Integral Vector Calculus

Transcription

1 ontents 29 Integral Vector alculus 29.1 Line Integrals Involving Vectors Surface and Volume Integrals Integral Vector Theorems 54 Learning outcomes In this Workbook you will learn how to integrate functions involving vectors. You will learn how to evaluate line integrals i.e. where a scalar or a vector is summed along a line or contour. You will be able to evaluate surface and volume integrals where a function involving vectors is summed over a surface or volume. You will learn about some theorems relating to line, surface or volume integrals i.e Stokes' theorem, Gauss' theorem and Green's theorem.

2 Line Integrals Involving Vectors 29.1 Introduction 28 considered the differentiation of scalar and vector fields. Here we consider how to integrate such fields along a line. Firstly, integrals involving scalars along a line will be considered. Subsequently, line integrals involving vectors will be considered. These can be scalar or vector depending on the form of integral used. Of particular interest are the integrals of conservative vector fields. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... have a thorough understanding of the basic techniques of integration be familiar with the operators div, grad and curl integrate a scalar or vector quantity along a line 2 HELM (26): Workbook 29: Integral Vector alculus

3 1. Line integrals 28 was concerned with evaluating an integral over all points within a rectangle or other shape (or over a cuboid or other volume). In a related manner, an integral can take place over a line or curve running through a two-dimensional (or three-dimensional) shape. Line integrals may involve scalar or vector fields. Those involving scalar fields are dealt with first. Line integrals in two dimensions A line integral in two dimensions may be written as F (x, y)dw There are three main features determining this integral: F (x, y) This is the scalar function to be integrated e.g. F (x, y) x 2 + 4y 2. This is the curve along which integration takes place. e.g. y x 2 or x sin y or x t 1; y t 2 (where x and y are expressed in terms of a parameter t). dw The integral. The integrals This states the variable of the integration. Three main cases are dx, dy and ds. Here s is arc length and so indicates position along the curve. ( ) 2 dy ds may be written as ds (dx) 2 + (dy) 2 or ds 1 + dx. dx A fourth case is when F (x, y) dw has the form: F 1 dx+f 2 dy. This is a combination of the cases dx and dy. F (x, y) ds represents the area beneath the surface z F (x, y) but above the line F (x, y) dx and and yz planes respectively. A particular case of the integral the length along a curve i.e. an arc length. F (x, y) dy represent the projections of this area onto the xz F (x, y) ds is the integral 1 ds. This is a means of calculating z f(x, y)dy y f(x, y)ds curve f(x, y)dx Figure 1: Representation of a line integral and its projections onto the xz and yz planes x HELM (26): Section 29.1: Line Integrals Involving Vectors 3

4 The technique with a line integral is to express all quantities in an integral in terms of a single variable. Often, if the integral is with respect to x or y, the curve and the function F may be expressed in terms of the relevant variable. If the integral is carried out with respect to ds, normally everything is expressed in terms of x. If x and y are given in terms of a parameter t, normally everything is expressed in terms of t. Example 1 Find x (1 + 4y) dx where is the curve y x 2, starting from x, y c and ending at x 1, y 1. Solution As this integral concerns only points along and the integration is carried out with respect to x, y may be replaced by x 2. The limits on x will be to 1. So the integral becomes x(1 + 4y) dx 1 x [ x 2 x ( 1 + 4x 2) dx 2 + x4 ] 1 1 ( ) x ( x + 4x 3 ) dx () 3 2 Example 2 Find x (1 + 4y) dy where is the curve y x 2, starting from c x, y and ending at x 1, y 1. This is the same as Example 1 other than dx being replaced by dy. Solution As this integral concerns only points along and the integration is carried out with respect to y, everything may be expressed in terms of y, i.e. x may be replaced by y 1/2. The limits on y will be to 1. So the integral becomes x(1 + 4y) dy 1 y y 1/2 (1 + 4y) dx [ 2 3 y3/ x5/2 ] 1 1 y ( ) 5 ( y 1/2 + 4y 3/2) dx () HELM (26): Workbook 29: Integral Vector alculus

