2-5 The Calculus of Scalar and Vector Fields (pp.33-55)

Transcription

1 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 1/9 2-5 The Calculus of Scalar and Vector Fields (pp.33-55) Fields are functions of coordinate variables (e.g.,, ρ, θ) Q: How can we integrate or differentiate vector fields?? A: There are many ways, we will study: A. The Integration of Scalar and Vector Fields 1. The Line Integral

2 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 2/9 Q1: A C ( r ) c d A1: HO: Differential Displacement Vectors HO: The Differential Displacement Vectors for Coordinate Systems Q2: A2: HO: The Line Ingtegral Q3: A3: HO: The Contour C HO: Line Integrals with Comple Contours Q4:

3 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 3/9 A4: HO: Steps for Analyzing Line Integrals Eample: The Line Integral 2. The Surface Integral Another important integration is the surface integral: A S ( r ) s ds Q1: A1: HO: Differential Surface Vectors HO: The Differential Surface Vectors for Coordinate Systems Q2: A2: HO: The Surface Integral

4 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 4/9 Q3: A3: HO: The Surface S HO: Integrals with Comple Surfaces Q4: A4: HO: Steps for Analyzing Surface Integrals Eample: The Surface Integral 3. The Volume Integral The third important integration is the volume integral it s the easiest of the 3! V ( ) g r dv Q1:

5 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 5/9 A1: HO: The Differential Volume Element HO: The Volume V Eample: The Volume Integral B. The Differentiation of Vector Fields 1. The Gradient The Gradient of a scalar field g ( r ) is epressed as: g ( r ) Q: ( ) ( r ) g r = A A: HO: The Gradient

6 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 6/9 Q: A: HO: The Gradient Operator in Coordinate Systems Q: The gradient of every scalar field is a vector field does this mean every vector field is the gradient of some scalar field? A: HO: The Conservative Field Eample: Integrating the Conservative Field 2. Divergence The Divergence of a vector field A ( r ) is denoted as: A ( r ) ( r ) g ( r ) A =

7 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 7/9 Q: A: HO: The Divergence of a Vector Field Q: A: HO: The Divergence Operator in Coordinate Systems HO: The Divergence Theorem 3. Curl The Curl of a vector field A ( r ) is denoted as: A ( r ) ( r ) ( r ) A = B

8 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 8/9 Q: A: HO: The Curl Q: A: HO: The Curl Operator in Coordinate Systems HO: Stoke s Theorem HO: The Curl of a Conservative Vector Field 4. The Laplacian 2 g ( r )

9 9/6/2005 section_2_5_the_calculus_of_vector_fields_empty.doc 9/9 C. Helmholtz s Theorems Q: ( r ) and/or ( r ) A A A: HO: Helmholtz s Theorems

10 9/6/2005 Differential Line Vector Elements.doc 1/4 Differential Displacement Vectors The derivative of a position vector r, with respect to, y, z, ρ, φ, r, θ ) is epressed as: coordinate value (where { } dr d = + y + z d d ( aˆ ˆ ˆ ay az ) ( aˆ ) d ( yaˆ y ) ( ˆ d zaz ) d = + + d d d d dy dz ˆ ˆ = ˆ + y + z d a d a d a Q: Immediately tell me what this incomprehensible result means or I shall be forced to pummel you! A: The vector above describes the change in position vector r due to a change in coordinate variable. This change in position vector is itself a vector, with both a magnitude and direction.

11 9/6/2005 Differential Line Vector Elements.doc 2/4 For eample, if a point moves such that its coordinate changes from to +, then the position vector that describes that point changes from r to r +. r r + In other words, this small vector is simply a directed distance between the point at coordinate and its new location at coordinate +! This directed distance is related to the position vector derivative as: dr = d d dy dz ˆ a ˆ = + a + ˆ a d d d y z As an eample, consider the case when and y = ρ sinφ we find that: = ρ. Since = ρ cosφ

12 9/6/2005 Differential Line Vector Elements.doc 3/4 dr d dy dz = ˆ a + ˆ a + ˆ a d ρ d ρ d ρ d ρ ( ρ cosφ) ( ρ sinφ) d d dz = ˆ a + ˆ a + ˆ a d ρ d ρ d ρ = cosφ ˆ a + sinφ ˆ a = ˆ a ρ y z y z y A change in position from coordinates ρ, φ,z to ρ + ρφ,,z results in a change in the position vector from r to r +. The vector is a directed distance etending from point ρ, φ,z to point ρ + ρφ,,z, and is equal to: y d r = ρ d ρ = ρ cosφ ˆ a + ρ sinφ ˆ a = ρ ˆ a ρ y = ρa ˆ p r If is really small (i.e., as it approaches zero) we can define something called a differential displacement vector d :

13 9/6/2005 Differential Line Vector Elements.doc 4/4 d lim 0 dr = lim 0 d dr = d d For eample: d r d ρ = d ρ = ˆ aρ d ρ d ρ Essentially, the differential line vector d is the tiny directed distance formed when a point changes its location by some tiny amount, resulting in a change of one coordinate value by an equally tiny (i.e., differential) amount d. The directed distance between the original location (at coordinate value ) and its new location (at coordinate value + d ) is the differential displacement vector d. d r r + d We will use the differential line vector when evaluating a line integral.

