Department of Mathematics and Statistics

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1 Applied eading University of eading Department of Mathematics and Statistics Useful reading: Calculus: A Complete Course 6 th Ed.,.A. Adams, Pearson Addison Wesley. Mathematical Methods in the Physical Sciences 3 rd Ed., M. Boas, Wiley. Vector algebra Let u, v n then the scalar or dot product u v is the scalar number given by: u v { u v cos θ where θ is the angle between the vectors n k u kv k u v + u v + + u n v n Geometrically, if v is a unit vector then one can interpret the dot product u v as a measure of how much of u is projected onto v i.e., their overlap in some sense. If u v then u and v are said to be orthogonal some authors prefer the term perpendicular but this concept does not generalise beyond 3 and as such we shall not use it. Example. Let u 3, + 3 T and v 3, 3 T, as shown in figure, then u v 6 3. We verify the assertions about the geometric nature of the scalar product by additionally computing: Figure : The scalar product of two vectors u and v 3. Therefore, the angle θ between the vectors satisfies 6 3 cos θ and is consequently found to be θ π/6. Thus, u cos θ this adjacent component of u is precisely the amount of u projected onto v is u cos θ cosπ/6 3/ 6. To see that this is precisely the overlap of u and v we rotate u by π/4 thereby bringing v in line with the horizontal axis whereon its components become 6, T which readily confirms our earlier assertion. Let u, v 3 then the vector or cross product u v is the vector given by: u v u v 3 u 3 v u 3 v u v 3 u v u v u v 3 u 3 v î + u3 v u v 3 ĵ + u v u v ˆk î ĵ ˆk u u u 3 v v v 3 where we formally evaluate the determinant by expanding along the first row. The vector product is constructed so as to produce a vector u v that is orthogonal to both u and v.

2 Applied eading Example. Let u,, T and v π,, T then u v î ĵ ˆk π î ĵ π + ˆk π î ĵ + πˆk. Hence, u v,, π T. We check the previous assertion that u v is orthogonal to both u and v by computing u v u π π + + ; and u v v π π π π π + + π. It follows from elementary row properties of the determinant that the vector product is an antisymmetric operation; i.e., u v v u. A further corollary of this observation is that u u u 3. Unlike the scalar product, the vector product does not generalise straightforwardly to higher dimensions its generalisation, the wedge product, is beyond the scope of this discussion. We close this section with some standard identities of vector algebra and a note on their proof: Let u, v, w 3 then u v w v w u w u v note the cyclical nature and u v w u w v u v w. The standard strategy for establishing vector identities is to consider their components. For example, the first component, [ u v w ], of u v w is [ ] u v w u [ ] v w u [ ] 3 3 v w u v w v w u3 v3 w v w 3 u w + u 3 w 3 v u v + u 3 w 3 w u w u w v u v u v w u w v u v w [ u w v u v w ] and to complete the proof one must establish the same result for all the other components individually. The Levi-Civita symbol The Levi-Civitia or permutation symbol ε ijk can greatly simplify vector calculations involving the vector product, it is defined to be Consequently, for example, ε ijk ε kij. + if ijk is a cyclic permutation of 3 ε ijk if ijk is an anti-cyclic permutation of 3 otherwise Example. ε 3 ε 3 ε 3 +, ε 3 ε 3 ε 3 and all other symbols vanish.