5 Example 3 Find x (1 + 4y) ds where is the curve y x 2, starting from x, y c and ending at x 1, y 1. Once again, this is the same as the previous two examples other than the integration being carried out with respect to s, the coordinate along the curve. Solution As this integral is with respect to x, all parts of the integral can be expressed in terms of x, Along ( ) 2 dy y x 2, ds 1 + dx 1 + (2x) 2 dx 1 + 4x dx 2 dx So, the integral is 1 x (1 + 4y) ds x ( 1 + 4x 2) x 2 dx x ( 1 + 4x 2) 3/2 dx c x This can be evaluated using the transformation U 1 + 4x 2 so du 8xdx i.e. x dx du 8. When x, U 1 and when x 1, U 5. The integral therefore equals 1 x x ( 1 + 4x 2) 3/2 1 5 dx 8 U [ U 3/2 du x ] 5 U 5/ [ 5 5/2 1 ] Note that the results for Examples 1,2 and 3 are all different: Example 3 is the area between a curve and a surface above; Examples 1 and 2 give projections of this area onto other planes. Example 4 Find xy dx where, on, x and y are given by x 3t 2, y t 3 1 for t starting at t and progressing to t 1. Solution Everything can be expressed in terms of t, the parameter. Here x 3t 2 so dx 6t dt. The limits on t are t and t 1. The integral becomes xy dx 1 t 3t 2 (t 3 1) 6t dt [ 18 7 t t4 ] 1 1 t (18t 6 18t 3 ) dt HELM (26): Section 29.1: Line Integrals Involving Vectors 5

6 Key Point 1 A line integral is normally evaluated by expressing all variables in terms of one variable. In general f(x, y) ds f(x, y) dy f(x, y) dx Task For F (x, y) 2x + y 2, find F (x, y) dx, F (x, y) dy and where is the line y 2x from (, ) to (1, 2). F (x, y) ds Express each integral as a simple integral with respect to a single variable and hence evaluate each integral: Your solution Answer 1 x (2x + 4x 2 ) dx, 7 3, 2 y (y + y 2 ) dy, 14 3, 1 x (2x + 4x 2 ) 5 dx, Task Find F (x, y) dx, F (x, y) dy and F (x, y) ds where F (x, y) 1 and is the curve y 1 2 x2 1 ln x from (1, 1) to (2, 2 1 ln 2) Your solution 6 HELM (26): Workbook 29: Integral Vector alculus

7 Answer dx 1, 2 1/4 ln 2 1/2 (x + 1 4x ) dx ln dy ln 2, 4 Task Find F (x, y) dx, F (x, y) dy and F (x, y) ds where F (x, y) sin 2x and is the curve y sin x from (, ) to ( π 2, 1). Your solution Answer π/2 π/2 sin 2x dx 1, π/2 2 sin x cos 2 x dx 2 3 sin 2x 1 + cos 2 x dx Using the substitution u 1 + cos 2 x gives 2 3 (2 2 1). 2. Line integrals of scalar products Integrals of the form such as the following. F dr, referred to at the end of the previous sub-section, occur in applications B A T δr dr S (current position) v Figure 2: Schematic for cyclist travelling from A to B into a head wind HELM (26): Section 29.1: Line Integrals Involving Vectors 7

8 onsider a cyclist riding along the road from A to B (Figure 2). Suppose it is necessary to find the total work the cyclist has to do in overcoming a wind of velocity v. On moving from S to T, the work done is given by Force distance F δr cos θ where F, the force, is directly proportional to v, but in the opposite direction, and δr cos θ is the component of the distance travelled in the direction of the wind. So, the work done travelling δr is kv δr. Letting δr become infinitesimally small, the work done becomes kv dr and the total work is k B A v dr. This is an example of the integral along a line of the scalar product of a vector field and a vector describing the line. The term scalar line integral is often used for integrals of this form. The vector dr may be considered to be dxi + dyj + dxk. Multiplying out the scalar product, in three dimensions, the scalar line integral of the vector F along contour is given by F dr and equals {F x dx + F y dy + F z dz} in three dimensions ( {F x dx + F y dy} in two dimensions.) If the contour has its start and end points in the same positions i.e. it represents a closed contour, the symbol rather than is used, i.e. F dr. As before, to evaluate the line integral, express the path and the function F in terms of either x, y and z, or in terms of a parameter t. Note that in examples t often represents time. Example 5 Find {2xy dx 5x dy} where is the curve y x 3 with x varying from x to x 1. [This is the integral F dr where F 2xyi 5xj and dr dxi + dyj.] Solution It is possible to split this integral into two different integrals and express the first term as a function of x and the second term as a function of y. However, it is also possible to express everything in terms of x. Note that on, y x 3 so dy 3x 2 dx and the integral becomes {2xy dx 5x dy} 1 x ( 2x x 3 dx 5x 3x 2 dx ) 1 (2x 4 15x 3 ) dx [ 2 5 x x4 ] HELM (26): Workbook 29: Integral Vector alculus

