About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project

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2 About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project undertaken by a consortium of five English universities led by Loughborough University, funded by the Higher Education Funding ouncil for England under the Fund for the Development of Teaching and Learning for the period October eptember 5. HELM aims to enhance the mathematical education of engineering undergraduates through a range of flexible learning resources in the form of Workbooks and web-delivered interactive segments. HELM supports two AA regimes: an integrated web-delivered implementation and a D-based version. HELM learning resources have been produced primarily by teams of writers at six universities: Hull, Loughborough, Manchester, Newcastle, Reading, underland. HELM gratefully acknowledges the valuable support of colleagues at the following universities and colleges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston, Bournemouth & Poole ollege, ambridge, ity, Glamorgan, Glasgow, Glasgow aledonian, Glenrothes Institute of Applied Technology, Harper Adams University ollege, Hertfordshire, Leicester, Liverpool, London Metropolitan, Moray ollege, Northumbria, Nottingham, Nottingham Trent, Oxford Brookes, Plymouth, Portsmouth, Queens Belfast, Robert Gordon, Royal Forest of Dean ollege, alford, ligo Institute of Technology, outhampton, outhampton Institute, urrey, Teesside, Ulster, University of Wales Institute ardiff, West Kingsway ollege (London), West Notts ollege. HELM ontacts: Post: HELM, Mathematics Education entre, Loughborough University, Loughborough, LE 3TU. helm@lboro.ac.uk Web: HELM Workbooks List Basic Algebra 6 Functions of a omplex Variable Basic Functions 7 Multiple Integration 3 Equations, Inequalities & Partial Fractions 8 Differential Vector alculus 4 Trigonometry 9 Integral Vector alculus 5 Functions and Modelling 3 Introduction to Numerical Methods 6 Exponential and Logarithmic Functions 3 Numerical Methods of Approximation 7 Matrices 3 Numerical Initial Value Problems 8 Matrix olution of Equations 33 Numerical Boundary Value Problems 9 Vectors 34 Modelling Motion omplex Numbers 35 ets and Probability Differentiation 36 Descriptive tatistics Applications of Differentiation 37 Discrete Probability Distributions 3 Integration 38 ontinuous Probability Distributions 4 Applications of Integration 39 The Normal Distribution 5 Applications of Integration 4 ampling Distributions and Estimation 6 equences and eries 4 Hypothesis Testing 7 onics and Polar oordinates 4 Goodness of Fit and ontingency Tables 8 Functions of everal Variables 43 Regression and orrelation 9 Differential Equations 44 Analysis of Variance Laplace Transforms 45 Non-parametric tatistics z-transforms 46 Reliability and Quality ontrol Eigenvalues and Eigenvectors 47 Mathematics and Physics Miscellany 3 Fourier eries 48 Engineering ase tudies 4 Fourier Transforms 49 tudent s Guide 5 Partial Differential Equations 5 Tutor s Guide opyright Loughborough University, 6

3 ontents 9 Integral Vector alculus 9. Line Integrals 9. urface and Volume Integrals Integral Vector Theorems 55 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 viz tokes' theorem, Gauss' divergence theorem and Green's theorem.

4 urface and Volume Integrals 9. Introduction A vector or scalar field - including one formed from a vector derivative (div, grad or curl) - can be integrated over a surface or volume. This ection shows how to carry out such operations. Prerequisites Before starting this ection you should... Learning Outcomes On completion you should be able to... be familiar with vector derivatives be familiar with double and triple integrals carry out operations involving integration of scalar and vector fields 34 HELM (8): Workbook 9: Integral Vector alculus

5 . urface integrals involving vectors The unit normal For the surface of any three-dimensional shape, it is possible to find a vector lying perpendicular to the surface and with magnitude. The unit vector points outwards from a closed surface and is usually denoted by ˆn. Example 7 If is the surface of the sphere x + y + z a find the unit normal ˆn. olution The unit normal at the point P (x, y, z) points away from the centre of the sphere i.e. it lies in the direction of xi + yj + zk. To make this a unit vector it must be divided by its magnitude x + y + z i.e. the unit vector is ˆn x x + y + z i + y x + y + z j + z x + y + z k x a i + y a j + z a k where a x + y + z is the radius of the sphere. z n P (x, y, z) a x y k j i Figure 6: A unit normal ˆn to a sphere HELM (8): ection 9.: urface and Volume Integrals 35

6 Integral Vector Theorems 9.3 Introduction Various theorems exist relating integrals involving vectors. Those involving line, surface and volume integrals are introduced here. They are the multivariable calculus equivalent of the fundamental theorem of calculus for single variables ( integration and differentiation are the reverse of each other ). Use of these theorems can often make evaluation of certain vector integrals easier. This ection introduces the main theorems which are Gauss divergence theorem, tokes theorem and Green s theorem. Prerequisites Before starting this ection you should... Learning Outcomes On completion you should be able to... be able to find the gradient of a scalar field and the divergence and curl of a vector field be familiar with the integration of vector functions use vector integral theorems to facilitate vector integration HELM (8): ection 9.3: Integral Vector Theorems 55

7 . tokes theorem This is a theorem that equates a line integral to a surface integral. For any vector field F and a contour which bounds an area, ( F ) d F dr d Notes Figure 6: A surface for tokes theorem (a) d is a vector perpendicular to the surface and dr is a line element along the contour. The sense of d is linked to the direction of travel along by a right hand screw rule. (b) Both sides of the equation are scalars. (c) The theorem is often a useful way of calculating a line integral along a contour composed of several distinct parts (e.g. a square or other figure). (d) F is a vector field representing the curl of the vector field F and may, alternatively, be written as curl F. Justification of tokes theorem Imagine that the surface is divided into a set of infinitesimally small rectangles ABD where the axes are adjusted so that AB and D lie parallel to the new x-axis i.e. AB δx and B and AD lie parallel to the new y-axis i.e. B δy. Now, F dr is calculated, where is the boundary of a typical such rectangle. The contributions along AB, B, D and DA are Thus, F (x, y, ) δx F x (x, y, z)δx, F (x + δx, y, ) δy F y (x + δx, y, z)δy, F (x, y + δy, ) ( δx) F x (x, y + δy, z)δx F (x, y, ) ( δx) F y (x, y, z)δy. F dr (F x (x, y, z) F x (x, y + δy, z))δx +(F y (x + δx, y, z) F y (x, y, z))δy F y x δxδy F x y δxδy ( F ) z δ ( F ) d as d is perpendicular to the x- and y- axes. 56 HELM (8): Workbook 9: Integral Vector alculus

8 Thus, for each small rectangle, F dr ( F ) d When the contributions over all the small rectangles are summed, the line integrals along the inner parts of the rectangles cancel and all that remains is the line integral around the outside of the surface. The surface integrals sum. Hence, the theorem applies for the area bounded by the contour. While the above does not constitute a formal proof of tokes theorem, it does give an appreciation of the origin of the theorem. ontribution does not cancel ontributions cancel Figure 7: Line integral cancellation and non-cancellation Key Point 8 tokes Theorem F dr ( F ) d The closed contour integral of the scalar product of a vector function with the vector along the contour is equal to the integral of the scalar product of the curl of that vector function and the unit normal, over the corresponding surface. HELM (8): ection 9.3: Integral Vector Theorems 57

9 Example 3 Verify tokes theorem for the vector function F y i (x + z)j + yzk and the unit square x, y, z. olution If F y i (x + z)j + yzk then F (z +)i +( y)k i +( y)k (as z ). Note that d dxdyk so that ( F ) d ( y)dydx Thus To evaluate ( F ) d x x y ( y)dydx ( y y ) x dx y + F dr, we must consider the four sides separately. x ( )dx When y, F xj and dr dxi so F dr i.e. the contribution of this side to the integral is zero. When x, F y i j and dr dyj so F dr dy so the contribution to the integral is ( dy) y. y When y, F i xj and dr dxi so F dr dx so the contribution to the integral is ( dx) x. x When x, F y i and dr dyj so F dr so the contribution to the integral is zero. The integral F dr is the sum of the contributions i.e. +. Thus ( F ) d F dr i.e. tokes theorem has been verified. 58 HELM (8): Workbook 9: Integral Vector alculus

10 Example 33 Using cylindrical polar coordinates verify tokes theorem for the function F ρ ˆφ the circle ρ a, z and the surface ρ a, z. olution Firstly, find F dr. This can be done by integrating along the contour ρ a from φ to φ π. Here F a ˆφ (as ρ a) and dr a dφ ˆφ (remembering the scale factor) so F dr a 3 dφ and hence π F dr a 3 dφ πa 3 As F ρ ˆφ, F 3ρẑ and ( F ) d 3ρ as d ẑ. Thus π π a ( F ) d 3ρ ρdρdφ Hence φ π φ ρ ρ 3 a ρ dφ F dr ( F ) d πa 3 π φ ρ a 3 dφ πa 3 3ρ dρdφ Example 34 Find the closed line integral F dr for the vector field F y i+(x z)j+xyk and for the contour ABDEF GHA in Figure 8. y F (, 7) E(5, 7) H(, 4) G(, 4) D(, 4) (6, 4) A(, ) B(6, ) x Figure 8: losed contour ABDEF GHA HELM (8): ection 9.3: Integral Vector Theorems 59