3 Applied eading The advantage of the Levi-Civita symbol is that one can use it to give a concise expression for the i-th component of u v in the following way: [ u v ]i j k ε ijk u j v k. Example. ecall that to compute [ u v ] u v 3 u 3 v we must formally evaluate the determinant of a 3 3 matrix and then isolate the appropriate component. Consider [ u v ] j k ε jk u j v k εj u j v + ε j u j v + ε j3 u j v 3 j ε u v + ε u v + ε 3 u 3 v + ε u v + ε u v + ε 3 u 3 v + ε 3 u v 3 + ε 3 u v 3 + ε 33 u 3 v 3 ε 3 u 3 v + ε 3 u v 3 u v 3 u 3 v You are entitled to make the objection that this isn t a significantly easier question but it does enable for the direct manipulation of components of a vector product as we shall soon establish especially, the curl of a vector field. Scalar and vector fields Throughout this section we let O n denote a subset of n possibly the whole of n. Let f : O be a scalar field, i.e., a scalar valued function. For convenience, we denote the set of scalar fields by SO,. Example. The mapping f : 3 defined by x, y, z xy + z is a scalar function; hence, f S 3,. The set S n, is easily shown to be a linear space. Example. The mapping f : is defined by x, y xy is a scalar function; hence, f S,. The mapping g : is defined by x, y x + y is a scalar function; hence, g S,. As both f, g S, they may be added and their sum f + g :, defined by x, y fx, y + gx, y xy + x + y, is also a scalar field. Let F : O n be a vector field, i.e., a vector valued function. For convenience, we denote the set of vector fields by V O, n. The components of a vector field are themselves scalar fields. Example. The mapping F : 3 3 defined by x, y, z x,, x + y T is a vector field; hence, F V 3, 3. The set V O, n is easily shown to be a linear space. Example. The mapping F : is defined by x, y x, y T is a vector field; that is, F V,. The mapping G : is defined by x, y x, x + y T is a vector field; that is, G V,. As both F, G V, they may be added and their sum F + G :, defined by x, y F x, y + Gx, y x, x + y T, is also a vector field.

4 Applied eading grad, div and curl Throughout this section we let O n denote a subset of n possibly the whole of n. The gradient operator : SO, V O, n is defined by f f, f,..., f x x x n for all sufficiently differentiable f SO,. Example. Let f : be defined by x, y xy + cos x then f : is given by x, y y sin x, x T. Typically, we write fx, y y sin x, x T or fx, y y sin xî + xĵ. Let σ : 3 then σx, y, z defines a two-dimensional surface in 3 and the vector σ is everywhere normal orthogonal to the surface defined by σx, y, z. Example. Let σ : 3 be defined by x, y, z x + y 3z + 5 then σx, y, z defines a two-dimensional surface in 3 indeed, x + y 3z + 5 is an equation of a plane whose normal vector is easily inspected to be,, 3 T. We compute T σ x + y 3z + 5, x + y 3z + 5, x + y 3z + 5,, 3 T x y z as expected. Example. Let σ : 3 be defined by x, y, z x + y + z 9 then σx, y, z defines a two-dimensional surface in 3 indeed, x + y + z 9 defines the surface of the sphere of radius 3 centred at the origin. We compute σ x, y, z T which is certainly an outward normal vector consider the various octants for confirmation. Let F : n n be a vector field whose components F T F,..., F n are each sufficiently differentiable scalar fields. The divergence of the vector field F, denoted F, is the scalar field given by n F k F. x k One can think of the divergence operator : V O, SO, should one wish to. Example. Consider the 3 vector field F given by x, y, z x, yz sin x, T then F is given by k T x, y, z x x + y yz sin x + x + z sin x ; z alternatively written F x, y, z x + z sin x. Let F : 3 3 be a vector field as before. The curl of the vector field F, denoted F, is the 3 vector field given by F 3 y F z F F z F3 F3 x y F F î+ z z F 3 F ĵ+ x x F î ĵ ˆk ˆk y x y z. F F F 3 F x F y Note that again when we write the curl of a field in terms of a determinant that this is pure formalism. Again, one can think of the curl operator : V O, V O, should one wish to. It is at this point that we have the opportunity to adopt a more powerful notation. Consider the arbitrary 3 vector x x, y, z T ; rather than using the usual Cartesian labels we instead write x x, x, x 3 T. The advantage of this notation is that one can now write i-th component i of the gradient operator is i / x i.

