vector calculus 2 Learning outcomes

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1 29 ontents vector calculus 2 1. Line integrals involving vectors 2. Surface and volume integrals 3. Integral vector theorems 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. Time allocation You are expected to spend approximately thirteen hours of independent study on the material presented in this workbook. However, depending upon your ability to concentrate and on your previous experience with certain mathematical topics this time may vary considerably. 1

2 Line Integrals involving Vectors 29.1 Introduction The previous section considered the differentiation of scalar and vector fields. The current section considers how to integrate such fields along a contour of integration. Firstly, integrals along a line will be considered in a general (non-vector) context. Subsequently, line integrals involving vectors will be considered. These can integrate either to a scalar or to a 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... 1 have a thorough understanding of the basic techniques of integration 2 be familiar with the operators div, grad and curl

3 1. Line Integrals Workbook 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- (or three- ) dimensional shape. 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 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 dw or x t 1; y t 2. The last case is where x and y are expressed in terms of a parameter t. This states the variable of the integration. Three main cases are dx, dy and ds. Here s isarc length and so indicates position along the curve. ds may be written as ds (dx) 2 +(dy) 2 or ds 1+ ( dy 2dx. 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. In the case where F (x, y) and the contour lies in the Oxy plane, the integral F (x, y) ds gives the area under the surface z F (x, y) and above the curve. The integrals F (x, y) dx and F (x, y) dy give the areas of the projections of this area onto the planes above the x and y axes (see Figure 1). Figure 1. 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. Ifx and y are given in terms of a parameter t, normally everything is expressed in terms of t. Example Find c x (1 + 4y) dx where is the curve y x2, starting from x,y and ending at x 1,y 1. 3 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

4 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 x ( 1+4x 2) ( dx ) x +4x 3 dx x [ x x4 ] 1 ( ) x () 3 2 Example Find c x (1 + 4y) dy where is the curve y x2, starting from x,y and ending at x 1,y 1. This is the same as the previous Example 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 y 1/2 (1 + 4y) dx y 1/2 +4y 3/2) dx y [ 2 3 y3/ x5/2 ] 1 y ( ) 5 () Example Find c x (1 + 4y) ds where is the curve y x2, starting from x,y 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. HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 4

5 Solution As this integral is with respect to x, all parts of the integral can be expressed in terms of x, Along y x 2,ds 1+ ( ) dy 2dx dx 1+(2x) 2 dx 1+4x 2 dx So, the integral is x (1 + 4y) ds x ( 1+4x 2) 1+4x 2 dx c x x x ( 1+4x 2) 3/2 dx This can be evaluated using the transformation U 1+4x 2 so du 8xdx i.e. dx du 8. When x, U 1 and when x 1, U 5. The integral therefore equals x 5 x ( 1+4x 2) 3/2 1 dx U 3/2 du 8 U1 1 2 [ ] U 5/ [ 1 5 5/2 1 ] Example Find xy dx where, on, x and y are given by x 3t2, 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 6tdt. The limits on t are t and t 1. The integral becomes xy dx 3t 2 (t 3 1) 6t dt (18t 6 18t 3 ) dt t [ 18 7 t t4 ] 1 t Key Point 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. 5 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

6 1. For F (x, y) 2x+y 2, find F (x, y) dx, F (x, y) dy and F (x, y) ds where is the line y 2x from (, ) to (1, 2). (a) Express each integral as a simple integral with respect to a single variable. (b) Hence evaluate each integral. 2. For F (x, y) 1, find F (x, y) dx, F (x, y) dy and F (x, y) ds where is the curve y 1 2 x2 1 ln x from (1, 1)to(2, 2 1 ln 2) For F (x, y) sin 2x, find F (x, y) dx, F (x, y) dy and F (x, y) ds where is the curve y sin x from (, ) to ( π, 1). 2 Your solution 1.) x (2x +4x 2 )dx, 7 3, 2 y (y + y 2 )dy, 14 3, x (2x +4x 2 ) 5dx, Your solution 2.) 1, ln 2, ln 2. HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 6