11 olution To find the line integral directly would require eight line integrals i.e. along AB, B, D, DE, EF, FG, GH and HA. It is easier to carry out a surface integral to find ( F ) d which is equal to the required line integral F dr by tokes theorem. i j k As F y i +(x z)j +xyk, F x y z (x +)i yj + (x y)k y x z xy As the contour lies in the x-y plane, the unit normal is k and d dxdyk Hence ( F ) d (x y)dxdy. To work out ( F ) d, it is necessary to divide the area inside the contour into two smaller areas i.e. the rectangle ABDGH and the trapezium DEFG. On ABDGH, the integral is 4 y 6 x (x y)dxdy On DEFG, the integral is 7 y4 y x (x y)dxdy x xy dy (36 y)dy y x y 4 36y 6y y4 y x xy 3 y3 + y +3y dy x o the full integral is, ( F ) d By tokes theorem, F dr 3 y4 ( y +y +3)dy From tokes theorem, it can be seen that surface integrals of the form ( F ) d depend only on the contour bounding the surface and not on the internal part of the surface. 6 HELM (8): Workbook 9: Integral Vector alculus

12 Task Verify tokes theorem for the vector field F x i +xyj + zk and the triangle with vertices at (,, ), (3,, ) and (3,, ). First find the normal vector d: Your solution Answer dxdyk Then find the vector F : Your solution Answer yk Now evaluate the double integral ( F ) d over the triangle: Your solution Answer HELM (8): ection 9.3: Integral Vector Theorems 6

13 Finally find the integral the tokes theorem are equal: Your solution F dr along the 3 sides of the triangle and so verify that the two sides of Answer 9+3, Both sides of tokes theorem have value. Exercises. Using plane-polar coordinates (or cylindrical polar coordinates with z ), verify tokes πρ theorem for the vector field F ρˆρ + ρ cos ˆφ and the semi-circle ρ, π φ π.. Verify tokes theorem for the vector field F xi +(y z)j + xzk and the contour around the rectangle with vertices at (,, ),(,, ), (,, ) and (,, ). 3. Verify tokes theorem for the vector field F yi + xj + zk (a) Over the triangle (,, ), (,, ), (,, ). (b) Over the triangle (,, ), (,, ), (,, ). 4. Use tokes theorem to evaluate the integral F dr where F sin( x +)+5y i + (x e y )j and is the contour starting at (, ) and going to (5, ), (5, ), (6, ), (6, 5), (3, 5), (3, ), (, ) and returning to (, ). Answers. Both integrals give,. Both integrals give 3. (a) Both integrals give (b) Both integrals give (as F is perpendicular to d) 4. 57, [ F 3 k]. 6 HELM (8): Workbook 9: Integral Vector alculus

14 . Gauss theorem This is sometimes known as the divergence theorem and is similar in form to tokes theorem but equates a surface integral to a volume integral. Gauss theorem states that for a volume V, bounded by a closed surface, any well-behaved vector field F satisfies F d F dv Notes: V (a) d is a unit normal pointing outwards from the interior of the volume V. (b) Both sides of the equation are scalars. (c) The theorem is often a useful way of calculating a surface integral over a surface composed of several distinct parts (e.g. a cube). (d) F is a scalar field representing the divergence of the vector field F and may, alternatively, be written as div F. (e) Gauss theorem can be justified in a manner similar to that used for tokes theorem (i.e. by proving it for a small volume element, then summing up the volume elements and allowing the internal surface contributions to cancel.) Key Point 9 Gauss Theorem F d F dv The closed surface integral of the scalar product of a vector function with the unit normal (or flux of a vector function through a surface) is equal to the integral of the divergence of that vector function over the corresponding volume. V HELM (8): ection 9.3: Integral Vector Theorems 63

15 Example 35 Verify Gauss theorem for the unit cube x, y, z and the function F xi + zj olution To find F d, the integral must be evaluated for all six faces of the cube and the results summed. On the face x, F zj and d i dydz so F d and F d dydz On the face x, F i + zj and d i dydz so F d dydz and F d dydz On the face y, F xi + zj and d j dxdz so F d z dxdz and F d z dxdz On the face y, F xi + zj and d j dxdz so F d z dxdz and F d z dxdz On the face z, F xi and d k dydz so F d dxdy and F d dxdy On the face z, F xi + j and d k dydz so F d dxdy and F d dxdy Thus, summing over all six faces, F d To find F dv note that F V x x + y z +. o F dv dxdydz. V o F d F dv hence verifying Gauss theorem. V Note: The volume integral needed just one triple integral, but the surface integral required six double integrals. Reducing the number of integrals is often the motivation for using Gauss theorem. 64 HELM (8): Workbook 9: Integral Vector alculus

16 Example 36 Use Gauss theorem to evaluate the surface integral F d where F is the vector field x yi +xyj + z 3 k and is the surface of the unit cube x, y, z. olution Note that to carry out the surface integral directly will involve, as in Example 35, the evaluation of six double integrals. However, by Gauss theorem, the same result comes from the volume integral F dv. As F xy +x +3z, we have the triple integral V (xy +x +3z ) dxdydz x y + x +3xz x dydz y + y +3yz 3 z + z3 5 dz y ( ++3z )dz (y ++3z )dydz ( 3 +3z )dz The six double integrals would also sum to 5 but this approach would require much more effort. Engineering Example 5 Gauss law Introduction From Gauss theorem, it is possible to derive a result which can be used to gain insight into situations arising in Electrical Engineering. Knowing the electric field on a closed surface, it is possible to find the electric charge within this surface. Alternatively, in a sufficiently symmetrical situation, it is possible to find the electric field produced by a given charge distribution. Gauss theorem states F d F dv V If F E, the electric field, it can be shown that, F E q ε HELM (8): ection 9.3: Integral Vector Theorems 65

17 where q is the amount of charge per unit volume, or charge density, and ε is the permittivity of free space: ε 9 /36π Fm 8.84 Fm. Gauss theorem becomes in this case q E d E dv dv q dv Q ε ε ε i.e. E d Q ε V which is known as Gauss law. Here Q is the total charge inside the surface. Note: this is one of the important Maxwell s Laws. Problem in words V A point charge lies at the centre of a cube. Given the electric field, find the magnitude of the charge, using Gauss law. Mathematical statement of problem onsider the cube x, y, z where the dimensions are in metres. A point charge Q lies at the centre of the cube. If the electric field on the top face (z ) is given by xi + yj + zk E (x + y + z ) 3 find the charge Q from Gauss law. V Hint : x y x + y + 3 4π dy dx 4 3 Mathematical analysis From Gauss law so E d Q ε Q ε E d 6ε (top) E d since, using the symmetry of the six faces of the cube, it is possible to integrate over just one of them (here the top face is chosen) and multiply by 6. On the top face and E xi + yj + k x + y + 43 d (element of surface area) (unit normal) dx dy k 66 HELM (8): Workbook 9: Integral Vector alculus

18 o E d x + y + 43 dy dx 5 x + y + 3 dy dx 4 Now (top) E d x y 5 x + y + 3 dy dx 4 5 4π 3 π 3 (using the hint) o, from Gauss law, Q 6ε π 3 4πε 9 Interpretation Gauss law can be used to find a charge from its effects elsewhere. The form of E xi + yj + k comes from the fact that E is radial and equals r x + y + r ˆr 43 3 r Example 37 Verify Gauss theorem for the vector field F y j xzk and the triangular prism with vertices at (,, ), (,, ), (,, ), (, 4, ), (, 4, ) and (, 4, ) (see Figure 9). z (, 4, ) y (,, ) (,, ) (, 4, ) (,, ) (, 4, ) x Figure 9: The triangular prism defined by six vertices HELM (8): ection 9.3: Integral Vector Theorems 67

19 olution As F y j xzk, F +y x y x. Thus 4 x/ F dv (y x)dzdydx V To work out x y 4 x x y z yz xz x/ z y xy xy + x y 6x 6x + 3 x3 4 3 dydx 4 x 4 dx y y (y xy x + x )dydx x (6 x +x )dx F d, it is necessary to consider the contributions from the five faces separately. On the front face, y, F xzk and d j thus F d and the contribution to the integral is zero. On the back face, y 4, F 6j xzk and d j thus F d 6 and the contribution to the integral is x x/ z 6dzdx x x/ 6z dx 6( x/)dx 6x 4x 6. z x On the left face, x, F y j and d i thus F d and the contribution to the integral is zero. On the bottom face, z, F y j and d k thus F d and the contribution to the integral is zero. On the top right (sloping) face, z x/, F y j+( x x)k and the unit normal ˆn 5 i+ 5 k Thus d 5 i + 5 k dydw where dw measures the distance along the slope for a constant y. As dw 5 dx, d i + k dydx thus F d 6 and the contribution to the integral is 4 ( x x)dydx (x 4x)dx 3 x3 x 8 3. x y Adding the contributions, Thus F d V x F d F dv 4 3 hence verifying Gauss divergence theorem. 68 HELM (8): Workbook 9: Integral Vector alculus