5 Applied eading Example. Let f : 3 be defined by x, x, x 3 x + x x then [ ] f [ ] f + x, f [ ] f x and f 3 3f. A consequence of this new notation is the following expression for the curl F of the vector field F : 3 3 with components F, F, F 3 T where F, F, F 3 S 3, : [ F ]i j k ε ijk j F k. Using this notation one can begin to establish those vector calculus identities involving the curl. For example, let F : 3 3 be a vector field as before then [ i F ]i i ε ijk j F k ε ijk i j F k. j k [ Summing the term i F over all admissible i one obtains ]i F i j k ε ijk i j F k j k ε 3 F 3 + ε 3 3 F + ε 3 3 F + ε 3 3 F + ε 3 3 F + ε 3 3 F F F F as partial derivatives commute i.e., F 3 F 3, etc. for sufficiently well behaved functions Clairaut s theorem. Thus we learn the famous vector calculus identity that div-curl vanishes. We close this section with the remark that vector calculus has many important applications in science outside of its interest in mathematics. Example Maxwell s equations. An electric field E with associated magnetic field B satisfy the Maxwell equations: D ρ, B, E B t and H j + D t where ρ is the charge density, j is the current and the auxiliary vectors D and H for a field propagating in an isotropic and homogeneous medium satisfy E D and B H. Any electromagnetic field must obey all of Maxwell s equations. Standard identities involving grad, div and curl In this brief section we cover the basic identities of vector calculus beginning with a familiar object from the study of differential equations. ecall the Laplacian which typically features in the Laplace, Poisson or heat equation; in two-dimensions is given by x + y or in n-dimensions x + x + + x. n Let f : n be a twice differentiable scalar field then f x x f x f f. x f x x n. f x n + f +... x x x x f x n f.

6 Applied eading Hence, we obtain the standard interpretation of the Laplacian as div-grad :. We may prove other vector calculus identities by adopting this strategy. V 3, 3 be suitably differentiable vector fields and consider F G i F G i i i i j k i j k i j k k ε ijk F j G k ε ijk i F j G k + F j i G k εkij i F j G k ε jik F j i G k F G k k F j G j j F G F G F G. For example, let F, G By continuing in this manner one can establish the following standard vector calculus identities: fg f g + g f F G F G F G ff f F + f F F G F G G F F G + G F ff f F + f F F G F G + G F f F f f for any scalar fields f, g and vector fields F, G. Line integrals + F G + G F F F F Let a < b, f : n be a scalar field and consider the curve γ : n. For convenience we define the set n by {γt t [a, b]}. The line integral b f ds fγt γ t dt a where ds is an infinitesimal piece of the curve γ. If γa γb then we say the line integral is a contour integral and denote thus f ds. Example. Consider the points, and, and the connecting path γ : defined by t t, t t [, ]. Then γ t, T, thus γ t ; hence, x + y ds t + t dt 5 3. Example. Consider the points, and, and the connecting path γ : defined by t sinπt/, sinπt/ t [, ]. Then γ t π/ cosπt/, T, thus γ t π/ cosπt/ ; hence, x + y ds sin πt/ + sinπt/ π cosπt/ dt 5 3.

7 Applied eading Let a < b, F : n n be a bounded vector field and consider the curve γ : n. For convenience we define the set n by {γt t [a, b]}. The line integral F dr b a F γt γ t dt where dr dx, dy, dz T. If γa γb then we say the line integral is a contour integral and denote it thus F dr. Contrary to intuition it is not a priori true that F dr ; in fact, it is generally not true. Example. Consider the points, and, and the connecting path γ : defined by t t, t t [, ]. Then γ t, T ; hence, t yî xĵ dr dt dt. t Example. Consider the points, and, and the connecting path γ : defined by t t, sinπt/ t [, ]. Then γ t, π/ cosπt/ T ; hence, sinπt/ yî xĵ dr dt t π/ cosπt/ sinπt/ π t cosπt/ dt 4 π π. Note that the results are different depending on the path taken between the end points of the curve. Conservative vector fields and scalar potentials A vector field F : n n is said to be conservative if there exists a f : n such that F f. The function f is referred to as a scalar potential. Some authors insist on F f; this alternative convention choice has considerable advantages in the theory of classical mechanics and electromagnetism. Example. The vector field F : defined by x, y x + y, x T is conservative; indeed, it arises from the scalar potential f : defined by x, y xy + x /. Example. Let f : 3 be given by x, x, x 3 x x x 3 then the vector field F : 3 3 defined by x, x, x 3 f, f, 3 f T x x 3, x x 3, x x T is conservative. Example. Consider the vector field F : defined by x, y y +, x T which is easily shown to be conservative by solving f x y +... f y x.... Integration w.r.t. x of equation implies that fx, y xy++ϕy for some arbitrary function ϕ. This proposed general solution must also satisfy equation, i.e., x + ϕ y x ϕ y ; the solution ϕ to this ode is y c where c is an arbitrary constant of integration. Thus, the general scalar potential solution f is given by x, y xy + + c, however, in practice it is typical to drop the constant as f only appears under a divergence or as a difference term when this cannot influence the answer. Therefore, fx, y xy +. Hence, the vector field is conservative.