7 Your solution 3.) 2, 2, 2(2 2 1) Line Integrals of Scalar Products Integrals of the form F dr, referred to at the end of the previous sub-section, occur in applications such as the following. B A v v T δr T S (current position) onsider a cyclist riding along the road from A to B. Suppose it is necessary to find the total work the cyclist has to do in overcoming the wind. The work done is proportional to the component of the wind speed in the direction travelled i.e. proportional to v δr The total work done over AB is approximately all δr v δr, where the summation is carried out over all small elements making up AB. B In the limit δr, the total work done over AB is lim all v δr v δr. δr A This is an example of the integral along a specific line of the scalar product of a vector field and avector describing the contour. The term scalar line integral is often used for integrals of this form but the role of the vector v should not be forgotten. The vector dr may be considered to be dx i + dy j + dz k. 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] inthree dimensions ( [F x dx + F y dy] intwo 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, orinterms of a parameter t. Note that in examples t often represents time. 7 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

8 Example Find (2xy dx 5x dy) where is the curve y x3 with x varying from x tox 1. 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) 2xx 3 dx +5x 3x 2 dx ) (2x 4 15x 3 )dx x [ 2 5 x x4 ] Key Point An integral of the form 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 Three paths from (, ) to (1, 2) are defined by (i) 1 : y 2x (ii) 2 : y 2x 2 (iii) 3 : y from (, ) to (1, ) and x 1from (1, ) to (1, 2) Sketch each path and find F dr, where F y 2 i + xyj, along each path. HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 8

9 2 y y 2x A(1, 2) 2 y y 2x 2 A(1, 2) 2 y x 1 A(1, 2) x 1 x y 1 x Solution (i) F dr (y 2 dx + xy dy ). Along y 2x, dy dx 1 F dr x ( (2x) 2 dx + x (2x)(2 dx) ) 2sody 2dx. Then ( 4x 2 +4x 2) dx 8x 2 dx [ ] x2 8 3 Solution (ii) F dr (y 2 dx + xy dy ). Along y 2x 2, dy dx 2 F dr x 4x so dy 4x dx. Then ( (2x 2 ) 2 dx + x ( 2x 2 ) (4x dx)) 12x 4 dx [ ] x HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

10 Solution (iii) 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 O B Along OB, y sody. Then B O F dr Along AB, x 1sodx. Then B A F dr 2 Hence F dr y x ( 2 dx + x ) dx ( y 2 +1 y dy ) 2 y dy [ ] y2 2. Key Point In general the value of the line integral depends on the path of integration as well as the end points. Example Find O A F dr, where F y 2 i + xyj (as in the previous example) and the path from A to O is the straight line from (1, 2) to (, ), that is the reverse of 1 in (i) above. 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 (, ). HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 1

11 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 F dr 8 A O 3 The integral along the parabola (calculated in (iii) above) evaluates to 12 5, then F dr F dr + F dr Example onsider the two vector fields F y 2 z 3 i +2xyz 3 j +3xy 2 z 2 k and 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, y t 2, z t 2 ( t 1) (a) Evaluate the scalar integral of each vector field along each path. (b) Find the value of F dr and F dr where is the closed path along 1 from O to A and back along 2 to O. 11 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

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 1and 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 1correspond to the start and end point of 1 and so these are the required limits of integration. Now 1 F dr F dr dt dt Substituting for x y z t in G we have 6t 5 dt [ t 6] 1 1 G ti +3tj and G dr t +3t 4t dt The limits of integration are again t and t 1,then 1 G dr G dr dt dt 4t dt [ 2t 2] 1 2 For the path 2 the parameterisation is x t 2, y t and z t 2 so dr dt 2ti+j+2tk. 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 F dr 1t 9 dt [ t 1] 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 ( G dr 2t 3 +4t 2 t ) [ 1 dt 2 2 t t3 1 ] 1 2 t2 4 3 (b) For the closed path F dr F dr F dr G dr G dr G dr HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 12

13 Summary Vector Field Path Line Integral F 1 1 F 2 1 F closed G 1 2 G 2 4/3 G closed 2/3 Note that the line integral of F is 1 for both paths. 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 divergence of a scalar field or the curl of a vector field. Example Find [ (x2 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 + x2 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 13 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