20 Engineering Example 6 Field strength around a charged line Problem in words Find the electric field strength at a given distance from a uniformly charged line. Mathematical statement of problem Determine the electric field at a distance r from a uniformly charged line (charge per unit length ρ L ). You may assume from symmetry that the field points directly away from the line. l r Mathematical analysis Figure : Field strength around a line charge Imagine a cylinder a distance r from the line and of length l (see Figure ). From Gauss law E d Q ε As the charge per unit length is ρ L, then the right-hand side equals ρ L l/ε. On the left-hand side, the integral can be expressed as the sum E d E d + E d (ends) (curved) Looking first at the circular ends of the cylinder, the fact that the field lines point radially away from the charged line implies that the electric field is in the plane of these circles and has no normal component. Therefore E d will be zero for these ends. Next, over the curved surface of the cylinder, the electric field is normal to it, and the symmetry of the problem implies that the strength of the electric field will be constant (here denoted by E). Therefore the integral Total curved surface area Field strength πrle. o, by Gauss law E d + E d Q (ends) (curved) ε or +πrle ρ Ll ε Interpretation Hence, the field strength E is given by E ρ L πε r HELM (8): ection 9.3: Integral Vector Theorems 69

21 Engineering Example 7 Field strength on a cylinder Problem in words Given the electric field E on the surface of a cylinder, use Gauss law to find the charge per unit length. Mathematical statement of problem On the surface of a long cylinder of radius a and length l, the electric field is given by E ρ L πε (a + b cos θ)ˆr b sin θ ˆθ (a +ab cos θ + b ) (using cylindrical polar co-ordinates) due to a line of charge a distance b (< a) from the centre of the cylinder. Using Gauss law, find the charge per unit length. Hint:- π Mathematical analysis a + b cos θ (a +ab cos θ + b ) dθ π a onsider a cylindrical section - as in the previous example, there are no contributions from the ends of the cylinder since the electric field has no normal component here. However, on the curved surface so d a dθ dz ˆr E d ρ L πε a + b cos θ (a +ab cos θ + b ) a dθ dz Integrating over the curved surface of the cylinder E d l z aρ Ll πε ρ Ll ε θπ θ π aρ L πε a + b cos θ (a +ab cos θ + b ) dθ a + b cos θ (a +ab cos θ + b ) dθ dz using the given result for the integral. Then, if Q is the total charge inside the cylinder, from Gauss law ρ L l ε Q ε so ρ L Q l Interpretation as one would expect. Therefore the charge per unit length on the line of charge is given by ρ L (i.e. the charge per unit length is constant). 7 HELM (8): Workbook 9: Integral Vector alculus

22 Task Verify Gauss theorem for the vector field F xi yj + zk and the unit cube x, y, z. (a) Find the vector F. (b) Evaluate the integral z y x F dxdydz. (c) For each side, evaluate the normal vector d and the surface integral F d. (d) how that the two sides of the statement of Gauss theorem are equal. Your solution Answer (a) + (b) (c) dxdyk, ; dxdyk, ; dxdzj, ; dxdzj, ; dydzi, ; dydzi, (d) Both sides are equal to. HELM (8): ection 9.3: Integral Vector Theorems 7

23 Exercises. Verify Gauss theorem for the vector field F 4xzi y j + yzk and the cuboid x, y 3, z 4.. Verify Gauss theorem, using cylindrical polar coordinates, for the vector field F ρ ˆρ over the cylinder ρ r, z for (a) r (b) r 3. If is the surface of the tetrahedron with vertices at (,, ), (,, ), (,, ) and (,, ), find the surface integral (xi + yzj) d (a) directly (b) by using Gauss theorem Hint :- When evaluating directly, show that the unit normal on the sloping face is 3 (i + j + k) and that d (i + j + k)dxdy Answers. Both sides are 56,. Both sides equal (a) 4π, (b) π, 3. (a) 5 4 [only contribution is from the sloping face] (b) 5 4 [by volume integral of ( + z)]. 7 HELM (8): Workbook 9: Integral Vector alculus

24 3. Green s Identities (3D) Like Gauss theorem, Green s identities relate surface integrals to volume integrals. However, Green s identities are concerned with two scalar fields u(x, y, z) and w(x, y, z). Two statements of Green s identities are as follows (u w) d u w + u w dv [] V and {u w v u} d V u w w u dv [] Proof of Green s identities Green s identities can be derived from Gauss theorem and a vector derivative identity. Vector identity () from subsection 6 of 8. states that (φa) ( φ) A + φ( A). Letting φ u and A w in this identity, (u w) ( u) ( w)+u( ( w)) ( u) ( w)+u w Gauss theorem states F d Now, letting F u w, (u w) d V F dv V V (u w)dv ( u) ( w)+u w dv This is Green s identity []. Reversing the roles of u and w, (w u) d ( w) ( u)+w u dv V ubtracting the last two equations yields Green s identity []. Key Point [] [] Green s Identities (u w) d u w + u w dv V {u w v u} d u w w u dv V HELM (8): ection 9.3: Integral Vector Theorems 73

25 Example 38 Verify Green s first identity for u (x x )y, w xy + z and the unit cube, x, y, z. olution As w xy + z, w yi + xj +zk. Thus u w (xy x y)(yi + xj +zk) and the surface integral is of this quantity (scalar product with d) integrated over the surface of the unit cube. On the three faces x, x, y, the vector u w and so the contribution to the surface integral is zero. On the face y, u w (x x )(i+xj +zk) and d dxdzj so (u w) d (x x 3 )dxdz and the contribution to the integral is x (x x 3 )dzdx (x x 3 3 )dx 3 x4 4. x z On the face z, u w (x x )y(yi + xj) and d dxdzk so (u w) d and the contribution to the integral is zero. On the face z, u w (x x )y(yi+xj+k) and d dxdyk so (u w) d y(x x )dxdy and the contribution to the integral is y(x x )dydx y (x x ) dx (x x )dx x y x y 6. Thus, (u w) d Now evaluate V u w + u w dv. Note that u ( x)yi +(x x )j and w so u w + u w ( x)y +(x x )x +(x x )y x x 3 +xy x y + y xy and the integral Hence V u w + u w dv (u w) d V z y z y z y x x 3 z( 4 )dz z 4 (x x 3 +xy x y + y xy )dxdydz 3 x4 4 + x y 3 x3 y + xy x y ( + y 3 )dydz z 4 u w + u w dv 4 z y + y dz 6 y dydz x and Green s first identity is verified. 74 HELM (8): Workbook 9: Integral Vector alculus

26 Green s theorem in the plane This states that Q (P dx + Qdy) x P dxdy y is a -D surface with perimeter ; P (x, y) and Q(x, y) are scalar functions. This should not be confused with Green s identities. Justification of Green s theorem in the plane Green s theorem in the plane can be derived from tokes theorem. ( F ) d F dr Now let F be the vector field P (x, y)i + Q(x, y)j i.e. there is no dependence on z and there are no components in the z direction. Now i j k F Q x y z x P k y P (x, y) Q(x, y) Q and d dxdyk giving ( F ) d x P dxdy. y Thus tokes theorem becomes Q x P dxdy F dr y and Green s theorem in the plane follows. Key Point Green s Theorem in the Plane Q (P dx + Qdy) x P dxdy y This relates a line integral around a closed path with a double integral over the region enclosed by. It is effectively a two-dimensional form of tokes theorem. HELM (8): ection 9.3: Integral Vector Theorems 75

27 Example 39 Evaluate the line integral (4x + y 3)dx + (3x +4y )dy around the rectangle x 3, y. olution The integral could be obtained by evaluating four line integrals but it is easier to note that [(4x + y 3)dx + (3x +4y )dy] is of the form P dx + Qdy with P 4x + y 3 and Q 3x +4y. It is thus of a suitable form for Green s theorem in the plane. Note that Q P 6x and x y. Green s theorem in the plane becomes {(4x + y 3)dx + (3x +4y )dy} y y 3 x 3x x (6x ) dxdy 3 dy x y 4 dy 4 76 HELM (8): Workbook 9: Integral Vector alculus

28 Example 4 Verify Green s theorem in the plane for the integral 4zdy +(y )dz and the triangular contour starting at the origin O (,, ) and going to A (,, ) and B (,, ) before returning to the origin. olution The whole of the contour is in the plane x and Green s theorem in the plane becomes Q (P dy + Qdz) y P dydz z (a) Firstly evaluate 4zdy +(y )dz. On OA, z and dz. As the integrand is zero, the integral will also be zero. On AB, z ( y ) and dz dy. The integral is (4 y)dy y (y )dy (5 y y )dy 5y y 6 y3 4 3 On BO, y and dy. The integral is ( )dz z. umming, 4zdy +(y )dz 8 3 Q (b) econdly evaluate y P dydz z In this example, P 4z and Q y. Thus P z Q y P dydz z Hence: (P dy + Qdz) y y y/ z yz 4z 4and Q y (y 4) dzdy y/ z 3 y3 +y 4y Q y P dydz 8 z 3 dy 8 3 y y. Hence, y +4y 4 dy and Green s theorem in the plane is verified. HELM (8): ection 9.3: Integral Vector Theorems 77