8 Applied eading Not all vector fields are conservative. To prove a given vector field is not conservative one must establish that there does not exist a scalar potential, typically one use a reductio ad absurdum argument. Example. The vector field F : defined by x, y y, x T is not conservative; indeed, suppose it were and that there exist a scalar potential f : satisfying f x y... f y x.... Integration w.r.t. x of equation implies that fx, y xy + ϕy for some arbitrary function ϕ. This proposed general solution must also satisfy equation, i.e., x + ϕ y x but no such function of a single variable can exist our desired contradiction. Hence, the vector field is not conservative. An 3 conservative vector field F : 3 3 satisfies F. This is a straightforward corollary of f. A vector field of vanishing curl is said to be irrotational. The fundamental theorem of vector calculus Let n define a closed path between the points p and q then the fundamental theorem of vector calculus states that: if F : n n is a conservative vector field then q F dr f fq fp p where f : n is a scalar potential associated to F i.e., F f. We can make the connection with the usual one variable situation by writing the result exclusively in terms of the scalar potential whereon it becomes f dr f q which is pleasingly reminiscent of the standard fundamental p theorem of calculus. Example. Consider the vector field F : defined by x, y y +, x T which we have previously established is conservative with associated scalar potential f : defined by x, y xy + and the points p, and q,. Then F dr fq fp xy + independently of the specific path joining the endpoints. To further reinforce this remarkable fact we specify a particular path by {t, t t [, ]} then as expected. F dr t + t dt,, t + dt Example. We have previously established that the vector field F : given by x, y y, x T is not conservative. The points p, and q, can be connected by a variety of paths, for definiteness we consider the paths {t, t t [, ]} ; {t, t t [, ]} ; 3 {, s s [, ]} {t, t [, ]}. Then F dr t t dt dt.

9 Applied eading Whereas, and, finally, 3 F dr F dr s t t ds + t t dt t dt 3, dt ds + Hence, F dr F dr F dr and F dr F dr 3 3 despite the fact that, and 3 all have the same endpoints. dt. Path independence of line integral quantities is not just a pleasing mathematical fact, it is of central importance of the field of classical physics and thermodynamics where it has profound consequences. For example, the work done by to resist a force F along the path is given by F dr; hence, if the force is not conservative one may have to work harder along certain paths between the endpoints of the motion. Multiple integration Let a < b, g : [a, b] be a bounded function and consider the expression b a gx dx which we typically interpret as the signed area between the curves x a, x b, y and y gx. This interpretation rests on the fact that gxdx can be identified as the area of the rectangle [x, x + dx] [, gx] which one sums continuously over all admissible x [a, b]. We wish to extend this notion and consider the integral of a function of two and later more variables. Suppose is a bounded region and f : is a bounded function of two variables. We wish to give meaning to expressions of the form fx, y da where the region of integration is and the area element is da. If we consider an infinitesimal element of area da in Cartesian coordinates of the rectangular region with vertices x, y, x, +dx, y, x+dx, y and x + dx, y + dy as shown in figure. This indicates that, in Cartesian coordinates, da dx dy and that we should be able to evaluate the area integral by a double integral. The question remains as to the limits of integration in the variables:???????? fx, yda fx, ydx dy or fx, yda fx, ydy dx?????? as, at least intuitively, the area dx dy is the same as dy dx. The answer is provided by considering the boundary of. The boundary is a curve in, that is a one dimensional object. Let us denote the equation of the curve by ϕx, y which one may solve for either variable we are at liberty to consider either as the independent one. Consider figure 3 which indicates that we may evaluate these double integrals using the following limit prescription: max y max xy fx, yda fx, ydx dy or fx, yda min y max x min x min xy?? max yx fx, ydy dx min yx as desired. In the examples which follow the reader is strongly advised to verify the asserted limits by means of a suitable sketch in the form of figure 3.????