9 An integral of the form Key Point 2 F dr may be expressed as {F x dx + F y dy + F z dz}. Knowing the expression for the path, every term in the integral can be further expressed in terms of one of the variables x, y or z or in terms of a parameter t and hence integrated. If an integral is two-dimensional there are no terms involving z. The integral F dr evaluates to a scalar. Example 6 Three paths from (, ) to (1, 2) are defined by (a) 1 : y 2x (b) 2 : y 2x 2 (c) 3 : y from (, ) to (1, ) and x 1 from (1, ) to (1, 2) Sketch each path and find F dr, where F y 2 i + xyj, along each path. Solution (a) F dr 1 F dr {y 2 dx + xydy }. Along y 2x, dy dx 1 x 1 { (2x) 2 dx + x (2x) (2dx) } ( 4x 2 + 4x 2) 1 dx 8x 2 dx 2 so dy 2dx. Then [ ] x y y 2x A(1, 2) 1 1 x Figure 3(a): Integration along path 1 HELM (26): Section 29.1: Line Integrals Involving Vectors 9

10 Solution (contd.) {y (b) F dr 2 dx + xydy }. Along y 2x 2, dy 4x so dy 4xdx. Then dx 1 { (2x F dr ) 2 2 ( ) 1 [ ] 1 dx + x 2x 2 12 (4xdx)} 12x 4 dx 2 x 5 x y A(1, 2) y 2x x Figure 3(b): Integration along path 2 Note that the answer is different to part (a), i.e., the line integral depends upon the path taken. (c) As the contour 3, has two distinct parts with different equations, it is necessary to break the full contour OA into the two parts, namely OB and BA where B is the point (1, ). Hence B A F dr F dr + F dr 3 Along OB, y so dy. Then B O F dr O 1 x Along AB, x 1 so dx. Then B A F dr 2 y Hence F dr B ( 2 dx + x ) 1 dx ( y y dy ) 2 ydy y [ ] y x 1 A(1, 2) 3 O y B 1 x Figure 3(c): Integration along path 3 Once again, the result is path dependent. 1 HELM (26): Workbook 29: Integral Vector alculus

11 Key Point 3 In general, the value of the line integral depends on the path of integration as well as the end points. Example 7 Find O A F dr, where F y 2 i + xyj (as in Example 6) and the path from A to O is the straight line from (1, 2) to (, ), that is the reverse of 1 in Example 6(a). Deduce F dr, the integral around the closed path formed by the parabola y 2x 2 from (, ) to (1, 2) and the line y 2x from (1, 2) to (, ). Solution Reversing the path swaps the limits of integration, this results in a change of sign for the value of the integral. O A F dr A O F dr 8 3 The integral along the parabola (calculated in (iii) above) evaluates to 12 5, then F dr F dr + F dr Example 8 onsider the vector field F y 2 z 3 i + 2xyz 3 j + 3xy 2 z 2 k Let 1 and 2 be the curves from O (,, ) to A (1, 1, 1), given by 1 : x t, y t, z t ( t 1) 2 : x t 2, y t, z t 2 ( t 1) (a) Evaluate the scalar integral of the vector field along each path. (b) Find the value of F dr where is the closed path along 1 from O to A and back along 2 from A to O. HELM (26): Section 29.1: Line Integrals Involving Vectors 11

12 Solution (a) The path 1 is given in terms of the parameter t by x t, y t and z t. Hence dx dt dy dt dz dt 1 and dr dt dx dt i + dy dt j + dz dt k i + j + k Now by substituting for x y z t in F we have F t 5 i + 2t 5 j + 3t 5 k Hence F dr dt t5 + 2t 5 + 3t 5 6t 5. The values of t and t 1 correspond to the start and end point of 1 and so these are the required limits of integration. Now 1 F dr 1 F dr dt dt 1 6t 5 dt [ t 6 ] 1 1 For the path 2 the parameterisation is x t 2, y t and z t 2 so dr 2ti + j + 2tk. dt Substituting x t 2, y t and z t 2 in F we have F t 8 i + 2t 9 j + 3t 8 k and F dr dt 2t9 + 2t 9 + 6t 9 1t 9 2 F dr 1 1t 9 dt [ t 1 ] 1 1 (b) For the closed path F dr F dr 1 F dr (Note that the line integral round a closed path is not necessarily zero - see Example 7.) Further points on Example 8 Vector Field Path Line Integral F 1 1 F 2 1 F closed Note that the line integral of F is 1 for both paths 1 and 2. In fact, this would hold for any path from (,, ) to (1, 1, 1). The field F is an example of a conservative vector field; these are discussed in detail in the next subsection. In F dr, the vector field F may be the gradient of a scalar field or the curl of a vector field. 12 HELM (26): Workbook 29: Integral Vector alculus

13 Task onsider the vector field G xi + (4x y)j Let 1 and 2 be the curves from O (,, ) to A (1, 1, 1), given by 1 : x t, y t, z t ( t 1) 2 : x t 2, y t, z t 2 ( t 1) (a) Evaluate the scalar integral G dr of each vector field along each path. (b) Find the value of O to A and back along 2 from A to O. G dr where is the closed path along 1 from Your solution HELM (26): Section 29.1: Line Integrals Involving Vectors 13

