Dr. Allen Back. Nov. 21, 2014

Transcription

1 Dr. Allen Back of Nov. 21, 2014

2 The most important thing you should know (e.g. for exams and homework) is how to setup (and perhaps compute if not too hard) surface integrals, triple integrals, etc. But occasionally symmetries can save a considerable amount of work. of

3 Problem: Calculate the following surface integral: y 2 z 2 ds S for S the part of the paraboloid x = y 2 + z 2 with 1 x 5. of

4 of In this case the symmetry of interchanging y and z is an orthogonal (and so distance/angle preserving transformation of R 3.) (Algebraically this is the transformation y z, z y, x x.) And this transformation preserves the region we are integrating over. At the Riemann sum level, for every little area S near (x, y, z) S there is a matching area near (x, z, y) whose contributions at the Riemann sum level cancel. So the value of the surface integral is 0.

5 If you use this argument on homework or an exam, please 1 Describe the symmetry you are using. 2 Mention that the region of integration is symmetric under this symmetry. 3 State clearly what cancellation or simplification results. (In other words, don t just say by symmetry this is 0 unless it really is obvious.) of

6 Using the same potion of a paraboloid, here are two more examples where we can use symmetry to argue for an answer of 0 : 1 (0, y, z) ˆn ds 2 S S yz ds of

7 Definition: A path c : [a, b] R 3 is said to be a flow line (or integral curve) of a vector field F (x, y, z) if for all t [a, b]. c (t) = F (c(t)) of

8 Flow lines always exist at least locally because of existence and uniqueness theorems for ordinary differential equations. of

9 At the end of Math 2210, you studied how to use eigenvalues/eigenvectors/diagonalization/(jordan Normal form) to solve linear differential equations. of

10 Show that c(t) = (x 0 e t, y 0 e t ) is a flow line of F (x, y) = (x, y). of

11 Show that c(t) = [ ] x(t) = y(t) [ cos t ] [ ] sin t x0 sin t cos t y 0 is a flow line of F (x, y) = ( y, x). of

12 We will discuss the meaning of these when we get to the 3d integral theorems (8.2 and 8.4.) For a vector field F (x, y, z) = (P, Q, R) div( F ) is a (scalar) function; i.e. a number at each point. curl( F ) is another vector field; i.e. a vector at each point. of

13 for F (x, y, z) = (P, Q, R). div( F ) = P x + Q y + R z ( F ). of

14 for F (x, y, z) = (P, Q, R). î ĵ ˆk curl( F ) = x y z P Q R ( F ). of

15 Please take a look at the table of vector analysis identities on page 255. Many are just the 1-variable product or chain rules. 1 (fg) 2 div(f F ) 3 curl(f F ) 4 curl( G F ) would be more complicated. of

16 (There is a more complicated identity a ( b c) = ( a c) b ( a b) c useful in E & M, but the related curl identity (useful in some proofs of Stokes theorem is tricky to get right.) of

17 very important ones are 1 div(curl F ) = 0. 2 curl( f ) = 0. (for C 2 functions and vector fields.) They all are consequences of f xy = f yx for such functions. of

18 Problem: Let r = (x, y, z), and so r = r = x 2 + y 2 + z 2. Show that a vector field of the form f (r) r satisfies F = 0. (In physics, this is called a central force field.) of

19 of Green s theorem says that for simple closed (piecewise smooth) curve C whose inside is a region R, we have Q P(x, y) dx + Q(x, y) dy = x P dx dy y C as long as the vector field F (x, y) = (P(x, y), Q(x, y)) is C 1 on the set R and C is given its usual inside to the left orientation. R

20 For a y-simple region P y dy dx = R is fairly easily justified. C= R P dx. of

21 For an x-simple region Q x dx dy = R C= R Q dy. of

22 So for a region that is both y-simple and x-simple we have Green: P(x, y) dx + Q(x, y) dy = Q y P x dx dy. R R of

23 Intuitively, why the different signs? +Q x yet P y. And why this combination of Q x P y? of

24 Intuitively, why the different signs? +Q x yet P y. And why this combination of Q x P y? Think about the line integral around a small rectangle with sides x and y. If you assume (or justify) that evaluating the vector field in the middle of each edge gives a good approximation in the line integral, then Q x P y emerges quite naturally. of

25 If you can cut a region into two pieces where we know Green s holds on each piece (e.g. a ring shaped region), then Green also holds for the entire region. (Because the line integrals over the cuts show up twice with opposite signs (think inside to the left ) and cancel. of

26 So in the end, Green s theorem holds for regions whose boundaries include several closed curves (multiply-connected regions) as long as we orient each boundar curve according to the inside to the left rule. of

27 1 2 C x dy y dx oriented counterclockwise. for C the boundary of the ellipse x y = 1 of

28 Let F = 1 x 2 ( y, x). + y 2 If C 1 and C 2 are two simple closed curves enclosing the origin (and oriented with the usual inside to the left), can you say whether one of C 1 F d s and C 2 F d s is bigger than the other? of

29 Integration of a conservative vector field cartoon. of

30 Green s Theorem cartoon. of

31 Stokes Theorem cartoon. of

32 of Both sides of Stokes involve integrals whose signs depend on the orientation, so to have a chance at being true, there needs to be some compatibility between the choices. The rule is that, from the positive side of the surface, (i.e. the side chosen by the orientation), the positive direction of the curve has the inside of the surface to the left. As with all orientations, this can be expressed in terms of the sign of some determinant. (Or in many cases in terms of the sign of some combination of dot and cross products.)