29 One very useful, special case of Green s theorem in the plane is when Q x and P y. The theorem becomes { ydx + xdy} ( ( )) dxdy The right-hand side becomes A {xdy ydx} dxdy i.e. A where A is the area inside the contour. Hence This result is known as the area theorem. It gives us the area bounded by a curve in terms of a line integral around. Example 4 Verify the area theorem for the segment of the circle x + y 4lying above the line y. olution Firstly, the area of the segment ADB can be found by subtracting the area of the triangle OADB from the area of the sector OAB. The triangle has area 3 3. The sector has area π 3 4π. Thus segment ADB has area 4π Now, evaluate the integral {xdy ydx} around the segment. Along the line, y, dy so the integral 3 3 ( dx) 3. {xdy ydx} becomes 3 3 (x dx) Along the arc of the circle, y 4 x (4 x ) / so dy x(4 x ) / dx. The integral {xdy ydx} becomes o, 3 3 { x (4 x ) / (4 x ) / }dx {xdy ydx} 3 3 π/3 8 3 π π 3. π/3 π/3 π/3 4 4 x dx 4 cos θ dθ (letting x sin θ) cos θ 4dθ 8 3 π Hence both sides of the area theorem equal 4 3 π 3 thus verifying the theorem. 78 HELM (8): Workbook 9: Integral Vector alculus

30 Task Verify Green s theorem in the plane when applied to the integral {(5x +y 7)dx + (3x 4y +5)dy} where represents the perimeter of the trapezium with vertices at (, ), (3, ), (6, ) and (, ). First let P 5x +y 7 and Q 3x 4y +5and find Q x P y : Your solution Answer Now find Your solution Q x P dxdy over the trapezium: y Answer 4 (by elementary geometry) Now find (P dx + Qdy) along the four sides of the trapezium, beginning with the line from (, ) to (3, ), and then proceeding anti-clockwise. Your solution Answers.5, 66, 6.5, whose sum is 4. HELM (8): ection 9.3: Integral Vector Theorems 79

31 Finally show that the two sides of the statement of Green s theorem are equal: Your solution Answer Both sides are 4. Exercises. Verify Green s identity [] (page 73) for the functions u xyz, w y and the unit cube x, y, z.. Verify the area theorem for Answers (a) The area above y, but below y x. (b) The segment of the circle x + y, to the upper left of the line y x.. Both integrals in [] equal. (a) both sides give a value of 4 3, (b) both sides give a value of π 4. 8 HELM (8): Workbook 9: Integral Vector alculus

32 Example 8 For the cube x, y, z, find the unit outward normal ˆn for each face. olution On the face given by x, the unit normal points in the negative x-direction. Hence the unit normal is i. imilarly :- On the face x the unit normal is i. On the face y the unit normal is j. On the face y the unit normal is j. On the face z the unit normal is k. On the face z the unit normal is k. d and the unit normal The vector d is a vector, being an element of the surface with magnitude du dv and direction perpendicular to the surface. If the plane in question is the Oxy plane, then d ˆn du dv k dx dy. u d v du dv Figure 7: The vector d as an element of a surface, with magnitude du dv If the plane in question is not one of the three coordinate planes (Oxy, Oxz, Oyz), appropriate adjustments must be made to express d in terms of two of dx and dy and dz. Example 9 The rectangle OAB lies in the plane z y (Figure 8). The vertices are O (,, ), A (,, ), B (,, ) and (,, ). Find a unit vector ˆn normal to the plane and an appropriate vector d expressed in terms of dx and dy. z y B O E A D x Figure 8: The plane z y passing through OAB 36 HELM (8): Workbook 9: Integral Vector alculus

33 olution Note that two vectors in the rectangle are OA i and O j + k. A vector perpendicular to the plane is i (j + k) j + k. However, this vector is of magnitude so the unit normal vector is ˆn ( j + k) j + k. The vector d is therefore ( j + k) du dv where du and dv are increments in the plane of the rectangle OAB. Now, one increment, say du, may point in the x-direction while dv will point in a direction up the plane, parallel to O. Thus du dx and (by Pythagoras) dv (dy) +(dz). However, as z y, dz dy and hence dv dy. Thus, d ( j + k) dx dy ( j + k) dx dy. Note :- the factor of could also have been found by comparing the area of rectangle OAB, i.e., with the area of its projection in the Oxy plane i.e. OADE with area. Integrating a scalar field A function can be integrated over a surface by constructing a double integral and integrating in a manner similar to that shown in 7. and 7.. Often, such integrals can be carried out with respect to an element containing the unit normal. Example Evaluate the integral A +x d over the area A where A is the square x, y, z. olution In this integral, d becomes k dx dy i.e. the unit normal times the surface element. Thus the integral is y x k dx dy k +x k π 4 k y y tan x dy ( π 4 ) dy π 4 k dy y HELM (8): ection 9.: urface and Volume Integrals 37

34 Example Find u d where u x + y + z and is the surface of the unit cube x, y, z. olution The unit cube has six faces and the unit normal vector ˆn points in a different direction on each face; see Example 8. The surface integral must be evaluated for each face separately and the results summed. On the face x, the unit normal ˆn i and the surface integral is y z ( + y + z )( i)dzdy i i y y y z + 3 z3 z dy On the face x, the unit normal ˆn i and the surface integral is y z ( + y + z )(i)dzdy i i y y y + dy i 3 3 y3 + 3 y 3 i z + y z + 3 z3 z dy y + 4 dy i 3 3 y y 5 3 i The net contribution from the faces x and x is i + 5i i. 3 3 Due to the symmetry of the scalar field u and the unit cube, the net contribution from the faces y and y is j while the net contribution from the faces z and z is k. Adding, we obtain ud i + j + k Key Point 4 A scalar function integrated with respect to a normal vector d gives a vector quantity. When the surface does not lie in one of the planes Oxy, Oxz, Oyz, extra care must be taken when finding d. 38 HELM (8): Workbook 9: Integral Vector alculus

35 Example Find ( F )d where F xi + yzj + xyk and is the surface of the triangle with vertices at (,, ), (,, ) and (,, ). olution Note that F +z as z everywhere along. As the triangle lies in the Oxy plane, the normal vector n k and d kdydx. Thus, ( F )d x x y dydxk x y dxk xdxk x k k Here the scalar function being integrated was the divergence of a vector function. Example 3 Find f d where f is the function x and is the surface of the triangle bounded by (,, ), (,, ) and (,, ). (ee Figure 9.) z Area 3 y Area x Figure 9: The triangle defining the area HELM (8): ection 9.: urface and Volume Integrals 39

36 olution The unit vector n is perpendicular to two vectors in the plane e.g. (j + k) and (i + k). The vector (j + k) (i + k) i + j k which has magnitude 3. Hence the unit normal vector ˆn 3 i + 3 j 3 k. As the area of the triangle is 3 and the area of its projection in the Oxy plane is, the vector 3/ d ˆn dydx (i + j + k)dydx. / Thus x fd (i + j + k) x dydx (i + j + k) (i + j + k) (i + j + k) x x x y xy x y dx (x x )dx x 3 x3 (i + j + k) 3 Task Evaluate the integral 4x d where represents the trapezium with vertices at (, ), (3, ), (, ) and (, ). (a) Find the vector d: Your solution Answer k dx dy (b) Write the surface integral as a double integral: Your solution 4 HELM (8): Workbook 9: Integral Vector alculus

37 Answer It is easier to integrate first with respect to x. This gives The range of values of y is y to y. For each value of y, x varies from x to x 3 y 3 y y x 4x dx dy k. (c) Evaluate this double integral: Your solution Answer 38 3 k HELM (8): ection 9.: urface and Volume Integrals 4

38 Exercises. Evaluate the integral xyd where is the triangle with vertices at (,, 4), (,, ) and (,, ).. Find the integral z. 3. Evaluate the integral at (,, ), (,, ), (,, ) and (,, ). xyzd where is the surface of the unit cube x, y, Answers. 3 i + 3 j + 6 k. 4 (x + y + z), 3. 5 i (x i + yzj + x yk) d where is the rectangle with vertices Integrating a vector field In a similar manner to the case ofa scalar field, a vector field may be integrated over a surface. Two common types of integral are F (r) d and F (r) d which integrate to a scalar and a vector respectively. Again, when d is expressed appropriately, the expression will reduce to a double integral. The form F (r) d has many important applications, e.g. the flux of a vector field such as an electric or magnetic field. Example 4 Evaluate the integral A (x yi + zj + (x + y)k) d over the area A where A is the square x, y, z. olution On A, the unit normal is dx dy k (x yi + zj + (x + y)k) (k dx dy) A y y x (x + y) dx dy x + xy dy y x y + y 3 ( + y)dy 4 HELM (8): Workbook 9: Integral Vector alculus