14 9/6/2005 The Differential Line Vector for Coordinate Systems.doc 1/3 The Differential Displacement Vector for Coordinate Systems Let s determine the differential displacement vectors for each coordinate of the Cartesian, cylindrical and spherical coordinate systems! Cartesian This is easy! dr d dy dz d d ˆ a ˆ = = ˆ ay az d d d + + d d = ˆ a d dr d dy dz dy = dy = ˆ a ˆ ˆ + ay + az dy dy dy dy dy = ˆ a dy y dr d dy dz dz dz ˆ a ˆ = = ˆ ay az dz dz dz + + dz dz = ˆ a dz z

15 9/6/2005 The Differential Line Vector for Coordinate Systems.doc 2/3 Cylindrical Likewise, recall from the last handout that: d ρ = ˆ a d ρ ρ Maria, look! I m starting to see a trend! dr d = d = ˆ a d d dr dy = dy = ˆ ay dy dy dr dz = dz = ˆ az dz dz dr d ρ = d ρ = ˆ aρ d ρ d ρ Q: It seems very apparent that: d ˆ a d = for all coordinates ; right? A: NO!! Do not make this mistake! For eample, consider d φ : Q: No!! dφ = ˆ a ρ dφ?!? How did the coordinate ρ get in there? φ d r dφ = dφ d φ d dy dz = ˆ a ˆ ˆ + ay + az dφ dφ dφ dφ d ρcosφ d ρsinφ dz = ˆ a ˆ ˆ + ay + az dφ dφ dφ dφ ( ρsinφ ˆ ρ sφ ˆ y ) ( sin ˆ sφ ˆ y ) = a + co a dφ = φa + co a ρ d φ = ˆ a ρ dφ φ

16 9/6/2005 The Differential Line Vector for Coordinate Systems.doc 3/3 The scalar differential value ρ d φ makes sense! The differential displacement vector is a directed distance, thus the units of its magnitude must be distance (e.g., meters, feet). The differential value dφ has units of radians, but the differential value ρ d φ does have units of distance. The differential displacement vectors for the cylindrical coordinate system is therefore: dr d ρ = d ρ = ˆ ap d ρ d ρ dr dφ = dφ = ˆ aφ ρdφ d φ dr dz = dz = ˆ az dz dz Likewise, for the spherical coordinate system, we find that: dr dr = dr = ˆ ar dr dr dr dθ = dθ = ˆ aθ r dθ d θ dr dφ = dφ = ˆ aφ r sinθdφ d φ

17 09/06/05 The Line Integral.doc 1/6 The Line Integral This integral is alternatively known as the contour integral. The reason is that the line integral involves integrating the projection of a vector field onto a specified contour C, e.g., A C ( r ) c d Some important things to note: * The integrand is a scalar function. * The integration is over one dimension. * The contour C is a line or curve through threedimensional space. * The position vector r c denotes only those points that lie on contour C. Therefore, the value of this integral only depends on the value of vector field A ( r ) at the points along this contour.

18 09/06/05 The Line Integral.doc 2/6 Q: What is the differential vector d, and how does it relate to contour C? A: The differential vector d is the tiny directed distance formed when a point moves a small distance along contour C. d C Contour C rc r + d c origin As a result, the differential line vector d is always tangential to every point of the contour. In other words, the direction of d always points down the contour. Q: So what does the scalar integrand A( r ) d mean? What is it that we are actually integrating? A: Essentially, the line integral integrates (i.e., adds up ) the values of a scalar component of vector field A ( r ) at each and every point along contour C. This scalar component of vector field A ( r ) is the projection of A ( ) onto the direction of the contour C. c r c

19 09/06/05 The Line Integral.doc 3/6 First, I must point out that the notation A ( r c ) is nonstandard. Typically, the vector field in the line integral is denoted simply as A ( r ). I use the notation A ( r c ) to emphasize that we are integrating the values of the vector field A ( r ) only at point that lie on contour C, and the points that lie on contour C are denoted as position vector r c. In other words, the values of vector field A ( r ) at points that do not lie on the contour (which is just about all of them!) have no effect on the integration. The integral only depends on the value of the vector field as we move along contour C we denote these values as A ( ). r c Moreover, the line integral depends on only one scalar component of A ( )! r c Q: On just what component of A ( r c ) does the integral depend? A: Look at the integrand A( rc ) d we see it involves the dot product! Thus, we find that the scalar integrand is simply the scalar projection of A ( r c ) onto the differential vector d. As a result, the integrand depends only the component of A ( r c ) that lies in the direction of d and d always points in the direction of the contour C!

20 09/06/05 The Line Integral.doc 4/6 To help see this, first note that A ( r c ), the value of the vector field along the contour, can be written in terms of a vector component tangential to the contour (i.e, A ( rc ) ˆ a ), and a vector component that is normal (i.e., orthogonal) to the contour (i.e, A ( r ) ˆ a ): n c n A ( r ) = A ( r ) ˆ a + A ( r ) ˆ a c c n c n A ( r ) ˆ a n c n ( ) A r c A ( rc ) ˆ a ˆ a ˆ a n = 0 C r c We likewise note that the differential line vector d, like any and all vectors, can be written in terms of its magnitude ( d ) and direction ( â ) as: d ˆ a d = For eample, for ˆ a ˆ = a φ. dφ = ρdφ ˆ a φ, we can say d = ρdφ and

21 09/06/05 The Line Integral.doc 5/6 As a result we can write: ( ) d A( r ) ˆ = a + A ( r ) ˆ a d A rc n n C C = A( r ) ˆ a A ( ) ˆ ˆ + n r an a d C = A( r ) ˆ a ˆ A ( ) ˆ ˆ a + n r an a d = C C A( r ) d In other words, the line integral is simply an integration along contour C, of the scalar component of vector field A ( r ) that lies in the direction tangential to the contour C! Note if vector field A ( r ) is orthogonal to the contour at every point, then the resulting line integral will be zero. A d C A ( ) r d = 0

22 09/06/05 The Line Integral.doc 6/6 Although C represents any contour, no matter how comple or convoluted, we will study only basic contours. In other words, d will correspond to one of the differential line vectors we have previously determined for Cartesian, cylindrical, and spherical coordinate systems.