14 Answer (a) The path 1 is given in terms of the parameter t by x t, y t and z t. Hence dx dt dy dt dz dt 1 and dr dt dx dt i + dy dt j + dz dt k i + j + k Substituting for x y z t in G we have G ti + 3tj and G dr dt t + 3t 4t The limits of integration are t and t 1, then 1 G dr G dr 1 [ ] 1 1 dt dt 4tdt 2t 2 2 For the path 2 the parameterisation is x t 2, y t and z t 2 so dr dt Substituting x t 2, y t and z t 2 in G we have G t 2 i + ( 4t 2 t ) j and G dr dt 2t3 + 4t 2 t 1 ( G dr 2t 3 + 4t 2 t ) [ 1 dt 2 2 t t3 1 ] 1 2 t2 4 3 (b) For the closed path G dr G dr G dr (Note that the integral around the closed path is non-zero, unlike Example 8.) 2ti + j + 2tk. Example 9 { Find (x 2 y) } dr where is the contour y 2x x 2 from (, ) to (2, ). Solution { Note that (x 2 y) 2xyi + x 2 j so the integral is 2xy dx + x 2 dy }. On y 2x x 2, dy (2 2x) dx so the integral becomes { 2xy dx + x 2 dy } 2 { 2x(2x x 2 ) dx + x 2 (2 2x) dx } x 2 (6x 2 4x 3 ) dx [ 2x 3 x 4 ] 2 14 HELM (26): Workbook 29: Integral Vector alculus

15 Task Evaluate paths F dr, where F (x y)i + (x + y)j along each of the following (a) 1 : from (1, 1) to (2, 4) along the straight line y 3x 2: (b) 2 : from (1, 1) to (2, 4) along the parabola y x 2 : (c) 3 : along the straight line x 1 from (1, 1) to (1, 4) then along the straight line y 4 from (1, 4) to (2, 4). Your solution Answer (a) (b) (c) (1x 4) dx 11, (x + x 2 + 2x 3 ) dx 35, (this differs from (a) showing path dependence) 3 (1 + y) dy (x 4) dx 8 Task For the function F and paths in the last Task, deduce F dr for the closed paths (a) 1 followed by the reverse of 2. (b) 2 followed by the reverse of 3. (c) 3 followed by the reverse of 1. Your solution HELM (26): Section 29.1: Line Integrals Involving Vectors 15

16 Answer (a) 1 3 1, (b), (c) 3. (note that all these are non-zero.) 3 1. onsider Exercises F dr, where F 3x 2 y 2 i + (2x 3 y 1)j. Find the value of the line integral along each of the paths from (, ) to (1, 4). (a) y 4x (b) y 4x 2 (c) y 4x 1/2 (d) y 4x 3 2. onsider the vector field F 2xi + (xz 2)j + xyk and the two curves between (,, ) and (1, 1, 2) defined by 1 : x t 2, y t, z 2t for t 1. 2 : x t 1, y 1 t, z 2t 2 for 1 t 2. (a) Find F dr, 1 F dr 2 (b) Find F dr where is the closed path from (,, ) to (1, 1, 2) along 1 and back to (,, ) along onsider the vector field G x 2 zi + y 2 zj (x3 + y 3 )k and the two curves between (,, ) and (1, 1, 2) defined by 1 : x t 2, y t, z 2t for t 1. 2 : x t 1, y 1 t, z 2t 2 for 1 t 2. (a) Find G dr, 1 G dr 2 (b) Find G dr where is the closed path from (,, ) to (1, 1, 2) along 1 and back to (,, ) along Find F dr) along y 2x from (, ) to (2, 4) for Answers (a) F (x 2 y) (b) F ( 1 2 x2 y 2 k) 1. All are 12, and in fact the integral would be 12 for any path from (,) to (1,4). 5 2 (a) 2,, (b) (a),, (b) , HELM (26): Workbook 29: Integral Vector alculus

