i 1,,1 = x+ y 8+ x+ y = x+ y 4 ( ) ( ) ( 8 y) 4 8 y ( ) Φ = x + y 4 dxdy 4 8 y x = + xy 4x dy x = 4 = + ( 8 y) y 4( 8 y) ( + + ) dy 4 ( y y y y y) = dy 3 4 y 64 = 4 ( 4y y ) dy = 4 y = 4 3 ( ) 3 3 Therefore the net flux is Φ = 18 3 The iteration could be taken in the other order: x/ ( ) Φ = x + y 4 dy dx 4 x/ = x y + xy 4y dx 8 y = 1 18 = = ( 4x x ) dx = = 3

32 ENGI urface Integrals - urface Method Page.3 Example.5.4 Find the total flux Φ of the vector field F= z kˆ through the simple closed surface + + = 1 x y z a b c Use the parametric grid ( θ, φ ), such that the displacement vector to any point on the ellipsoid is r = asinθ cos φ, bsinθ sin φ, ccosθ This grid is a generalisation of the spherical polar coordinate grid and covers the entire surface of the ellipsoid for θ π, φ < π. One can verify that x= asinθ cos φ, y = bsinθ sin φ, z = ccosθ does lie on the ellipsoid x y z + + = 1 for all values of ( θ, φ ) : a b c x y z a sin θ cos φ b sin θsin φ c cos θ + + = + + a b c a b c = sin θ cos φ + sin θsin φ + cos θ = sin θ cos φ + sin φ + cos θ = sin θ + cos θ = 1 θ and φ ( ) The tangent vectors along the coordinate curves φ = constant and θ = constant are dr acosθ cos φ, bcosθ sin φ, csin dθ = θ and d r asinθ sin φ, bsinθ cos φ, dφ =. The normal vector at every point on the ellipsoid follows: ˆi ˆj kˆ dr dr N = = acosθ cosφ bcosθ sinφ csinθ dθ dφ asinθ sinφ bsinθ cosφ ( φ) = bc θ φ ac θ φ ab θ θ φ+ sin cos, sin sin, sin cos cos sin (and this vector points away from the origin).

33 ENGI urface Integrals - urface Method Page.33 Example.5.4 (continued) On the ellipsoid, F= zkˆ = ccosθ k ˆ FN i = c cosθ absinθ cosθ = abcsinθ cos θ ( ) The total flux of F through the surface is therefore π π π π Φ = Fd i = FN i dθ dφ = abc sinθ cos θ dθ dφ Let u = cos θ, then du = sin θ dθ and θ = u = +1, θ = π u = π 1 π u abc 1 dφ u du abc φ Fd i = = = abc( π ) Φ = 4π abc 3

34 ENGI urface Integrals - urface Method Page.34 For vector fields F(r), Line integral: Fdr i urface integral: ( ) ( ) r r F r id = F r ind = FiNdu dv = ± Fi du dv u v On a closed surface, take the sign such that N points outward. ome ommon Parametric Nets o o 1) The circular plate ( x x ) ( y y ) + a in the plane z = z o. Let the parameters be r, θ where < r a, θ < π x = x o + r cos θ, y = y o + r sin θ, z = z o ˆi ˆj kˆ r r N = ± = ± cosθ sinθ = ± r ˆ k r θ r sinθ r cosθ o o ) The circular cylinder ( x x ) ( y y ) + = a with z o z z 1. Let the parameters be z, θ where z o z z 1, θ < π x = a cos θ, y = a sin θ, z = z ˆi ˆj kˆ r r N = ± = ± 1 = ± acosθ ˆ asinθ ˆ i j z θ asinθ acosθ Outward normal: N = acosθ ˆi + asinθ ˆj ( ) 3) The frustrum of the circular cone w w = a ( u u ) + ( v v ) w w w and w w 1 o 1 o o o. Let the parameters here be r, θ where where w1 wo w w r o, a a θ < π x = u = uo + r cos θ, y = v = vo + rsin θ, z = w = wo + ar ˆi ˆj kˆ r r N = ± = ± cosθ sinθ a r θ r sinθ r cosθ ( cosθ) ( sinθ) = ± ar ˆ + ar ˆ + r ˆ i j k Outward normal: N = ar cosθ ˆi + ar sinθ ˆj r kˆ