23 09/06/05 The Contour C.doc 1/11/ The Contour C In this class, we will limit ourselves to studying only those contours that are formed when we change the location of a point by varying just one coordinate parameter. In other words, the other two coordinate parameters will remain fied. Mathematically, therefore, a contour is described by: 2 equalities (e.g., =2, y =-4; r =3, φ =π/4) AND 1 inequality (e.g., -1 < z < 5; 0 < θ < π/2) Likewise, we will need to eplicitly determine the differential displacement vector d for each contour. Recall we have studied seven coordinate parameters (, y, z, ρ, φ, r, θ ). As a result, we can form seven different contours C! Cartesian Contours Say we move a point from P( =1, y =2, z =-3) to P( =1, y =2, z =3) by changing only the coordinate variable z from z =-3 to z =3. In other words, the coordinate values and y remain constant at = 1 and y = 2.

24 09/06/05 The Contour C.doc 2/11/ We form a contour that is a line segment, parallel to the z- ais! z P(1,2,3) C y P(1,2,-3) Note that every point along this segment has coordinate values =1 and y =2. As we move along the contour, the only coordinate value that changes is z. Therefore, the differential directed distance associated with a change in position from z to z +dz, is d = dz = a dz. ˆ z z P(1,2,3) C dz y P(1,2,-3)

25 09/06/05 The Contour C.doc 3/11/ Similarly, a line segment parallel to the -ais (or y-ais) can be formed by changing coordinate parameter (or y), with a resulting differential displacement vector of d = d = a d ˆ y (or d = dy = a dy ). ˆ The three Cartesian contours are therefore: 1. Line segment parallel to the z-ais = c y = c c z c y z1 z2 d = a dz ˆz 2. Line segment parallel to the y-ais = c c y c z = c y1 y2 z d = a dy ˆy 3. Line segment parallel to the -ais c c y = c z = c 1 2 y z d = a d ˆ

26 09/06/05 The Contour C.doc 4/11/ Cylindrical Contours Say we move a point from P(ρ =1, φ = 45, z =2) to P(ρ =3, φ = 45, z =2) by changing only the coordinate variable ρ from ρ =1 to ρ =3. In other words, the coordinate values φ and z remain constant at φ = 45 and z =2. We form a contour that is a line segment, parallel to the -y plane (i.e., perpendicular to the z-ais). z P(1,45,2) C P(3,45,2) y Note that every point along this segment has coordinate valuesφ = 45 and z =2. As we move along the contour, the only coordinate value that changes is ρ.

27 09/06/05 The Contour C.doc 5/11/ Therefore, the differential directed distance associated with a change in position from ρ to ρ +dρ, is d = d ρ = a d ρ. ˆ ρ z P(1,45,2) d ρ P(3,45,2) y Alternatively, say we move a point from P(ρ =3, φ = 0, z =2) to P(ρ =3, φ = 90, z =2) by changing only the coordinate variable φ from φ=0 to φ = 90. In other words, the coordinate values ρ and z remain constant at ρ =3 and z =2. We form a contour that is a circular arc, parallel to the -y plane. z P(3,90,2) C P(3,0,2) y

28 09/06/05 The Contour C.doc 6/11/ Note: if we move from φ = 0 to φ = 360, a complete circle is formed around the z-ais. Every point along the arc has coordinate values ρ = 3 and z =2. As we move along the contour, the only coordinate value that changes is φ. Therefore, the differential directed distance associated with a change in position from φ to φ +dφ, is: ˆ φ d = dφ = a ρdφ. z P(3,90,2) C P(3,0,2) d φ y Finally, changing coordinate z generates the third cylindrical contour but we already did that in Cartesian coordinates! The result is again a line segment parallel to the z-ais. The three cylindrical contours are therefore described as:

29 09/06/05 The Contour C.doc 7/11/ 1. Line segment parallel to the z- ais. ρ = c φ = c c z c ρ φ z1 z2 d = a dz ˆz 2. Circular arc parallel to the -y plane. ρ = c c φ c z = c ρ φ1 φ2 z d = ˆ a ρdφ φ 3. Line segment parallel to the -y plane. c ρ c φ = c z = c ρ1 ρ2 φ z d = ˆ a dρ ρ

30 09/06/05 The Contour C.doc 8/11/ Spherical Contours Say we move a point from P(r =0, θ = 60, φ = 45 ) to P(r =3, θ = 60, φ = 45 ) by changing only the coordinate variable r from r=0 to r =3. In other words, the coordinate values θ and φ remain constant atθ = 60 and φ = 45. We form a contour that is a line segment, emerging from the origin. z P(3,60,45 ) C P(0,60,45 ) y Every point along the line segment has coordinate values θ = 60 and φ = 45. As we move along the contour, the only coordinate value that changes is r.

31 09/06/05 The Contour C.doc 9/11/ Therefore, the differential directed distance associated with a change in position from r to r +dr, is d = dr = a dr. ˆ r z P(0,60,45 ) dr P(3,60,45 ) y Alternatively, say we move a point from P(r =3, θ = 0, φ = 45 ) to P(r =3, θ = 90, φ = 45 ) by changing only the coordinate variable θ from θ = 0 to θ =90. In other words, the coordinate values θ and φ remain constant atθ = 60 and φ = 45. We form a circular arc, whose plane includes the z-ais. P(3,0,45 ) z y C P(3,90,45 )

32 09/06/05 The Contour C.doc 10/11/ Every point along the arc has coordinate values r = 3 and φ = 45. As we move along the contour, the only coordinate value that changes is θ. Therefore, the differential directed distance associated with a change in position from θ to θ +dθ, is d = dθ = ˆ a rdθ. θ P(3,0,45 ) z d θ y C P(3,90,45 ) Finally, we could fi coordinates r and θ and vary coordinate φ only but we already did this in cylindrical coordinates! We again find that a circular arc is generated, an arc that is parallel to the -y plane. The three spherical contours are therefore:

