29.3. Integral Vector Theorems. Introduction. Prerequisites. Learning Outcomes

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1 Integral ector Theorems 9. Introduction arious 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.: Integral ector Theorems 55

2 1. 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 16: 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 ector alculus

3 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 17: 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.: Integral ector Theorems 57

4 Example erify tokes theorem for the vector function F y i (x + z)j + yzk and the unit square x 1, y 1, z. olution If F y i (x + z)j + yzk then F (z + 1)i + ( 1 y)k i + ( 1 y)k (as z ). Note that d dxdyk so that ( F ) d ( 1 y)dydx Thus To evaluate ( F ) d x x y ( 1 y)dydx ( y y ) x ] 1 ] 1 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 1, F y i j and dr dyj so F dr dy so the contribution to the integral is ] 1 ( dy) y 1. y When y 1, F i xj and dr dxi so F dr dx so the contribution to the integral is ] 1 ( dx) x 1. 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 ector alculus

5 Example 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 dφ and hence π F dr a dφ πa As F ρ ˆφ, F ρẑ and ( F ) d ρ as d ẑ. Thus π π a ( F ) d ρ ρdρdφ Hence φ π φ ρ ρ ] a ρ dφ F dr ( F ) d πa π φ ρ a dφ πa ρ dρdφ Example 4 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 18. y F (1, 7) E(5, 7) H(, 4) G(1, 4) D(, 4) (6, 4) A(, ) B(6, ) x Figure 18: losed contour ABDEF GHA HELM (8): ection 9.: Integral ector Theorems 59

6 olution To find the line integral directly would require eight line integrals i.e. along AB, B, D, DE, EF, F G, 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 + 1)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 DEF G. On ABDGH, the integral is 4 y 6 x (x y)dxdy On DEF G, the integral is 7 y4 y x1 (x y)dxdy 4 ] 6 4 x xy dy (6 1y)dy y x y ] 4 6y 6y y4 ] y x xy 1 y + y + y dy x1 4 o the full integral is, ( F ) d By tokes theorem, F dr ] y4 ( y + y + )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 ector alculus

7 Task erify tokes theorem for the vector field F x i + xyj + zk and the triangle with vertices at (,, ), (,, ) and (, 1, ). 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 1 HELM (8): ection 9.: Integral ector Theorems 61

8 Finally find the integral the tokes theorem are equal: Your solution F dr along the sides of the triangle and so verify that the two sides of Answer , Both sides of tokes theorem have value 1. Exercises 1. 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 1, π φ π.. erify tokes theorem for the vector field F xi + (y z)j + xzk and the contour around the rectangle with vertices at (,, ),(,, ), (,, 1) and (,, 1).. erify tokes theorem for the vector field F yi + xj + zk (a) Over the triangle (,, ), (1,, ), (1, 1, ). (b) Over the triangle (1,, ), (1, 1, ), (1, 1, 1). 4. Use tokes theorem to evaluate the integral F dr where F (sin( 1x ) + 1) + 5y i + (x e y )j and is the contour starting at (, ) and going to (5, ), (5, ), (6, ), (6, 5), (, 5), (, ), (, ) and returning to (, ). Answers 1. Both integrals give,. Both integrals give 1. (a) Both integrals give 1 (b) Both integrals give (as F is perpendicular to d) 4. 57, F k]. 6 HELM (8): Workbook 9: Integral ector alculus

9 . 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, bounded by a closed surface, any well-behaved vector field F satisfies F d F d Notes: (a) d is a unit normal pointing outwards from the interior of the volume. (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 d 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. HELM (8): ection 9.: Integral ector Theorems 6

10 Example 5 erify Gauss theorem for the unit cube x 1, y 1, z 1 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 1, F i + zj and d i dydz so F d 1 dydz and F d 1 dydz 1 On the face y, F xi + zj and d j dxdz so F d z dxdz and F d z dxdz 1 On the face y 1, F xi + zj and d j dxdz so F d z dxdz and F d z dxdz 1 On the face z, F xi and d k dydz so F d dxdy and F d dxdy On the face z 1, 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 d note that F x x + y z o F d 1 dxdydz 1. o F d F d 1 hence verifying Gauss theorem. 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 ector alculus

11 Example 6 Use Gauss theorem to evaluate the surface integral F d where F is the vector field x yi + xyj + z k and is the surface of the unit cube x 1, y 1, z 1. olution Note that to carry out the surface integral directly will involve, as in Example 5, the evaluation of six double integrals. However, by Gauss theorem, the same result comes from the volume integral F d. As F xy + x + z, we have the triple integral (xy + x + z ) dxdydz x y + x + xz ] 1 x dydz 1 y + y + yz z + z ] 1 5 ] 1 dz y ( z )dz (y z )dydz ( + z )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 d If F E, the electric field, it can be shown that, F E q ε HELM (8): ection 9.: Integral ector Theorems 65