35 ENGI urface Integrals - urface Method Page.35 4) The portion of the elliptic paraboloid ( ) ( ) o o o with o 1 z z = a x x + b y y z z z z Let the parameters here be r, θ where z1 zo z zo r, θ < π a cos θ + b sin θ a cos θ + b sin θ x = x o + r cos θ, y = y o + r sin θ, z = z o + r (a cos θ + b sin θ) ˆi ˆj kˆ r r N = ± = ± cosθ sinθ r( a cos θ + b sin θ) r θ rsinθ r cosθ r b a sinθ cosθ ( ) = ± ( ar cosθ) ˆ + ( br sinθ) ˆ + rˆ i j k Outward normal: = ( cosθ) + ( sinθ) j N ar ˆi br ˆ rkˆ. 5) The surface of the sphere ( x x ) + ( y y ) + ( z z ) = a Let the parameters here be θ, φ where θ π, φ < π x = x o + a sin θ cos φ, y = y o + a sin θ sin φ, z = z o + a cos θ ˆ ˆ ˆ i j k r r N = ± = ± acosθ cosφ acosθ sinφ asinθ θ φ asinθ sinφ asinθ cosφ a ( ) i ( ) j ( ) sinθ sinθ cosφ ˆ sinθ sinφ ˆ cosθ ˆ k N = a sinθ sinθ cosφ ˆ + sinθ sinφ ˆ + cosθ) ˆ i j k = ± + + Outward normal: ( ) ( ) ( 6) The part of the plane ( x x ) + B( y y ) + ( z z ) = A in the first octant with A, B, > and Ax o + By o + z o >. Let the parameters be x, y where x Ax By z By ; y Ax By z A B ˆi ˆj kˆ r r A 1 A ˆ Bˆ ˆ N = ± = ± = ± + + x y i j k 1 B

36 ENGI Gauss, tokes; Potential Functions Page.36.6 Theorems of Gauss and tokes; Potential Functions Gauss Divergence Theorem Let be a piecewise-smooth closed surface enclosing a volume V in ú 3 and let F be a vector field. Then the net flux of F out of V is Fd i = F d N. But the divergence of F is a flux density, or an outflow per unit volume at a point. Integrating div F over the entire enclosed volume must match the net flux out through the boundary of the volume V. Gauss divergence theorem then follows: Fd i = ifdv V Example.6.1 (Example.5.4 repeated) Find the total flux Φ of the vector field F= z kˆ through the simple closed surface + + = 1 x y z a b c Use Gauss Divergence Theorem: Fd i = V div FdV F is differentiable everywhere in 3, so Gauss divergence theorem is valid. ( ˆ div F = i zk) =,, i,, z = = 1 x z 4π abc div F dv = 1 dv = V = the volume of the ellipsoid! 3 V V Therefore Φ = Fd i = 4π abc 3 In fact, the flux of by that surface. F= z kˆ through any simple closed surface is just the volume enclosed

37 ENGI Gauss, tokes; Potential Functions Page.37 Example.6. Archimedes Principle Gauss divergence theorem may be used to derive Archimedes principle for the buoyant force on a body totally immersed in a fluid of constant density ρ (independent of depth). Examine an elementary section of the surface of the immersed body, at a depth z < below the surface of the fluid: The pressure at any depth z is the weight of fluid per unit area from the column of fluid above that area. Therefore pressure = p = ρ gz ρ g is the weight of the column z is the height of the column (note z < ). The normal vector N to is directed outward, but the hydrostatic force on the surface (due to the pressure p) acts inward. The element of hydrostatic force on Δ is ( pressure) ( area) ( direction) = ( ρgz)( Δ)( N ˆ ) = ( + ρgzδ) N ˆ The element of buoyant force on Δ is the component of the hydrostatic force in the direction of k (vertically upwards): ( ρ gz ˆ ) + Δ N i kˆ Define F= ρ gzkˆ and d = N ˆ d. umming over all such elements Δ, the total buoyant force on the immersed object is ρ g z kn ˆiˆ d = Fd i = i FdV (by the Gauss Divergence Theorem) V