33 09/06/05 The Contour C.doc 11/11/ 1. A circular arc parallel to the -y plane. r = c θ = c c φ c r θ φ1 φ2 d = ˆ aφ r sin θdφ 2. A circular arc in a plane that includes the z-ais. r = c c θ c φ = c φ r θ 1 θ 2 d = ˆ a rdθ θ 3. A line segment directed toward the origin. c r c θ = c φ = c r1 r2 θ φ d = a dr ˆr

34 09/06/05 Line Integrals with Comple Contours.doc 1/4 Line Integrals with Comple Contours Consider a more comple contour, such as: C y Q: What s this flim-flam?! This contour can neither be epressed in terms of single coordinate inequality, nor with single differential line vector! A: True! But we can still easily evaluate a line integral over this contour C. The trick is to divide C into two contours, denoted as C 1 and C 2 : y C 1 C 2

35 09/06/05 Line Integrals with Comple Contours.doc 2/4 We can denote contour C as C = C 1 + C 2. It can be shown that: ( ) ( ) ( ) A r d = A r d + A r d c c c C C C 1 2 Note for the eample given, we can evaluate the integral over both contour C 1 and contour C 2. The first is a circular arc around the z-ais, and the second is a line segment parallel to the y-ais. Q: Does the direction of the contour matter? A: YES! Every contour has a starting point and an end point. Integrating along the contour in the opposite direction will result in an incorrect answer! For eample, consider the two contours below: y y C 1 C 2

36 09/06/05 Line Integrals with Comple Contours.doc 3/4 In this case, the two contours are identical, with the eception of direction. In other words the beginning point of one is the end point of the other, and vice versa. For this eample, we would relate the two contours by saying: C 1 = -C 2 and/or C 2 = -C 1 Just like vectors, the negative of a contour is an otherwise identical contour with opposite direction. We find that: C ( ) A( ) A r d = r d c C c Q: Does the shape of the contour really matter, or does the result of line integration only depend on the starting and end points?? A: Generally speaking, the shape of the contour does matter. Not only does the line integral depend on where we start and where we finish, it also depends on the path we take to get there!

37 09/06/05 Line Integrals with Comple Contours.doc 4/4 For eample, consider these two contours: y C 1 C 2 Generally speaking, we find that: C ( ) A( ) A r d r d c C 1 2 c An eception to this is a special category of vector fields called conservative fields. For conservative fields, the contour path does not matter the beginning and end points of the contour are all that are required to evaluate a line integral! Remember the name conservative vector fields, as we will learn all about them later on. You will find that a conservative vector field has many properties that make it well EXCELLENT!

38 09/06/05 Steps for Analyzing Line Integrals.doc 1/1 Steps for Analyzing Line Integrals You wish to evaluate an integral of the form: A C ( ) r c d To successfully accomplish this, simply follow these steps: Step 1: Determine the 2 equalities, 1 inequality, and d for the contour C. Step 2: Evaluate the dot product ( r ) A d. Step 3: Transform all coordinates of the resulting scalar field to the same system as C. Step 4: Evaluate the scalar field using the two coordinate equalities that describe contour C. Step 5: Determine the limits of integration from the inequality that describes contour C (be careful of order!). Step 6: Integrate the remaining function of one coordinate variable.

39 09/06/05 Eample The Line Integral.doc 1/4 Eample: The Line Integral Consider the vector field: ( r ) = z ˆ ˆ A a a c y Integrate this vector field over contour C, a straight line that P r = 4, θ=60, φ=45. begins at the origin and ends at point ( ) z C P (4,60,45 ) y Step 1: Determine the two equalities, one inequality, and proper d for the contour C. This contour is formed as the coordinate r changes from r =0 to r =4, where θ = 60 and φ = 45 for all points. The two equalities and one inequality that define this contour are thus: 0 r 4 θ = 60 φ = 45

40 09/06/05 Eample The Line Integral.doc 2/4 and the differential displacement vector for this contour is therefore: d = dr = a dr Step 2: Evaluate the dot product ( r ) A ˆr A d. ( r ) ( ˆ ˆ ) ˆ c d = z a ay ar dr = ( ˆ ˆ ˆ ˆ r y r ) z a a a a dr ( sinθcosφ sinθsinφ) = z dr c Step 3: Transform all coordinates of the resulting scalar field to the same system as C. The contour is a spherical contour. Recall that z = r cosθ and = r sin θcosφ, therefore: A ( rc ) d = ( z sinθcosφ sinθsinφ) dr = ( cosθ sinθcosφ sinθcosφ sinθsinφ) = r sinθcosφ( cosθ sinθsinφ) dr r r dr Step 4: Evaluate the scalar field using the two coordinate equalities that describe contour C. Recall that θ =60 and φ=45 at every point along the contour we are integrating over. Thus, functions of θ or φ are constants with respect to the integration! For eample, cos θ = cos 45 = 0.5. Therefore:

41 09/06/05 Eample The Line Integral.doc 3/4 A ( c ) = sin60 cos45 ( cos60 sin60 sin45 ) r d r dr ( ) = r dr = r dr = rdr 8 Step 5: Determine the limits of integration from the inequality that describes contour C (be careful of order!). We note the contour is described as: 0 r 4 and the contour C moves from r = 0 to r = 4. Thus, we integrate from 0 to 4 : C A( rc ) d = r dr 8 0 Note: if the contour ran from r = 4 to r = 0 the limits of integration would be flipped! I.E., rdr 8

42 09/06/05 Eample The Line Integral.doc 4/4 It is readily apparent that the line integral from r = 0 to r =4 is the opposite (i.e., negative) of the integral from r = 4 to r =0. Step 6: Integrate the remaining function of one coordinate variable. C A( rc ) d = r dr = = = rdr

43 09/06/05 Differential Surface Vectors.doc 1/3 Differential Surface Vectors Consider a rectangular surface, oriented in some arbitrary direction: B C=AB A We can describe this surface using vectors! One vector (say A), is a directed distance that denotes the length (i.e, magnitude) and orientation of one edge of the rectangle, while another directed distance (say B) denotes the length and orientation of the other edge. Say we take the cross-product of these two vectors (AB=C). Q: What does this vector C represent? A: Look at the definition of cross product!