14 1. Evaluate F dr, where F (x y)i +(x + y)j along each of the following paths from (1, 1) to (2, 4). (a) 1 : the straight line y 3x 2. (b) 2 : the parabola y x 2. (c) 3 : the straight line x 1from (1, 1) to (1, 4) followed by the straight line y 4from (1, 4) to (2, 4). 2. For the function F and paths in question 1, 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 onsider F dr, where F 3x2 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 4. onsider the two vector fields F 2xi +(xz 2)j + xyk and G x 2 zi + y 2 zj (x3 + y 3 )k and the two curves between (,, ) and (1, 1, 2) 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 1 F dr, 2 F dr, 1 G dr, 2 G dr (b) Find F dr and G dr where is the closed path from (,, ) to (1, 1, 2) along 1 and back to (,, ) along Find F dt along y 2x from (, ) to (2, 4) for (a) F (x 2 y) (b) F ( 1 2 x2 y 2 k) HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 14

15 Your solution 1.) 11, 34 3,8 Your solution 2.) 1 3, 1 3, 3. Your solution 3.) All are HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

16 Your solution 4.) 2, 5 3,,, 1 3,. Your solution 5.) 16, onservative vector fields For certain of the line integrals in the previous section, the integral depended only on the vector field F and the start and end points of the contour but not on the actual path of the contour between the start and end points. However, for other line integrals, the result depended on the actual details of the path of the contour. 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 conservative vector field. Since it is impossible to check the value of every line integral over every path, it is possible to use these five properties (and in particular property P 3 below to determine whether a vector field is conservative. They are also used to simplify calculations with conservative vector fields. P1 The line integral B of the actual path taken. A F dr depends only on the end points A and B and is independent HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 16

17 P2 The line integral around any closed curve is zero. That is F dr for all. P3 The curl of a conservative vector field F is zero i.e. F P4 Forany conservative vector field F,itispossible 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 All gradient fields are conservative. That is F φ is a conservative vector field for any scalar field φ. Example The following vector fields were considered in the examples of the previous subsection. 1. F 1 y 2 i + xyj 2. F 2 2xi +2yj 3. F 3 y 2 z 3 i +2xyz 3 j +3xy 2 z 2 k 4. F 4 xi +(4x y) j 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 P4 to verify the line integrals found. Solution 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 by finding that F yk ]. 17 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

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,P3gives 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 Thus y Using P4: (4,2) (,) φ x 2x φ x2 + f(y) φ y 2y φ y2 + g(x) F 2 dr (4,2) (,) φ x 2 + y 2 (+ constant) ( φ) dr φ(4, 2) φ(, ) ( ) ( )2. 3 The fact that line integrals by two different paths between the same start and end points is consistent with F 3 being a conservative field. So too is the fact that the integral around a closed path is zero. 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,P3gives that F 3 is a conservative field. To find φ that satisfies φ F 3,itisnecessary to satisfy φ x y2 z 3 φ xy 2 z 3 + f(y, z) φ y 2xyz3 φ xy 2 z 3 + g(x, z) φ xy 2 z 3 φ z 3xy2 z 2 φ xy 2 z 3 + h(x, y) Using P4: (1,1,1) (,,) F 3 dr φ(1, 1, 1) φ(,, ) 1 1. HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 18

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 3, F 4 is NOT a conservative field. Note that F 4 4k so this is an independent conclusion that F 4 is NOT conservative. Example 1. Show that I (2,1) (,) [(2xy +1)dx +(x2 2y)dy] isindependent of the path taken 2. Find I using property P1. 3. Find I using property P4. 4. Find I [(2xy +1)dx +(x2 2y)dy] where is (a) the circle x 2 + y 2 1 (b) the square with vertices (, ), (1, ), (1, 1), (, 1). 19 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

20 Solution 1. The integral I (2,1) [(2xy +1)dx (,) +(x2 2y)dy] may be re-written F dr where F (2xy +1)i +(x 2 2y)j. i j k Now F x y z i +j +k 2xy +1 x 2 2y 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 2 2 integral I becomes I (2,1) (,) 2 [ (2xy +1)dx +(x 2 2y)dy ] 2 ( 3 2 x2 1 2 x +1)dx [ 1 2 x3 1 4 x2 + x x ] 2 [(2x 12 x +1)dx +(x2 4x) 12 dx ] If F φ then φ x 2xy +1 φ x2 y + x + f(y) φ xy 2 z 3. φ y x2 2y 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 set equal to zero). So I φ(2, 1) φ(, ) (4+2 1) 5 4. As F is a conservative field, all integrals around a closed contour are zero. HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 2