39 Example 5 Evaluate r d where A represents the surface of the unit cube A x, y, z and r represents the vector xi + yj + zk. olution The vector d (in the direction of the normal vector) will be a constant vector on each face, but will be different for each face. On the face x, d dy dz i and the integral on this face is z y (i + yj + zk) ( dy dz i) z y dy dz imilarly on the face y, d dx dz j and the integral on this face is z x (xi +j + zk) ( dx dz j) z x dx dz Furthermore on the face z, d dx dy k and the integral on this face is x y (xi + yj +k) ( dx dy k) x y dx dy On these three faces, the contribution to the integral is thus zero. However, on the face x, d +dy dz i and the integral on this face is z y (i + yj + zk) (+dy dz i) z y dy dz imilarly, on the face y, d +dx dz j and the integral on this face is z x (xi +j + zk) (+dx dz j) z x dx dz Finally, on the face z, d +dx dy k and the integral on this face is y x (xi + yj +k) (+dx dy k) Adding together the contributions gives A y x dx dy r d HELM (8): ection 9.: urface and Volume Integrals 43

40 Engineering Example 3 Magnetic flux Introduction The magnetic flux through a surface is given by B is the magnetic field and d is the vector normal to the surface. Problem in words B d where is the surface under consideration, The magnetic field generated by an infinitely long vertical wire on the z-axis, carrying a current I, is given by: B µ I yi + xj π x + y Find the flux through a rectangular region (with sides parallel to the axes) on the plane y. Mathematical statement of problem Find the integral B d over the surface, x x x, z z z. (see Figure which shows part of the plane y for which the flux is to be found and a single magnetic field line. The strength of the field is inversely proportional to the distance from the axis.) z z x x x z y Mathematical analysis Figure : The surface defined by x x x,z z z On y, B µ I πx j and d dx dz j so B d µ I πx The flux is given by the double integral: z zz x xx µ I πx dx dz µ I π µ I π µ I π z zz z zz x ln x dz x ln x ln x dz z ln x ln x z dx dz zz µ I π (z z ) ln x x 44 HELM (8): Workbook 9: Integral Vector alculus

41 Interpretation The magnetic flux increases in direct proportion to the extent of the side parallel to the axis (i.e. along the z-direction) but logarithmically with respect to the extent of the side perpendicular to the axis (i.e. along the x-axis). Example 6 If F x i + y j + z k, evaluate F d where is the part of the plane z bounded by x ±, y ±. olution i j k Here d dx dy k and hence F d x y z y dx dy i x dx dy j dx dy F d y dx dy i x dx dy j The first integral is imilarly Thus y x y y y dx dy x x y x dx dy 4 3. F d 4 3 i 4 3 j y x y x dy y dy x y 3 y3 4 3 Key Point 5 (a) An integral of the form F (r) d evaluates to a scalar. (b) An integral of the form F (r) d evaluates to a vector. The vector function involved may be the gradient of a scalar or the curl of a vector. HELM (8): ection 9.: urface and Volume Integrals 45

42 Example 7 Integrate ( φ).d where φ x +yz and is the area between y and y x for x and z. (ee Figure.) y x olution Figure : The area between y and y x,for x and z Here φ xi +zj +yk and d k dydx. Thus ( φ).d ydydx and ( φ).d For integrals of the form x x x 5 x5 y y x 5 y dydx dx y x x 4 dx F d, non-artesian coordinates e.g. cylindrical polar or spherical polar coordinates may be used. Once again, it is necessary to include any scale factors along with the unit normal. Example 8 Using cylindrical polar coordinates, (see 8.3), find the integral F (r) d for F ρz ˆρ + z sin φẑ and being the complete surface (including ends) of the cylinder ρ a, z. (ee Figure.) z z y ρ a x Figure : The cylinder ρ a, z 46 HELM (8): Workbook 9: Integral Vector alculus

43 olution The integral F (r) d must be evaluated separately for the curved surface and the ends. For the curved surface, d ˆρadφdz (with the a coming from ρ the scale factor for φ and the fact that ρ a on the curved surface.) Thus, F d a z dφdz and F (r) d z π φ a z dφdz z πa z dz πa πa z On the bottom surface, z so F and the contribution to the integral is zero. On the top surface, z and d ẑρ dρdφ and F d ρz sin φ dφdρ ρ sin φ dφdρ and o F (r) d π a π ρ φ a ρ ρ sin φ dφdρ ρ dρ πa F (r) d πa + πa 3 πa Engineering Example 4 The current continuity equation Introduction When an electric current flows at a constant rate through a conductor, then the current continuity equation states that J d where J is the current density (or current flow per unit area) and is a closed surface. The equation is an expression of the fact that, under these conditions, the current flow into a closed volume equals the flow out. Problem in words A person is standing nearby when lightning strikes the ground. Find the potential difference between the feet of that person. HELM (8): ection 9.: urface and Volume Integrals 47

44 Figure 3: Lightning: a current dissipating into the ground Mathematical statement of problem The current from the lightning dissipates radially (see Fig 3). (a) Find a relationship between the current I and current density J at a distance r from the strike by integrating the current density over the hemisphere I J d (b) Find the field E from the equation E ρi where E E and I is the current. π r (c) Find V from the integral Mathematical analysis R R E dr Imagine a hemisphere of radius r level with the surface of the ground so that the point of lightning strike is at its centre. By symmetry, the pattern of current flow from the point of strike will be uniform radial lines, and the magnitude of J will be a constant, i.e. over the curved surface of the hemisphere J J ˆr. ince the amount of current entering the hemisphere is I, then it follows that the current leaving must be the same i.e. I J d (where c is the curved surface of the hemisphere) c (J ˆr) (d ˆr) c J d c πr J [ surface area (πr ) flux (J)] since the surface area of a sphere is 4πr. Therefore J I πr Note that if the current density J is uniformly radial over the curved surface, then so must be the electric field E, i.e. E Eˆr. Using Ohm s law J σe or E ρj where σ conductivity /ρ. 48 HELM (8): Workbook 9: Integral Vector alculus

45 Hence E ρi πr The potential difference between two points at radii R and R from the lightning strike is found by integrating E between them, so that V R R R ρi π ρi π ρi π E dr E dr R R R r dr r R R R R ρi π R R R R Interpretation uppose the lightning strength is a current I, A, that the person is m away with feet.35 m apart, and that the resistivity of the ground is 8 Ω m. learly, the worst case (i.e. maximum voltage) would occur when the difference between R and R is greatest, i.e. R m and R.35 m which would be the case if both feet were on the same radial line. The voltage produced between the person s feet under these circumstances is V ρi π R R 8 π 3 V.35 Task For F (x + y )i +(x + z )j +xzk and the square bounded by (,, ), (,, ), (,, ) and (,, ) find the integral F d Your solution Answer d dxdzj (x + z ) dxdz 8 3 HELM (8): ection 9.: urface and Volume Integrals 49

46 Task For F (x + y )i +(x + z )j +xzk and being the rectangle bounded by (,, ), (,, ), (,, ) and (,, ) (i.e. the same F and as in the previous Task), find the integral F d Your solution Answer ( xz)i + (x +)k dxdz 4 3 k. Evaluate the integral Exercises φ d for φ x z sin y and being the rectangle bounded by (,, ), (,, ), (,π,) and (,π,).. Evaluate the integral ( F ) d where F xe y i + ze y j and represents the unit square x, y. 3. Using spherical polar coordinates (r, θ, φ), evaluate the integral and is the curved surface of the top half of the sphere r a. Answers.,. (e )j, 3. πa3 3 F d where F r cos θˆr. Volume integrals involving vectors Integrating a scalar function of a vector over a volume involves essentially the same procedure as in 7.3. In 3D cartesian coordinates the volume element dv is dxdydz. The scalar function may be the divergence of a vector function. 5 HELM (8): Workbook 9: Integral Vector alculus

47 Example 9 Integrate F over the unit cube x, y, z where F is the vector function x yi +(x z)j +xz k. olution F x (x y)+ y (x z)+ z (xz )xy +4xz The integral is x y x x y 3xdx z (xy +4xz)dzdydx (xy +x) dydx 3 x 3 x x y xy +xy xyz +xz dydx dx Key Point 6 The volume integral of a scalar function (including the divergence of a vector) is a scalar. Task Using spherical polar coordinates (r, θ, φ) and the vector field F r ˆr+r sin θ ˆθ, evaluate the integral F dv over the sphere given by r a. V Your solution Answer F 4r +r cos θ, a r π θ π φ {(4r +r cos θ)r sin θ} dφdθdr 4πa 4 The r sin θ term comes from the Jacobian for the transformation from spherical to cartesian coordinates (see 7.4 and 8.3). HELM (8): ection 9.: urface and Volume Integrals 5

