three

Appendix B VectorCalculusinThreeDimensions In this appendix, we review the fundamentals of three-dimen sional vector calculus. The student is expected to have already encountered most of t hese topics in an introduc- tory multi-variable calculus course. We shall be dealing wi th calculus on curves, surfaces and solid bodies in three-dimensional space. The three meth ods of integration — line, surface and volume (triple) integrals — and the fundamental vector differential operators — gradient, curl and divergence — are intimately related. Th e differential operators and integrals underlie the multivariate versions of the fundam ental theorem of calculus, known as Stokes’ Theorem and the Divergence Theorem. All of these topics will be reviewed in rapid succession, wit h most details being rel- egated to the exercises. A more detailed development can be f ound in any reasonable multi-variable calculus text, including [ 9,46,70]. B.1. Dot and Cross Product. We begin by reviewing the basic algebraic operations betwee n vectors in three-dim- ensional space R3. We shall continue to use column vector notation v= v1 v2 v3 = (v1,v2,v3)T∈R3. The standard basis vectors of R3are e1=i= 1 0 0 , e2=j= 0 1 0 , e3=k= 0 0 1 .(B.1) We prefer the former notation, as it easily generalizes to n-dimensional space. Any vector v= v1 v2 v3 =v1e1+v2e2+v3e3 isa linearcombinationofthe basisvectors. The coefficients hev1,v2,v3arethecoordinates of the vector with respect to the standard basis. 12/11/12 1230 c/ci∇cleco√y∇t2012 Peter J. Olver Space comes equipped with an orientation — either right- or l eft-handed. One cannot alter†the orientation by physical motion, although looking in a mi rror — or, mathemat- ically, performing a reflection — reverses the orientation. The standard basis vectors are graphed with a right-hand orientation, as in Figure rhr . When you point with your right hand,e1lies in the direction of your index finger, e2lies in the direction of your middle finger, and e3is in the direction of your thumb. In general, a set of three li nearly inde- pendent vectors v1,v2,v3is said to have a right-handed orientation if they have the same orientation as the standard basis. It is not difficult to prove that this is the case if and only if the determinant of the 3 ×3 matrix whose columns are the given vectors is positive: det(v1,v2,v3)>0. Interchanging the order of the vectors may switch their or ientation; for example if v1,v2,v3are right-handed, then v2,v1,v3is left-handed. We have already made extensive use of the Euclidean dot product v·w=v1w1+v2w2+v3w3,where v= v1 v2 v3 ,w= w1 w2 w3 ,(B.2) along with the Euclidean norm /ba∇dblv/ba∇dbl=/radicalbig v·v=/radicalbig v2 1+v2 2+v2 3. (B.3) As in Definition 3.1, the dot product is bilinear, symmetric: v·w=w·v, and positive. The Cauchy–Schwarz inequality |v·w| ≤ /ba∇dblv/ba∇dbl/ba∇dblw/ba∇dbl. (B.4) implies that the dot product can be used to measure the angle θbetween the two vectors vandw: v·w=/ba∇dblv/ba∇dbl/ba∇dblw/ba∇dblcosθ. (B.5) (See also (3.17).) Remark: In this chapter, we will only use the Euclidean dot product a nd its associ- ated norm. Adapting the constructions to more general norms and inner products is an interesting exercise, but will not concern us here. Also of great importance — but particular to three-dimensio nal space — is the cross productbetween vectors. While the dot product produces a scalar, th e three-dimensional cross product produces a vector, defined by the formula v×w= v2w3−v3w2 v3w1−v1w3 v1w2−v2w1 where v= v1 v2 v3 ,w= w1 w2 w3 ,(B.6) †This assumes that space is identified with the three-dimensional E uclidean space R3, or, more generally, an oriented three-dimensional manifold, [ 21]. 12/11/12 1231 c/ci∇cleco√y∇t2012 Peter J. Olver We have chosen to employ the more modern wedge notation rathe r the more traditional cross symbol, v×w, for this quantity. The cross product formula is most easily memorized as a formal 3 ×3 determinant v×w= det v1w1e1 v2w2e2 v3w3e3 = (v2w3−v3w2)e1+(v3w1−v1w3)e2+(v1w2−v2w1)e3, (B.7) involving the standard basis vectors (B.1). We note that, li ke the dot product, the cross product is a bilinear function, meaning that (cu+dv)×w=c(u×w)+d(v×w),u×(cv+dw) =c(u×v)+d(u×w), (B.8) for any vectors u,v,w∈R3and any scalars c,d∈R. On the other hand, unlike the dot product, the cross product is an anti-symmetric quantity v×w=−w×v, (B.9) which changes its sign when the two vectors are interchanged . In particular, the cross product of a vector with itself is automatically zero: v×v=0. Geometrically, the cross product vector u=v×wis orthogonal to the two vectors v andw: v·(v×w) =0=w·(v×w). Thus, when vandware linearly independent, their cross product u=v×w/\e}atio\slash=0defines a normal direction to the plane spanned by vandw. The direction of the cross product is fixed by the requirement that v,w,u=v×wform a right-handed triple. The length of the cross product vector is equal to the area of the paralle logram defined by the two vectors, which is /ba∇dblv×w/ba∇dbl=/ba∇dblv/ba∇dbl/ba∇dblw/ba∇dbl|sinθ| (B.10) whereθis than angle between the two vectors, as in Figure para . Consequently, the cross product vector is zero, v×w=0, if and only if the two vectors are collinear (linearly dependent) and hence only span a line. Thescalar triple product u·(v×w) between three vectors u,v,wis defined as the dot product between the first vector with the cross product of the second and third vec- tors. The parenthesis is often omitted because there is only one way to make sense of u·v×w. Combining (B.2), (B.7), shows that one can compute the trip le product by the determinantal formula u·v×w= det u1v1w1 u2v2w2 u3v3w3 . (B.11) By the properties of the determinant, permuting the order of the vectors merely changes the sign of the triple product: u·v×w=−v·u×w= +v·w×u=···. 12/11/12 1232 c/ci∇cleco√y∇t2012 Peter J. Olver The triple product vanishes, u·v×w= 0, if and only if the three vectors are linearly dependent, i.e., coplanar or collinear. The triple product is positive, u·v×w>0 if and only if the three vectors form a right-handed basis. Its magn itude|u·v×w|measures the volume of the parallelepiped spanned by the three vector su,v,w, as in Figure ppp . B.2. Curves. Aspace curve C⊂R3is parametrized by a vector-valued function x(t) = x(t) y(t) z(t) ∈R3, a ≤t≤b, (B.12) that depends upon a single parameter tthat varies over some interval. We shall always assume that x(t) is continuously differentiable. The curve is smoothprovided its tangent vectoris continuous and everywhere nonzero: dx dt=/squaresmallsolidx= /squaresmallsolidx /squaresmallsolidy /squaresmallsolidz /\e}atio\slash=0. (B.13) As in the planar situation, the smoothness condition (B.13) precludes the formulation of corners, cusps or other singularities in the curve. Physically, we can think of a curve as the trajectory describ ed by a particle moving in space. At each time t, the tangent vector/squaresmallsolidx(t) represents the instantaneous velocity of the particle. Thus, as long as the particle moves with nonzero sp eed,/ba∇dbl/squaresmallsolidx/ba∇dbl=/radicalbig/squaresmallsolidx2+/squaresmallsolidy2+/squaresmallsolidz2> 0, its trajectory is necessarily a smooth curve. Example B.1. Acharged particleinaconstant magneticfield movesalongth ecurve x(t) = ρcost ρsint ct , (B.14) wherec>0 andρ>0 are positive constants. The curve describes a circular helixofradius ρspiraling up the zaxis. The parameter cdetermines the pitchof the helix, indicating how tightly its coils are wound; the smaller cis, the closer the winding. See Figure B.1 for an illustration. DNA is, remarkably, formed in the shape of a (bent and twisted) double helix. The tangent to the helix at a point x(t) is the vector /squaresmallsolidx(t) = −ρsint ρcost c . Note that the speed of the particle, /ba∇dbl/squaresmallsolidx/ba∇dbl=/radicalig ρ2sin2t+ρ2cos2t+c2=/radicalbig ρ2+c2, (B.15) remains constant, although the velocity vector/squaresmallsolidxtwists around. 