44 09/06/05 Differential Surface Vectors.doc 2/3 C = AB = ˆa AB n = ˆa AB n sinθ AB Note that: C = A B The magnitude of vector C is therefore product of the lengths of each directed distance the area of the rectangle! Likewise, C A = 0 and C B = 0, therefore vector C is orthogonal (i.e., normal ) to the surface of the rectangle. As a result, vector C indicates both the size and the orientation of the rectangle. The differential surface vector For eample, consider the very small rectangular surface resulting from two differential displacement vectors, say d and dm. dm ds = d dm d

45 09/06/05 Differential Surface Vectors.doc 3/3 For eample, consider the situation if d = d and dm = dy : ds = d dy = ( ˆ ˆ y ) = ˆ a a a d dy z d dy In other words the differential surface element has an area equal to the product d dy, and a normal vector that points in the ˆa z direction. The differential surface vector ds specifies the size and orientation of a small (i.e., differential) patch of area, located on some arbitrary surface S. We will use the differential surface vector in evaluating surface integrals of the type: A S ( ) r s ds

46 09/06/05 The Differential Surface Vector for Coordinate Systems.doc 1/2 The Differential Surface Vector for Coordinate Systems Given that ds = d dm, we can determine the differential surface vectors for each of the three coordinate systems. dm ds = d dm d Cartesian ds = dy dz = ˆ a dy dz ds = dz d = ˆ a d dz y ds = d dy = ˆ a d dy z y z We shall find that these differential surface vectors define a small patch of area on the surface of flat plane.

47 09/06/05 The Differential Surface Vector for Coordinate Systems.doc 2/2 Cylindrical ds = d φ dz = ˆ a ρ d φ dz ρ ds = dz dp = ˆ a d ρ dz φ ds = d ρ d φ = ˆ a ρ d ρ d φ z φ ρ z We shall find that ds ρ describes a small patch of area on the surface of a cylinder, ds φ describes a small patch of area on the surface of a half-plane, and ds z again describes a small patch of area on the surface of a flat plane. Spherical 2 ds ˆ r = dθ d φ = ar r sinθ dθ d φ ds d dr ˆ θ = φ = aθ r sinθ dr d φ ds = dr dθ = ˆ a r dr dθ φ φ We shall find that ds r describes a small patch of area on the surface of a sphere, ds θ describes a small patch of area on the surface of a cone, and ds φ again describes a small patch of area on the surface of a half plane.

48 9/6/2005 The Surface Integral.doc 1/5 The Surface Integral An important type of vector integral that is often quite useful for solving physical problems is the surface integral: A S ( r ) s ds Some important things to note: * The integrand is a scalar function. * The integration is over two dimensions. * The surface S is an arbitrary two-dimensional surface in a three-dimensional space. * The position vector r s denotes only those points that lie on surface S. Therefore, the value of this integral only depends on the value of vector field A ( r ) at the points on this surface.

49 9/6/2005 The Surface Integral.doc 2/5 Q: How are differential surface vector ds and surface S related? A: The differential vector ds describes a differential surface area at every point on S. ds S As a result, the differential surface vector ds is normal (i.e., orthogonal) to surface S at every point on S. Q: So what does the scalar integrand A ( rs ) ds mean? What is it that we are actually integrating? A: Essentially, the surface integral integrates (i.e., adds up ) the values of a scalar component of vector field A ( r ) at each and every point on surface S. This scalar component of vector field A ( r ) is the projection of A ( r s ) onto a direction perpendicular (i.e., normal) to the surface S.

50 9/6/2005 The Surface Integral.doc 3/5 First, I must point out that the notation A ( r s ) is nonstandard. Typically, the vector field in the surface integral is denoted simply as A ( r ). I use the notation A ( r s ) to emphasize that we are integrating the values of the vector field A ( r ) only at points that lie on surface S, and the points that lie on surface S are denoted by position vector r s. In other words, the values of vector field A ( r ) at points that do not lie on the surface (which is just about all of them!) have no effect on the integration. The integral only depends on the value of the vector field as we move over surface S we denote these values as A ( ). r s Moreover, the surface integral depends on only one component of A ( )! r s Q: On just what component of A ( r s ) does the integral depend? A: Look at the integrand A ( rs ) ds --we see it involves the dot product! Thus, we find that the scalar integrand is simply the scalar projection of A ( r s ) onto the differential vector ds. As a result, the integrand depends only the component of A ( r s ) that lies in the direction of ds --and ds always points in the direction orthogonal to surface S!

51 9/6/2005 The Surface Integral.doc 4/5 To help see this, first note that every vector A ( r s ) can be written in terms of a component tangential to the surface (i.e, A ( rs ) ˆ a ), and a component that is normal (i.e., orthogonal) to the surface (i.e, A ( r ) ˆ a ): n s n A ( r ) = A ( r ) ˆ a + A ( r ) ˆ a s s n s n A ( r ) ˆ a n n A( r ) A ( r ) ˆ a S ˆ a ˆ a n = 0 We note that the differential surface vector ds can be written in terms of its magnitude ( ds ) and direction ( a ˆn ) as: ds = aˆn ds 2 For eample, for ds = ˆ a r sin θd θd φ, we can say ds = r 2 sin θd θd φ and ˆ a = ˆ a. r r r n r

52 9/6/2005 The Surface Integral.doc 5/5 As a result we can write: S A ( ) = ˆ + ˆ d s r ds A ( r ) a An( r ) an S = A ( r ) ˆa + An( r ) ˆan S = A ( r ) ˆa ˆan + An( r ) ˆan ˆan ds = S S A( r ) n ds Note if vector field A ( r ) is tangential to the surface at every point, then the resulting surface integral will be zero. ˆa n ds ds A S Although S represents any surface, no matter how comple or convoluted, we will study only basic surfaces. In other words, ds will correspond to one of the differential surface vectors from Cartesian, cylindrical, or spherical coordinate systems.

