EE2 Mathematics : Vector Calculus

Transcription

1 EE2 Mathematics : Vector alculus J. D. Gibbon Professor J. D Gibbon, Dept of Mathematics) j.d.gibbon@ic.ac.uk jdg These notes are not identical word-for-word with m lectures which will be given on a BB/WB. Some of these notes ma contain more eamples than the corresponding lecture while in other cases the lecture ma contain more detailed working. I will NOT be handing out copies of these notes ou are therefore advised to attend lectures and take our own.. The material in them is dependent upon the Vector Algebra ou were taught at A-level and our st ear. A summar of what ou need to revise lies in Handout : Things ou need to recall about Vector Algebra which is also of this document. 2. Further handouts are : a) Handout 2 : The role of grad, div and curl in vector calculus summarizes most of the material in 3. b) Handout 3 : hanging the order in double integration is incorporated in 5.5. c) Handout 4 : Green s, Divergence & Stokes Theorems plus Mawell s Equations summarizes the material in 6, 7 and 8. Technicall, while Mawell s Equations themselves are not in the sllabus, three of the four of them arise naturall out of the Divergence & Stokes Theorems and the connect all the subsequent material with that given from lectures on e/m theor given in our own Department. Do not confuse me with Dr J. Gibbons who is also in the Mathematics Dept.

2 4th// EE2Ma-V.pdf) ontents evision : Things ou need to recall about Vector Algebra 2 2 Scalar and Vector Fields 3 3 The vector operators : grad, div and curl 3 3. Definition of the gradient operator Definition of the divergence of a vector field div B Definition of the curl of a vector field curl B Five vector identities Irrotational and solenoidal vector fields Line path) integration 8 4. Line integrals of Tpe : ψ,, z) ds Line integrals of Tpe 2 : F,, z) dr Independence of path in line integrals of Tpe Double and multiple integration 5 5. How to evaluate a double integral Applications Eamples of multiple integration hange of variable and the Jacobian hanging the order in double integration Green s Theorem in a plane D Divergence and Stokes Theorems 25 8 Mawell s Equations 29

3 4th// EE2Ma-V.pdf) 2 evision : Things ou need to recall about Vector Algebra Notation: a = a î + a 2 ĵ + a 3ˆk a, a 2, a 3 ).. The magnitude or length of a vector a is a = a = a 2 + a a 2 3) /2..) 2. The scalar dot) product of two vectors a = a, a 2, a 3 ) & b = b, b 2, b 3 ) is given b a b = a b + a 2 b 2 + a 3 b 3..2) Since a b = ab cos θ, where θ is the angle between a and b, then a and b are perpendicular if a b =, assuming neither a nor b are null. 3. The vector cross) product 2 between two vectors a and b is î ĵ ˆk a b = a a 2 a 3..3) b b 2 b 3 ecall that a b can also be epressed as a b = ab sin θ) ˆn.4) where ˆn is a unit vector perpendicular to both a and b in a direction determined b the right hand rule. If a b = then a and b are parallel if neither vector is null. 4. The scalar triple product between three vectors a, b and c is a a 2 a 3 a a b c) = b b 2 b 3 c c 2 c 3 c b According to the cclic rule clic ule: clockwise +ve.5) a b c) = b c a) = c a b).6) One consequence is that if an two of the three vectors are equal or parallel) then their scalar product is zero: e.g. a b a) = b a a) =. If a b c) = and no pair of a, b and c are parallel then the three vectors must be coplanar. 5. The vector triple product between three vectors is a b c) = ba c) ca b)..7) The placement of the brackets on the LHS is important: the HS is a vector that lies in the same plane as b and c whereas a b) c = bc a) ac b) lies in the plane of a and b. Thus, a b c without brackets is a meaningless statement! 2 It is acceptable to use the notation a b as an alternative to a b.