12/11/12 1233 c/ci∇cleco√y∇t2012 Peter J. Olver -1-0.500.51 -1-0.500.51 -10010-0.500.51 Figure B.1. A Helix. Most of the terminology introduced in Chapter A for planar cu rves carries over to space curves without significant alteration. In particular , a curve is simpleif it never crosses itself, and closedif its ends meet, x(a) =x(b). In the plane, simple closed curves are all topologically equivalent, meaning one can be contin uously deformed to the other. In space, this is no longer true. Closed curves can be knotted , and thus have nontrivial topology. Example B.2. The curve x(t) = (2+cos3t)cos2t (2+cos3t)sin2t sin3t for 0 ≤t≤2π, (B.16) describes a closedcurve that isintheshape of a trefoilknot , asdepicted inFigureB.2. The trefoil is a genuine knot, meaning it cannot be deformed into an unknotted circle without cutting and retying. (However, a rigorous proof of this fact is not easy.) The trefoil is the simplest of the “toroidal knots”, investigated in more deta il in Exercise B.2.6. The study and classification of knots is a subject of great his torical importance. In- deed, they were first considered from a mathematical viewpoi nt in the nineteenth century, when the English applied mathematician William Thompson (l ater Lord Kelvin) proposed a theory of atoms based on knots! In recent years, knot theory has witnessed a tremendous revival, owing to its great relevance to modern day mathemat ics and physics. We refer the 12/11/12 1234 c/ci∇cleco√y∇t2012 Peter J. Olver Figure B.2. Two Views of a Trefoil Knot. interested reader to the advanced text [ 115] for details. B.3. Line Integrals. In Section A.5, we encountered three different types of line i ntegrals along plane curves. Two of these — integrals with respect to arc length, ( A.35), and circulation integrals, (A.37) — are directly applicable to space curves . On the other hand, for three- dimensional flows, the analog of the flux line integral (A.42) is a surface integral, and will be discussed later in the chapter. Arc Length Thelengthof the space curve x(t) over the parameter range a≤t≤bis computed by integrating the norm of its tangent vector: L(C) =/integraldisplayb a/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt=/integraldisplayb a/radicalbig/squaresmallsolidx2+/squaresmallsolidy2+/squaresmallsolidz2dt. (B.17) It is not hard to show that the length of the curve is independe nt of the parametrization — as it should be. Starting at the endpoint x(a), thearc length parameter sis given by s=/integraldisplayt a/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledtand sods=/ba∇dbl/squaresmallsolidx/ba∇dbldt=/radicalbig/squaresmallsolidx2+/squaresmallsolidy2+/squaresmallsolidz2dt. (B.18) The arc length smeasures the distance along the curve starting from the init ial point x(a). Thus, the length of the part of the curve between s=αands=βis exactlyβ−α. It is often convenient to reparametrize the curve by its arc len gth,x(s). This has the same effect as moving along the curve at unit speed, since, by the ch ain rule, dx ds=dx dtdt ds=/squaresmallsolidx /ba∇dbl/squaresmallsolidx/ba∇dbl,so that/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx ds/vextenddouble/vextenddouble/vextenddouble/vextenddouble= 1. 12/11/12 1235 c/ci∇cleco√y∇t2012 Peter J. Olver Thereforedx/dsis the unit tangent vector pointing in the direction of motio n along the curve. Example B.3. The length of one turn of a helix (B.14) is, using (B.15), L(C) =/integraldisplay2π 0/vextenddouble/vextenddouble/vextenddouble/vextenddoubledx dt/vextenddouble/vextenddouble/vextenddouble/vextenddoubledt=/integraldisplay2π 0/radicalbig ρ2+c2dt= 2π/radicalbig ρ2+c2. The arc length parameter, measured from the point x(0) = (r,0,0)Tis merely a rescaling, s=/integraldisplayt 0/radicalbig ρ2+c2dt=/radicalbig ρ2+c2t, of the original parameter t. When the helix is parametrized by arc length, x(s) =/parenleftigg ρcoss/radicalbig ρ2+c2, ρsins/radicalbig ρ2+c2,cs/radicalbig ρ2+c2/parenrightiggT , we move along it with unit speed. It now takes time s= 2π/radicalbig ρ2+c2to complete on turn of the helix. Example B.4. To compute the length of the trefoil knot (B.16), we begin by c om- puting the tangent vector dx dt= −2(2+cos3 t)sin2t−3 sin3tcos2t 2(2+cos3 t)cos2t−3 sin3tsin2t 3 cos3t . After some algebra involving trigonometric identities, we find /ba∇dbl/squaresmallsolidx/ba∇dbl=√ 27+16 cos3 t+2 cos6t, which is never 0. Unfortunately, the resulting arc length in tegral /integraldisplay2π 0/ba∇dbl/squaresmallsolidx/ba∇dbldt=/integraldisplay2π 0√ 27+16 cos3 t+2 cos6t dt cannot be completed in elementary terms. Numerical integra tion can be used to find the approximate value 31 .8986 for the length of the knot. Thearc length integral of a scalar field u(x) =u(x,y,z) along a curve Cis /integraldisplay Cuds=/integraldisplayℓ 0u(x(s))ds=/integraldisplayℓ 0u(x(s),y(s),z(s))ds, (B.19) whereℓis the total length of the curve. For example, if ρ(x,y,z) represents the density at position x= (x,y,z) of a wire bent in the shape of the curve C, then/integraldisplay Cρdsrepresents the total mass of the wire. In particular, the integral /integraldisplay Cds=/integraldisplayℓ 0ds=ℓ 12/11/12 1236 c/ci∇cleco√y∇t2012 Peter J. Olver recovers the length of the curve. If it isnot convenient to work directly with the arc length pa rametrization, we can still compute the arc length integral in terms of the original para metrization x(t) fora≤t≤b. Using the change of parameter formula (B.18), we find /integraldisplay Cuds=/integraldisplayb au(x(t))/ba∇dbl/squaresmallsolidx/ba∇dbldt=/integraldisplayb au(x(t),y(t),z(t))/radicalbig/squaresmallsolidx2+/squaresmallsolidy2+/squaresmallsolidz2dt. (B.20) Example B.5. The density of a wire that is wound in the shape of a helix is proportional to its height. Let us compute the mass of one ful l turn of the helical wire. Thus, the density is given by ρ(x,y,z) =az, whereais the constant of proportionality, and we are assuming z≥0. Substituting into (B.20), the total mass of the wire is L(C) =/integraldisplay Cazds=/integraldisplay2π 0act/radicalbig r2+c2dt= 2π2ac/radicalbig r2+c2. Line Integrals of Vector Fields As in the two-dimensional situation (A.37), the line integral of a vector field valong a parametrized curve x(t) isobtained by integrationofitstangential component wit hrespect to the arc length. The tangential component of vis given by v·t,where t=dx ds is the unit tangent vector to the curve. Thus, the line integr al ofvis written as /integraldisplay Cv·dx=/integraldisplay Cv1(x,y,z)dx+v2(x,y,z)dy+v3(x,y,z)dz=/integraldisplay Cv·tds.(B.21) We can evaluate the line integral in terms of an arbitrary par ametrization of the curve by the general formula /integraldisplay Cv·dx=/integraldisplayb av(x(t))·dx dtdt (B.22) =/integraldisplayb a/bracketleftbigg v1(x(t),y(t),z(t))dx dt+v2(x(t),y(t),z(t))dy dt+v3(x(t),y(t),z(t))dz dt/bracketrightbigg dt. Line integrals in three dimensions enjoy all of the properti es of their two-dimensional siblings: Reversing the direction of parameterization alo ng the curve changes the sign; also, the integral can be decomposed into sums over componen ts: /integraldisplay −Cv·dx=−/integraldisplay Cv·dx,/integraldisplay Cv·dx=/integraldisplay C1v·dx+/integraldisplay C2v·dx, C=C1∪C2. (B.23) Iff(x) represents a force field, e.g., gravity, electromagnetic f orce, etc., then its line integral/integraldisplay Cf·dxrepresents the workdone by moving along the curve. As in two dimen- sions, work is independent of the parametrization of the cur ve, i.e., the particle’s speed of traversal. 12/11/12 1237 c/ci∇cleco√y∇t2012 Peter J. Olver Example B.6. Our goal is to move a mass through the force field f= (y,−x,1)T startingfromtheinitialpoint(1 ,0,1)Tandmovingverticallytothefinalpoint(1 ,0,2π)T. Question : does it require more work to move in a straight line x(t) = (1,0,t)Tor along the spiral helix x(t) = (cost,sint,t)T, where, in both cases, 0 ≤t≤2π? The work line integral has the form /integraldisplay Cf·dx=/integraldisplay Cydx−xdy+dz=/integraldisplay2π 0/bracketleftbigg ydx dt−xdy dt+dz dt/bracketrightbigg dt. Along the straight line, the amount of work is /integraldisplay Cf·dx=/integraldisplay2π 0dt= 2π. As for the spiral helix, /integraldisplay Cf·dx=/integraldisplay2π 0/bracketleftbig −sin2t−cos2t+1/bracketrightbig dt= 0. Thus, although we travel a more roundabout route, it takes no work to move along the helix! Thereasonforthesecondresultisthattheforcevectorfield fiseverywhereorthogonal to the tangent to the curve: f·t= 0, and so there is no tangential force exerted upon the motion. In such cases, the work line integral /integraldisplay Cf·dx=/integraldisplay Cf·tds= 0 automatically vanishes. In other words, it takes no work wha tsoever to move in any direction which is orthogonal to the given force vector. B.4. Surfaces. Curves are one-dimensional, and so can be traced out by a sing le parameter. Surfaces are two-dimensional, and hence require two distinct parame ters. Thus, a surfaceS⊂R3 is parametrized by a vector-valued function x(p,q) = (x(p,q),y(p,q),z(p,q))T(B.24) that depends on two variables. As the parameters ( p,q)∈Ω range over a prescribed plane domain Ω ⊂R2, the locus of points x(p,q) traces out the surface in space. See Figure surf for an illustration. The parameters are often thought of as d efining a system oflocal coordinates on the curved surface. We shall always assume that the surface is simple, meaning that it does not intersect itself, so x(p,q) =x(/tildewidep,/tildewideq) if and only if p=/tildewidepandq=/tildewideq. In practice, this condition can be quite hard to check! The boundary ∂S={x(p,q)|(p,q)∈∂Ω} (B.25) of a simple surface consists of one or more simple curves, as i n Figure surf . If the underlying parameter domain Ω is bounded and simply connect ed, then∂Ω is a simple closed plane curve, and so ∂Sis also a simple closed curve. 