17 3. onservative vector fields For some line integrals in the previous section, the integral depended only on the vector field F and the start and end points of the line but not on the actual path of the line between the start and end points. However, for other line integrals, the result depended on the actual details of the path of the line. Vector fields are classified according to whether the line integrals are path dependent or path independent. Those vector fields for which all line integrals between all pairs of points are path independent are called conservative vector fields. There are five properties of a conservative vector field (P1 to P5). It is impossible to check the value of every line integral over every path, but instead it is possible to use any one of these five properties (and in particular property P3 below) to determine whether a vector field is conservative. They are also used to simplify calculations with conservative vector fields. P1 P2 The line integral B A F dr depends only on the end points A and B and is independent of the actual path taken. The line integral around any closed curve is zero. That is P3 The curl of a conservative vector field F is zero i.e. F. F dr for all. P4 For any conservative vector field F, it is possible to find a scalar field φ such that φ F. Then, F dr φ(b) φ(a) where A and B are the start and end points of contour. P5 [This is sometimes called the Fundamental Theorem of Line Integrals and is comparable with the Fundamental Theorems of alculus.] All gradient fields are conservative. That is, F φ is a conservative vector field for any scalar field φ. Solution Example 1 The following vector fields were considered in the Examples of the previous subsection. 1. F 1 y 2 i + xyj (Example 6) 2. F 2 2xi + 2yj (Example 7) 3. F 3 y 2 z 3 i + 2xyz 3 j + 3xy 2 z 2 k (Example 8) 4. F 4 xi + (4x y) j (Task on page 13) Determine which of these vector fields are conservative e.g. by referring to the answers given in the solution. For those that are conservative find a scalar field φ such that F φ and use property P4 to verify the line integrals found. 1. Two different values were obtained for line integrals over the paths 1 and 2. Hence, by P1, F 1 is not conservative. [It is also possible to reach this conclusion from P3 by finding that F yk.] HELM (26): Section 29.1: Line Integrals Involving Vectors 17

18 Solution 2. Both line integrals from (, ) to (4, 2) had the same value i.e. 2 and for the closed path the line integral was. This alone does not mean that F 2 is conservative as there could be other untried paths giving different values. So by using P3 i j k F 2 x y z 2x 2y i( ) j( ) + k( ) As F 2, P3 gives that F 2 is a conservative vector field. Now, find a φ such that F 2 φ. Then φ x i + φ j 2xi + 2yj. y Thus Using P4: φ x 2x φ x2 + f(y) φ y 2y φ y2 + g(x) (4,2) (,) F 2 dr (4,2) (,) φ x 2 + y 2 (+ constant) ( φ) dr φ(4, 2) φ(, ) ( ) ( ) The fact that line integrals along two different paths between the same start and end points is consistent with F 3 being a conservative field according to P1. So too is the fact that the integral around a closed path is zero according to P2. However, neither fact can be used to conclude that F 3 is a conservative field. This can be done by showing that F 3. i j k Now, x y z (6xyz 2 6xyz 2 )i (3y 2 z 2 3y 2 z 2 )j + (2yz 3 2yz 3 )k. y 2 z 3 2xyz 3 3xy 2 z 2 As F 3, P3 gives that F 3 is a conservative field. To find φ that satisfies φ F 3, it is necessary to satisfy φ x y2 z 3 φ xy 2 z 3 + f(y, z) Using P4: φ y 2xyz3 φ xy 2 z 3 + g(x, z) φ z 3xy2 z 2 φ xy 2 z 3 + h(x, y) (1,1,1) (,,) φ xy 2 z 3 F 3 dr φ(1, 1, 1) φ(,, ) HELM (26): Workbook 29: Integral Vector alculus

19 Solution 4. As the integral along 1 is 2 and the integral along 2 (same start and end points but different intermediate points) is 4, F 3 4 is not a conservative field using P1. Note that F 4 4k so, using P3, this is an independent conclusion that F 4 is not conservative. Engineering Example 1 Work done moving a charge in an electric field Introduction If a charge, q, is moved through an electric field, E, from A to B, then the work required is given by the line integral W AB q B A Problem in words E dl ompare the work done in moving a charge through the electric field around a point charge in a vacuum via two different paths. Mathematical statement of problem An electric field E is given by E Q 4πε r ˆr 2 Q xi + yj + zk 4πε (x 2 + y 2 + z 2 ) x2 + y 2 + z 2 Q(xi + yj + zk) 4πε (x 2 + y 2 + z 2 ) 3 2 where r is the position vector with magnitude r and unit vector ˆr, and constants of proportionality, where ɛ 1 9 /36π F m πɛ is a combination of Given that Q 1 8, find the work done in bringing a charge of q 1 1 from the point A (1, 1, ) to the point B (1, 1, ) (where the dimensions are in metres) (a) by the direct straight line y x, z (b) by the straight line pair via (1, 1, ) HELM (26): Section 29.1: Line Integrals Involving Vectors 19

20 y A a b B O b x Figure 4: Two routes (a and b) along which a charge can move through an electric field The path comprises two straight lines from A (1, 1, ) to B (1, 1, ) via (1, 1, ) (see Figure 4). Mathematical analysis (a) Here q/(4πε ) 9 so E 9[xi + yj] (x 2 + y 2 ) 3 2 as z over the region of interest. The work done B W AB q E dl A B 1 1 A 9 (x 2 + y 2 ) 3 2 [xi + yj] [dxi + dyj] Using y x, dy dx 1 W AB {x dx + x dx} x1 (2x 2 ) (2 2) x 3 2x dx x 2 dx [ ] x [ ] x [1.1] J 2 HELM (26): Workbook 29: Integral Vector alculus