4 4th// EE2Ma-V.pdf) 3 2 Scalar and Vector Fields L) Our first aim is to step up from single variable calculus that is, dealing with functions of one variable to functions of two, three or even four variables. The phsics of electromagnetic e/m) fields requires us to deal with the three co-ordinates of space,, z) and also time t. There are two different tpes of functions of the four variables :. A scalar field 3 is written as ψ = ψ,, z, t). 2.) Note that one cannot plot ψ as a graph in the conventional sense as ψ takes values at ever point in space and time. A good eample of a scalar field is the temperature of the air in a room. If the bo-shape of a room is thought of as a co-ordinate sstem with the origin in one corner, then ever point in that room can be labelled b a co-ordinate,, z). If the room is poorl air-conditioned the temperature in different parts ma var widel : moving a thermometer around will measure the variation in temperature from point to point spatiall) and also in time temporall). Another eample of a scalar field is the concentration of salt or a de dissolved in a fluid. 2. A vector field B,, z, t) must have components B, B 2, B 3 ) in terms of the three unit vectors î, ĵ, ˆk) B = îb,, z, t) + ĵb 2,, z, t) + ˆkB 3,, z, t). 2.2) These components can each be functions of,, z, t). vector fields are : Three phsical eamples of a) An electric field : E,, z, t) = îe,, z, t) + ĵe 2,, z, t) + ˆkE 3,, z, t), 2.3) b) A magnetic field : H,, z, t) = îh,, z, t) + ĵh 2,, z, t) + ˆkH 3,, z, t), 2.4) c) The velocit field u,, z, t) in a fluid. A classic illustration of a three-dimensional vector field in action is the e/m signal received b a mobile phone which can be received anwhere in space. 3 The vector operators : grad, div and curl 3. Definition of the gradient operator The gradient operator grad) is denoted b the smbol and is defined as = î + ĵ + ˆk z. 3.) 3 The convention is to use Greek letters for scalar fields and bold oman for vector fields.

5 4th// EE2Ma-V.pdf) 4 As such it is a vector form of partial differentiation because it has spatial partial derivatives in each of the three directions. On its right, can operate on a scalar field ψ,, z) ψ = î ψ + ĵ ψ ψ + ˆk z. 3.2) Note that while ψ is a scalar field, ψ itself is a vector. Eample ) : With ψ = z 3) ψ = î2 + ĵ2 + ˆkz ) As eplained above, the HS is a vector whereas ψ is a scalar. Eample 2) : With ψ = z the vector ψ is ψ = îz + ĵz + ˆk. 3.4) End of L 3.2 Definition of the divergence of a vector field div B L2 Because vector algebra allows two forms of multiplication the scalar and vector products) there are two was of operating on a vector B = îb,, z, t) + ĵb 2,, z, t) + ˆkB 3,, z, t). 3.5) The first is through the scalar or dot product div B = B = î + ĵ + ˆk ) îb + z ĵb 2 + ˆkB ) ) ecalling that î î = ĵ ĵ = ˆk ˆk = but î ĵ = î ˆk = ˆk ĵ =, the result is div B = B = B + B 2 + B 3 z. 3.7) Note that div B is a scalar because div is formed through the dot product. The best phsical eplanation that can be given is that div B is a measure of the compression or epansion of a vector field through the 3 faces of a cube. If div B = the the vector field B is incompressible ; If div B > the the vector field B is epanding ; If div B < the the vector field B is compressing.

6 4th// EE2Ma-V.pdf) 5 Eample : Let r be the straight line vector r = î + ĵ + ˆkz then div r = + + = 3 3.8) Eample 2 : Let B = î2 + ĵ2 + ˆkz 2 then div B = z 3.9) Note : the usual rule in vector algebra that a b = b a that is, a and b commute) doesn t hold when one of them is an operator. Thus B = B + B 2 + B 3 z B 3.) 3.3 Definition of the curl of a vector field curl B The alternative in vector multiplication is to use in a cross product with a vector B : î ĵ ˆk curl B = B = z. 3.) B B 2 B 3 The best phsical eplanation that can be given is to visualize in colour the intensit of B then curl B is a measure of the curvature in the field lines of B. Eample ) : Take the vector denoting a straight line from the origin to a point,, z) denoted b r = î + ĵ + ˆkz. Then î ĵ ˆk curl r = r = z =. 3.2) z Eample 2) : hoose B = 2î 2 + ĵ2 + ˆkz 2) then curl B = B = î ĵ ˆk z z2 =. 3.3) Eample 3) : Take the vector denoted b B = î2 z 2 + ĵ2 z 2 + ˆk 2 2. Then î ĵ ˆk curl B = B = z = 2 2 î z) 2 2 ĵ z)+2z 2ˆk ). 3.4) 2 z 2 2 z 2 2 2