48 . Evaluate V Exercises F dv when F is the vector field yzi + xyj and V is the unit cube x, y, z. For the vector field F (x y+sin z)i+(xy +e z )j+(z +x y )k, find the integral V F dv where V is the volume inside the tetrahedron bounded by x, y, z and x+y+z. Answers. F x,. 7 6 Integrating a vector function over a volume integral is similar, but less common. are should be taken with the various components. It may help to think in terms of a separate volume integral for each component. The vector function may be of the form f or F. Example 3 Integrate the function F x i+j over the prism given by x, y, z ( x). (ee Figure 4.) z y x olution The integral is Figure 4: The prism bounded by x, y, z ( x) x y x x 6 i +j y x z (x i +j)dzdydx x y x ( x)i + ( x)j dydx (x x 3 )i + (4 4x)j dx x x zi +zj x y z dydx ( 3 x3 x4 )i + (4x x )j (x x 3 )i + ( x)j dydx 5 HELM (8): Workbook 9: Integral Vector alculus

49 Example 3 For F x yi + y j evaluate ( F )dv where V is the volume under the V plane z x + y +(and above z ) for x, y. olution i j k F x y z x k x y y so ( F )dv V x y x y x x x y x+y+ z ( x )zk ( x )k dzdydx x+y+ z dydx x 3 x y x dydx k x 3 y x y x y x 3 4x dx k z dx k y x4 4 3 x3 k 8 3 k (,, 4) y (,, ) (,, ) x (,, ) (,, ) Figure 5: The plane defined by z x + y + z, for z>, x, y HELM (8): ection 9.: urface and Volume Integrals 53

50 Key Point 7 The volume integral of a vector function (including the gradient of a scalar or the curl of a vector) is a vector. Task Evaluate the integral F dv for the case where F xi + y j + zk and V is the cube x, y, z. V Your solution Answer x y z (xi + y j + zk)dzdydx 8 3 j Exercises. For f x + yz, and V the volume bounded by y, x + y and x + y for z, find the integral ( f)dv.. Evaluate the integral V V ( F )dv for the case where F xzi +(x 3 + y 3 )j 4yk and V is the cube x, y, z. Answers. (xi + zj + yk)dv 3 k,. V V ( 4i + xj +3x k)dv 3i +8k 54 HELM (8): Workbook 9: Integral Vector alculus

51 Line Integrals 9. Introduction workbook 8 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. ubsequently, line integrals involving vectors will be considered. These can give scalar or vector answers depending on the form of integral involved. Of particular interest are the integrals of conservative vector fields. Prerequisites Before starting this ection 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 HELM (8): Workbook 9: Integral Vector alculus

52 . Line integrals 8 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) region. 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 +4y. : This is the curve along which integration takes place. e.g. y x or x sin y or x t ; y t (where x and y are expressed in terms of a parameter t). dw: The integral. The integrals This gives 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. dy ds may be written as ds (dx) +(dy) or ds + dx. dx A fourth case is when F (x, y) dw has the form: F dx+f 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 curve 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 ds. This is a means of calculating z f(x, y)dy y f(x, y)ds curve f(x, y)dx Figure : Representation of a line integral and its projections onto the xz and yz planes x HELM (8): ection 9.: Line Integrals 3

53 The technique for evaluating a line integral is to express all quantities in the integral in terms of a single variable. If the integral is with respect to x or y, then the curve and the function F may be expressed in terms of the relevant variable. If the integral is with respect to ds, normally all quantities are expressed in terms of x. If x and y are given in terms of a parameter t, then t is used as the variable. Example Find x ( + 4y) dx where is the curve y x, starting from x,y c and ending at x,y. olution As this integral concerns only points along and the integration is carried out with respect to x, y may be replaced by x. The limits on x will be to. o the integral becomes x( + 4y) dx x x x +4x dx + x4 + x x +4x 3 dx () 3 Example Find x ( + 4y) dy where is the curve y x, starting from c x,y and ending at x,y. This is the same as Example other than dx being replaced by dy. olution 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 /. The limits on y will be to. o the integral becomes x( + 4y) dy y y / ( + 4y) dx 3 y3/ x5/ y y / +4y 3/ dx () HELM (8): Workbook 9: Integral Vector alculus

54 Example 3 Find x ( + 4y) ds where is the curve y x, starting from x,y c and ending at x,y. This is the same integral and curve as the previous two examples but the integration is now carried out with respect to s, the arc length parameter. olution As this integral is with respect to x, all parts of the integral can be expressed in terms of x, Along dy y x, ds + dx +(x) dx +4x dx dx o, the integral is x ( + 4y) ds x +4x +4x dx x +4x 3/ dx c x This can be evaluated using the transformation u +4x so du 8xdx i.e. x dx du 8. When x, u and when x, u 5. Hence, x +4x 3/ 5 dx u 3/ du x 8 u 8 5 u 5/ 5 x 5 5/.745 Note that the results for Examples, and 3 are all different: Example 3 is the area between a curve and a surface above; Examples and give projections of this area onto other planes. Example 4 Find xy dx where, on, x and y are given in terms of a parameter t by x 3t, y t 3 for t varying from to. olution Everything can be expressed in terms of t, the parameter. Here x 3t so dx 6t dt. The limits on t are t and t. The integral becomes xy dx t 3t (t 3 ) 6t dt 8 7 t7 8 4 t4 t (8t 6 8t 3 ) dt HELM (8): ection 9.: Line Integrals 5

55 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 Task For F (x, y) x + y, find (i) F (x, y) dx, (ii) F (x, y) dy, (iii) F (x, y) ds where is the line y x from (, ) to (, ). Express each integral as a simple integral with respect to a single variable and hence evaluate each integral: Your solution Answer (i) x (x +4x ) dx 7 3, (ii) y (y + y ) dy 4 3, (iii) +4x x(x ) 5 dx HELM (8): Workbook 9: Integral Vector alculus

56 Task Find (i) F (x, y) dx, (ii) F (x, y) dy, (iii) F (x, y) ds where F (x, y) and is the curve y x ln x from (, ) to (, ln ). 4 4 Your solution Answer (i) dx, (ii) ds (/4) ln / +(x 4x ) dx dy 3 4 ln, (iii) y x 4 x + + 6x dx ln x dy dx x 4x (x + 4x ) dx ln. Task Find (i) F (x, y) dx, (ii) F (x, y) dy, (iii) F (x, y) ds where F (x, y) sin x and is the curve y sin x from (, ) to ( π, ). Your solution Answer (i) (iii) π/ π/ sin x dx, (ii) π/ sinx cos x dx 3 sin x + cos x dx 3 ( ), using the substitution u + cos x. HELM (8): ection 9.: Line Integrals 7

57 . Line integrals of scalar products Integrals of the form F dr occur in applications such as the following. B A T δr dr (current position) v Figure : chematic for cyclist travelling from A to B into a head wind onsider a cyclist riding along the road from A to B (Figure ). uppose it is necessary to find the total work the cyclist has to do in overcoming a wind of velocity v. On moving from to T, along an element δr of road, 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. o, 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, with a vector element of the line. The term scalar line integral is often used for integrals of this form. The vector dr may be considered to be dx i + dy j + dz k. Multiplying out the scalar product, 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, and {F x dx + F y dy} in two dimensions, where F x, F y, F z are the components of F. 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 t often represents time. Example 5 Find {xy dx 5x dy} where is the curve y x 3 x. [This is the integral F dr where F xyi 5xj and dr dx i + dy j.] 8 HELM (8): Workbook 9: Integral Vector alculus

58 olution 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 dx and the integral becomes {xy dx 5x dy} x xx 3 dx 5x 3x dx (x 4 5x 3 ) dx 5 x5 5 4 x An integral of the form Key Point 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 (, ) are defined by (a) : y x (b) : y x (c) 3 : y from (, ) to (, ) and x from (, ) to (, ) ketch each path, and along each path find F dr, where F y i + xyj. HELM (8): ection 9.: Line Integrals 9

59 olution (a) F dr F dr y dx + xydy. Along y x, dy dx x (x) dx + x (x) (dx) 4x +4x dx 8x dx so dy dx. Then 8 3 x 8 3 y y x A(, ) x (b) Figure 3(a): Integration along path y F dr dx + xydy. Along y x, dy 4x so dy 4xdx. Then dx x F dr dx + x x (4xdx) x 4 dx x 5 x5 5 y A(, ) y x x Figure 3(b): Integration along path 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 (, ). Hence B A F dr F dr + F dr 3 O B HELM (8): Workbook 9: Integral Vector alculus

60 olution (contd.) Along OB, y so dy. Then B O F dr x Along AB, x so dx. Then B A F dr y Hence F dr + 3 dx + x dx y + y dy ydy y y. x A(, ) 3 O y B x Figure 3(c): Integration along path 3 Once again, the result is path dependent. Key Point 3 In general, the value of a line integral depends on the path of integration as well as upon the end points. HELM (8): ection 9.: Line Integrals