10 Applied eading Figure : The area element da in Cartesian coordinates Example. Let [, ] [, ] and f : be defined by x, y x + x cos πy then fx, y da x + x cos πy dx dy 3 + cos πy dy 3. Example. Let be the triangle with vertices,,, and, 3 and f : be given by x, y sinh y then 3 3 fx, y da sinh y dx dy sinh y y dy + y/3 3 3 sinh 3. Example. Let {x, y x + y } and f : be defined by x, y xy then y fx, y da x + y dx dy y y dy π y 4. We remark that this integral is remarkably symmetric in the following sense. Suppose we perform the change of variables y ỹ y then y x + y ỹ dx dy x + ỹ dx dỹ y ỹ ỹ Therefore, y y x + y dx dy which may be equivalently expressed x + y da D ỹ x + ỹ dx y y x + y dx x + y da where D is the upper unit semi-circle centred at the origin. Thus, D dỹ. dy x + y da π/8.

11 Applied eading Figure 3: The limits of a double integral Polar coordinates Some situations admit particular symmetry which may be exploited to enable easier computation. In this section we briefly consider planar circular symmetry where the polar coordinates r, θ [, [, π are related to the Cartesian coordinates x, y through x r cos θ and y r sin θ. By considering an infinitesimal area da element in planar polar coordinates one can show that da rdr dθ; see figure 4. Figure 4: Area element in polar coordinates Example. Let be the upper half of the unit circle centred at the origin in and f : be given by x, y x + y then π fx, y da r cos θ + r sin θ π dθ rdr r rdr π 8 as expected.

12 Applied eading Multiple integrals and the Jacobian determinant One may extend the idea of double integrals to higher dimensional settings. For the purposes of this discussion we shall restrict ourselves to a brief remark on higher integrals of the form fx,..., x n dv where n, f : is a bounded function and the volume element dv in Cartesian coordinates is given by dv n k dx k dx... dx n. Example. Let a < b and [a, b] [a, b] n-times then b b dv... dx... dx n b a n as one would expect for the volume of the n-cube of side length b a. a a Example. Let h, l >, {x, y, z 3 x + y l and z [, h]} [, h] where {x, y x + y l } and consider dv which we expect, on geometric grounds, to be equal to πl h. Indeed, one may verify this by writing dv [,h] dv da h dz h da h l π dθ rdr h l π rdr πl h. We turn our attention briefly to the subject of the change of variables in multiple integration. Suppose we wish to change from Cartesian variables x,..., x n to another set of variables t,..., t n then one can show that dx... dx n x,..., x n t,..., t n dt... dt n where is the Jacobian determinant of the change of variables transform. x,..., x n t,..., t n x x t n t x n t... Example. Consider the change from Cartesian to planar polar coordinates via x r cos θ and y r sin θ then x,..., x n t,..., t n x/ r y/ r x/ θ y/ θ cos θ sin θ x n t n r sin θ r cos θ r cos θ + r sin θ r. Hence, as before. dx dy rdr dθ Example Spherical polar coordinates. Let l >, {x, y, z 3 x + y + z l} and define f : by x, y, z x + y then fx, y, z dv may be evaluated by the following change of variables x r sin θ cos ϕ r y r sin θ sin ϕ where θ [, π]. z r cos θ ϕ [, π