38 ENGI Gauss, tokes; Potential Functions Page.38 Example.6. Archimedes Principle (continued) ( ˆ = i ρgzk) dv =,, i,, ρgz dv x z V = ρgdv provided ( ρg) z V = weight of fluid displaced V Therefore the total buoyant force on an object fully immersed in a fluid equals the weight of the fluid displaced by the immersed object (Archimedes principle).

39 ENGI Gauss, tokes; Potential Functions Page.39 Gauss Law A point charge q at the origin O generates an electric field q q E = r = r ˆ 3 4πε r 4πε r If is a smooth simple closed surface not enclosing the charge, then the total flux through is Ed i = iedv (Gauss divergence theorem) V 1 But Example showed that i = 3 r r. r Therefore Ed i =. There is no net outflow of electric flux through any closed surface not enclosing the source of the electrostatic field. If does enclose the charge, then one cannot use Gauss divergence theorem, because ie is undefined at the origin. Remedy: onstruct a surface 1 identical to except for a small hole cut where a narrow tube T connects it to another surface, a sphere of radius a centre O and entirely inside. Let * = 1 T (which is a simple closed surface), then O is outside *! Applying Gauss divergence theorem to *, Ed i = iedv = V * * Ed i + Ed i + Ed i = 1 T

40 ENGI Gauss, tokes; Potential Functions Page.4 Gauss Law (continued) As the tube T approaches zero thickness, Ed i and therefore T Ed i Ed i 1 But is a sphere, centre O, radius a. Using parameters ( θ, φ ) on the sphere, r = a sinθ cos φ, sinθ sin φ, cosθ r r Finding, as before leads to θ φ N =±asinθ r. But the outward normal to actually points towards O. N = a sinθ r on the sphere q and E = 3 r everywhere on. 4πε a Also r = arˆ r i r = a q qsinθ qsinθ EN i = ri 3 ( asinθ r) = rr i = 4πε a 4πε a 4πε Recall that d = Nd = N N dθ dφ = N dθ dφ q π π Ed i = EN i dθ dφ = sinθ dθ dφ 4πε q q π [ ] [ ] π cosθ φ ( 1 1)( π ) = = + + = 4πε 4πε ε q Ed i = Ed i = + ε 1 q

41 ENGI Gauss, tokes; Potential Functions Page.41 Gauss Law (continued) But, as ( ) a O, 1 The surface 1 looks more and more like the surface as the tube T collapses to a line and the sphere collapses into a point at the origin. Gauss law then follows. Gauss law for the net flux through any smooth simple closed surface, in the presence of a point charge q at the origin, then follows: Ed i q if encloses = ε otherwise O

42 ENGI Gauss, tokes; Potential Functions Page.4 Example.6.3 Poisson s Equation The exact location of the enclosed charge is immaterial, provided it is somewhere inside the volume V enclosed by the surface. The charge therefore does not need to be a concentrated point charge, but can be spread out within the enclosed volume V. Let the charge density be ρ (x, y, z), then the total charge enclosed by is q = ρ dv V Gauss law q Ed i = ε Apply Gauss divergence theorem to the left hand side, substitute for q on the right hand side and assume that the permittivity ε is constant throughout the volume: ie dv = V V ρ dv ε E ρ dv = V i ε V This identity will hold for all volumes V only if the integrand is zero everywhere. Poisson s equation then follows: ie = ρ ε ρ E = V and i V = V V = ε This reduces to Laplace s equation V = when ρ.