21 (b) The first part of the path is A to where x 1, dx and y goes from 1 to 1. W A Q E dl A y u2 9 (1 + y 2 ) 3 2 9y dy (1 + y 2 ) du u u [ 2u 1 2 [xi + yj] [i + dyj] substituting du ] ( ) J u 1 + y 2, du 2y dy The second part is to B, where y 1, dy and x goes from 1 to 1. 1 W B [xi + yj] [dxi + j] x1 (x 2 + 1) x dx 1 (x 2 + 1) du substituting u x 2 + 1, du 2x dx u u 3 2 u ] [ 2u 1 2 du ( ) J The sum of the two components W A and W B is J. Therefore the work done over the two routes is identical. Interpretation In fact, the work done is independent of the route taken as the electric field E around a point charge in a vacuum is a conservative field. HELM (26): Section 29.1: Line Integrals Involving Vectors 21

22 Example Show that I path taken. (2,1) (,) { (2xy + 1)dx + (x 2 2y)dy } is independent of the 2. Find I using property P1. 3. Find I using property P4. { 4. Find I (2xy + 1)dx + (x 2 2y)dy } where is (a) the circle x 2 + y 2 1 (b) the square with vertices (, ), (1, ), (1, 1), (, 1). Solution 1. The integral I (2,1) (,) { (2xy + 1)dx + (x 2 2y)dy } may be re-written F (2xy + 1)i + (x 2 2y)j. i j k Now F x y z i + j + k 2xy + 1 x 2 2y F dr where As F, F is a conservative field and I is independent of the path taken between (, ) and (2, 1). 2. As I is independent of the path taken from (, ) to (2, 1), it can be evaluated along any such path. One possibility is the straight line y 1x. On this line, dy 1 dx. The integral 2 2 I becomes I (2,1) (,) 2 x 2 { (2xy + 1)dx + (x 2 2y)dy } {(2x 12 x + 1)dx + (x2 4x) 12 dx } ( 3 2 x2 1 2 x + 1)dx [ 1 2 x3 1 4 x2 + x ] HELM (26): Workbook 29: Integral Vector alculus

23 Solution (contd.) 3. If F φ then φ x 2xy + 1 φ x2 y + x + f(y) φ y x2 2y φ x 2 y + x y 2 +. φ x 2 y y 2 + g(x) These are consistent if φ x 2 y + x y 2 (plus a constant which may be omitted since it cancels). So I φ(2, 1) φ(, ) ( ) 5 4. As F is a conservative field, all integrals around a closed contour are zero. Exercises 1. Determine whether the following vector fields are conservative (a) F (x y)i + (x + y)j (b) F 3x 2 y 2 i + (2x 3 y 1)j (c) F 2xi + (xz 2)j + xyk (d) F x 2 zi + y 2 zj (x3 + y 3 )k 2. onsider the integral F dr with F 3x 2 y 2 i + (2x 3 y 1)j. From Exercise 1(b) F is a conservative vector field. Find a scalar field φ so that φ F. Hence use P4 to evaluate the integral F dr where is an integral with start-point (, ) and end point (1, 4). 3. For the following conservative vector fields F, find a scalar field φ such that φ F and hence evaluate the I F dr for the contours indicated. (a) F (4x 3 y 2x)i + (x 4 2y)j; any path from (, ) to (2, 1). (b) F (e x + y 3 )i + (3xy 2 )j; closed path starting from any point on the circle x 2 + y 2 1. (c) F (y 2 + sin z)i + 2xyj + x cos zk; any path from (1, 1, ) to (2,, π). (d) F 1 x i + 4y3 z 2 j + 2y 4 zk; any path from (1, 1, 1) to (1, 2, 3). Answers 1. (a) No, (b) Yes, (c) No, (d) Yes 2. x 3 y 2 y +, (a) x 4 y x 2 y 2, 11; (b) e x + xy 3, ; (c) xy 2 + x sin z, 1; (d) ln x + y 4 z 2,143 HELM (26): Section 29.1: Line Integrals Involving Vectors 23

24 4. Vector line integrals It is also possible to form the less commonly used integrals f(x, y, z) dr and F (x, y, z) dr. Each of these integrals evaluates to a vector. Remembering that dr dx i + dy j + dz k, an integral of the form f(x, y, z) dr becomes f(x, y, z)dx i + f(x, y, z) dy j + f(x, y, z)dz k. The first term can be evaluated by expressing y and z in terms of x. Similarly the second and third terms can be evaluated by expressing all terms as functions of y and z respectively. Alternatively, all variables can be expressed in terms of a parameter t. If an integral is two-dimensional, the term in z will be absent. Example 12 Evaluate the integral xy 2 dr where represents the contour y x 2 from (, ) to (1, 1). Solution This is a two-dimensional integral so the term in z will be absent. I xy 2 dr xy 2 (dxi + dyj) xy 2 dx i + xy 2 dy j 1 x 1 [ 1 6 x6 x(x 2 ) 2 dx i + x 5 dx i + ] i j i + 1 [ 2 7 x7/2 1 y y 5/2 dy j ] 1 j y 1/2 y 2 dy j 24 HELM (26): Workbook 29: Integral Vector alculus