12/11/12 1238 c/ci∇cleco√y∇t2012 Peter J. Olver Example B.7. The simplest instance of a surface is a graphof a function. The parameters are the x,ycoordinates, and the surface coincides with the portion of t he graph of the function z=u(x,y) that lies over a fixed domain ( x,y)∈Ω⊂R2, as illustrated in Figure gsurf . Thus, a graphical surface has the parametric form x(p,q) = (p,q,u(p,q))T, (p,q)∈Ω. Thus, theparametrizationidentifies x=pandy=q, whilez=u(p,q) =u(x,y)represents the height of the surface above the point ( x,y)∈Ω. For example, the upper hemisphere S+ rof radiusrcentered at the origin can be parametrized as a graph z=/radicalbig r2−x2−y2, x2+y2<r2, (B.26) sitting over the disk Dr={x2+y2<r2}of radiusr. The boundary of the hemisphere is the image of the circle Cr=∂Dr={x2+y2=r2}of radiusr, and is itself a circle of radiusrsitting in the x,yplane:∂S+ r={x2+y2=r,z= 0}. Remark: One can interpret the Dirichlet problem (15.1), (15.4) for the two-dimen- sional Laplace equation as the problem of finding a surface Sthat is the graph of a harmonic function with a prescribed boundary ∂S={z=h(x,y) for (x,y)∈∂Ω}. Example B.8. AsphereSrofradiusrcanbeexplicitlyparametrizedby twoangular variablesϕ,θin the form x(ϕ,θ) = (rsinϕcosθ,rsinϕsinθ,rcosϕ),0≤θ<2π,0≤ϕ≤π.(B.27) The reader can easily check that /ba∇dblx/ba∇dbl2=r2, as it should be. As illustrated in Fig- ure sangle ,θmeasures the azimuthal angle orlongitude , whileϕmeasures the zenith angleorlatitude. Thus, the upper hemisphere S+ ris obtained by restricting the zenith parameter to the range 0 ≤ϕ≤1 2π. Each parameter value ϕ,θcorresponds to a unique point on the sphere, exceptwhenϕ= 0 orπ. All points ( θ,0)are mapped to the north pole (0,0,r), while all points ( θ,π) are mapped to the south pole (0 ,0,−r). Away from the poles, the spherical angles provide bona fide coordinates on the sphere. Fortunately, the polar singularities do not interfere with the overall smoot hness of the sphere. Nevertheless, one must always be careful at or near these two distinguished points. Remark: In terrestrial cartography and navigation, the latitudeis measured from the equator, and equals1 2π−ϕ, with postive values referring to the northern hemishpere a nd negative the southern henisphere (or, vice versa, if you are an anitpodean). The longitude istaken withrespect to the prime meridian, throughGreenwi ch, England, and equals π−θ, with positive vlaues referring to the western hemisphere. O f course, in practice, both are measured in degrees rather than radians. The curves {ϕ=c}where the zenith angle takes a prescribed constant value are the circular parallels of co nstant latitude — except for the north andsouth poles which aremerely points. The equatoris atϕ=1 2π, whilethe tropics of Cancer and Capricorn are 231 2◦≈0.41 radians above and below the equator. The curves {θ=c}where the meridial angle is constant are the semi-circular m eridians of constant longitude stretching from north to south pole. Note that θ= 0 andθ= 2πdescribe the same meridian. In terrestrial navigation, latitude is the a ngle, in degrees, measured from the equator, while longitude is the angle measured from the G reenwich meridian. 12/11/12 1239 c/ci∇cleco√y∇t2012 Peter J. Olver Example B.9. Atorusis a surface of the form of an inner tube. One convenient parametrization of a particular toroidal surface is x(ψ,θ) = ((2+cos ψ)cosθ,(2+cosψ)sinθ,sinψ)Tfor 0 ≤ψ,θ≤2π.(B.28) Note that the parametrization is 2 πperiodic in both ψandθ. If we introduce cylindrical coordinates x=rcosθ, y =rsinθ, z, then the torus is parametrized by r= 2+cosψ, z = sinψ. Therefore, the relevant values of ( r,z) all lie on the circle (r−2)2+z2= 1 (B .29) of radius 1 centered at (2 ,0). As the polar angle θincreases from 0 to 2 π, the circle rotates around the zaxis, and thereby sweeps out the torus. Remark: The sphere and the torus are examples of closed surfaces . The requirements for a surface to be closed are that it be simple and bounded, an d, moreover, have no boundary. In general, a subset S⊂R3isbounded provided it does not stretch off infinitely far away. More precisely, boundedness is equivalent to the e xistence of a fixed number R>0 which bounds the norm /ba∇dblx/ba∇dbl<Rof all points x∈S. Tangents to Surfaces Consider a surface Sparameterized by x(p,q) where (p,q)∈Ω. Each parametrized curve(p(t),q(t))intheparameter domainΩwillbemappedtoaparametrizedc urveC⊂S contained in the surface. The curve Cis parametrized by the composite map x(t) =x(p(t),q(t)) = (x(p(t),q(t)),y(p(t),q(t)),z(p(t),q(t)))T. The tangent vector dx dt=∂x ∂pdp dt+∂x ∂qdq dt(B.30) to such a curve will be tangent to the surface. The set of all po ssible tangent vectors to curves passing through a given point in the surface traces ou t thetangent plane to the surface at that point, as in Figure tp . Note that the tangent vector (B.30) is a linear combination of the two basis tangent vectors xp=∂x ∂p=/parenleftbigg∂x ∂p,∂y ∂p,∂z ∂p/parenrightbiggT , xq=∂x ∂q=/parenleftbigg∂x ∂q,∂y ∂q,∂z ∂q/parenrightbiggT ,(B.31) which therefore span the tangent plane to the surface at the p ointx(p,q)∈S. The first basis vector is tangent to the curves where q= constant, while the second is tangent to the curves where p= constant. 12/11/12 1240 c/ci∇cleco√y∇t2012 Peter J. Olver Example B.10. Consider the torus Tparametrized as in (B.28). The basis tangent vectors are ∂x ∂ψ= −(2+cosθ)sinψ (2+cosθ)cosψ 0 ,∂x ∂θ= −sinθcosψ −sinθsinψ cosθ .(B.32) They serve to span the tangent plane to the torus at the point x(θ,ψ). For example, at the point x(0,0) = (3,0,0)Tcorresponding to the particular parameter values θ=ψ= 0, the basis tangent vectors are xψ(0,0) = (0,3,0)T= 3e2, xθ(0,0) = (0,0,1)T=e3, andsothetangentplaneatthisparticularpointisthe( y,z)–planespanned bythestandard basis vectors e2,e3. The tangent to any curve contained within the torus at the giv en point will be a linear combination of these two vectors. For instance, the toroida l knot (B.16) corresponds to the straight line ψ(t) = 2t, 0≤t≤2π, θ (t) = 3t, in the parameter space. Its tangent vector dx dt= −(4+2cos3 t)sin2t−3 sin3tcos2t (4+2cos3 t)cos2t−3 sin3tsin2t 3 cos3t  lies in the tangent plane to the torus at each point. In partic ular, att= 0, the knot passes through the point x(0,0) = (3,0,0)T, and has tangent vector dx dt= 0 6 3 = 2xψ(0,0)+3xθ(0,0) sincedψ dt= 2,dθ dt= 3. A pointx(p,q)∈Son the surface is said to be nonsingular provided the basis tangent vectorsxp(p,q),xq(p,q) arelinearly independent. Thus the point is nonsingular if and only if the tangent vectors span a full two-dimensional subspace ofR3— the tangent plane to the surface at the point. Nonsingularity ensures the smooth ness of the surface at each point, which is a consequence of the general Implicit Functi on Theorem, [ 159]. Singular points, where the tangent vectors are linearly dependent, c an take the form of corners, cusps and folds in the surface. From now on, we shall always as sume that our surface is nonsingular meaning every point is a nonsingular point. Linear independence of the tangent vectors is equivalent to the requirement that their cross product is a nonzero vector: N=∂x ∂p×∂x ∂q=/parenleftbigg∂(y,z) ∂(p,q),∂(z,x) ∂(p,q),∂(x,y) ∂(p,q)/parenrightbiggT /\e}atio\slash=0. (B.33) 12/11/12 1241 c/ci∇cleco√y∇t2012 Peter J. Olver In this formula, we have adopted the standard notation ∂(x,y) ∂(p,q)= det/parenleftbigg xpxq ypyq/parenrightbigg =∂x ∂p∂y ∂q−∂x ∂q∂y ∂p(B.34) for theJacobian determinant of the functions x,ywith respect to the variables p,q, which we already encountered in the change of variables formula (A .52) for double integrals. The cross-product vector Nin (B.33) is orthogonal to both tangent vectors, and hence orthogonal to the entire tangent plane. Therefore, Ndefines a normalvector to the surface at the given (nonsingular) point. Example B.11. Consider a surface Sparametrized as the graph of a function z= u(x,y), and so, as in Example B.7 x(x,y) = (x,y,u(x,y))T, (x,y)∈Ω. The tangent vectors ∂x ∂x=/parenleftbigg 1,0,∂u ∂x/parenrightbiggT ,∂x ∂y=/parenleftbigg 0,1,∂u ∂y/parenrightbiggT , span the tangent plane sitting at the point ( x,y,u(x,y) onS. The normal vector is N=∂x ∂x×∂x ∂y=/parenleftbigg −∂u ∂x,−∂u ∂y,1/parenrightbiggT , and points upwards, as in Figure graphN . Note that every point on the graph is nonsin- gular. Theunit normal to the surface at the point is a unit vector orthogonal to the t angent plane, and hence given by n=N /ba∇dblN/ba∇dbl=xp×xq /ba∇dblxp×xq/ba∇dbl. (B.35) In general, the direction of the normal vector Ndepends upon the order of the two param- etersp,q. Computing the cross product in the reverse order, xq×xp=−N, reverses the sign of the normal vector, and hence switches its direction. Thus, there are two possible unit normals to the surface at each point, namely nand−n. For a closed surface, one normal points outwards and one points inwards. When possible, a consistent (meaning continuously varying ) choice of a unit normal serves to define an orientation of the surface. All closed surfaces, and most other surfaces can be oriented. The usual convention for closed surfaces is to choose the orientation defined by the outward normal. The simplest example of a non-o rientable surface is the M¨ obius strip obtained by gluing together the ends of a twist ed strip of paper; see Exercise . 