43 ENGI Gauss, tokes; Potential Functions Page.43 tokes Theorem Let F be a vector field acting parallel to the xy-plane. components by F = f ˆ ˆ,, 1i + f j = f1 f. Then ˆi ˆj kˆ 1 ( ) ˆ f f F = F i k = x y z x y f1 f Green s theorem can then be expressed in the form ˆ Fdr i = Fk i da D Represent its artesian Now let us twist the simple closed curve and its enclosed surface out of the xy-plane, so that the normal vector k is replaced by a more general normal vector N. If the surface (that is bounded in ú 3 by the simple closed curve ) can be represented by z = f (x, y), then a normal vector at any point on is N z z =,, x 1 is oriented coherently with respect to if, as one travels along with N pointing from one s feet to one s head, is always on one s left side. The resulting generalization of Green s theorem is tokes theorem: Fdr i = FN i d = ( curl F ) i d This can be extended further, to a non-flat surface with a non-constant normal vector N. Example.6.4 Find the circulation of F = xyz, xz, e xy around : the unit square in the xz-plane. Because of the right-hand rule, the positive orientation around the square is OGHJ (the y axis is directed into the page). In the xz plane y = F =, xz,1

44 ENGI Gauss, tokes; Potential Functions Page.44 Example.6.4 (continued) omputing the line integral around the four sides of the square: OG :,, t ( t 1) d r r = =,,1 dt and,,1 d r F = F i =,,1 i,,1 = 1 dt Fdr i = dt = = 1 = 1 OG [] t Fdr =, Fdr i i = 1 and Fdr i = In a similar way (Problem et 6 Question 6), it can be shown that GH HJ JO Fdr i = = OR use tokes theorem: On D F ˆi ˆj kˆ = = x,, z x z xz 1 da = ˆ j da FdA i = x,, z i,1, da = da FdA i = Fdr i = D Note that this vector field F is not conservative, because F.

45 ENGI Gauss, tokes; Potential Functions Page.45 Domain A region Ω of ú 3 is a domain if and only if 1) For all points P o in Ω, there exists a sphere, centre P o, all of whose interior points are inside Ω; and ) For all points P o and P 1 in Ω, there exists a piecewise smooth curve, entirely in Ω, from P o to P 1. A domain is simply connected if it has no holes. Example.6.5 Are these regions simply-connected domains? The interior of a sphere. The interior of a torus. The first octant. YE NO YE On a simply-connected domain the following statements are either all true or all false: F is conservative. F φ F Fdr i = φ( Pend ) φ( Pstart ) - independent of the path between the two points. Fdr i = Ω Example.6.6 Find a potential function φ (x, y, z) for the vector field F = x, y,z. First, check that a potential function exists at all: ˆi ˆj kˆ curl F = F = =,, = x z x y z Therefore F 3 is conservative on.

46 ENGI Gauss, tokes; Potential Functions Page.46 Example.6.6 (continued) F = φ = φ, φ, φ x z φ x = = + ( ) x φ x g y, z φ g = + =, = + ( ) φ = x + y + h z y g( y z) y h( z) φ dh = + + = z h( z) = z + c z dz φ = x + y + z +c We have a free choice for the value of the arbitrary constant c. hoose c =, then ( x, yz, ) x y z r φ = + + =

47 ENGI Gauss, tokes; Potential Functions Page.47 Maxwell s Equations (not examinable in this course) We have seen how Gauss and tokes theorems have led to Poisson s equation, relating the electric intensity vector E to the electric charge density ρ : ρ ie = ε Where the permittivity is constant, the corresponding equation for the electrical flux density D is one of Maxwell s equations: id = ρ. Another of Maxwell s equations follows from the absence of isolated magnetic charges (no magnetic monopoles): ih = ib =, where H is the magnetic intensity and B is the magnetic flux density. Faraday s law, connecting electric intensity with the rate of change of magnetic flux density, is = Edr i t Bd i. Applying tokes theorem to the left side produces B E = t Ampère s circuital law, I = Hdl i, leads to H = J + J d, where the current density is J = σ E= ρv v, σ is the conductivity, ρ V is the volume charge D density; and the displacement charge density is Jd = t The fourth Maxwell equation is D H = J + t The four Maxwell s equations together allow the derivation of the equations of propagating electromagnetic waves. END OF HAPTER