7 4th// EE2Ma-V.pdf) 6 Eample 4) : For the curl of a two-dimensional vector B = îb, ) + ĵb 2, ) we have î ĵ ˆk curl B = B = z = ˆk B2 B ), 3.5) B, ) B 2, ) which points in the vertical direction onl because there are no î and ĵ components. End of L2 3.4 Five vector identities L3 There are 5 useful vector identities see hand out No 2). The proofs of ), 4) and 5) are obvious : for No ) use the product rule. 2) and 3) can be proved with a little effort.. The gradient of the product of two scalars ψ and φ φψ) = ψ φ + φ ψ. 3.6) 2. The divergence of the product of a scalar ψ with a vector b 3. The curl of the product of a scalar ψ with a vector B 4. The curl of the gradient of an scalar ψ because the cross product is zero. 5. The divergence of the curl of an vector B div ψb) = ψ div B + ψ) B. 3.7) curl ψb) = ψ curl B + ψ) B. 3.8) curl ψ) = ψ =, 3.9) div curl B) = B) =. 3.2) The cclic rule for the scalar triple product in 3.2) shows that this is zero for all vectors B because two vectors ) in the triple are the same. 3.5 Irrotational and solenoidal vector fields onsider identities 4) & 5) above see also Handout 2 The role of grad, div & curl... ) curl φ) = φ =, 3.2) div curl B) =. 3.22) 3.2) sas that if an vector B,, z) can be written as the gradient of a scalar φ,, z) which can t alwas be done) B = φ 3.23)

8 4th// EE2Ma-V.pdf) 7 then automaticall curl B =. Such vector fields are called irrotational vector fields. It is equall true that for a given field B, then if it is found that curl B =, then we can write 4 B = ± φ 3.24) φ is called the scalar potential. Note that not ever vector field has a corresponding scalar potential, but onl those that are curl-free. Likewise we now turn to 3.22) : vector fields B for which div B = are called solenoidal, in which case B can be written as B = curl A 3.25) where the vector A is called a vector potential. Note that onl vectors those that are div-free have a corresponding vector potential 5. Eample : The Newtonian gravitational force between masses m and M with gravitational constant G) is where r = î + ĵ + ˆkz and r 2 = z 2.. Let us first calculate curl F F = GmM r r 3, 3.26) curl F = GmM curl ψr), where ψ = r ) The 3rd in the list of vector identities gives curl ψr) = ψ curl r + ψ) r 3.28) and we alread know that curl r =. It remains to calculate ψ : { ψ = z 2) } 3/2 3î + ĵ + ˆkz) = = 3r z 2 ) 5/2 r. 3.29) 5 Thus, from 3.26), curl F = GmM 3r ) 2r r =. 3.3) 5 Thus the Newton gravitational force field is curl-free, which is wh a gravitational potential eists. We can write 6 F = φ. 3.3) One can find φ b inspection : it turns out that φ = GmM /r. 4 Whether we use + or in 3.24) depends on convention : in e/m theor normall uses a minus sign whereas fluid dnamics uses a plus sign. 5 The last lecture on Mawell s Equations stresses that all magnetic fields in the universe are div-free as no magnetic monopoles have et been found : thus all magnetic fields are solendoidal vector fields. 6 In this case a negative sign is adopted on the scalar potential.

9 4th// EE2Ma-V.pdf) 8 2. Now let us calculate div F div F = GmM div ψr) where ψ = r ) The 2nd in the list of vector identities gives div ψr) = ψ div r + ψ) r 3.33) and we alread know that div r = 3 and we have alread calculated ψ in 3.29). 3 div F = GmM r 3r ) 3 r r =. 3.34) 5 Thus the Newton gravitational force field is also div-free, which is wh a gravitational vector potential A also eists. As a final remark it is noted that with F satisfing both F = φ and div F =, then div φ) = 2 φ =, 3.35) which is known as Laplace s equation. End of L3. 4 Line path) integration L4) In single variable calculus the idea of the integral b a f) d = N f i )δ i 4.) i= is a wa of epressing the sum of values of the function f i ) at points i multiplied b the area of small strips δ i : correctl it is often epressed as the area under the curve f). Pictoriall the concept of an area sits ver well in the plane with = f) plotted against. However, the idea of area under a curve has to be dropped when line integration is considered because we now wish to place our curve in 3-space where a scalar field ψ,, z) or a vector field F,, z) take values at ever point in this space. Instead, we consider a specified continuous curve in 3-space known as the path 7 of integration and then work out methods for summing the values that either ψ or F take on that curve. It is essential to realize that the curve sitting in 3-space and the scalar/vector fields ψ or F that take values at ever point in this space are wholl independent quantities and must not be conflated. 7 The same idea arises in comple integration where, b convention, a closed is known as a contour.