53 9/6/2005 The Surface S.doc 1/10 The Surface S In this class, we will limit ourselves to studying only those surfaces that are formed when we change the location of a point by varying two coordinate parameters. In other words, the other coordinate parameters will remain fied. Mathematically, therefore, a surface is described by: 1 equality (e.g., =2 or r =3) AND 2 inequalities (e.g., -1 < y < 5 and -2 < z < 7, or 0 < θ < π/2 and 0 < φ < π) Likewise, we will need to eplicitly determine the differential surface vector ds for each contour. We will be able to describe a surface for each of the coordinate values we have studied in this class!

54 9/6/2005 The Surface S.doc 2/10 Cartesian Coordinate Surfaces The single equation z =3 specifies all points P(,y,z) with a coordinate value z =3. These points form a plane that is parallel to the -y plane. z ds z y * As we move across this plane, the coordinate values of and y will vary. Thus, the size of this rectangular plane is defined by two inequalities -- c c and c y c. 1 2 y1 y2 * Note the differential surface vector ds z (or orthogonal to every point on this plane. dsz ) is * Similarly, the equations y =-2 or =6 describe planes orthogonal to the -z plane and the y-z plane, respectively. Likewise, the differential surface vectors ds y and ds are orthogonal to each point on these planes.

55 9/6/2005 The Surface S.doc 3/10 Summarizing the Cartesian surfaces: 1. Flat plane parallel to the y-z plane. = c c y c c z c y 1 y2 z1 z2 ds =± ds =±ˆ a dy dz 2. Flat plane parallel to the -z plane. c c y = c c z c 1 2 y z1 z2 ds =± ds =±ˆ a dz d y y 3. Flat plane parallel to the -y plane. c c c y c z = c 1 2 y1 y2 z ds =± ds =±ˆ a dy d z z Cylindrical Coordinate Surfaces With cylindrical coordinates, we can define surfaces such as φ = 45 or ρ = 4. These surfaces, however, are more comple than simply planes.

56 9/6/2005 The Surface S.doc 4/10 For eample, the surface denoted by ρ=4 is formed by all points with coordinate ρ=4. In other words, this surface is formed by all points that are a distance of 4 units from the z-ais a cylinder! z ds ρ y * As we move across this cylinder, the coordinate values of φ and z will vary. Thus, the size of this cylinder is defined by two inequalities--cφ1 φ cφ2 and cz1 z cz2. * Note a cylinder that completely surrounds the z-ais is described by the inequality 0 φ 2π. However, the cylinder does not have to be complete! For eample, the inequality 0 φ π defines a half-cylinder, * We note the differential surface vector ds ρ (or ds ρ ) is orthogonal to this surface at all points.

57 9/6/2005 The Surface S.doc 5/10 Another surface is defined by the equation φ = 45. This surface is formed only from points with coordinate value φ = 45. The surface is a half-plane that etends outward from the z-ais. I see. Sort or like a big door with a z-ais hinge! ds φ z y Note the differential surface vector ds φ is orthogonal to this surface at every point. The final cylindrical surface that we will consider the type formed by the equality z = 2. We know that this forms a flat plane that is parallel to the -y plane. * Using the inequalities of Cartesian coordinates, this flat plane is rectangular in shape. However, using cylindrical coordinates inequalities, this plane will be shaped like a ring or a disk. * For eample, the surface z = 0, 0 ρ 2, 0 φ 2π describes a circular disk of radius 2, lying on the -y plane, and centered at the z-ais:

58 9/6/2005 The Surface S.doc 6/10 y 2 Now let s see if you ve been paying attention! Determine the two inequalities that define this flat surface. y z = 0 1 ρ φ 2

59 9/6/2005 The Surface S.doc 7/10 Summarizing our cylindrical surface results: 1. Circular cylinder centered around the z-ais. ρ = c c φ c c z c ρ φ1 φ2 z1 z2 ds =± dsρ =±ˆp a ρd φdz 2. Vertical half-plane etending from the z-ais. c ρ c φ = c c z c ρ1 ρ2 φ z1 z2 ds =± dsφ =±ˆ aφ dz d ρ 3. Flat plane parallel to the -y plane. c ρ c c φ c z = c ρ1 ρ2 φ1 φ2 z ds =± ds =±ˆ a ρd φd ρ z z

60 9/6/2005 The Surface S.doc 8/10 Spherical Coordinate Surfaces The surface defined by θ = 30 is formed only from points with coordinate θ = 30. This surface is a cone! The ape of the cone is centered at the origin, and its ais of rotation is the z- ais. z ds θ y * Note that the differential surface vector ds θ is normal to this surface at every point. * Just like a cylinder, a complete cone is defined by the inequality 0 φ 2π. Alternatively, for eample, the equation π φ 3π 2 defines a quarter cone.

61 9/6/2005 The Surface S.doc 9/10 Say instead our equality equation is r =3. This defines a surface formed from all points a distance of 3 units from the origin a sphere of radius 3! * This sphere is centered at the origin. * The differential surface vector ds r is normal to this sphere at all points on the surface. * If we wish to define a complete sphere, our inequalities must be: 0 θ < π and 0 φ < 2π otherwise, we will be defining some subsection of a spherical surface (e.g., the Northern Hemisphere.). Finally, we know that the equation φ = 45 defines a vertical half-plane, etending from the z-ais. However, using spherical inequalities, this vertical plane will be in the shape of a semi-circle (or some section thereof), as opposed to rectangular (with cylindrical inequalities). z φ = π 2 y