12/11/12 1242 c/ci∇cleco√y∇t2012 Peter J. Olver Example B.12. For the sphere of radius rparametrized by the spherical angles as in (B.27), the tangent vectors are ∂x ∂ϕ= rcosϕcosθ rsinϕcosθ −rsinϕ ,∂x ∂θ= −rsinϕsinθ rsinϕcosθ 0 . These vectors are tangent to, respectively, the meridians o f constant longitude, and the parallels of constant latitude. The normal vector is N=∂x ∂ϕ×∂x ∂θ= r2sin2ϕcosθ r2sin2ϕsinθ r2cosϕsinϕ =rsinϕx. (B.36) ThusNis a non-zero multiple of the radial vector x, except at the north or south poles whenϕ= 0 orπ. This reconfirms our earlier observation that the poles are p roblematic points for the spherical angle parametrization. The unit no rmal n=N /ba∇dblN/ba∇dbl=x r determined by the spherical coordinates ϕ,θis the outward pointing normal. Reversing the order of the angles, θ,ϕ, would lead to the outwards normal −n=−x/r. Remark: As we already saw in the example of the hemisphere, a given su rface can be parametrized in many different ways. In general, to change pa rameters p=g(/tildewidep,/tildewideq), q =h(/tildewidep,/tildewideq), requires a smooth, invertible map between the two parameter domains/tildewideΩ→Ω. Many interesting surfaces, particularly closed surfaces, cann ot be parametrized in a single con- sistent manner that satisfies the smoothness constraint (B. 33) on the entire surface. In such cases, one must assemble the surface out of pieces, each parametrized in the proper manner. The key problem in cartography is to find convenient p arametrizations of the globe that do not significantly distort the geographical fea tures of the planet. A surface is piecewise smooth if it can be constructed by gluing together a finite number of smooth parts, joined along piecewise smooth curve s. For example, a cube is a piecewise smooth surface, consisting of squares joined al ong straight line segments. We shall rely on the reader’s intuition to formalize these idea s, leaving a rigorous development to a more comprehensive treatment of surface geometry, e.g. , [60]. B.5. Surface Integrals. As with spatial line integrals, there are two important type s of surface integral. The first is the integration of a scalar field with respect to surfa ce area. A typical application is to compute the area of a curved surface or the mass and cente r of mass of a curved shell of possibly variable density. The second type is the surface integral that computes the flux 12/11/12 1243 c/ci∇cleco√y∇t2012 Peter J. Olver associated with a vector field through an oriented surface. A pplications appear in fluid mechanics, electromagnetism, thermodynamics, gravitati on, and many other fields. Surface Area According to (B.10), the length of the cross product of two ve ctors measures the area of the parallelogram they span. This observation underlies the proof that the length of the normal vector to a surface (B.35), namely /ba∇dblN/ba∇dbl=/ba∇dblxp×xq/ba∇dbl, is a measure of the infinitesimal element of surface area, den oted dS=/ba∇dblN/ba∇dbldpdq=/ba∇dblxp×xq/ba∇dbldpdq. (B.37) The total area of the surface is found by summing up these infin itesimal contributions, and is therefore given by the double integral areaS=/integraldisplay/integraldisplay SdS=/integraldisplay/integraldisplay Ω/ba∇dblxp×xq/ba∇dbldpdq =/integraldisplay/integraldisplay Ω/radicaligg/parenleftbigg∂(y,z) ∂(p,q)/parenrightbigg2 +/parenleftbigg∂(z,x) ∂(p,q)/parenrightbigg2 +/parenleftbigg∂(x,y) ∂(p,q)/parenrightbigg2 dpdq.(B.38) The surface’s area does not depend upon the parametrization used to compute the integral. In particular, if the surface is parametrized by x,yas the graph z=u(x,y) of a function over a domain ( x,y)∈Ω, then the surface area integral reduces to the familiar for m areaS=/integraldisplay/integraldisplay SdS=/integraldisplay/integraldisplay Ω/radicaligg 1+/parenleftbigg∂u ∂x/parenrightbigg2 +/parenleftbigg∂u ∂y/parenrightbigg dxdy. (B.39) A detailed justification of these formulae can be found in [ 9,70]. Example B.13. The well-known formula for the surface area of a sphere is a si mple consequence of the integral formula (B.38). Using the param etrization by spherical angles (B.27) and the formula (B.36) for the normal, we find areaSr=/integraldisplay/integraldisplay SrdS=/integraldisplay2π 0/integraldisplayπ 0r2sinϕ dϕdθ= 4πr2. (B.40) Fortunately, the problematic poles do not cause any difficult y in the computation, since they contribute nothing to the surface area integral. Alternatively, we can compute the area of one hemisphere S+ rby realizing it as a graph z=/radicalbig r2−x2−y2forx2+y2≤1, over the disk of radius r, and so, by (B.39), areaS+ r=/integraldisplay/integraldisplay Ω/radicaligg 1+x2 r2−x2−y2+y2 r2−x2−y2dxdy =/integraldisplay/integraldisplay Ωr/radicalbig r2−x2−y2dxdy=/integraldisplayr 0/integraldisplay2π 0rρ/radicalbig r2−ρ2dθdρ= 2πr2, 12/11/12 1244 c/ci∇cleco√y∇t2012 Peter J. Olver where we used polar coordinates x=ρcosθ,y=ρsinθto evaluate the final integral. The area of the entire sphere is twice the area of the hemisphere. Example B.14. Similarly, to compute the surface area of the torus Tparametrized in (B.28), we use the tangent vectors in (B.32) to compute the normal to the torus: N=xψ×xθ= (2+cosψ)cosψcosθ (2+cosψ)cosψsinθ (2+cosψ)sinψ ,with /ba∇dblxψ×xθ/ba∇dbl= 2+cosψ. Therefore, areaT=/integraldisplay2π 0/integraldisplay2π 0(2+cosψ)dψdθ= 8π2. IfS⊂R3is a surface with finite area, the meanoraverage of a scalar function f(x,y,z) overSis given by MS[f] =1 areaS/integraldisplay/integraldisplay SfdS. (B.41) For example, the mean of a function over a sphere Sr={/ba∇dblx/ba∇dbl=r}of radiusris explicitly given by MSr[f] =1 4πr2/integraldisplay/integraldisplay /bardblx/bardbl=rf(x)dS=1 4π/integraldisplay2π 0/integraldisplayπ 0F(r,ϕ,θ)sinϕdϕdθ, (B.42) whereF(r,ϕ,θ) is the spherical coordinate expression for the scalar func tionf. As usual, the mean lies between the maximum and minimum values of the fu nction on the surface: min Sf≤MS[f]≤max Sf. In particular, the center of mass Cof a surface (assuming it has constant density) is equal to the mean of the coordinate functions x= (x,y,z)T, so C= (MS[x],MS[y],MS[z])T=1 areaS/parenleftbigg/integraldisplay/integraldisplay SxdS,/integraldisplay/integraldisplay SydS,/integraldisplay/integraldisplay SzdS/parenrightbiggT .(B.43) More generally, the integral of a scalar field u(x,y,z) over the surface is given by /integraldisplay/integraldisplay SudS=/integraldisplay/integraldisplay Ωu(x(p,q),y(p,q),z(p,q))/ba∇dblxp×xq/ba∇dbldpdq. (B.44) IfSrepresents a thin curved shell, and u=ρ(x) the density of the material at position x∈S, then the surface integral (B.44) represents the total mass of the shell. For example, the integral of u(x,y,z) over a hemisphere S+ rof radiusrcan be evaluated by either of the formulae /integraldisplay/integraldisplay S+ rudS=/integraldisplay2π 0/integraldisplayπ/2 0u(rcosθsinϕ,rsinθsinϕ,rcosϕ)r2sinϕ dϕdθ =/integraldisplay/integraldisplay x2+y2≤r2r/radicalbig r2−x2−y2u(x,y,/radicalbig r2−x2−y2)dxdy,(B.45) 12/11/12 1245 c/ci∇cleco√y∇t2012 Peter J. Olver depending upon whether one prefers spherical or graphical c oordinates. Flux Integrals Now assumethat Sisanoriented surface withchosen unit normal n. Ifv= (u,v,w)T is a vector field, then the surface integral /integraldisplay/integraldisplay Sv·ndS=/integraldisplay/integraldisplay Ω/parenleftbig v·xp×xq/parenrightbig dpdq=/integraldisplay/integraldisplay Ωdet u xpxq v ypyq w zpzq dpdq (B.46) of the normal component of vover the entire surface measures its fluxthrough the surface. An alternative common notation for the flux integral is /integraldisplay/integraldisplay Sv·ndS=/integraldisplay/integraldisplay Sudydz+vdzdx+wdxdy (B.47) =/integraldisplay/integraldisplay Ω/parenleftbigg u(x,y,z)∂(y,z) ∂(p,q)+v(x,y,z)∂(z,x) ∂(p,q)+w(x,y,z)∂(x,y) ∂(p,q)/parenrightbigg dxdy, Note how the Jacobian determinant notation (B.34) seamless ly interacts with the integra- tion. In particular, if the surface is the graph of a function z=h(x,y), then the surface integral reduces to the particularly simple form /integraldisplay/integraldisplay Sv·ndS=/integraldisplay/integraldisplay Ω/parenleftbigg u(x,y,z)∂z ∂x+v(x,y,z)∂z ∂y+w(x,y,z)/parenrightbigg dpdq (B.48) The flux surface integral relies upon the consistent choice o f an orientation or unit normal on the surface. Thus, flux only makes sense through an o riented surface — it doesn’t make sense to speak of “flux through a M¨ obius band”. I f we switch normals, using, say, the inward instead of the outward normal, then th e surface integral changes sign — just like a line integral if we reverse the orientation of a curve. Similarly, if we decompose a surface into the union of two or more parts, with o nly their boundaries in common, then the surface integral similarly decomposes int o a sum of surface integrals. Thus, /integraldisplay/integraldisplay −Sv·ndS=−/integraldisplay/integraldisplay Sv·ndS, /integraldisplay/integraldisplay Sv·ndS=/integraldisplay/integraldisplay S1v·ndS+/integraldisplay/integraldisplay S2v·ndS, S =S1∪S2.