10 4th// EE2Ma-V.pdf) 9 z B A Figure 3. : The curve in 3-space starts at the point A and ends at B. B δs δr r + δr A r O Figure 3.2 : On a curve, small elements of arc length δs and the chord δr, where O is the origin. δ δs δ Figure 3.3 : In 2-space we can use Pthagoras Theorem to epress δs in terms of δ and δ. Pthagoras Theorem in 3-space see Fig 3.3 for a 2-space version) epress δs in terms of δ, δ and δz : δs) 2 = δ) 2 + δ) 2 + δz) 2. There are two tpes of line integral : Tpe : The first concerns the integration of a scalar field ψ along a path ψ,, z) ds 4.2)

11 4th// EE2Ma-V.pdf) Tpe 2 : The second concerns the integration of a vector field F along a path F,, z) dr 4.3) emark : In either case, if the curve is closed then we use the designations ψ,, z) ds and F,, z) dr 4.4) 4. Line integrals of Tpe : ψ,, z) ds How to evaluate these integrals is best shown b a series of eamples keeping in mind that, where possible, one should alwas draw a picture of the curve : Eample ) : Show that 2 ds = /3 where is the circular arc in the first quadrant of the unit circle., ), ) δθ θ δs, ), ) is the arc of the unit circle = represented in polars b = cos θ and = sin θ for θ π/2. Thus the small element of arc length is δs =.δθ. 2 ds = π/2 cos 2 θ sin θdθ) = /3. 4.5) Eample 2) : Show that 3 ds = 54 /5 where is the line = 3 from =. = 3 is an element on the line = 3. δ = 3δ so Thus δs) 2 = δ) 2 + 9δ) 2 = δ) 2 3 ds = 27 4 d = 54 /5. Eample 3) : Show that 2 ds = 2/2 where is the closed triangle in the figure.

12 4th// EE2Ma-V.pdf), ) 2 On : = so ds = d and = because = ). On 2 : = so d = d and so ds) 2 = 2ds) 2. On 3 : = so ds = d and 3 = because = ). 3, ) Therefore = = + 2 ) 2 d + = 2/2. 4.6) Note that we take the positive root of ds) 2 = 2d) 2 but use the fact that following the arrows on 2 the variable goes from. Eample 4) : see Sheet 2). Find z 2 ) ds where is the heli r = î cos θ + ĵ sin θ + ˆkθ, 4.7) with one turn : that is θ running from 2π. From 4.7) we note that = cos θ, = sin θ and z = θ. d = cos θdθ and dz = dθ. Thus Therefore d = sin θdθ, Therefore we can write the integral as z 2 ) ds = 2 End of L4 ds) 2 = sin 2 θ + cos 2 θ + ) dθ) 2 = 2dθ) ) 2π 4.2 Line integrals of Tpe 2 : F,, z) dr δs + θ 2 ) dθ = 2π 2 + 4π 2 /3 ). 4.9) B A r δr r + δr F O Figure 3.4 : A vector F and a curve with the chord δr : O is the origin. L5)

13 4th// EE2Ma-V.pdf) 2. Think of F as a force on a particle being drawn through the path of the curve. Then the work done δw in pulling the particle along the curve with arc length δs and chord δr is δw = F δr. Thus the full work W is W = F dr. 4.) 2. The second eample revolves around taking F as an electric field E,, z). Then the mathematical epression of Farada s Law sas that the electro-motive force on a particle of charge e travelling along is precisel e E,, z) dr. Eample : Evaluate F dr given that F = î2 + ĵ z) + ˆkz and the path is the parabola = 2 in the plane z = 2 from,, 2),, 2)., ) 2 is along the curve = 2 in the plane z = 2 in which case dz = and d = 2 d. 2 is along the straight line = in which case d = d. F dr = = = F d + F 2 d + F 3 dz) 2 d + z) d + z dz ) 2 d + 2) d ) 4.) Using the fact that d = 2 d we have I = = 4 d + 2)2 d ) ) d = 7/5. 4.2) Finall, with the same integrand but along 2 we have F dr = 3 d + 2) d ) 2 = /4 + /2 2 = 5/4. 4.3) This eample illustrates the point that with the same integrand and start/end points the value of an integral can differ when the route between these points is varied.

14 4th// EE2Ma-V.pdf) 3 Eample 2 : Evaluate I = 2 d 2 2 d) given that F = 2, 2 2 ) with the path taken as = 2 in the z = plane from,, ) 2, 4, ). 2, 4) is the curve = 2 in the plane z =, in which case dz = and d = 2 d. The starting point has co-ordinates,, ) and the end point 2, 4, ). I = = 2 d 2 2 d) ) d = 48/5. 4.4) Eample 3 : Given that F = î ĵz + 2ˆk, where is the curve z = 4 in the = plane, show that from,, ),, ) the value of the line integral is 7/5. On we have dz = 4 3 d and in the plane = we also have d =. Therefore F dr = d z d + 2 dz) End of L5 = 4.3 Independence of path in line integrals of Tpe ) d = 7/5. 4.5) L6) Are there circumstances in which a line integral F dr, for a given F, takes values which are independent of the path? To eplore this question let us consider the case where F dr is an eact differential : that is F dr = dφ for some scalar 8 φ. For starting and end co-ordinates A and B of we have F dr = dφ = φ[a] φ[b]. 4.6) This result is independent of the route or path taken between A and B. Thus we need to know what F dr = dφ means. Firstl, from the chain rule dφ = φ d + φ 8 As stated earlier, the choice of sign is b convention. φ d + dz = φ dr. 4.7) z