61 Example 7 Find O A F dr, where F y i + xyj (as in Example 6) and the path 4 from A to O is the straight line from (, ) to (, ), that is the reverse of in Example 6(a). Deduce F dr, the integral around the closed path formed by the parabola y x from (, ) to (, ) and the line y x from (, ) to (, ). olution Reversing the path interchanges the limits of integration, which 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 Example 6(b)) evaluates to 5, then F dr F dr + F dr Example 8 onsider the vector field F y z 3 i +xyz 3 j +3xy z k Let and be the curves from O (,, ) to A (,, ), given by : x t, y t, z t ( t ) : x t, y t, z t ( t ) (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 from O to A and back along from A to O. HELM (8): Workbook 9: Integral Vector alculus

62 olution (a) The path is given in terms of the parameter t by x t, y t and z t. Hence dx dt dy dt dz dt 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 +t 5 j +3t 5 k Hence F dr dt t5 +t 5 +3t 5 6t 5. The values of t and t correspond to the start and end point of and so these are the required limits of integration. Now F dr F dr dt dt 6t 5 dt t 6 For the path the parameterisation is x t, y t and z t so dr ti + j +tk. dt ubstituting x t, y t and z t in F we have F t 8 i +t 9 j +3t 8 k and F dr dt t9 +t 9 +6t 9 t 9 F dr t 9 dt t (b) For the closed path F dr F dr F dr (Note: A line integral round a closed path is not necessarily zero - see Example 7.) Further points on Example 8 Vector Field Path Line Integral F F F closed Note that the value of the line integral of F is for both paths and. In fact, this result would hold for any path from (,, ) to (,, ). 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. HELM (8): ection 9.: Line Integrals 3

63 Task onsider the vector field G xi + (4x y)j Let and be the curves from O (,, ) to A (,, ), given by : x t, y t, z t ( t ) : x t, y t, z t ( t ) (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 from A to O. G dr where is the closed path along from Your solution 4 HELM (8): Workbook 9: Integral Vector alculus

64 Answer (a) The path is given in terms of the parameter t by x t, y t and z t. Hence dx dt dy dt dz dt and dr dt dx dt i + dy dt j + dz dt k i + j + k ubstituting 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, then G dr G dr dt dt 4tdt t For the path the parameterisation is x t, y t and z t so dr dt ubstituting x t, y t and z t in G we have G t i + 4t t j and G dr dt t3 +4t t G dr t 3 +4t t dt t t3 t 4 3 (b) For the closed path G dr G dr G dr ti + j +tk. (Note: The value of the integral around the closed path is non-zero, unlike Example 8.) Example 9 Find (x y) dr where is the contour y x x from (, ) to (, ). Here, refers to the gradient operator, i.e. φ grad φ olution Note that (x y)xyi + x j so the integral is xy dx + x dy. On y x x, dy ( x) dx so the integral becomes xy dx + x dy x(x x ) dx + x ( x) dx x (6x 4x 3 ) dx x 3 x 4 HELM (8): ection 9.: Line Integrals 5

65 Example Two paths from (, ) to (4, ) are defined by (a) : y x x 4 (b) : The straight line y from (, ) to (4, ) followed by 3 : The straight line x 4from (4, ) to (4, ) For each path find F dr, where F xi +yj. olution (a) For the straight line y x we have dy dx 4 Then, F dr x dx +y dy x + x dx (b) For the straight line from (, ) to (4, ) we have F dr For the straight line from (4, ) to (4, ) we have F dr 3 Adding these two results gives F dr x dx x dx 6 y dy 4 Task Evaluate paths F dr, where F (x y)i +(x + y)j along each of the following (a) : from (, ) to (, 4) along the straight line y 3x : (b) : from (, ) to (, 4) along the parabola y x : (c) 3 : along the straight line x from (, ) to (, 4) then along the straight line y 4from (, 4) to (, 4). Your solution 6 HELM (8): Workbook 9: Integral Vector alculus

66 Answer (a) (b) (c) 4 (x 4) dx, (x + x +x 3 ) dx 34, (this differs from (a) showing path dependence) 3 ( + 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) followed by the reverse of. (b) followed by the reverse of 3. (c) 3 followed by the reverse of. Your solution Answer (a) 3, (b), (c) 3. (note that all these are non-zero.) 3 HELM (8): ection 9.: Line Integrals 7

67 . onsider Exercises F dr, where F 3x y i + (x 3 y )j. Find the value of the line integral along each of the paths from (, ) to (, 4). (a) y 4x (b) y 4x (c) y 4x / (d) y 4x 3. onsider the vector field F xi +(xz )j + xyk and the two curves between (,, ) and (,, ) defined by : x t, y t, z t for t. : x t, y t, z t for t. (a) Find F dr, F dr (b) Find F dr where is the closed path from (,, ) to (,, ) along and back to (,, ) along. 3. onsider the vector field G x zi + y zj + 3 (x3 + y 3 )k and the two curves between (,, ) and (,, ) defined by : x t, y t, z t for t. : x t, y t, z t for t. (a) Find G dr, G dr (b) Find G dr where is the closed path from (,, ) to (,, ) along and back to (,, ) along. 4. Find F dr) along y x from (, ) to (, 4) for Answers (a) F (x y) (b) F ( x y k) [Here f represents the curl of f]. All are, and in fact the integral would be for any path from (,) to (,4). (a), 5 3 (b) 3. 3 (a), (b). 4. (a) xy dx + x dy 6, (b) x y dx xy dy 4. 8 HELM (8): Workbook 9: Integral Vector alculus

68 3. onservative vector fields For some line integrals in the previous section, the value of the integral depended only on the vector field F and the start and end points of the line but not on the actual path 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 (P to P5 below). It is impossible to check the value of every line integral over every path, but it is possible to use any one of these five properties (particularly property P3 below) to determine whether or not a vector field is conservative. These properties are also used to simplify calculations with conservative vector fields over non-closed paths. P P 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 Theorem of alculus.] All gradient fields are conservative. That is, F φ is a conservative vector field for any scalar field φ. olution Example onsider the following vector fields.. F y i + xyj (Example 6). F xi +yj (Example ) 3. F 3 y z 3 i +xyz 3 j +3xy z k (Example 8) 4. F 4 xi + (4x y) j (Task on page 4) Determine which of these vector fields are conservative where possible 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 values of the line integrals.. Two different values were obtained for line integrals over the paths and. Hence, by P, F is not conservative. [It is also possible to reach this conclusion from P3 by finding that F yk.] HELM (8): ection 9.: Line Integrals 9

69 olution (contd.). For the closed path consisting of and 3 from (, ) to (4, ) and back to (, ) along we obtain the value + ( ). This alone does not mean that F is conservative as there could be other paths giving different values. o by using P3 i j k F x y z x y i( ) j( ) + k( ) As F, P3 gives that F is a conservative vector field. Now, find a φ such that F φ. Then φ x i + φ j xi +yj. y Thus Using P4: φ x x φ x + f(y) φ y y φ y + g(x) (4,) (,) F dr (4,) (,) φ x + y (+ constant) ( φ) dr φ(4, ) φ(, ) (4 + ) ( + ). 3. The fact that line integrals along two different paths between the same start and end points have the same value is consistent with F 3 being a conservative field according to P. o too is the fact that the integral around a closed path is zero according to P. 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 6xyz )i (3y z 3y z )j + (yz 3 yz 3 )k. y z 3 xyz 3 3xy z 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 y z 3 φ xy z 3 + f(y, z) Using P4: φ y xyz3 φ xy z 3 + g(x, z) φ z 3xy z φ xy z 3 + h(x, y) (,,) (,,) φ xy z 3 F 3 dr φ(,, ) φ(,, ) in agreement with Example 8(a). HELM (8): Workbook 9: Integral Vector alculus

70 olution (contd.) 4. As the integral along is and the integral along (same start and end points but different intermediate points) is 4, F 3 4 is not a conservative field using P. Note that F 4 4k so, using P3, this is an independent conclusion that F 4 is not conservative. Engineering Example 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 dr 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 Q xi + yj + zk 4πε (x + y + z ) x + y + z Q(xi + yj + zk) 4πε (x + y + z ) 3 where r is the position vector with magnitude r and unit vector ˆr, and constants of proportionality, where 9 /36π Fm. 4π is a combination of Given that Q 8, find the work done in bringing a charge of q from the point A (,, ) to the point B (,, ) (where the dimensions are in metres) (a) by the direct straight line y x, z (b) by the straight line pair via (,, ) HELM (8): ection 9.: Line Integrals

71 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 (,, ) to B (,, ) via (,, ) (see Figure 4). Mathematical analysis (a) Here Q/(4πε ) 9 so E 9[xi + yj] (x + y ) 3 as z over the region of interest. The work done B W AB q E dr A B A 9 (x + y ) 3 [xi + yj] [dxi + dyj] Using y x, dy dx W AB 9 {x dx + x dx} x (x ) 3 9 ( ) x 3 x dx 9 x dx x x 9 9 [.] J HELM (8): Workbook 9: Integral Vector alculus