(B.49) In the first formula, −Sdenotes the surface Swith the reverse orientation. In the second formula,S1andS2are only allowed to intersect along their boundaries; moreo ver, they must be oriented in the same manner as S, i.e., have the same unit normal direction. Example B.15. LetSdenote the triangular surface given by that portion of the planex+y+z= 1 that lies inside the positive orthant {x≥0, y≥0, z≥0}, as in 12/11/12 1246 c/ci∇cleco√y∇t2012 Peter J. Olver Figure tri3 . The flux of the vector field v= (y,xz,0)TthroughSequals the surface integral/integraldisplay/integraldisplay Sydydz+xzdzdx, where we orient Sby choosing the upwards pointing normal. To compute, we note that Scan be identified as the graph of the function z= 1−x−ylying over the triangle T={0≤x≤1,0≤y≤1−x}. Therefore, by (B.47), /integraldisplay/integraldisplay Sydydz+xzdzdx =/integraldisplay/integraldisplay T/bracketleftbigg y∂(y,1−x−y) ∂(x,y)+x(1−x−y)∂(1−x−y,x) ∂(x,y)/bracketrightbigg dxdy =/integraldisplay1 0/integraldisplay1−x 0(1−x)(y+x)dydx=/integraldisplay1 0/parenleftbig1 2+1 2x−1 2x2+1 2x3/parenrightbig dx=17 24. Ifvrepresents the velocity vector field for a steady state fluid fl ow, then its flux integral (B.46) tells us the total volume of fluid passing thr oughSper unit time. Indeed, at each point on S, the volume fluid that flows across a small part the surface in u nit time will fill a thin cylinder whose base is the surface area elemen tdSand whose height v·n is the normal component of the fluid velocity v, as pictured in Figure fluxs . Summing (integrating)allthesefluxcylindervolumesoverthesurfa ceresultsinthefluxintegral. The choice of orientation or unit normal specifies the conventio n for measuring the direction of positive flux through the surface. If Sis a closed surface, and we choose nto be the unit outward normal, then the flux integral (B.46) represents the net amount of fluid flowing outof the solid region bounded by Sper unit time. Example B.16. The vector field v= (0,0,1)Trepresents a fluid moving with con- stant velocity in the vertical direction. Let us compute the fluid flux through a hemisphere S+ r=/braceleftig z=/radicalbig r2−x2−y2/vextendsingle/vextendsingle/vextendsinglex2+y2≤1/bracerightig , sitting over the disk Drof radiusrin thex,yplane. The flux integral over S+ ris computed using (B.48), so /integraldisplay/integraldisplay S+ rv·ndS=/integraldisplay/integraldisplay S+ rdx×dy=/integraldisplay/integraldisplay Drdxdy=πr2. The resulting double integral is just the area of the disk. In deed, in this case, the value of the flux integral is the samefor all surfaces z=h(x,y) sitting over the disk Dr. This example provides a particular case of a surface-indepe ndent flux integral, which aredefined inanalogywiththepath-independent lineintegr alsthat weencountered earlier. In general, a flux integral is called surface-independent if /integraldisplay/integraldisplay S1v·ndS=/integraldisplay/integraldisplay S2v·ndS (B.50) whenever the surfaces S1andS2have a common boundary ∂S1=∂S2. In other words, the value of the integral depends only upon the boundary of th e surface. The veracity of 12/11/12 1247 c/ci∇cleco√y∇t2012 Peter J. Olver (B.50) requires that the surfaces be oriented in the “same ma nner”. For instance, if they do not cross, then the combined surface S=S1∪S2is closed, and one uses the outward pointing normal on one surface and the inward pointing norma l on the other. In more complex situations, one checks that the two surfaces induce the same orientation on their commonboundary. (Wedefer adiscussion oftheboundaryorie ntationuntil later.) Finally, applying (B.49) to the closed surface S=S1∪S2and using the prescribed orientations, we deduce an alternative characterization of surface-inde pendent vector fields. Proposition B.17. A vector field leads to a surface-independent flux integral if and only if/integraldisplay/integraldisplay Sv·ndS= 0 (B .51) for every closed surface Scontained in the domain of definition of v. A fluid is incompressible when its volume is unaltered by the flow. Therefore, in the absence of sources or sinks, there cannot be any net inflow or o utflow across a simple closed surfaceboundingaregionoccupiedbythefluid. Thus, theflux integraloveraclosedsurface must vanish:/integraldisplay/integraldisplay Sv·ndS= 0. Proposition B.17 implies that the fluid velocity vector field defines a surface-independent flux integral. Thus, the fl ux of an incompressible fluid flow through any surface depends only on the (oriented) bound ary curve of the surface! B.6. Volume Integrals. Volume or triple integrals take place over domains Ω ⊂R3representing solid three- dimensional bodies. A simple example of such a domain is a ball Br(a) ={x| /ba∇dblx−a/ba∇dbl<r} (B.52) of radiusr>0 centered at a point a∈R3. Other examples of domains include solid cubes, solid cylinders, solid tetrahedra, solid tori (doughnuts a nd bagels), solid cones, etc. In general, a subset Ω ⊂R3isopenif, for every point x∈Ω, a small open ball Bε(a)⊂Ω centered at aof radiusε=ε(a)>0, which may depend upon a, is also contained in Ω. In particular, the ball (B.52) is open. The boundary∂Ω of an open subset Ω consists of all limit points which are not in the subset. Thu s, the boundary of the open ball Br(a) is the sphere Sr(a) ={/ba∇dblx−a/ba∇dbl=r}of radiusrcentered at the point a. An open subset is called a domainif its boundary ∂Ω consists of one or more simple, piecewise smooth surfaces. We are allowing corners and edges in the bou nding surfaces, so that an open cube will be a perfectly valid domain. A subset Ω ⊂R3isbounded provided it fits inside a sphere of some (possibly large) radius. For example, the solid ball Br={/ba∇dblx/ba∇dbl<R}is bounded, while its exterior Er= {/ba∇dblx/ba∇dbl>R}is an unbounded domain. The sphere SR={/ba∇dblx/ba∇dbl=R}is the common boundary of the two domains: SR=∂Br=∂ER. Indeed, any simple closed surface separates R3into two domains that have a common boundary — its interior, which is bounded, and its unbounded exterior. 12/11/12 1248 c/ci∇cleco√y∇t2012 Peter J. Olver The boundary of a bounded domain consists of one or more close d surfaces. For instance, the solid annular domain Ar,R=/braceleftbig 0<r</ba∇dblx/ba∇dbl<R/bracerightbig (B.53) consisting of all points lying between two concentric spher es of respective radii randR has boundary given by the two spheres: ∂Ar,R=Sr∪SR. On the other hand, setting r= 0 in (B.53) leads to a punctured ball of radiusRwhose center point has been removed. A punctured ball is nota domain, since the center point is part of the boundary, but i s not abona fide surface. If the domain Ω ⊂R3represents a solid body, and the scalar field ρ(x,y,z) represents its density at a point ( x,y,z)∈Ω, then the triple integral /integraldisplay/integraldisplay/integraldisplay Ωρ(x,y,z)dxdydz (B.54) equals the total mass of the body. In particular, the volume o f Ω is equal to volΩ =/integraldisplay/integraldisplay/integraldisplay Ωdxdydz. (B.55) Triple integrals can be directly evaluated when the domain h as the particular form Ω =/braceleftbig ξ(x,y)<z<η(x,y), ϕ(x)<y<ϕ(x), a<x<b/bracerightbig (B.56) where thezcoordinate lies between two graphical surfaces sitting ove r a common domain in the (x,y)–plane that is itself of the form of (A.47) used to evaluate d ouble integrals; see Figure triple . In such cases we can evaluate the triple integral by iterate d integration first with respect to z, then with respect to yand, finally, with respect to x: /integraldisplay/integraldisplay/integraldisplay Ωu(x,y,z)dxdydz =/integraldisplayb a/parenleftigg/integraldisplayψ(x) ϕ(x)/parenleftigg/integraldisplayη(x,y) ξ(x,y)u(x,y,z)dz/parenrightigg dy/parenrightigg dx. (B.57) A similar result holds for other orderings of the coordinate s. Fubini’s Theorem, [ 159,158], assures us that the result of iterated integration does not depend upon the order in which the variables are integrat ed. Of course, the domain must be of the requisite type in order to write the volume inte gral as repeated single integrals. More general triple integrals can be evaluated b y chopping the domain up into disjoint pieces that have the proper form. Example B.18. The volume of a solid ball BRof radiusRcan be computed as follows. We express the domain of integration x2+y2+z2<R2in the form −R<x<R, −/radicalbig R2−x2<y</radicalbig R2−x2,−/radicalbig R2−x2−y2<z</radicalbig R2−x2−y2. 12/11/12 1249 c/ci∇cleco√y∇t2012 Peter J. Olver Therefore, in accordance with (B.57), /integraldisplay/integraldisplay/integraldisplay BRdxdydz =/integraldisplayR −R/parenleftigg/integraldisplay√ R2−x2 −√ R2−x2/parenleftigg/integraldisplay√ R2−x2−y2 −√ R2−x2−y2dz/parenrightigg dy/parenrightigg dx =/integraldisplayR −R/parenleftigg/integraldisplay√ R2−x2 −√ R2−x22/radicalbig R2−x2−y2dy/parenrightigg dx =/integraldisplayR −R/parenleftbigg y/radicalbig R2−x2−y2+(R2−x2)sin−1y√ R2−x2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle√ R2−x2 y=−√ R2−x2dx =/integraldisplayR −Rπ(R2−x2)dx=π/parenleftbigg R2x−x3 3/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleR x=−R=4 3πR3, recovering the standard formula, as it should. Change of Variables Sometimes, an inspired change of variables can be used to sim plify a volume integral. If x=f(p,q,r), y =g(p,q,r), z =h(p,q,r), (B.58) is an invertible change of variables — meaning that each poin t (x,y,z) corresponds to a unique point ( p,q,r) — then /integraldisplay/integraldisplay/integraldisplay Ωu(x,y,z)dxdydz =/integraldisplay/integraldisplay/integraldisplay DU(p,q,r)/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂(x,y,z) ∂(p,q,r)/vextendsingle/vextendsingle/vextendsingle/vextendsingledpdqdr. (B.59) Here U(p,q,r) =u(x(p,q,r),y(p,q,r),z(p,q,r)) is the expression for the integrand in the new coordinates, w hileDis the domain consisting of all points ( p,q,r) that