15 4th// EE2Ma-V.pdf) 4 Hence, if F dr = dφ we have F = φ, which means that F must be a curl-free vector field curl F =, 4.8) where φ is the scalar potential. A curl-free F is also known as a conservative vector field. esult : The integral F dr is independent of path onl if curl F =. Moreover, when is closed then F dr =. 4.9) Eample : Is the line integral 22 d d) independent of path? We can see that F is epressed as F = 2 2 î ĵ + ˆk. Then î ĵ ˆk curl F = z = 4 4)ˆk =. 4.2) Thus the integral is independent of path and we should be able to calculate φ from F = φ. φ = 22, φ = 22, φ z =. 4.2) Partial integration of the st equation gives φ = A) where A) is an arbitrar function of onl, whereas from the second φ = B). Thus A) = B) = const =. Hence φ = ) Eample 2 : Find the work done F dr b the force F = z, z, ) moving from,, ) 3, 3, 2). Is this line integral independent of path? We check to see if curl F =. î ĵ ˆk curl F = z =. 4.23) z z Thus the integral is independent of path and φ must eist. In fact φ = z, φ = z and φ z =. Integrating the first gives φ = z + A, z), the second gives φ = z + B, z) and the third φ = z +, ). Thus A = B = = const and φ = z + const. 4.24) 3,3,2) W = dφ = [ z ] 3,3,2) = 8 = ),,),,)

16 4th// EE2Ma-V.pdf) 5 Eample 3 : Find F dr on ever path between,, ) and, π/4, 2) where F = 2z 2, 2 z 2 + z cos z), 2 2 z + cos z) ). 4.26) We check to see if curl F =. î ĵ ˆk curl F = z =. 4.27) 2z 2 2 z 2 + z cos z) 2 2 z + cos z) Thus the integral is independent of path and φ must eist. Then φ = 2z 2, φ = 2 z 2 + z cos z) and φ z = 2 2 z + cos z). Integration of the first gives of the second and the third φ = 2 z 2 + A, z), 4.28) φ = 2 z 2 + sin z) + B, z), 4.29) φ = 2 z 2 + sin z) +, ). 4.3) The onl wa these are compatible is if A, z) = sin z + c and B = = c = const. and so φ = 2 z 2 + sin z) + c. In consequence F dr = [ φ ],π/4, 2) = π +. End of L6 4.3),,) 5 Double and multiple integration L7) Now we move on to et another different concept of integration : that is, summation over an area instead of along a curve. onsider the value of a scalar function ψ i, i ) at the co-ordinate point i, i ) at the lower left hand corner of the square of area δa i = δ i δ i. Then N i= M ψ i, i ) δa i ψ, ) dd } {{ } j= double integral as δ, δ. 5.) We sa that the HS is the double integral of ψ over the region. Note : do not confuse this with the area of itself, which is Area of = dd. 5.2)

17 4th// EE2Ma-V.pdf) 6 5. How to evaluate a double integral = h) = g) a b Figure 4.2 : The region is bounded between the upper curve = h), the lower curve = g) and the vertical lines = a and = b. { b } =h) ψ, ) dd = ψ, ) d d 5.3) a =g) The inner integral is a partial integral over holding constant. Thus the inner integral is a function of and so Moreover the area of itself is 5.2 Applications Area of = b a =h) =g) ψ, ) d = P ) 5.4) ψ, ) dd = b { } =h) d d = =g) a P ) d. 5.5) b a {h) g)} d 5.6). Area under a curve : For a function of a single variable = f) between limits = a and = b { b } f) b Area = d d = f) d. 5.7) a 2. Volume under a surface : A surface is 3-space ma be epressed as z = f, ) { } f,) Volume = dddz = dz dd = V a f, ) dd 5.8)

