ENGI 4430 Surface Integrals Page and 0 2 r

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1 ENGI 4430 Surface Integrals Page Surface Integrals - Projection Method Surfaces in 3 In 3 a surface can be represented by a vector parametric equation r x u, v ˆi y u, v ˆj z u, v k ˆ where u, v are parameters. Example 9.01 The unit sphere, centre O, can be represented by sin cos, sin sin 0 and 0 2 r cos If every vertical line (parallel to the z-axis) in 3 meets the surface no more than once, then the surface can also be parameterized as x r x, y y or as z f x, y f x, y Example 9.02 z x y x y x y ,, 4 is a A simple surface does not cross itself. If the following condition is true: r u, v r u, v u, v 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 (2, 0, 0) and (2, 0, ) both represent the same Cartesian point (0, 0, 2). Yet a sphere remains simple at its z-intercepts.

2 ENGI 4430 Surface Integrals Page 9.02 Tangent and Normal Vectors to Surfaces A surface S is represented by uv, u, v r 0 0. Hold parameter v constant at 0 r. Examine the neighbourhood of a point P 0 at v (its value at P 0 ) and allow the other parameter u to vary. This generates a slice through the two-dimensional surface, namely a one-dimensional curve C containing P 0 and represented by a vector parametric equation uv, o u r r with only one freely-varying parameter (u). If, instead, u is held constant at u 0 and v is allowed to vary, we obtain a different slice containing 0 P, the curve C : u, v v r. On each curve a unique tangent vector can be defined. o At all points along C u, a tangent vector is defined by Tu r uv, o. u [Note that this is not necessarily a unit tangent vector.] At 0 T u u v u v Po u r u r. P the tangent vector becomes, o, o Similarly, along the other curve C v, the tangent vector at P o is Tv r uo, vo. Po 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 0. Po

3 ENGI 4430 Surface Integrals Page 9.03 A normal vector to the tangent plane is y, z z, x x, y u, v u, v u, v T uo, vo, x x xy, u v where is the Jacobian det. uv, y y u v Cartesian parameters u x, v y, z f x, y, the components of the normal vector With N N ˆi N ˆj N k ˆ are: N 1 yz, xy, N 2 zx, xy, N 3 xy, xy, a normal vector to the surface z f x, y at,, x y z is N f f 1 x y T xo, yo

4 ENGI 4430 Surface Integrals Page 9.04 If the normal vector N is continuous and non-zero over all of the surface S, then the surface is said to be smooth. Example 9.03 A sphere is smooth. A cube is A cone is Surface Integrals (Projection Method) This method is suitable mostly for surfaces which can be expressed easily in the Cartesian form z = f (x, y). The plane region D is the projection (or shadow) of the surface plane (usually the xy-plane) in a 1:1 manner. S : f r c onto a The plane containing D has a constant unit normal ˆn. N is any non-zero normal vector to the surface S.

5 ENGI 4430 Surface Integrals Page 9.05 Surface Integrals (Projection Method) (continued) For z f x, y and D = a region of the xy-plane, z z N 1 and ˆ ˆ x y n k Nn ˆ 1 and S T 2 2 z z g r ds g r 1 da x y D which is the projection method of integration of g x, y, z over the surface z f x, y Advantage:. Disadvantage:

6 ENGI 4430 Surface Integrals Page 9.06 Example 9.04 Evaluate S z ds, where the surface S is the section of the cone octant, between z = 2 and z = z x y in the first

7 ENGI 4430 Surface Integrals Page 9.07 Example 9.04 (continued) Flux through a Surface (Projection Method) g r F (the normal component of vector field F, that is, F resolved in the Set N direction of the normal N to the surface S), then proceed as before: where N z z x y 1 T

8 ENGI 4430 Surface Integrals Page 9.08 Example 9.05 Find the flux due to the vector field origin. F r r through the sphere S, radius 2, centre the

9 ENGI 4430 Surface Integrals Page 9.09 Example 9.05 (continued)

10 ENGI 4430 Surface Integrals Page 9.10 Surface Integrals - Surface Method When a surface S is defined in a vector parametric form uv, coordinate grid (u, v) down on the surface S. r r A normal vector everywhere on S is N. u v r r, one can lay a Advantage: only one integral to evaluate S r r g r ds g r du dv u v Disadvantage: it is often difficult to find optimal parameters (u, v). S The total flux of a vector field F through a surface S is ˆ r r F ds F N ds F du dv u v S S r r (which involves the scalar triple product F ). u v S

11 ENGI 4430 Surface Integrals Page 9.11 Example 9.06: (same as Example 9.04, but using the surface method). Evaluate S z ds, where the surface S is the section of the cone octant, between z = 2 and z = z x y in the first Choose a convenient parametric net:

12 ENGI 4430 Surface Integrals Page 9.12 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 9.07 Find the centre of mass of the part of the unit sphere (of constant surface density) that lies in the first octant.