map to points ( x,y,z)∈Ω in the original domain. Invertibility requires that each point in Dcorresponds to a unique point in Ω. The change in volume is governed by the absolute value of the three-dimensional Jacobian determinant ∂(x,y,z) ∂(p,q,r)= det xpxqxr ypyqyr zpzqzr =xp·xq×xr (B.60) for the change of variables. The identification of the vector triple product (B.60) with an (infinitesimal) volume element lies behind the justificat ion of the change of variables formula; see [ 9,70] for a detailed proof. By far, the two most important cases are cylindrical and sphe rical coordinates. Cylin- drical coordinates correspond to replacing the xandycoordinates by their polar counter- parts, whileretainingthe vertical zcoordinateunchanged. Thus, thechange of coordinates has the form x=rcosθ, y =rsinθ, z =z. (B.61) 12/11/12 1250 c/ci∇cleco√y∇t2012 Peter J. Olver The Jacobian determinant for cylindrical coordinates is ∂(x,y,z) ∂(r,θ,z)= det xrxθxz yryθyz zrzθzz = det cosθ−rsinθ0 sinθ rcosθ0 0 0 1 =r. (B.62) Therefore, the general change of variables formula (B.59) t ells us the formula for a triple integral in cylindrical coordinates: /integraldisplay/integraldisplay/integraldisplay f(x,y,z)dxdydz =/integraldisplay/integraldisplay/integraldisplay f(rcosθ,rsinθ,z)rdrdθdz. (B.63) Example B.19. For example, consider an ice cream cone Ch=/braceleftbig x2+y2<z2,0<z <h/bracerightbig =/braceleftbig r<z,0<z <h/bracerightbig of heighthplotted in Figure cone . To compute its volume, we express the domain in terms of the cylindrical coordinates, leading to /integraldisplay/integraldisplay/integraldisplay Chdxdydz =/integraldisplayh 0/integraldisplay2π 0/integraldisplayz 0rdrdθdz =/integraldisplayh 0πz2dz=1 3πh3. Spherical coordinates are denoted by r,ϕ,θ, where x=rsinϕcosθ, y =rsinϕsinθ, z =rcosϕ. (B.64) Herer=/ba∇dblx/ba∇dbl=/radicalbig x2+y2+z2represents the radius, 0 ≤ϕ≤πis the zenith angle or latitude, while 0 ≤θ<2πis the azimuthal angle or longitude. The reader may recall th at we already encountered these coordinates in our parametriz ation (B.27) of the sphere. It is important to distinguish between the spherical r,θand the cylindrical r,θ— even though the same symbols are used, they represent different quantities. Warning : In many books, particularly those in physics, the roles of θandϕare reversed, leading to much confusion when perusing the literature. We prefer the mathe- matical convention as the azimuthal angle θagrees with its cylindrical counterpart. You need to be very careful to determine which convention is bein g used when consulting any reference! A short computation proves that the spherical coordinate Ja cobian determinant is ∂(x,y,z) ∂(r,ϕ,θ)= det xrxϕxθ yryϕyθ zrzϕzθ  = det sinϕcosθ rcosϕcosθ−rsinϕsinθ sinϕsinθ rcosϕsinθ rsinϕcosθ cosϕ−rsinϕ 0 =r2sinϕ.(B.65) Therefore, a triple integral is evaluated in spherical coor dinates according to the formula /integraldisplay/integraldisplay/integraldisplay f(x,y,z)dxdydz =/integraldisplay/integraldisplay/integraldisplay F(r,ϕ,θ)r2sinϕ drdϕdθ, (B.66) 12/11/12 1251 c/ci∇cleco√y∇t2012 Peter J. Olver where we rewrite the integrand F(r,ϕ,θ) =f(rsinϕcosθ,rsinϕsinθ,rcosϕ) (B .67) as a function of the spherical coordinates. Example B.20. The integration required in Example B.18 to compute the volu me of a ballBRof radiusRcan be considerably simplified by switching over to spherica l coordinates. The ball is given by BR=/braceleftbig 0≤r<R,0≤ϕ≤π,0≤θ<2π/bracerightbig . Thus, using (B.66), we compute /integraldisplay/integraldisplay/integraldisplay BRdxdydz =/integraldisplayR 0/integraldisplayπ 0/integraldisplay2π 0r2sinϕ dθdϕdr =/integraldisplayR 04πr2dr=4 3πR3.(B.68) The reader may note that the next-to-last integrand represe nts the surface area of the sphere of radius R. Thus, we are, in effect, computing the volume by summing up (i .e., integrating) the surface areas of concentric thin spherica l shells. Remark: Sometimes, we will be sloppy and use the same letter for a fun ction in an alternative coordinate system. Thus, we may use f(r,ϕ,θ) to represent the spherical coordinate form (B.67) of a function f(x,y,z). Technically, this is not correct! However, the clarity and intuitionsometimes outweighs the pedantic use of a new letter each time we change coordinates. Moreover, in geometry and modern physi cal theories, [ 21], the symbol “f” represents an intrinsic scalar field, and f(x,y,z) andf(r,ϕ,θ) merely its incarnations in two different coordinate charts on R3. Hopefully, this will be clear from the context. B.7. Gradient, Divergence, and Curl. There are three important vector differential operators tha t play a ubiquitous role in three-dimensional vector calculus, known as the gradient, divergence and curl. We have already encountered their two-dimensional counterparts i n Chapter A. The Gradient We begin with the three-dimensional version of the gradient operator ∇u= ux uy uz . (B.69) The gradient defines a linear operator that maps a scalar func tionu(x,y,z) to the vector fieldwhosecomponentsareitspartialderivativeswithresp ect totheCartesiancoordinates. Ifx(t) = (x(t),y(t),z(t))Tis any parametrized curve, then the rate of change in the functionuas we move along the curve is given by the inner product d dtu(x(t),y(t),z(t))=∂u ∂xdx dt+∂u ∂ydy dt+∂u ∂zdz dt=∇u·/squaresmallsolidx (B.70) between the gradient and the tangent vector to the curve. The refore, as we reasoned earlier in the planar case, the gradient ∇upoints in the direction of steepest increase in 12/11/12 1252 c/ci∇cleco√y∇t2012 Peter J. Olver the function u, while its negative −∇upoints in the direction of steepest decrease. For example, if u(x,y,z) represents the temperature at a point ( x,y,z) in space, then ∇u points in the direction in which temperature is getting the h ottest, while −∇upoints in the directionit gets the coldest. Therefore, ifone wants to cool down asrapidly as possible, one should move in the direction of −∇uat each instant, which is the direction of the flow of heat energy. Thus, the path x(t) to be followed for the fastest cool down will be a solution to the gradient flow equations /squaresmallsolidx=−∇u, (B.71) or, explicitly, dx dt=−∂u ∂x(x,y,z),dy dt=−∂u ∂y(x,y,z),dz dt=−∂u ∂z(x,y,z). A solution x(t) to such a system of ordinary differential equations will exp erience continu- ously decreasing temperature. In Chapter 19, we will learn h ow to use such gradient flows to locate and numerically approximate the minima of functio ns. The set of all points where a scalar field u(x,y,z) has a given value, u(x,y,z) =c (B.72) for some fixed constant c, is known as a level set ofu. Ifumeasures temperature, then its level sets are the isothermal surfaces of equal temperat ure. Ifuis sufficiently smooth, most of its level sets are smooth surfaces. In fact, if ∇u/\e}atio\slash= 0 at a point, then one can prove that all nearby level sets are smooth surfaces near the point in question. This important fact is a consequence of the general Implicit Function Theor em, [159]. Thus, if ∇u/\e}atio\slash= 0 at all points on a level set, then the level set is a smooth surfac e, and, if bounded, a simple closed surface. (On the other hand, finding an explicit param etrization of a level set may be quite difficult!) Theorem B.21. If nonzero, the gradient vector ∇u/\e}atio\slash=0defines the normal direction to the level set {u=c}at each point. Proof: Indeed, suppose x(t) is any curve contained in the level set, so that u(x(t),y(t),z(t)) =cfor allt. Sincecis constant, the derivative with respect to tis zero, and hence, by (B.70), d dtu(x(t),y(t),z(t))=∇u·/squaresmallsolidx= 0, which implies that the gradient vector ∇uis orthogonal to the tangent vector/squaresmallsolidxto the curve. Since this holds for all such curves contained within the level set, the gradient must be orthogonal to the entire tangent plane at the point, and he nce, if nonzero, defines a normal direction to the level surface. Q.E.D. 12/11/12 1253 c/ci∇cleco√y∇t2012 Peter J. Olver Physically,Theorem B.21tellsusthatthedirectionofstee pest increaseintemperature is perpendicular to the isothermal surfaces at each point. C onsequently, the solutions to the gradient flow equations (B.71) form an orthogonal system of curves to the level set surfaces of u, and one should follow these curves to minimize the temperat ure as rapidly as possible. Similarly, in a steady state fluid flow, the fluid p otential is represented by a scalar field ϕ(x,y,z). Its gradient v=∇ϕdetermines the fluid velocity at each point. The streamlines followed by the fluid particles are the solution s to the gradient flow equations /squaresmallsolidx=v=∇ϕ, while the level sets of ϕare the equipotential surfaces. Thus, fluid particles flow in a direction orthogonal to the equipotential surfaces . Example B.22. The level sets of the radial function u=x2+y2+z2are the concentric spheres centered at the origin. Its gradient ∇u= (2x,2y,2z)T= 2xpoints in the radial direction, orthogonal to each spherical level se t. Note that ∇u=0only at the origin, which is a level set, but not a smooth surface. The radial vector also specifies the direction of fastest inc rease (decrease) in the func- tionu. Indeed, the solution to the