25 Example 13 Find I xdr for the contour given parametrically by x cos t, y sin t, z t π starting at t and going to t 2π, i.e. (1,, π) and finishes at (1,, π). the contour starts at Solution The integral becomes x(dxi + dyj + dzk). Now, x cos t, y sin t, z t π so dx sin t dt, dy cos t dt and dz dt. So I 2π 2π cos t( sin t dti + cos t dtj + dtk) 2π cos t sin t dt i + 2π 2π cos 2 t dt j + 2π 1 sin 2t dt i + 1 [ (1 + cos 2t) dt j [ ] 2π cos 2t i + 1 [t + 12 ] 2π 4 2 sin 2t j + k i + π j πj cos t dt k ] 2π sin t k Integrals of the form and dr dx i + dy j + dz k so i j k F dr F 1 F 2 F 3 dx dy dz F dr can be evaluated as follows. The vector field F F 1 i + F 2 j + F 3 k (F 2 dz F 3 dy)i + (F 3 dx F 1 dz)j + (F 1 dy F 2 dx)k (F 3 j F 2 k)dx + (F 1 k F 3 i)dy + (F 2 i F 1 j)dz There are a maximum of six terms involved in one such integral; the exact details may dictate which form to use. HELM (26): Section 29.1: Line Integrals Involving Vectors 25

26 Example 14 Evaluate the integral (x 2 i + 3xyj) dr where represents the curve y 2x 2 from (, ) to (1, 2). Solution Note that the z component of F and dr are both zero. i j k So F dr x 2 3xy (x 2 dy 3xydx)k dx dy and (x 2 i + 3xyj) dr (x 2 dy 3xydx)k Now, on, y 2x 2 so dy 4xdx and (x 2 i + 3xyj) dr {x 2 dy 3xydx}k 1 x k { x 2 4xdx 3x 2x 2 dx } k 2x 3 dxk [ 1 2 x4 ] 1 k 26 HELM (26): Workbook 29: Integral Vector alculus

27 Engineering Example 2 Force on a loop from a magnetic field Introduction A current I in a magnetic field B is subject to a force F given by F I dl B where the current can be regarded as having magnitude I and flowing (positive charge) in the direction given by the vector dl. The force is known as the Lorentz force and is responsible for the workings of an electric motor. If current flows around a loop, the total force on the loop is given by the integral of F around the loop, i.e. F (I dl B) I (B dl) where the closed path of the integral represents one circuit of the loop. z B y x I dl Figure 5: The magnetic field through a loop of current Problem in words A current of 1 amp flows around a circuit in the shape of the unit circle in the Oxy plane. A field of 1 gauss in the positive z-direction is present. Find the total force on the circuit loop. Mathematical statement of problem hoose an origin at the centre of the circuit and use polar coordinates to describe the position of any point on the circuit and the length of a small element. alculate the line integral around the circuit representing the force using the given values of current and magnetic field. Mathematical analysis The circuit is described parametrically by with x cos θ y sin θ z dl sin θ dθ i + cos θ dθ j B B k HELM (26): Section 29.1: Line Integrals Involving Vectors 27

28 since B is constant. Therefore, the force on the circuit is given by F IB k dl k dl (since I 1 A and B 1 G) where k dl i j k 1 sin θ dθ cos θ dθ ( cos θ i sin θ j ) dθ So 2π ( ) F cos θ i sin θ j dθ θ [ ] 2π sin θ i cos θ j θ ( ) i (1 1) j Hence there is no net force on the loop. Interpretation At any given point of the circle, the force on the point opposite is of the same magnitude but opposite direction, and so cancels, leaving a zero net force. Tip: Use symmetry argument to avoid detailed calculations whenever possible! 28 HELM (26): Workbook 29: Integral Vector alculus