33 Problem: Let S be the portion of the unit sphere x 2 + y 2 + z 2 = 1 with z 0. Orient the hemisphere with an upward unit normal. Let F (x, y, z) = (y, x, z). Calculate the value of the surface integral F ˆn ds. S of

34 Theorem field cartoon. of

35 of The surface integral side of depends on the orientation, so there needs to be a choice making the theorem true. The rule is that the normal to the surface should point outward from the inside of the region. (For the 2d analogue of (really an application of ) F ˆn = (P x + Q y ) dx dy C inside we also use an outward normal, where here C must of course be a closed curve.

36 Problem: Let W be the solid cylinder x 2 + y 2 3 with 1 z 5. Let F (x, y, z) = (x, y, z). Find the value of the surface integral F ˆn ds. W of

37 of This is not worth memorizing! If one rotates about the z-axis the path (curve) z = f (x) in the xz-plane for 0 a x b, one obtains a surface of revolution with a parametrization of and ds =? Φ(u, v) = (u cos v, u sin v, f (u))

38 of ds = u 1 + (f (u)) 2 du dv. of

39 This is not worth memorizing! For the graph parametrization of z = f (x, y), and ds =? Φ(u, v) = (u, v, f (u, v)) of

40 ds = 1 + fu 2 + fv 2 du dv. of

41 of For such a graph, the normal to the surface at a point (x, y, f (x, y)) (this is the level set z f (x, y) = 0) is so we can see that ( f x, f y, 1) 1 cos γ = 1 + fx 2 + fy 2 determines the angle γ of the normal with the z-axis. And so at the point (u, v, f (u, v)) on a graph, ds = 1 cos γ du dv. (Note that du dv is essentially the same as dx dy here.)

42 Picture of T u, T v for a Lat/Long Param. of the Sphere. of

43 Basic Parametrization Picture of

44 of Parametrization Φ(u, v) = (x(u, v), y(u, v), z(u, v)) Tangents T u = (x u, y u, z u ) T v = (x v, y v, z v ) Area Element ds = T u T v du dv Normal N = T u T v Unit normal ˆn = ± T u T v T u T v (Choosing the ± sign corresponds to an orientation of the surface.)

45 of Two Kinds of Integral of a scalar function f (x, y, z) : f (x, y, z) ds Integral of a vector field F (x, y, z) : F (x, y, z) ˆn ds. S S

46 of Integral of a scalar function f (x, y, z) calculated by f (x, y, z) ds = f (Φ(u, v)) T u T v du dv S D where D is the domain of the parametrization Φ. Integral of a vector field F (x, y, z) calculated by F (x, y, z) ˆn ds S ( Tu = ± F (Φ(u, v)) ) T v D T u T T u T v du dv v where D is the domain of the parametrization Φ.

47 3d Flux Picture of

48 of The preceding picture can be used to argue that if F (x, y, z) is the velocity vector field, e.g. of a fluid of density ρ(x, y, z), then the surface integral S ρ F ˆn ds (with associated Riemann Sum ρ(x i, yj, zk ) F (xi, yj, zk ) ˆn(x i, yj, zk ) S ijk) represents the rate at which material (e.g. grams per second) crosses the surface.

49 From this point of view the orientation of a surface simple tells us which side is accumulatiing mass, in the case where the value of the integral is positive. of

50 2d Flux Picture of There s an analagous 2d Riemann sum and interp of F ˆn ds. C

51 of

52 Problem: Calculate S F (x, y, z) ˆn ds for the vector field F (x, y, z) = (x, y, z) and S the part of the paraboloid z = 1 x 2 y 2 above the xy-plane. Choose the positive orientation of the paraboloid to be the one with normal pointing downward. of

53 Problem: Calculate the surface area of the above paraboloid. of

54 Problem: Find S F ˆn ds for F (x, y, z) = (0, yz, z 2 ) and S the portion of the cylinder y 2 + z 2 = 1 with 0 x 1, z 0, and the positive orientation chosen to be a radial outward (from the axis of the cylinder) normal. of

55 Problem: Letting r denote the vector field (x, y, z), find the value of the surface integral r ˆn ds r 3 S where S is the sphere of radius R about the origin, oriented with an outward normal. of

56 Spherical coordinates (letting ρ = R be constant) gives a natural parametrization of the sphere of radius R centered at the origin. It is geometrically to be expected (and you ve done some verifications like this at least in look at problems) that the Jacobian result for spherical coordinates translates to the area element on such a sphere to be ds = R 2 sin φ dφ dθ. of

57 Paraboloid z = x 2 + 4y 2 The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. of

58 Paraboloid z = x 2 + 4y 2 of The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Parameters u and v just different names for x and y resp.