72 (b) The first part of the path is A to where x, dx and y goes from to. W A q E dr A y u 9 ( + y ) 3 9y dy ( + y ) 3 45 du u 3 45 u 3 45 u [xi + yj] [i + dyj] (substituting du J u + y, du y dy) The second part is to B, where y, dy and x goes from to. W B 9 [xi + yj] [dxi +j] x (x +) 3 9x dx (x +) 3 45 du (substituting u x +, du x dx) u 45 u 3 u 3 45 u du J The sum of the two components W A and W B is J. Therefore the work done over the two paths (a) and (b) 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 (8): ection 9.: Line Integrals 3

73 Example. how that I path taken. (,) (,) (xy +)dx +(x y)dy is independent of the. Find I using property P. (Page 9) 3. Find I using property P4. (Page 9) 4. Find I (xy +)dx +(x y)dy where is (a) the circle x + y (b) the square with vertices (, ), (, ), (, ), (, ). olution. The integral I (,) (,) (xy +)dx +(x y)dy may be re-written F (xy +)i +(x y)j. i j k Now F x y z i +j +k xy + x y F dr where As F, F is a conservative field and I is independent of the path taken between (, ) and (, ).. As I is independent of the path taken from (, ) to (, ), it can be evaluated along any such path. One possibility is the straight line y x. On this line, dy dx. The integral I becomes I (,) (,) x (xy +)dx +(x y)dy (x x +)dx +(x x) dx ( 3 x x +)dx x3 4 x + x HELM (8): Workbook 9: Integral Vector alculus

74 olution (contd.) 3. If F φ then φ x xy + φ x y + x + f(y) φ y x y φ x y + x y +. φ x y y + g(x) These are consistent if φ x y + x y (plus a constant which may be omitted since it cancels). o I φ(, ) φ(, ) (4 + ) 5 4. As F is a conservative field, all integrals around a closed contour are zero. Exercises. Determine whether the following vector fields are conservative (a) F (x y)i +(x + y)j (b) F 3x y i + (x 3 y )j (c) F xi +(xz )j + xyk (d) F x zi + y zj + 3 (x3 + y 3 )k. onsider the integral F dr with F 3x y i + (x 3 y )j. From Exercise (b) F is a conservativevector field. Find a scalar field φ so that φ F. Use property P4 to evaluate the integral F dr where is an integral with start-point (, ) and end point (, 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 x)i +(x 4 y)j; any path from (, ) to (, ). (b) F (e x + y 3 )i + (3xy )j; closed path starting from any point on the circle x + y. (c) F (y + sin z)i +xyj + x cos zk; any path from (,, ) to (,,π). (d) F x i +4y3 z j +y 4 zk; any path from (,, ) to (,, 3). Answers. (a) No, (b) Yes, (c) No, (d) Yes. x 3 y y +, 3. (a) x 4 y x y, ; (b) e x + xy 3, ; (c) xy + x sin z, ; (d) ln x + y 4 z,43 HELM (8): ection 9.: Line Integrals 5

75 4. Vector line integrals It is also possible to form less commonly used integrals of the types: 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. imilarly 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 3 Evaluate the integral xy dr where represents the contour y x from (, ) to (, ). olution This is a two-dimensional integral so the term in z will be absent. I xy dr xy (dxi + dyj) xy dx i + xy dy j x 6 x6 x(x ) dx i + x 5 dx i + 6 i + 7 j i + 7 x7/ y y 5/ dy j j y / y dy j 6 HELM (8): Workbook 9: Integral Vector alculus

76 Example 4 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 π, i.e. (,, π) and finishes at (,,π). the contour starts at olution The integral becomes x(dx i + dy j + dz k). Now, x cost, y sin t, z t π so dx sin t dt, dy cost dt and dz dt. o I π π cos t( sin t dt i + cos t dt j + dt k) π cos t sin t dt i + π π cos t dt j + π sin t dt i + ( + cos t) dt j + π cos t i + t + π 4 sin t j + k i + πj πj cos t dt k π sin t k Integrals of the form and dr dx i + dy j + dz k then: i j k F dr F F F 3 dx dy dz F dr can be evaluated as follows. If the vector field F F i + F j + F 3 k (F dz F 3 dy)i +(F 3 dx F dz)j +(F dy F dx)k (F 3 j F k)dx +(F k F 3 i)dy +(F i F j)dz There are thus a maximum of six terms involved in one such integral; the exact details may dictate which method to use. HELM (8): ection 9.: Line Integrals 7

77 Example 5 Evaluate the integral (x i +3xyj) dr where represents the curve y x from (, ) to (, ). olution Note that the z components of both F and dr are zero. i j k F dr x 3xy (x dy 3xydx)k and dx dy (x i +3xyj) dr (x dy 3xydx)k Now, on, y x dy 4xdx and (x i +3xyj) dr {x dy 3xydx}k x k x 4xdx 3x x dx k x 3 dxk x4 k 8 HELM (8): Workbook 9: Integral Vector alculus

78 Engineering Example Force on a loop due to a magnetic field Introduction A current I in a magnetic field B is subject to a force F given by F I dr B where the current can be regarded as having magnitude I and flowing (positive charge) in the direction given by the vector dr. 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 dr B) I (B dr) where the closed path of the integral represents one circuit of the loop. Problem in words Figure 5: The magnetic field through a loop of current A current of amp flows around a circuit in the shape of the unit circle in the Oxy plane. A magnetic field of tesla (T) 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 to give the force required using the given values of current and magnetic field. Mathematical analysis The circuit is described parametrically by with x cosθ y sin θ z dr sin θ dθ i + cos θ dθ j B Bk HELM (8): ection 9.: Line Integrals 9

79 since B is constant. Therefore, the force on the circuit is given by F IB k dr k dr (since I A and B T) where k dr i j k sin θ dθ cos θ dθ cos θi sin θj dθ o π F cos θi sin θj dθ θ π sin θi cos θj θ ( ) i ( ) 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 arguments to avoid detailed calculations whenever possible! 3 HELM (8): Workbook 9: Integral Vector alculus

80 A scalar or vector involved in a vector line integral may itself be a vector derivative as this next Example illustrates. Example 6 Find the vector line integral ( F ) dr where F is the vector x i+xyj +xzk and is the curve y x, z x 3 from x to x i.e. from (,, ) to (,, ). Here F is the (scalar) divergence of the vector F. olution As F x i +xyj +xzk, F x +x +x 6x. The integral ( F ) dr 6x(dx i + dy j + dz k) 6x dx i + 6x dy j + 6x dz k The first term is 6x dx i x 6x dx i 3x i 3i In the second term, as y x on, dy may be replaced by x dx so 6x dy j 6x x dx j x dx j 4x 3 j 4j x In the third term, as z x 3 on, dz may be replaced by 3x dx so 9 6x dz k 6x 3x dx k 8x 3 dx k x x4 k 9 k On summing, ( F ) dr 3i +4j + 9 k. Task Find the vector line integral fdr where f x and is (a) the curve y x / from (, ) to (9, 3). (b) the line y x/3 from (, ) to (9, 3). HELM (8): ection 9.: Line Integrals 3

81 Your solution Answer (a) 9 (x i + x3/ j)dx 43i j, (b) 9 (x i + 3 x j)dx 43i +8j. Task Evaluate the vector line integral F dr when represents the contour y 4 4x, z x from (, 4, ) to (,, ) and F is the vector field (x z)j. Your solution Answer {(4 6x)i + ( 3x)k} i + k 3 HELM (8): Workbook 9: Integral Vector alculus

82 . Evaluate the vector line integral Exercises ( F ) dr in the case where F xi + xyj + xy k and is the contour described by x t, y t, z t for t starting at t and going to t.. When is the contour y x 3, z, from (,, ) to (,, ), evaluate the vector line integrals (a) { (xy)} dr (b) (x i + y k) dr Answers. ( + x)(dx i + dy j + dz k) 4i j k,. (a) k y dy x dx k, (b) k y dy k k HELM (8): ection 9.: Line Integrals 33

83 Index for Workbook 9 Area theorem 78 onservative vector field 3, 9 urrent continuity equation 47 urrent in line 69 urrent in loop 9 ylinder 7 ylindrical polar coordinates 46, 59 Divergence theorem ee Gauss Electric current 47 Electric field, 65, 69, 7 Electric motor 9 Field strength 69, 7 Magnetic Field 9 Magnetic Flux 44 Ohm s law 48 Pythagoras theorem 37 calar line integral 8 calar product 8 tokes theorem 56-6 urface integrals 35-5 Unit normal 35 Volume integrals 5-54 Work Gauss law 65, 67 Gauss theorem 63-7 Green s theorem 73-8 Integral vector theorems - Gauss Green s tokes 56-6 Integrating - scalar field 37 - vector field 4 Lightning strike 47 Line integrals - scalar products 8 - vector 6 Lorentz force 9 EXERIE 8, 5, 33, 4, 5, 5, 54, 6, 7, 8 ENGINEERING EXAMPLE Work done moving a charge in an electric field Force on a loop due to a magnetic field 9 3 Magnetic flux 44 4 The current continuity equation Gauss law 65 6 Field strength around a charged line 69 7 Field strength on a cylinder 7

84