associated gradient flow system (B.71), namely /squaresmallsolidx=−2x is x(t) =x0e−2t, wherex0=x(0) is the initial position. Therefore, to decrease the func tionuas rapidly as possible, one should follow a radial ray into the origin. Example B.23. An implicit equation for the torus (B.28) is obtained by repl acing r=/radicalbig x2+y2in(B.29). In thismanner, we areledto consider the levelset s of thefunction u(x,y,z) =x2+y2+z2−4/radicalbig x2+y2=c, (B.73) with the particular value c=−3 corresponding to (B.28). The gradient of the function is ∇u(x,y,z) =/parenleftigg 2x−4x/radicalbig x2+y2,2y−4y/radicalbig x2+y2,2z/parenrightiggT , (B.74) which is well-define except on the zaxis, where x=y= 0. Note that ∇F/\e}atio\slash=0unlessz= 0 andx2+y2= 4. Therefore, the level sets of uare smooth, toroidal surfaces except for z axis and the circle of radius 2 in the ( x,y) plane. Divergence and Curl The second important vector differential operator is the divergence , divv=∇·v=∂v1 ∂x+∂v2 ∂y+∂v3 ∂z. (B.75) The divergence maps a vector field v= (v1,v2,v3)Tto a scalar field f=∇ ·v. For example, the radial vector field v= (x,y,z)Thas constant divergence ∇·v= 3. In fluid mechanics, the divergence measures the local, insta ntaneous change in the volume of a fluid packet as it moves. Thus, a steady state fluid fl ow isincompressible , with 12/11/12 1254 c/ci∇cleco√y∇t2012 Peter J. Olver unchanging volume, if and only if its velocity vector field is divergence-free: ∇·v≡0. The connection between incompressibility and the earlier zero -flux condition will be addressed in the Divergence Theorem B.35 below. As in the two-dimensional situation, the composition of div ergence and gradient pro- duces the Laplacian operator: ∇·∇u= ∆u=uxx+uyy+uzz. (B.76) Indeed, as we shall see, except for the missing minus sign and the all-important boundary conditions, this is effectively the same as the self-adjoint form of the three-dimensional Laplacian: ∇∗◦∇u=−∇·∇u=−∆u. Thethirdimportantvectordifferentialoperatoristhe curl, which, inthreedimensions, maps vector fields to vector fields. It is most easily memorize d in the form of a (formal) 3×3 determinant curlv=∇×v= ∂v3 ∂y−∂v2 ∂z ∂v1 ∂z−∂v3 ∂x ∂v2 ∂x−∂v1 ∂y = det ∂xv1e1 ∂yv2e2 ∂zv3e3 , (B.77) in analogy with the determinantal form (B.6) of the cross pro duct. For instance, the radial vector field v= (x,y,z)Thas zero curl: ∇×v= det ∂xxe1 ∂yye2 ∂zze3 =0. This is indicative of the lack or any rotational effect of the i nduced flow. Ifvrepresents the velocity vector field of a steady state fluid flo w, its curl ∇ ×v measures the instantaneous rotation of the fluid flow at a poin t, and is known as the vorticity of the flow. When non-zero, the direction of the vorticity vec tor represents the axis of rotation, while its magnitude /ba∇dbl∇×v/ba∇dblmeasures the instantaneous angular velocity of the swirling flow. Physically, if we place a microscopic tu rbine in the fluid so that its shaft points in the direction specified by a unit vector n, then its rate of spin will be proportional to component of the vorticity vector ∇×vin the direction of its shaft. This is equal to the dot product n·(∇×v) =/ba∇dbl∇×v/ba∇dblcosϕ, whereϕis the angle between nand the curl vector. Therefore, the maximal rate of spin will occur when ϕ= 0, and so the shaft of the turbine lines up with the direction of the vorticity vector ∇ ×v. In this orientation, the angular velocity of the turbine wi ll be proportional to its magnitude /ba∇dbl∇×v/ba∇dbl. On the other hand, if the axis of the turbine is 12/11/12 1255 c/ci∇cleco√y∇t2012 Peter J. Olver orthogonal to the direction of the vorticity, then it will no t rotate. If ∇×v≡0, then there is no net motion of a turbine, not matter which orientation it is placed in the fluid flow. Thus, a flow with zero curl is irrotational . The precise connection between this definition and the earlier zero circulation condition will be explaine d shortly. Example B.24. Consider a helical fluid flow with velocity vector v= (−y,x,1)T. Integrating the ordinary differential equations/squaresmallsolidx=v, namely /squaresmallsolidx=−y,/squaresmallsolidy=x,/squaresmallsolidz= 1, with initial conditions x(0) =x0,y(0) =y0,z(0) =z0gives the flow x(t) =x0cost−y0sint, y (t) =x0sint+y0cost, z (t) =z0+t.(B.78) Therefore, the fluid particles move along helices spiraling up thezaxis, as illustrated in Figure hel . The divergence of the vector field vis ∇·v=∂ ∂x(−y)+∂ ∂yx+∂ ∂z1 = 0, and hence the flow is incompressible. Indeed, any fluid packet will spiral up the zaxis unchanged in shape, and so its volume does not change. The vorticity or curl of the velocity is ∇×v= ∂ ∂y1−∂ ∂zx ∂ ∂z(−y)−∂ ∂x1 ∂ ∂xx−∂ ∂y(−y) = 0 0 2 , which points along the z-axis. This reflects the fact that the flow is spiraling up the z-axis. If a turbine is placed in the fluid at an angle ϕwith thez-axis, then its rate of rotation will be proportional to 2cos ϕ. Example B.25. Any planar vector field v= (v1(x,y),v2(x,y))Tcan be identified with a three-dimensional vector field v= (v1(x,y),v2(x,y),0)T that has no vertical component. If vrepresents a fluid velocity, then the fluid particles remain on horizontal planes {z=c}, and the individual planar flows are identical. Its three-dimensional curl ∇×v=/parenleftbigg 0,0,∂v2 ∂x−∂v1 ∂y/parenrightbiggT 12/11/12 1256 c/ci∇cleco√y∇t2012 Peter J. Olver is a purely vertical vector field, whose third component agre es with the scalar two-dimen- sional curl (A.18) of v. This provides the direct identification between the two- an d three- dimensional versions of the curl operation. Indeed, our ana lysis of flows around airfoils in Chapter 16 directly relied upon this identification betwe en two- and three-dimensional flows. Interconnections and Connectedness The three basic vector differential operators — gradient, cu rl and divergence — are intimately inter-related. The proof of the key identities r elies on the equality of mixed par- tial derivatives, which in turn requires that the functions involved are sufficiently smooth. We leave the explicit verification of the key result to the rea der. Proposition B.26. Ifuis a smooth scalar field, then ∇×∇u≡0. Ifvis a smooth vector field, then ∇·(∇×v)≡0. Therefore, thecurlofanygradientvectorfieldisautomatic allyzero. Asaconsequence, allgradientvectorfieldsrepresent irrotationalflows. Als o,thedivergenceofanyvectorfield thatisa curl isalsoautomaticallyzero. Thus, allcurl vect orfields represent incompressible flows. On the other hand, the divergence of a gradient vector fi eld is the Laplacian of the underlying potential, as we previously noted, and hence is z ero if and only if the potential is a harmonic function. The converse statements are almost true. As in the two-dimen sional case, the precise statement of this result depends upon the topology of the und erlying domain. In two dimensions, we only had to worry about whether or not the doma in contained any holes, i.e., whether or not the domain was simply connected. Simila r concerns arise in three dimensions. Moreover, there are two possible classes of “ho les” in a solid domain — called tunnels and voids – and so there are two different types of conn ectivity. For lack of a better terminology, we introduce the following definition. Definition B.27. A domain Ω ⊂R3is said to be (a) 0–connected orpathwise connected if there is a curve C⊂Ω connecting any two points x0,x1∈Ω, so that†∂C={x0,x1}. (b) 1–connected if every unknotted simple closed curve C⊂Ω is the boundary, C=∂S of an oriented surface S⊂Ω. (c) 2–connected if every simple closed surface S⊂Ω is the boundary, S=∂Dof a subdomain D⊂Ω. Remark: The unknotted condition is to avoid considering “wild” cur ves that fail to bound any oriented surface S⊂R3whatsoever. For example, R3is both 0, 1 and 2–connected, as are all solid balls, cubes, te trahedra, solid cylinders, and so on. A disjoint union of balls is not 0– connected, although it does remainboth1and2–connected. ThedomainΩ =/braceleftbig 0≤r</radicalbig x2+y2<R/bracerightbig lyingbetween †We use the notation ∂Cto denote the endpoints of a curve C. 12/11/12 1257 c/ci∇cleco√y∇t2012 Peter J. Olver two cylinders is not 1–connected since it has a “one-dimensi onal” hole drilled through it. Indeed, ifC⊂Ω is any closed curve that encircles the inner cylinder, then every bounding surfaceSwith∂S=Cmust pass across the inner cylinder and hence will not lie ent irely withinthedomain. Ontheotherhand, thiscylindricaldomai nΩisboth0and2–connected — even an annular surface that encircles the inner cylinder w ill bound a solid annular domain contained inside Ω. Similarly, the domain Ω = {0≤r</ba∇dblx/ba∇dbl<R}between two concentric spheres is 0 and 1–connected, but not 2–connecte d owing to the spherical cavity inside. Any closed curve C⊂Ω will bound a surface S⊂Ω; for instance, a circle going around the equator of the inner sphere will still bound a hemi spherical surface that does not pass through the spherical cavity. On the other hand, a sp here that lies between the inner and outer spheres will not bound a solid domain contain ed within the domain. A full discussion of the topology underlying the various type s of connectivity, the nature of tunnels and voids or cavities, and