29 Engineering Example 3 Magnetic field from a line current Introduction A charge Q, moving at a steady velocity v produces a magnetic field given by db Qµ (v r) 4πr2 where µ is the permeability of free space (4π 1 7 H m 1 ), r is the position vector from the point of interest, P, to the line current. If, instead of a single charge, a current is used, then it is necessary to integrate over all charges in the current. So, the total magnetic field due to the current is given by B D db D µ I 4π dl ˆr r 2 where Idl is the continuous form of Qv, ˆr is a unit vector along r and the current extends from to D. Note that the field is perpendicular to both the current and the line from the current to P. Problem in words Find the magnetic field strength (or magnetic flux density), measured in tesla (T), due to a current I directed vertically downwards, starting at z c and ending at z d. What is the field 1 m from the current when c 5 m, d 1 m and I 1 A? Mathematical statement of problem Here Idl Ikdz z 5 dl φ z z 1 r h h2 + z 2 Figure 6: The current element dl and point P where the field is calculated i.e. dl k dz (pointing downwards). Imagine (without loss of generality) a point P a distance h from the line current and a distance z below a typical line element of the current. The increment of field is given by db µ I 4π(h 2 + z 2 ) dl ˆr where h 2 + z 2 is the distance of P from the typical element. Since dl k dz and ˆr is a unit vector, the magnitude of the vector product is dl ˆr sin φ dz h dz h2 + z 2 P HELM (26): Section 29.1: Line Integrals Involving Vectors 29

30 and is in a direction which (by the right-hand-rule) points OUT of the page to the right of the line and IN to it on the left. Knowing the direction of the field, now calculate the magnitude: the increment of field is given by db µ I 4π(h 2 + z 2 ) so that the total field at a point is B c z d Mathematical analysis h dz h2 + z 2 µ I 4π h(h2 + z 2 ) 3/2 dz µ I 4π h(h2 + z 2 ) 3/2 dz This integral can be evaluated by means of the substitution z h tan u where z h tan u dz h sec 2 u du z c u tan 1 (c/h) u c z d u tan 1 (d/h) u d Substituting into the total field integral gives B µ I 4π µ I 4π µ [ I 4πh µ I 4πh µ I 4πh uc u d uc u d sin u ( ( h(h 2 sec 2 u) 3/2 h sec 2 u du cos u du h ] uc u d c/h 1 + (c/h) 2 + c (h2 + c 2 ) + ) d/h 1 + (d/h) 2 ) d (h2 + d 2 ) as sin(tan 1 y) y 1 + y 2 and B B ˆθ where ˆθ is a unit vector in a circumferential direction around the line current. Now if I 1 A, c 5 m, d 1 m and h 1 m the magnetic field becomes B 1 7 ( ) T 1.98 milli-gauss. Interpretation Note that if c and d then ( ) B µ I c 4πh (h2 + c 2 ) + d µ I (h2 + d 2 ) 4πh [2] µ I 2πh i.e. the field lines are circles around the line current and the magnetic field strength is inversely proportional to the distance of the point of interest P from the current. A scalar or vector involved in a vector line integral may itself be a vector derivative as this next Example illustrates. 3 HELM (26): Workbook 29: Integral Vector alculus

31 Example 15 Find the vector line integral ( F ) dr where F is the vector x 2 i+2xyj +2xzk and is the curve y x 2, z x 3 from x to x 1 i.e. from (,, ) to (1, 1, 1). Solution As F x 2 i + 2xyj + 2xzk, F 2x + 2x + 2x 6x. The integral ( F ) dr 6x(dxi + dyj + dzk) 6x dx i + 6x dy j + 6x dz k The first term is 6x dx i 1 x 6x dx i [ ] 1 3x 2 i 3i In the second term, as y x 2 on, dy may be replaced by 2x dx so 1 1 [ ] 1 6x dy j 6x 2x dx j 12x 2 dx j 4x 3 j 4j x In the third term, as z x 3 on, dz may be replaced by 3x 2 dx so 1 1 [ ] 1 9 6x dz k 6x 3x 2 dx k 18x 3 dx k x 2 x4 k 9 2 k On summing, ( F ) dr 3i + 4j k. HELM (26): Section 29.1: Line Integrals Involving Vectors 31

32 Task Find the vector line integral fdr where f x 2 and is (a) the curve y x 1/2 from (, ) to (9, 3). (b) the line y x/3 from (, ) to (9, 3). Your solution Answer (a) 9 (x 2 i x3/2 j)dx 243i j, (b) 9 (x 2 i x2 j)dx 243i j. Task Evaluate the vector line integral F dr when represents the contour y 4 4x, z 2 2x from (, 4, 2) to (1,, ) and F is the vector field (x z)j. Your solution Answer 1 {(4 6x)i + (2 3x)k} i k 32 HELM (26): Workbook 29: Integral Vector alculus

33 1. Evaluate the vector line integral Exercises ( F ) dr in the case where F xi + xyj + xy 2 k and is the contour described by x 2t, y t 2, z 1 t for t starting at t and going to t When is the contour y x 3, z, from (,, ) to (1, 1, ), evaluate the vector line integrals (a) { (xy)} dr { (b) (x 2 i + y 2 k) } dr Answers 1. 4i j 2k, 2. (a), (b) k HELM (26): Section 29.1: Line Integrals Involving Vectors 33