18 4th// EE2Ma-V.pdf) 7 B this process, a 3-integral has been reduced to a double integral. A specific eample would be the volume of an upper unit hemisphere z = ) f, ) centred at the origin. 3. Mass of a solid bod : Let ρ,, z) be the variable densit of the material in a solid bod. Then the mass δm of a small volume δv = δδδz is δm = ρ δv, End of L7 Mass of bod = ρ,, z) dv. 5.9) 5.3 Eamples of multiple integration L8 Eample : onsider the first quadrant of a circle of radius a. Show that : V, a), ) i) Area of = πa 2 /4 ii) dd = a 4 /8 iii) 2 2 dd = πa 6 /96, ) a, ) i) : The area of is given b A = a { } a 2 2 d d = a a2 2 d. 5.) Let = a cos θ and = a sin θ then A = 2 a2 π/2 cos 2θ) dθ = πa 2 /4. ii) : a ) a 2 2 dd = d d iii) : 2 2 dd = = 2 = 2 = 3 a a 2 2 ) d [ 2 2 a 2 4] a = 4 a4 /8. 5.) a a ) a d d 2 a 2 2 ) 3/2 d π/2 = 3 a6 cos 2 θ sin 4 θ dθ = 3 a6 I 2 I 3 ). 5.2)

19 4th// EE2Ma-V.pdf) 8 where I n = π/2 sin 2n θ dθ. Using a eduction Formula method we find that ) 2n I n = I n, I = π/2. 5.3) 2n Thus I 2 = 3π/6 and I 2 = 5π/96 and so from 5.2) the answer is πa 6 / hange of variable and the Jacobian In the last eample it might have been easier to have invoked the natural circular smmetr in the problem. Hence we must ask how δa = δδ would be epressed in polar co-ordinates. This suggests considering a more general co-ordinate change. In the two figures below we see a region in the plane that is distorted into in the plane of the new co-ordinates u = u, ) and v = v, ) v The transformation relating the two small areas δδ and δuδv is given here b the modulus sign is a necessit) : esult : where the Jacobian J u,v, ) is defined b Note that dd = J u,v, ) dudv, 5.4) J u,v, ) = u u u v v u 5.5) ) u 5.6) because u is computed at = const whereas u is computed at v = const. However, luckil there is an inverse relationship between the two Jacobians J, u, v) = [J u,v, )] 5.7) dudv = J, u, v) dd 5.8) Either 5.4) and 5.8) can therefore be used at one s convenience.

20 4th// EE2Ma-V.pdf) 9 Proof : not eaminable). onsider two sets of orthogonal unit vectors ; î, ĵ) in the - plane, and û, ˆv) in the u v-plane. Keeping in mind that the component of δ along û is δ/δu)δu v is constant along û), in vectorial notation we can write and so the cross-product is δ δ = δ = û δ δu δ = û δ δu û δ ) δ δu + ˆv δu δv δv δ δu + ˆv δv 5.9) δv δ δu + ˆv δv 5.2) δv û δ δ δu + ˆv δu δv δv ), 5.2) Therefore d d = δ δ = det u u v v ) û ˆv dudv = J u,v, ) dudv. 5.22) The inverse relationship 5.8) can be proved in a similar manner esult 2 : f, ) dd = fu, v), u, v)) J u,v, ) dudv. 5.23) Eample : For polar co-ordinates = r cos θ and = r sin θ we take u = r and v = θ J r,θ, ) = r θ = cos θ r sin θ sin θ r cos θ = r 5.24) Thus dd = r drdθ. r θ Eample 2 : To calculate the volume of a sphere of radius a we note that z = ± a Doubling up the two hemispheres we obtain Volume = 2 a2 2 2 dd 5.25) where is the circle = a 2 in the z = plane. Using a change of variable Volume = 2 a2 2 2 dd = 2 a2 r 2 rdrdθ = 2 a 2π a2 r 2 rdr dθ = 4πa 3 / )

21 4th// EE2Ma-V.pdf) 2 Eample 3 : From the eample in 5.3 over the quarter circle in the first quadrant it would be easier to compute in polars. Thus Area of = dd = rdrdθ = dd = = a rdr a π/2 dθ = πa 2 /4 5.27) r 3 cos θ sin θdrdθ r 3 dr = 4 a4 π/2 π/2 cos θ sin θdθ sin 2θdθ 2 = a 4 /6)[ cos 2θ] π/2 = a 4 / ) Likewise one ma show that 2 2 dd = = r 5 cos 2 θ sin 2 θ drdθ a r 5 dr π/2 cos 2 θ sin 2 θdθ = πa 6 / ) Eample 4 : Show that ) dd = 8/3 using u = + and v =. where has corners at, ),, ), 2, ) and, ) and rotates to a square with corners at, ), 2, ), 2, 2) and, 2). From = u + v) and = u v) it is found that 2 2 End of L8 J u,v, ) = 2 5.3) I = 2u 2 + v 2 ) dudv 4 2 { [ = u3] [v]2 + [ } v3] 2 3 [u]2 = 8/3. 5.3) 5.5 hanging the order in double integration In a handout, not in lectures. An eample is used to illustrate this : consider a a ) 2 d I = d. 5.32) 2 + 2