13 Applied eading The Jacobian determinant x, y, z/ r, θ, ϕ of the transform is x x x x, y, z r θ ϕ r, θ, ϕ y y y r θ ϕ z z z r θ ϕ sin θ cos ϕ r cos θ cos ϕ r sin θ sin ϕ sin θ sin ϕ r cos θ sin ϕ r sin θ cos ϕ cos θ r sin θ cos θ r cos θ cos ϕ r sin θ sin ϕ r cos θ sin ϕ r sin θ cos ϕ + r sin θ sin θ cos ϕ sin θ sin ϕ cos θ r cos θ sin θ + r sin θ r sin θ r sin θ ; dx dy dz r sin θ dr dθ dϕ r sin θ sin ϕ r sin θ cos ϕ as r and sin θ if θ [, π]. Hence, noting that x + y r sin θ, x + y l π dv l l π r sin θ dϕ sin θ dθ r dr π πr sin θ sin θ dθ r dr 8 3 πr 8 5 πl5. r dr For clarity for fixed r: in our conventions for spherical polar coordinates the angle θ governs the displacement from the origin in the z-direction whereas φ determined the distance from the origin in the x y plane. We present a series of plots of the surfaces swept out for a variety of angular regions in figure 5. We provide a summary of these findings and another common coordinate system in table. Table : Some common coordinate systems Planar Volume Coordinate system Coordinates Measure Notes Cartesian x, y dx dy x, y Polar x r cos θ r dx dy rdr dθ y r sin θ θ [, π Cartesian x, y, z dx dy dz x, y, z x r sin θ cos ϕ r Polar y r sin θ sin ϕ dx dy dz r sin θ dr dθ dϕ θ [, π] z r cos θ ϕ [, π x r cos θ r Cylindrical y r sin θ dx dy dz r dr dθ dz θ [, π z z

14 Applied eading Figure 5: Spherical polar coordinates. θ [, π, ϕ [, π] top left, θ [, π/, ϕ [, π top right, θ [, π], ϕ [, π] bottom left and θ [, π/], ϕ [, π] bottom right Surface integrals Table also enables us to infer some results about surface integrals. A surface integral is one of the form fx, y ds where is a surface bounding the region and ds is the surface element of. In general ds dx dy i.e., a small patch of area on is not always a rectangle. One can become convinced of this by considering the following well known phenomenon/brand of chocolate orange: it is impossible to smoothly wrap a sheet of paper around a ball the fact that the surface of the ball is curved means that you need to introduce a fold. ather than integrating over a planar region using Cartesian coordinates we could use planar polar coordinates to integrate a function over. Planar polar coordinates provide us with a surface circle of radius r which we can integrate our function over before subsequently integrating over all r and thereby sweeping out the entire plane as shown in figure 6. Figure 6 encourages the following interpretation of the area integral: {}}{ π fx, y da fr, θ rdθ first integrate f over the surface of a circle of radius r dr. } {{ } then sweep out the plane by integrating over the radial direction r Hence, we may interpret rdθ as the surface element of the circle and π fr, θ rdθ as the integral of

15 Applied eading Figure 6: Sweeping out the plane using the surfaces of a series of circles f over the surface of a circle of fixed radius r. Denoting this surface we write this integral as fx, y ds. Example. Let l > and {x, y x + y l }. Consider the surface integral x + y + π ds l cos θ + l sin θ + ldθ πl + l. Using a similar argument to that of figure 6 we may infer that r sin θ dθ dϕ is the appropriate surface element for the sphere of radius r. Example. Let l > and {x, y, z 3 x + y + z l} be the surface of the sphere of radius l. Consider the integral x ds. In spherical polar coordinates ds l sin θ dθ dϕ and the value of x, restricted to the surface, is given by x l sin θ cos ϕ. Hence, x ds π π l sin θ cos ϕ l sin θ dθ dϕ π πl3 cos ϕ dϕ. eaders interested in learning how to construct the surface element ds for other surfaces are directed to the literature we confess that the technique we have employed here takes advantage of the special geometry considered and does not generalise. A brief digression on dimensional analysis The Cartesian coordinates x, y, z each have the fundamental dimension of length L. Let f be a physical quantity, then we denote the dimension of f by {f} dim. For example, {length} dim {radius} dim L, {area} dim L and {volume} dim L 3. Scalars numbers are dimensionless which we write {scalar} dim ; for example, {angle} dim. Pertinent to our discussion are the observations that {dx} dim {dr} dim L, {dx dy} dim L, {dx dy dz} dim L 3 and {dθ} dim {dϕ} dim. If we consider the change of coordinates and the Jacobian determinant then { } x,..., x n {dx... dx n } dim t,..., t n dt... dt n dim { } x,..., x n t,..., t n {dt } dim... {dt n } dim dim