62 9/6/2005 The Surface S.doc 10/10 Summarizing the spherical surfaces: 1. Sphere centered at the origin. r = c c θ c c φ c r θ1 θ2 φ1 φ2 2 ds =± ds =± ˆ a r sin θd θd φ r r 2. A cone with ape at the origin and aligned with the z-ais. c r c θ = c c φ c r1 r2 θ φ1 φ2 ds =± dsθ =±ˆ aθ r sinθd φdr 3. Vertical half-plane etending from the z-ais. c r c c θ c φ = c r1 r2 θ1 θ2 φ ds =± dsφ =±ˆ aφ r dr d θ

63 9/6/2005 Integrals with Comple Surfaces.doc 1/3 Integrals with Comple Surfaces Similar to contours, we can form comple surfaces by combining any of the seven simple surfaces that can easily be formed with Cartesian, cylindrical or spherical coordinates. For eample, we can define 6 planes to form the surface of a cube centered at the origin: z y The cube surface S is thus described as the sum of the si sides: S = S + S + S + S + S + S Therefore, a surface integration over S can be evaluated as: ( ) ( ) ( ) ( ) A r ds = A r ds + A r ds + A r ds s s s s S S S S ( ) ( ) ( ) + A r ds + A r ds + A r ds s s s S S S 4 5 6

64 9/6/2005 Integrals with Comple Surfaces.doc 2/3 This is a great eample for considering the direction of differential surface vector ds. Recall there are two differential surface vectors that are orthogonal to every surface: the first is simply the opposite of the second. For eample, if we were performing a surface integration over the top surface of this cube (i.e., z=1 plane), we would typically use ds = ds z = ˆ a d dy. z However, we could also use the differential surface vector ds = ds = a d dy! z ˆz Q: How would the results of the two integrations differ? A: By a factor of 1!! We find that a surface integration using ds is related to the surface integration using ds as: S A( r ) ( ds) = A ( r ) ds s S s

65 9/6/2005 Integrals with Comple Surfaces.doc 3/3 The surface of a cube is an eample of a closed surface. A closed surface is a surface that completely surrounds some volume. You cannot get from one side of a closed surface to the other side without passing through the surface. In other words, if your beverage is surrounded by a closed surface, better go get your can opener! In electromagnetics, we often define ds as the direction pointing outward from a closed surface. So, for eample, the differential surface vector for the top surface (z=1) would be: aˆz ddy ds = ds = ˆ a d dy, z z while on the bottom (z=-1) we would use : aˆz ddy ds = ds = ˆ a d dy z z Similarly, we would use differential line vectors of opposite directions for each of the pair of side surfaces (left and right), as well as for the front and back surfaces. Regardless if the surface is open or closed, the direction of ds must remain consistent across an entire comple surface!

66 9/6/2005 Steps for Analyzing a Surface Integral.doc 1/1 Steps for Analyzing Surface Integrals We wish to evaluate an integral of the form: A S ( r ) s ds To successfully accomplish this, simply follow these steps: Step 1: Determine the 1 equality, 2 inequalities, and ds for the surface S (be careful of direction!). Step 2: Evaluate the dot product ( r ) A ds. s Step 3: Write the resulting scalar field using the same coordinate system as surface S. Step 4: Evaluate the scalar field using the coordinate equality that described surface S. Step 5: Determine the limits of integration from the inequalities that describe surface S. Step 6: Integrate the remaining function of two coordinate variables.

67 09/06/05 Eample The Surface Integral.doc 1/5 Eample: The Surface Integral Consider the vector field: A ( r ) = aˆ Say we wish to evaluate the surface integral: A S ( r ) s ds where S is a cylinder whose ais is aligned with the z-ais and is centered at the origin. This cylinder has a radius of 1 unit, and etends 1 unit below the -y plane and one unit above the -y plane. In other words, the cylinder has a height of 2 units. z 1 1 y 1

68 09/06/05 Eample The Surface Integral.doc 2/5 This is a comple, closed surface. We will define the top of the cylinder as surface S 1, the side as S 2, and the bottom as S 3. The surface integral will therefore be evaluated as: ( ) ( ) ( ) ( ) A r ds = A r ds + A r ds + A r ds s s 1 s 2 s 3 S S S S Step 1: Determine ds for the surface S. Let s define ds as pointing in the direction outward from the closed surface. S 1 is a flat plane parallel to the -y plane, defined as: 0 ρ 1 0 φ 2π z = 1 and whose outward pointing ds is: ds = ds = ˆa p dp d φ 1 z z S 2 is a circular cylinder centered on the z- ais, defined as: ρ = 1 0 φ 2π 1 z 1 and whose outward pointing ds is: ds = ds = ˆa p dz d 2 ρ ρ φ

69 09/06/05 Eample The Surface Integral.doc 3/5 S 3 is a flat plane parallel to the -y plane, defined as: 0 ρ 1 0 φ 2π z = 1 and whose outward pointing ds is: ds = ds = ˆa p dp d φ 3 z z Step 2: Evaluate the dot product ( r ) ( ) s 1 = = 0 A ds. A r ds = ˆ a ˆ a ρ d ρdφ A A ( ) ( 0) s z ( cosφ ) ρ d ρdφ r ˆ ˆ s ds2 = a ap ρ dzdφ = ρ dzdφ ( ) r ˆ ˆ s ds3 = a az ρ d ρdφ = ( 0) ρ d ρd φ = 0 Look! Vector field A( r ) is tangential to surface S 1 and S 3 for all points on surface S 1 and S 3! Therefore: ( ) ( ) ( ) ( ) A r ds = A r ds + A r ds + A r ds s s 1 s 2 s 3 S S S S ( r ) ( r ) = 0+ A ds + 0 S S s s = A ds 2 2

70 09/06/05 Eample The Surface Integral.doc 4/5 Step 3: Write the resulting scalar field using the same coordinate system as ds. The differential vector ds ρ is epressed in cylindrical coordinates, therefore we must write the scalar integrand using cylindrical coordinates. We know that: Therefore: = ρ cosφ ( s ) 2 = ( cosφ) ρ = cos ( cos ) A r ds dzdφ = ρ φ φ ρdz d φ cos dz d 2 2 ρ φ φ Step 4: Evaluate the scalar field using the coordinate equality that described surface S. Every point on S 2 has the coordinate value ρ = 1. Therefore: A ( ) 2 2 rs ds2 = cos dzd 2 2 ρ φ φ = 1cosφdz d φ 2 = cos φ dz d φ Step 5: Determine the limits of integration from the inequalities that describe surface S. For S 2 we know that 0 φ 2π 1 z 1.