22 4th// EE2Ma-V.pdf) 2 ) Performing the integration in this order is hard as it involves a term in tan a. To make evaluation easier we change the order of the -integration and the -integration; first, however, we must deduce what the area of integration is from the internal limits in 5.32). The internal integral is of the tpe =a = f, ) d. 5.33) Being an integration over means that we are summing horizontall so we have left and right hand limits. The left limit is = and the right limit is = a. This is shown in the drawing of the graph where the region of integration is labelled as : = = a = a Now we want to perform the integration over but in reverse order to that above: we integrate verticall first b reading off the lower limit as = and the upper limit as =. After integrating horizontall with limits = to = a, this reads as I = a = = 2 d ) d. 5.34) The inside integral can be done with ease. is treated as a constant because the integration is over. Therefore define = θ where θ is the new variable. We find that Hence 2 d 2 + = dθ 2 + θ = π ) I = π 4 a d = πa )

23 4th// EE2Ma-V.pdf) 22 6 Green s Theorem in a plane L9 Green s Theorem in a plane which will be quoted at the bottom of an eam question where necessar tells us how behaviour on the boundar of a close curve through a closed line integral in two dimensions is related to the double integral over the region inside. Theorem Let be a closed bounded region in the plane with a piecewise smooth boundar. Let P, ) and Q, ) be arbitrar, continuous functions within having continuous partial derivatives Q and P. Then P d + Qd) = Q P ) dd. 6.) 2 = h) = g) a b Figure : In the plane the boundar curve is made up from two counter-clockwise curves and 2 : denotes the region inside. Proof : is represented b the upper and lower boundaries as in the Figure above) g) h) 6.2) and so P dd = = b a b = a b { =h) =g) } P d d { P, h) ) P, g) )} d a a P, g) ) d P, h) ) d b } {{ } switched limits & sign = P, ) d P, ) d 2 = P, ) d. 6.3)

24 4th// EE2Ma-V.pdf) 23 The same method can be used the other wa round to prove that note the +ve sign) Q dd = Q, ) d 6.4) Both these results are true separatel but can be pieced together to form the final result. Eample : Use Green s Theorem to evaluate the line integral { ) d 2 d }, 6.5) 2 where and are given b , 2) 2 Using G.T. with P = and Q = 2 over the bo-like region we obtain { ) d 2 d } = 2) dd = 2 2 d 2) d = ) Direct valuation gives = where is = ; 2 is = 2 ; 3 is = 2 and 4 is =. Thus = 2 d 4 2 d + 2 2) d + = = ) Eample 2 : Using Green s Theorem over in the diagram, show that { 3 d ) d } = 3/2 6.8) 2, ) = = 2

25 4th// EE2Ma-V.pdf) 24 Using Green s Theorem { 3 d ) d } = 3 = 3 = 3 2 dd 2 = = 2 d ) d 3 4) d = 3/2. End of L9 6.9) L Eample 3 : Part of 25 eam) : With a suitable choice of P and Q and as in the figure, show that d d) = dd = /3. 6.) 2 2, 2) = = 2 2 dd = 2 Around the line integral we have = ) d = d ) d ) d = /3. 6.) d d) + 2 /2 = 7/2 /4 = /3. 6.2) Eample 4 : Part of 24 eam) : If Q = 2 and P = 2 and is as in the figure below, show that 2 d 2 d) = 2 + ) dd = 24/5. 6.3) d

26 4th// EE2Ma-V.pdf) 25 2 = 2 2, 2) = The HS is obvious in the sense that Q P = and so 2 ) ) dd = 2 = 2 = d ) 2 d + 2 d d 2 2 [ 2 2 2) )] d [ 2 3/ ] d = 24/5. 6.4) The sum of the three line integrals can be done to get the same answer. 7 2D Divergence and Stokes Theorems B δs δr A r r + δr O Figure 6. : On a curve, with starting and ending points A and B, small elements of arc length δs and the chord δr, where O is the origin.

27 4th// EE2Ma-V.pdf) 26 The chord δr and the arc δs in the figure show us how to define a unit tangent vector ˆT δr ˆT = lim δr δs = dr ds = îd ds + ĵ d ds. 7.) The unit normal ˆn must be perpendicular to this : that is ˆn ˆT =, giving ˆn = ± î d ds ĵ d ), 7.2) ds where ± refer to inner and outer normals. A) The Divergence Theorem : Define a 2D vector u = îq ĵp. Then u ˆn = P d ds + Qd ds div u = Q P 7.3) Thus Green s Theorem turns into a 2D version of Divergence Theorem div u dd = u ˆn ds 7.4) This line integral simpl epresses the sum of the normal component of u around the boundar. If u is a solenoidal vector div u = ) then automaticall u ˆn ds =. The 3D version 9 uses an arbitrar 3D vector field u,, z) that lives in some finite, simpl connected volume V whose surface is S: da is some small element of area on the curved surface S div u dv = u ˆn da 7.5) V S End of L B) Stokes Theorem : L Now define a 2D vector v = îp + ĵq. Then v dr = v ˆT ) ds = P d + Q d. 7.6) Moreover curl v = î ĵ ˆk z P Q = ˆk Q P ) 7.7)