12 where q is the amount of charge per unit volume, or charge density, and ε is the permittivity of free space: ε 1 9 /6π F m F m 1. Gauss theorem becomes in this case q E d E d d 1 q d Q ε ε ε i.e. E d Q ε 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 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 1 x 1, 1 y 1, 1 z 1 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 1 ) is given by xi + yj + zk E 1 (x + y + z ) find the charge Q from Gauss law. Hint : x 1 y 1 ( x + y + 1 ) ] 4π dy dx 4 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 1 xi + yj + 1 k ( x + y + 1 4) d (element of surface area) (unit normal) dx dy k 66 HELM (8): Workbook 9: Integral ector alculus

13 o E d 1 1 ( x + y + 1 4) dy dx ( 5 x + y + 1 ) dy dx 4 Now (top) E d x 1 y 1 ( 5 x + y + 1 ) dy dx 4 5 4π π (using the hint) o, from Gauss law, Q 6ε π 4πε 1 9 Interpretation Gauss law can be used to find a charge from its effects elsewhere. The form of E 1 xi + yj + 1k ( comes from the fact that E is radial and equals 1 r x + y + 1 r 1 ˆr 4) r Example 7 erify Gauss theorem for the vector field F y j xzk and the triangular prism with vertices at (,, ), (,, ), (,, 1), (, 4, ), (, 4, ) and (, 4, 1) (see Figure 19). z (, 4, 1) y (,, 1) (,, ) (, 4, ) (,, ) (, 4, ) x Figure 19: The triangular prism defined by six vertices HELM (8): ection 9.: Integral ector Theorems 67

14 olution As F y j xzk, F + y x y x. Thus 4 x/ F d (y x)dzdydx To work out x y 4 x x y z yz xz ] 1 x/ z y 1 xy xy + 1 x y 16x 6x + ] x 4 dydx ] 4 x 4 dx y y (y xy x + 1 x )dydx x (16 1x + 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 16j xzk and d j thus F d 16 and the contribution to the integral is x x/ z 16dzdx x ] 1 x/ ] 16z dx 16(1 x/)dx 16x 4x 16. 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 1 x/, F y j+( 1 x x)k and the unit normal ˆn 1 5 i+ 5 k ] 1 Thus d 5 i + 5 k dydw where dw measures the distance along the slope for a constant y. As dw 5 dx, d 1 i + k] dydx thus F d 16 and the contribution to the integral is 4 ( 1 ] x x)dydx (x 4x)dx x x 8. x y Adding the contributions, Thus F d x F d F d 4 hence verifying Gauss divergence theorem. 68 HELM (8): Workbook 9: Integral ector alculus

15 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.: Integral ector Theorems 69

16 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 ector alculus

17 Task erify Gauss theorem for the vector field F xi yj + zk and the unit cube x 1, y 1, z 1. (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) 1 (c) dxdyk, ; dxdyk, 1; dxdzj, ; dxdzj, 1; dydzi, ; dydzi, 1 (d) Both sides are equal to 1. HELM (8): ection 9.: Integral ector Theorems 71

18 Exercises 1. erify Gauss theorem for the vector field F 4xzi y j + yzk and the cuboid x, y, z 4.. erify Gauss theorem, using cylindrical polar coordinates, for the vector field F ρ ˆρ over the cylinder ρ r, 1 z 1 for (a) r 1 (b) r. If is the surface of the tetrahedron with vertices at (,, ), (1,, ), (, 1, ) and (,, 1), 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 1 (i + j + k) and that d (i + j + k)dxdy Answers 1. Both sides are 156,. Both sides equal (a) 4π, (b) π,. (a) 5 4 only contribution is from the sloping face] (b) 5 4 by volume integral of (1 + z)]. 7 HELM (8): Workbook 9: Integral ector alculus

19 . Green s Identities (D) 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 } d 1] and {u w v u} d { u w w u } d ] Proof of Green s identities Green s identities can be derived from Gauss theorem and a vector derivative identity. ector identity (1) 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 F d (u w)d { ( u) ( w) + u w } d This is Green s identity 1]. Reversing the roles of u and w, { (w u) d ( w) ( u) + w u } d ubtracting the last two equations yields Green s identity ]. Key Point 1 1] ] Green s Identities { (u w) d u w + u w } d { {u w v u} d u w w u } d HELM (8): ection 9.: Integral ector Theorems 7