71 09/06/05 Eample The Surface Integral.doc 5/5 Therefore: S 2π 1 2 rs ds = rs ds2 = cos φ dz d S 0 1 ( ) ( ) A A 2 φ Step 6: Integrate the remaining function of two coordinate variables. Using all the results determined above, the surface integral becomes: S A 2π 1 2 rs ds = cos φ dzd 0 1 ( ) = 2π 1 2 cos φ dφ 0 1 φ dz ( π 0)( 1 ( 1) ) = = 2π

72 09/06/05 The Differential Volume Element.doc 1/3 The Differential Volume Element Consider a rectangular cube, whose three sides can be defined by three directed distances, say A, B, and C. B C A It is evident that the lengths of each side of the rectangular cube are A, B, and C, such that the volume of this rectangular cube can be epressed as: V = ABC Consider now what happens if we take the triple product of these three vectors: ABC = A a ˆ BC n sinθ BC However, we note that sinθ = sin90 = 1. 0, and that ˆ a = ˆ a (i.e., BC vector B C points in the same direction as vector A! ). n A

73 09/06/05 The Differential Volume Element.doc 2/3 Using the fact that A = A ˆ a A, we then find the result: ABC = A aˆ BCsinθ n = A aˆ B C A = Aaˆ aˆ B C A = ABC A BC Look what this means, the volume of a cube can be epressed in terms of the triple product! V = AB C= ABC Consider now a rectangular volume formed by three orthogonal line vectors (e.g., d, dy, dz or d ρ, dφ, dz ). dm dn d The result is a differential volume, given as: dv = d dm dn

74 09/06/05 The Differential Volume Element.doc 3/3 For eample, for the Cartesian coordinate system: dv = d dy dz = d dy dz and for the cylindrical coordinate system: dv = d ρ d φ dz = ρ dρ dφ dz and also for the spherical coordinate system: dv = dr d θ d φ = r 2 sinθ dr dφ dθ

75 09/06/05 The Volume V.doc 1/3 The Volume V As we might epect from out knowledge about how to specify a point P (3 equalities), a contour C (2 equalities and 1 inequality), and a surface S (1 equality and 2 inequalities), a volume V is defined by 3 inequalities. Cartesian The inequalities: c c c y c c z c 1 2 y1 y2 z1 z2 define a rectangular volume, whose sides are parallel to the -y, y-z, and -z planes. The differential volume dv used for constructing this Cartesian volume is: dv = d dy dz Cylindrical The inequalities: c ρ c c φ c c z c ρ1 ρ2 φ1 φ2 z1 z2 defines a cylinder, or some subsection thereof (e.g. a tube!).

76 09/06/05 The Volume V.doc 2/3 The differential volume dv is used for constructing this cylindrical volume is: dv = ρ d ρ d φ dz Spherical The equations: c r c c θ c c φ c r1 r2 θ1 θ2 φ1 φ2 defines a sphere, or some subsection thereof (e.g., an orange slice!). The differential volume dv used for constructing this spherical volume is: dv = r 2 sin θdr d θd φ * Note that the three inequalities become the limits of integration for a volume integral. For eample, integrating over a spherical volume would result in an integral of the form: c c c φ2 θ2 r 2 g( r ) dv = g( r ) r sinθdrdθdφ V c c c φ1 θ1 r 1 2 For this eample, if the scalar field g( r ) is not epressed in terms of spherical coordinates, it must first be transformed before the integral can be eplicitly evaluated.

77 09/06/05 The Volume V.doc 3/3 * Note also that we can construct comple volumes by combining the simple volumes discussed here. V = V1 + V2 + V3 + V4 V 2 V 3 V 4 V 1

78 9/6/2005 Eample The Volume Integral.doc 1/3 Eample: The Volume Integral Let s evaluate the volume integral: V ( ) g r where g( r ) = 1 and the volume V is a sphere with radius R. dv In other words, the volume V is described as: 0 r R 0 θ π 0 φ 2π And thus we use for the differential volume dv: dv = dr d θ d φ = r 2 sinθ dr dθd φ Therefore:

79 9/6/2005 Eample The Volume Integral.doc 2/3 V 2π π R 2 g r dv = r sin drd d ( ) = π π R 2 dφ sinθdθ r dr R = 2 π (2) 3 θ θ φ = 4 3 πr 3 Hey look! The answer is the volume (e.g., in m 3 ) of a sphere! Now, this result provided the numeric volume of V only because g( r ) = 1. We find that the total volume of any space V can be determined this way: Volume of V (1) dv = V Typically though, we find that g( r ) 1, and thus the volume integral does not provide the numeric volume of space V. Q: So what s the volume integral even good for? A: Generally speaking, the scalar function g( r ) will be a density function, with units of things/unit volume. Integrating g( r ) with the volume integral provides us the number of things within the space V!

80 9/6/2005 Eample The Volume Integral.doc 3/3 For eample, let s say g( r ) describes the density of a big swarm of insects, using units of insects/m 3 (i.e., insects are the things). Note that g( r ) must indeed a function of position, as the density of insects changes at different locations throughout the swarm. Now say we want to know the total number of insects within the swarm, which occupies some space V. We can determine this by simply applying the volume integral! number of insects in swarm = V ( ) g r dv where space V completely encloses the insect swarm.