21 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 3x2 y 2 i +(2x 3 y 1)j. Noting that F is a conservative vector field, find a scalar field φ so that φ F. Hence evaluate the integral F dr where is an integral with startpoint (, ) 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; 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 1i x +4y3 z 2 j +2y 4 zk; any path from (1, 1, 1) to (1, 2, 3) Your solution 1.) No, Yes, No, Yes Your solution 2.) x 3 y 2 y, HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

22 Your solution 3.) x 4 y x 2 y 2,11;e x + xy 3,;xy 2 + x sin z,1; ln x + y 4 z 2, Vector Line Integrals It is also possible to form the 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, anintegral 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. Ifanintegral is two-dimensional, the term in z will be absent. Example Evaluate the integral xy2 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 + x ] 1 [ 1 6 x6 x(x 2 ) 2 dx i + i + [ 2 7 x7/2 y ] 1 y 1/2 y 2 dy j j 1 6 i j x 5 dx i + xy 2 dy j y 5/2 dy j HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 22

23 Example Find I xzdr 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. the contour starts at (1,, π) and finishes at (1,,π)). Solution The integral becomes x(dxi + dyj + dzk). Now, x cos t, y sin t, z t π so dx sin tdt, dy cos tdtand dz dt. So I 2π 2π cos t( sin tdti+ cos tdtj+ dtk) 2π cos t sin tdti+ 2π 2π cos 2 tdtj+ 1 sin 2t dti [cos 2t]2π i + 1 [t + 12 ] 2π 4 2 sin 2t j +k i + πj πj 2π cos tdtk (1 + cos 2t) dt j + [sin t] 2π k Integrals of the form F dr can be evaluated as follows. The vector field F F 1i+F 2 j +F 3 k and dr dx i + dy j + dz k so F dr i j k F 1 F 2 F 3 dx dy dz (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)i +(F 1 k F 3 i)j +(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. Example Evaluate the integral (x2 i+3xyj) dr where represents the curve y 2x 2 from (, ) to (1, 2). 23 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

24 Solution Note that the z component of F and dz are both zero. i j k So F dr x 2 3xy (x 2 dy 3xydx)k dx dy and (x2 i +3xyj) dr (x2 dy 3xydx)k Now, on, y 2x 2 so dy 4xdx and (x 2 i +3xyj) dr (x 2 dy 3xydx)k x ] 1 [ 1 2 x4 [ x 2 4xdx 3x 2x 2 dx ] k ( 2x 3 )dxk k 1 2 k A scalar or vector involved in a vector line integral may itself be a vector derivative. Example Find the vector line integral ( F ) dr where F is the vector x2 i+2xyj+2xzk and is the curve y x 2, z x 3 from x tox 1i.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 6xdxi+ 6xdyj+ 6xdzk. The first term is 6x dxi 6x dxi [ 3x 2] 1 i 3i In the second term, as y x 2 on, dy may be replaced by 2x dxso 6x dyj 6x 2x dxj 12x 2 dx j [ 4x 3] 1 j 4j x In the third term, as z x 3 on, dz may be replaced by 3x 2 dx so [ ] 1 9 6x dzk 6x 3x 2 dx k 18x 3 dx k x 2 x4 k 9 2 k On summing, ( F ) dr 3i +4j k. x HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 24

25 1. Find the vector line integral fdr where f x2 and is (a) the curve y x 1/2 from (, ) to (9, 3). (b) the line y x/3 from (, ) to (9, 3). 2. When represents the contour y 4 4x, z 2 2x from (, 4, 2) to (1,, ) and F is the vector field (x z)j, evaluate the vector line integral F dr. 3. Evaluate the vector line integral ( 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) [ (x2 i + y 2 k)] dr Your solution 1.) 243i j,243i + j. 2 Your solution 2.) i j Your solution 3.) 25 HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus

26 4i j 2k Your solution 4.), k HELM (VERSION 1: April 13, 24): Workbook Level : Vector alculus 26