16 Applied eading Suppose we did not know dx dy dz r sin θ dr dθ dϕ we could at least deduce, on purely dimensional grounds, that dx dy dz cθ, ϕ r dr dθ, dϕ where {cθ, ϕ} dim. Indeed, we may sensibly assume that dx dy dz kr, θ, ϕ dr dθ dϕ {dx dy dz} dim {kr, θ, ϕ} dim {dr} dim {dθ} dim {dϕ} dim L 3 {kr, θ, ϕ} dim L L {kr, θ, ϕ} dim {kr, θ, ϕ} dim {r cθ, ϕ} dim where {cθ, ϕ} dim as previously asserted. Astute readers will have noticed that this argument relies on the implicit assumption that we have not omitted a dimensionful quantity. Hence, had we erroneously concluded earlier that dx dy dz r sin θ dr dθ dϕ we would know immediately that we had blundered. Finally, we note that this technique cannot provide any insight into the dimensionless quantities but is a useful tool in our armory nonetheless. The divergence theorem Gauss divergence theorem is a generalisation of the fundamental theorem of calculus. We present the planar version of this theorem it s generalisation to higher dimensions requires a discussion of surface integrals and the interested reader is directed to the literature. Let be a bounded region with boundary having outward unit normal ˆn and F V, then F da F ˆn ds. Physically one may interpret F ˆn ds as a measure of the amount of F flowing through the boundary in the direction ˆn; consequently, one sometimes refers to such terms as flux integrals. We remark that some authors write ˆnds ds and the divergence theorem as F da F ds. ightly or wrongly we shall not adopt this convention for this discussion. Example. Let l >, {x, y x + y l } and consider da which we have already shown to be equal to πl. This may be verified using the divergence theorem in the following way: define F : by x, y x, y T / then F. Moreover, the boundary {x, y x + y l } and ˆn x, y T /l. Therefore, the divergence theorem asserts that da F ˆn ds x + y ds l ds l ds l π l dθ πl l l as expected. It is straightforward to generalise the divergence theorem to higher dimensions although we restrict ourselves to the physically meaningful three. Let 3 be a bounded region with boundary having outward unit normal ˆn and F V, 3 then F dv F ˆn da where dv is the volume element dv dx dy dz in Cartesian coordinates. Example. Let l >, 3 be the solid sphere of radius l, i.e. {x, y, z 3 x + y + z l },

17 Applied eading with normal n x + y + z l x, y, z T. Therefore, ˆn x, y, z T/l. Define the vector field F : 3 by x, y, z x, y, z T then the flux integral F ˆn ds l l x y z x/l y/l z/l x + y z ds ds x + y l ds. There are many ways to parametrise the surface of the sphere. Perhaps the most conventional is x l cos θ cos ϕ, y l cos θ sin ϕ and z l sin θ θ [, π], ϕ [, π. The surface element ds l sin θ dθ dϕ. Therefore, x + y l π π ds l F ˆn ds l sin 3 θ dθ dϕ 4l 3 Example. Let l >, 3 be the solid cylinder of radius l and unit height, i.e. {x, y, z 3 x + y l and z [, ]}, π dϕ 8πl 3 and define F : 3 3 by x, y, z xy, x y, z x + y T. Then the flux F ˆn ds of F flowing out of the cylinder, i.e. across, is easily calculated using the divergence theorem: F ˆn ds F dv x + y + z dv l π r + z dθ rdr dz πl 4. l πr + z rdr dz πl 4 + z dz. We close this section with a brief remark about partial differential equations. It is straightforward to establish the identity f g f g + f g for any suitably defined scalar fields f, g; an immediate corollary of this observation is that f g g f f g g f. Hence, it follows from the divergence theorem that f g g f dv f g g f ˆn ds. This result is known as Green s identity. Green s identity is of central importance to the study of partial differential equations where one routinely wishes to construct objects called Green s functions which one can think of as the inverse of pde ivp/bvp s.