20 Example 8 erify Green s first identity for u (x x )y, w xy + z and the unit cube, x 1, y 1, z 1. 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 1, y, the vector u w and so the contribution to the surface integral is zero. On the face y 1, u w (x x )(i+xj +zk) and d dxdzj so (u w) d (x x )dxdz and the contribution to the integral is ] x (x x )dzdx (x x 1 )dx x 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 1, 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 ] 1 y(x x )dydx y (x x ) dx (x x )dx 1 x y x y 6. Thus, (u w) d Now evaluate { u w + u w } d. Note that u (1 x)yi + (x x )j and w so u w + u w (1 x)y + (x x )x + (x x )y x x + xy x y + y xy and the integral Hence { u w + u w } d (u w) d z y z y z y x x z( 1 4 )dz z 4 (x x + xy x y + y xy )dxdydz x4 4 + x y x y + xy x y ( y )dydz ] 1 z 1 4 u w + u w ] d 1 4 z ] 1 y 1 + y dz 6 y ] 1 dydz x and Green s first identity is verified. 74 HELM (8): Workbook 9: Integral ector alculus

21 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 11 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.: Integral ector Theorems 75

22 Example 9 Evaluate the line integral (4x + y )dx + (x + 4y )dy ] around the rectangle x, y 1. olution The integral could be obtained by evaluating four line integrals but it is easier to note that (4x + y )dx + (x + 4y )dy] is of the form P dx + Qdy with P 4x + y and Q x + 4y. It is thus of a suitable form for Green s theorem in the plane. Note that Q P 6x and x y 1. Green s theorem in the plane becomes {(4x + y )dx + (x + 4y )dy} y y x x x (6x 1) dxdy ] dy x y 4 dy 4 76 HELM (8): Workbook 9: Integral ector alculus

23 Example 4 erify 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 (,, 1) 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 (1 y ) and dz 1 dy. The integral is ( (4 y)dy 1 ) y (y )dy (5 y 1 y )dy 5y y 1 ] 6 y 14 ] On BO, y and dy. The integral is ( )dz z. 1 1 ( umming, 4zdy + (y )dz ) 8 ( 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 4 and Q y (y 4) dzdy ] 1 y/ z ] 1 y + y 4y ( Q y P ) dydz 8 z dy 8 y y. Hence, ( y + 4y 4 ) dy and Green s theorem in the plane is verified. HELM (8): ection 9.: Integral ector Theorems 77

24 One very useful, special case of Green s theorem in the plane is when Q x and P y. The theorem becomes { ydx + xdy} (1 ( 1)) dxdy The right-hand side becomes A 1 {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 41 erify the area theorem for the segment of the circle x + y 4 lying above the line y 1. 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 1 1. The sector has area π 4π. Thus segment ADB has area 4π. Now, evaluate the integral {xdy ydx} around the segment. Along the line, y 1, dy so the integral ( dx). {xdy ydx} becomes (x 1 dx) Along the arc of the circle, y 4 x (4 x ) 1/ so dy x(4 x ) 1/ dx. The integral {xdy ydx} becomes o, 1 { x (4 x ) 1/ (4 x ) 1/ }dx {xdy ydx} 1 π/ ] 8 π 4 π. π/ π/ π/ 4 4 x dx 1 4 cos θ dθ (letting x sin θ) cos θ 4dθ 8 π Hence both sides of the area theorem equal 4 π thus verifying the theorem. 78 HELM (8): Workbook 9: Integral ector alculus

25 Task erify Green s theorem in the plane when applied to the integral {(5x + y 7)dx + (x 4y + 5)dy} where represents the perimeter of the trapezium with vertices at (, ), (, ), (6, 1) and (1, 1). First let P 5x + y 7 and Q x 4y + 5 and find Q x P y : Your solution Answer 1 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 (, ), and then proceeding anti-clockwise. Your solution Answers 1.5, 66, 6.5, 1 whose sum is 4. HELM (8): ection 9.: Integral ector Theorems 79

26 Finally show that the two sides of the statement of Green s theorem are equal: Your solution Answer Both sides are 4. Exercises 1. erify Green s identity 1] (page 7) for the functions u xyz, w y and the unit cube x 1, y 1, z 1.. erify the area theorem for Answers (a) The area above y, but below y 1 x. (b) The segment of the circle x + y 1, to the upper left of the line y 1 x. 1. Both integrals in 1] equal 1. (a) both sides give a value of 4, (b) both sides give a value of π HELM (8): Workbook 9: Integral ector alculus