59 Paraboloid z = x 2 + 4y 2 of The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Use this idea if you can t think of something better.

60 Paraboloid z = x 2 + 4y 2 The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. of

61 Paraboloid z = x 2 + 4y 2 The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. of Note the curves where u and v are constant are visible in the wireframe.

62 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. of

63 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of

64 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of

65 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Algebraically, we are rescaling the algebra behind polar coordinates where x = r cos θ y = r sin θ leads to r 2 = x 2 + y 2.

66 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Here we want x 2 + 4y 2 to be simple. So x = 2r cos θ y = r sin θ will do better.

67 Paraboloid z = x 2 + 4y 2 A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Here we want x 2 + 4y 2 to be simple. So x = y = 2r cos θ r sin θ will do better. Plug x and y into z = x 2 + 4y 2 to get the z-component.

68 Parabolic Cylinder z = x 2 Graph parametrizations are often optimal for parabolic cylinders. of

69 Parabolic Cylinder z = x 2 Φ(u, v) =< u, v, u 2 > of

70 Parabolic Cylinder z = x 2 Φ(u, v) =< u, v, u 2 > of

71 Parabolic Cylinder z = x 2 Φ(u, v) =< u, v, u 2 > of One of the parameters (v) is giving us the extrusion direction. The parameter u is just being used to describe the curve z = x 2 in the zx plane.

72 Elliptic Cylinder x 2 + 2z 2 = 6 The trigonometric trick is often good for elliptic cylinders of

73 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 3 2 cos v, u, 3 sin v >=< 6 cos v, u, 3 sin v > of

74 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > of

75 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > of

76 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > of

77 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > of What happened here is we started with the polar coordinate idea x = r cos θ z = r sin θ but noted that the algebra wasn t right for x 2 + 2z 2 so shifted to x = 2r cos θ z = r sin θ

78 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > of x = z = 2r cos θ r sin θ makes the left hand side work out to 2r 2 which will be 6 when r = 3.

79 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids. of

80 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids. Φ(u, v) =< 2 sin u cos v, 4 2 sin u sin v, 3 cos u > of

81 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 Φ(u, v) =< 2 sin u cos v, 2 sin u sin v, 4 3 cos u > of

82 Hyperbolic Cylinder x 2 z 2 = 4 You may have run into the hyperbolic functions of cosh x = sinh x = ex + e x 2 ex e x 2

83 Hyperbolic Cylinder x 2 z 2 = 4 of You may have run into the hyperbolic functions cosh x = sinh x = ex + e x 2 ex e x Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic version cosh 2 θ sinh 2 θ = 1 leads to the nicest hyperbola parameterizations. 2

84 Hyperbolic Cylinder x 2 z 2 = 4 Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic version cosh 2 θ sinh 2 θ = 1 leads to the nicest hyperbola parameterizations. of Φ(u, v) =< 2 sinh v, u, 2 cosh v >

85 Hyperbolic Cylinder x 2 z 2 = 4 Φ(u, v) =< 2 sinh v, u, 2 cosh v > of

86 Saddle z = x 2 y 2 The hyperbolic trick also works with saddles of

87 Saddle z = x 2 y 2 Φ(u, v) =< u cosh v, u sinh v, u 2 > of

88 Saddle z = x 2 y 2 Φ(u, v) =< u cosh v, u sinh v, u 2 > of

89 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 The spherical coordinate idea for ellipsoids with sin φ replaced by cosh u works well here. of

90 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 Φ(u, v) =< cosh u cos v, cosh u sin v, sinh u > of

91 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 of

92 Hyperboloid of 2-Sheets x 2 + y 2 z 2 = 1 Φ(u, v) =< sinh u cos v, sinh u sin v, cosh u > of

93 Hyperboloid of 2-Sheets x 2 + y 2 z 2 = 1 of

94 Top Part of Cone z 2 = x 2 + y 2 So z = x 2 + y 2. of

95 Top Part of Cone z 2 = x 2 + y 2 So z = x 2 + y 2. The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u > of

96 Top Part of Cone z 2 = x 2 + y 2 So z = x 2 + y 2. The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u > of

97 Mercator Parametrization of the Sphere For 0 v, 0 u 2π Φ(u, v) = (sech(v) cos u, sech(v) sin u, tanh(v)). (Note tanh 2 (v) + sech 2 (v) = 1) of

98 ctionn of