18 Applied eading Green s theorem Let be a bounded region with boundary then: Green s theorem in the plane for the field F : defined by x, y F x, y, F x, y T is where F, F : are scalar fields. F x F da F dr y Example. Let {x, y x +y } and F : be defined by x, y x+y, x T then F x F da x x + y da da. y x y Next we remark that {x, y x + y } is parametrised by t cosπt, sinπt T where t [,. Hence, F dr π π cosπt + sinπt cosπt π sinπt π cosπt dt sinπt cosπt sin πt + cos πt dt sin4πt + cos4πt thereby verifying the statement of Green s theorem in the plane. Example. Let {x, y x + y and y } and F : be defined by x, y x, xy T then F x F da y 3 xy x da x y y da y y y dx y y dy Next we remark that {x, y x + y and y } {x, x [, ]}. Hence, F dr π 3 π sinπt π cosπt cosπt cosπt sinπt sin4πt + cos πt sinπt dt + thereby verifying the statement of Green s theorem in the plane. dy dt + s ds s ds

19 Applied eading Stokes theorem Stokes theorem generalises Green s theorem in the following way: let 3 be a surface i.e. dimension with normal vector ˆn and boundary. Let F : then F ˆn ds F dr. In the interests of clarify: this result is more correctly known as Kelvin s formulation of Stokes theorem or the Kelvin-Stokes theorem; the full Stokes theorem requires the machinery of forms from differential geometry and is beyond the scope of this discussion. Example. Let l > and {x, y, z 3 x + y + z l and z } be the surface of the upper hemi-sphere of radius l and outward normal ˆn x, y, z T /l. The boundary,, of is the circle given by {x, y, 3 x + y l }. The vector field F : 3 3 given by x, y, z z, y, x T T. has curl F,, Therefore, F ˆn y and π π/ F ˆn ds l sin θ sin ϕ l sin θ dθ dϕ Additionally, we compute F dr using the parametrisation: π πl3 sin ϕ dϕ. x l cos θ, y l sin θ and z r l cos θ, sin θ, T r l sin θ, cos θ, T Therefore, r l cos θ, sin θ, T from which we readily compute F dr π F r dθ l π sin θ cos θ thereby verifying the statement of Stokes theorem. sin θ cos θ dθ l π sin θ cos θ dθ Example. Let l > and {x, y, z 3 x + y + z l and z } be the surface of the upper hemi-sphere of radius l and outward normal ˆn. The boundary,, of is given by {x, y, 3 x + y l } note that this boundary is also the boundary of the disc D {x, y x + y l } whose normal vector is ˆk. Define F : 3 3 by x, y, z y cosxz, x 3 e yz, e xyz T then, by two applications of the Kelvin-Stokes theorem, F ˆn ds F dr F ˆk da. We evaluate this right most integrand: D F xze xyz x 3 ye yz, xy sinxz yze xyz, 3x e yz y cosxz T F ˆk 3x e yz y cosxz F ˆk 3x y recall that z on D. D Hence, F ˆn ds Acknowledgements D 3x y da l π 3r cos θ r sin θ dθ rdr l 3πr 3 dr 3 4 πl4. We are grateful to Prof Mark Matsen, Prof Peter Grindrod, Dr Alex Lukyanov, Dr Peter Chamberlain & Dr Nick Biggs for their contributions to this document. Any errors which remain are, of course, the fault of the author.