13 ENGI 4430 Surface Integrals Page 9.13 Example 9.07 (continued)

14 ENGI 4430 Surface Integrals Page 9.14 Example 9.08 (same as Example 9.05, but using the surface method) Find the flux due to the vector field origin. F r r through the sphere S, radius 2, centre the

15 ENGI 4430 Surface Integrals Page 9.15 Example 9.09 Find the flux of the field T the first octant. F x y z across that part of x + 2y + z = 8 that lies in

16 ENGI 4430 Surface Integrals Page 9.16 Example 9.09 (continued)

17 ENGI 4430 Surface Integrals Page 9.17 Example 9.10 Find the total flux of the vector field F z k ˆ through the simple closed surface S x y z a b c Use the parametric grid,, such that the displacement vector to any point on the ellipsoid is asin cos r, bsin sin c cos This grid is a generalisation of the spherical polar coordinate grid and covers the entire surface of the ellipsoid for 0, 0 2. One can verify that x asin cos, y bsin sin, z ccos does lie on the ellipsoid x y z a b c for all values of, : x y z a sin cos b sin sin c cos a b c a b c sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2 sin 2 cos sin cos 1 and The tangent vectors along the coordinate curves = constant and = constant are

18 ENGI 4430 Surface Integrals Page 9.18 Example 9.10 (continued)

19 ENGI 4430 Surface Integrals Page 9.19 For vector fields Line integral: Fr, F dr C Surface integral: F r ds F r N ds F N du dv r r F du dv u v S S S S On a closed surface, take the sign such that N points outward. Some Common Parametric Nets o o 1) The circular plate x x y y a in the plane z zo. Let the parameters be r, where 0 r a, 0 2 x x r cos, y y r sin, z z o o o ˆi cos r sin r r N ˆ sin r cos r ˆ j k r kˆ o o x x y y a with z0 z z1. Let the parameters be z, where z z z, x acos, y asin, z z ˆi 0 a sin r r N ˆ 0 a cos a cos ˆ a sin ˆ j i j z kˆ 1 0 2) The circular cylinder Outward normal: N acos ˆi asin ˆj 2 2 3) The frustum of the circular cone w w a u u v v 1 2 o 1 where o o o w w w and w w. Let the parameters here be r, where w1 wo w2 w r o, 0 2 a a x u u r cos, y v v r sin, z w w a r o o o r r N r ˆi cos r sin ˆj sin r cos kˆ a 0 cos sin a r ˆ a r ˆ r ˆ i j k Outward normal: N ar cos ˆi ar sin ˆj r k ˆ

20 ENGI 4430 Surface Integrals Page ) The portion of the elliptic paraboloid o o o with o 1 2 z z a x x b y y z z z z Let the parameters here be r, where z z z z a cos b sin a cos b sin 1 o 2 o r, x x r y y r z z r a b o cos, o sin, o cos sin r r N r k r a b r b a a r cos ˆ 2b r sin ˆ r ˆ i j k Outward normal: N 2a 2 r 2 cos ˆi 2b 2 r 2 sin ˆj r k ˆ ˆi cos r sin ˆj sin r cos ˆ 2 cos sin 2 sin cos y y z z a. x x Let the parameters here be, where 0, 0 2 x x asin cos, y y asin sin, z z acos 5) The surface of the sphere 2 o o o ˆi acos cos asin sin r r N ˆ j acos sin asin cos kˆ a sin 0 2 a sin sin cos ˆ sin sin ˆ cos ˆ i j k Outward normal: N a 2 sin sin cos ˆ sin sin ˆ cos ˆ i j k r r N a ˆ asin ˆ a sin rˆ a sin r OR 2 6) The part of the plane x x By y Cz z 0 A in the first octant with A, B, C 0 and Axo Byo Czo 0. Let the parameters be x, y where 0 Ax By Cz x By ; 0 y Ax By Cz A B ˆi 1 0 r r ˆ A 0 1 ˆ Bˆ ˆ N j x y i j k C C kˆ A B C C [End of Chapter 9]