28 4th// EE2Ma-V.pdf) 27 Thus Green s Theorem turns into a 2D version of Stokes Theorem ) ˆk curl v dd = v dr 7.8) The line integral on the HS is called the circulation. Note that if v is an irrotational vector then v dr = which means there is no circulation. The 3D-version of Stokes Theorem for an arbitrar 3D vector field v,, z) in a volume V is given b v dr = ˆn curl v da. 7.9) ˆn is the unit normal vector to the surface S of V. is a closed circuit circumscribed on the surface of the volume and S is the surface of the cap above. S Eample : part of 22 eam) If v = î2 + ĵ2 and is as in the figure below, b evaluating the line integral, show that v dr = 5/48. 7.) = 3 = 2 Firstl 2 v dr = = /4) i) On we have = /4) and so d = d/4 2 ). Therefore v dr = 6 ) d + 2 d ). 7.) d = ) ii) On 2 we have = and so d =. Therefore v dr = d = ) 4

29 4th// EE2Ma-V.pdf) 28 iii) On 3 we have = and so d = d. Therefore 3 v dr = 2 2 Summing these three results gives = Eample 2 : part of 23 eam) If 2 d = ) u = î ĵ [ ln + 2 ) ] 7.5) and is as in the figure below, show that div u dd = 2 ln ) 3 = 2 Firstl we calculate div u and so End of L div u = div u dd = 4 = 4 = 4 = ) = + dd 2 = d + 2 [ 2 ln + 2 ) ] d ln + 2 ) d ) d = 2 ln 2 7.8)

30 4th// EE2Ma-V.pdf) 29 8 Mawell s Equations L2 Mawell s equations are technicall not in m sllabus but I give this lecture to tie everthing together in a phsical contet. onsider a 3D volume with a closed circuit drawn on its surface. Let s use the 3D versions of the Divergence 7.5) and Stokes Theorems 7.9) to derive some relationships between various electro-magnetic variables.. Using the charge densit ρ, the total charge within the volume V must be equal to surface area integral of the electric flu densit D through the surface S recall that D = ɛe where E is the electric field). ρ dv = D ˆn da. 8.) V Using a 3D version of the Divergence Theorem 7.5) above on the HS of 8.) ρ dv = div D dv. 8.2) Hence we have the first of Mawell s equations This is also known as Gauss s Law. V S V div D = ρ 8.3) 2. Following the above in the same manner for the magnetic flu densit B recall that B = μh where H is the magnetic field) but noting that there are no magnetic sources so ρ mag = ), we have the 2nd of Mawell s equations div B = 8.4) 3. Farada s Law sas that the rate of change of magnetic flu linking a circuit is proportional to the electromotive force in the negative sense). Mathematicall this is epressed as d B ˆn da = E dr 8.5) dt S Using a 3D version of Stokes Theorem 7.9) on the HS of 8.4), and taking the time derivative through the surface integral thereb making it a partial derivative) we have the 3rd of Mawell s equations curl E + B t = 8.6)

31 4th// EE2Ma-V.pdf) 3 4. Ampère s Biot-Savart) Law epressed mathematicall the line integral of the magnetic field around a circuit is equal to the current enclosed) is J ˆn da = H dr, 8.7) S where J is the current densit and H is the magnetic field. Using 3D-Stokes Theorem 7.9) on the HS we find that we have curl H = J and therefore div J =, which is inconsistent with the first three of Mawell s equations. Wh? The continuit equation for the total charge is d ρ dv = J ˆn da. 8.8) dt V S Using the Divergence Theorem on the LHS we obtain div J + ρ t =. 8.9) If div J = then ρ would have to be independent of t. To get round this problem we use Gauss s Law ρ = div D epressed above to get { div J + D } = 8.) t and so this motivates us to replace J in curl H = J b J + D/ t giving the 4th of Mawell s equations curl H = J + D 8.) t 5. We finall note that because div B = then B is a solenoidal vector field: there must eist a vector potential A such that B = curl A. Using this in the 3rd of Mawell s equations, we have [ curl E + A ] =. 8.2) t This means that there must also eist a scalar potential φ that satisfies E = A t End of L2 and the Vector alculus part of EE2 Maths. φ. 8.3)