their connection with the existence of scalar and vector potentials, must be deferred to a more advanced course in diff erential topology, [ 23,86]. We cannow statethebasic theorem relatingthe connectivity of domainsto thekernels of the fundamental vector differential operators; see [ 23] for details. Theorem B.28. LetΩ⊂R3be a domain. (a)IfΩis0–connected, then a scalar field u(x,y,z)defined on all of Ωhas vanishing gradient, ∇u≡0, if and only if u(x,y,z) = constant . (b)IfΩis1–connected, then a vector field v(x,y,z)defined on all of Ωhas vanishing curl,∇×v≡0, if and only if there is a scalar field ϕ, known as a scalar potential forv, such that v=∇ϕ. (c)IfΩis2–connected, then a vector field v(x,y,z)defined on all of Ωhas vanishing divergence, ∇ ·v≡0, if and only if there is a vector field w, known as a vector potential forv, such that v=∇×w. Ifvrepresents the velocity vector field of a steady-state fluid fl ow, then the curl-free condition ∇×v≡0corresponds to an irrotational flow. Thus, on a 2–connected d omain, every irrotational flow field vhas a scalar potential ϕwith∇ϕ=v. The divergence-free condition ∇·v≡0corresponds to an incompressible flow. If the domain is 1–con nected, every incompressible flow field vhas a vector potential wthat satisfies ∇×w=v. The vector potential can be viewed as the three-dimensional ana log of the stream function for planar flows. If the fluid is both irrotational and incompress ible, then its scalar potential satisfies 0 =∇·v=∇·∇ϕ= ∆ϕ, which is Laplace’s equation! Thus, just as in the two-dimens ional case, the scalar potential to an irrotational, incompressible fluid flow is a harmonic fu nction. This fact is used in modeling many problems arising in physical fluids, includin g water waves, [ 126]. Unfortu- nately, in three dimensions there is no counterpart of compl ex function theory to represent the solutions of the Laplace equation, or to connect the vect or and scalar potentials. Example B.29. The vector field v= (−y,x,1)T 12/11/12 1258 c/ci∇cleco√y∇t2012 Peter J. Olver that generates the helical flow (B.78) satisfies ∇·v= 0, and so is divergence-free, recon- firming our observation that the flow is incompressible. Sinc evis defined on all of R3, Theorem B.28 assures us that there is a vector potential wthat satisfies ∇×w=v. One candidate for the vector potential is w=/parenleftbig y,0,1 2x2+1 2y2/parenrightbigT. The helical flow is not irrotational, and so it does not admit a scalar potential. Remark: The construction of a vector potential is not entirely stra ightforward, but we will not dwell on this problem. Unlike a scalar potential whi ch, when it exists, is uniquely defined up to a constant, there is, in fact, quite a bit of ambig uity in a vector potential. Adding in anygradient, /tildewidew=w+∇ϕ will give an equally valid vector potential. Indeed, using P roposition B.26, we have ∇×/tildewidew=∇×w+∇×∇ϕ=∇×w. Thus, any vector field of the form w=/parenleftbigg y+∂ϕ ∂x,∂ϕ ∂y,x2 2+y2 2+∂ϕ ∂z/parenrightbiggT , whereϕ(x,y,z) is anarbitrary function, is also a valid vector potential for the helical vector field v= (−y,x,1)T. B.8. The Fundamental Integration Theorems. In three-dimensional vector calculus there are 3 fundament al differential operators — gradient, curl and divergence. There are also 3 types of int egration — line, surface and volume integrals. And, not coincidentally, there are 3 b asic theorems that general- ize the Fundamental Theorem of Calculus to line, surface and volume integrals in three- dimensional space. In all three results, the integral of som e differentiated quantity over a curve, surface, or domain is related to an integral of the qua ntity over its boundary. The first theorem relates the line integral of a gradient over a cu rve to the values of the function at the boundary or endpoints of the curve. Stokes’ Theorem re lates the surface integral of the curl of a vector field to the line integral of the vector fiel d around the boundary curve of the surface. Finally, the Divergence Theorem, also known as Gauss’ Theorem, relates the volume integral of the divergence of a vector field to the s urface integral of that vector field over the boundary of the domain. The Fundamental Theorem for Line Integrals We begin with the Fundamental Theorem for line integrals. Th is is identical to the planar version, as stated earlier in Theorems A.20 and A.22. We do not need to reproduce its proof again here. 12/11/12 1259 c/ci∇cleco√y∇t2012 Peter J. Olver Theorem B.30. LetC⊂R3be a curve that starts at the endpoint aand goes to the endpoint b. Then the line integral of a gradient of a function along Cis given by /integraldisplay C∇u·dx=u(b)−u(a). (B.79) Since its value only depends upon the endpoints, the line int egral of a gradient is independent of path. In particular, if Cis a closed curve, then a=b, and so the endpoint contributions cancel out: /contintegraldisplay C∇u·dx= 0. Conversely, if vis any vector field with the property that its integral around any closed curve vanishes, /contintegraldisplay Cv·dx= 0, (B.80) thenv=∇ϕadmits a potential. Indeed, as long as the domain is 0–connec ted, one can construct a potential ϕ(x) by integrating over any convenient curve Cconnecting a fixed pointa∈Ω to the point x ϕ(x) =/integraldisplayx av·dx. The proof that this is a well-defined potential is similar to t he planar version discussed in Chapter A. Ifvrepresents the velocity vector field of a three-dimensional steady state fluid flow, then its line integral around a closed curve C, namely /contintegraldisplay Cv·dx=/contintegraldisplay Cv·tds is the integral of the tangential component of the velocity v ector field. This represents the circulation of the fluid around the curve C. In particular, if the circulation line integral is 0 for every closed curve, then the fluid flow will be irrotational because∇×v=∇×∇ϕ≡0. Stokes’ Theorem The second of the three fundamental integration theorems is known as Stokes’ The- orem. This important result relates the circulation line integr al of a vector field around a closed curve with the integral of its curl over any bounding surface. Stokes’ Theorem first appeared in an 1850 letter from Lord Kelvin (William Tho mpson) written to George Stokes, who made it into an undergraduate exam question for t he Smith Prize at Cam- bridge University in England. Theorem B.31. LetS⊂R3be an oriented, bounded surface whose boundary ∂S consists of one or more piecewise smooth simple closed curve s. Letvbe a smooth vector field defined on S. Then /contintegraldisplay ∂Sv·dx=/integraldisplay/integraldisplay S(∇×v)·ndS. (B.81) 12/11/12 1260 c/ci∇cleco√y∇t2012 Peter J. Olver To make sense of Stokes’ formula (B.81), we need to assign a co nsistent orientation to the surface — meaning a choice of unit normal n— and to its boundary curve — meaning a direction to go around it. The proper choice is described by the following left hand rule: If we walk along the boundary ∂Swith the normal vector nonSpointing upwards, then the surface should be on our left hand side; see Figure St okes. For example, if S⊂ {z= 0}is a planar domain and we choose the upwards normal n= (0,0,1)T, then Cshould be oriented in the usual, counterclockwise directio n. Indeed, in this case, Stokes’ Theorem B.31 reduces to Green’s Theorem A.26! Stokes’ formula (B.81) can be rewritten using the alternati ve notations (B.21), (B.47), for surface and line integrals in the form /contintegraldisplay ∂Sudx+vdy+wdz= /integraldisplay/integraldisplay S/parenleftbigg∂w ∂y−∂v ∂z/parenrightbigg dydz+/parenleftbigg∂u ∂z−∂w ∂x/parenrightbigg dzdx+/parenleftbigg∂v ∂x−∂u ∂y/parenrightbigg dxdy.(B.82) Recall that a closed surface is one without boundary: ∂S=∅. In this case, the left hand side of Stokes’ formula (B.81) is zero, and we find that in tegrals of curls vanish on closed surfaces. Proposition B.32. If the vector field v=∇×wis a curl, then/integraldisplay/integraldisplay Sv·ndS= 0 for every closed surface S. Thus, every curl vector field defines a surface-independent i ntegral. Example B.33. LetS={x+y+z= 1, x>0, y>0, z>0}denote the triangu- lar surface considered in Example B.15. Its boundary ∂S=Lx∪Ly∪Lzis a triangle composed of three line segments Lx={x= 0, y+z= 1, y≥0, z≥0}, Ly={y= 0, x+z= 1, x≥0, z≥0}, Lz={z= 0, x+y= 1, x≥0, y≥0}. To compute the line integral /contintegraldisplay ∂Sv·dx=/contintegraldisplay ∂Sy2dx+xz2dy of the vector field v=/parenleftbig y2,xz2,0/parenrightbigT, we could proceed directly, but this would require evaluatingthreeseparateintegralsoverthethreesidesof thetriangle. Asanalternative, we can use Stokes formula (B.81), and compute the integral of it s curl∇×v= (2y,2xz,0)T over the triangle, which is /contintegraldisplay ∂Sv·dx=/integraldisplay/integraldisplay S(∇×v)·ndS=/integraldisplay/integraldisplay S2ydydz+2xzdzdx =17 12, where this particular surface integral was already compute d in Example B.15. 12/11/12 1261 c/ci∇cleco√y∇t2012 Peter J. Olver We remark that Stokes’ Theorem B.31 is consistent with Theor em B.28. Suppose thatvis a curl-free vector field, so ∇×v=0, which is defined on a 1–connected domain Ω⊂R3. Since every simple (unknotted) closed curve C⊂Ω bounds a surface, C=∂S, withS⊂Ω also contained inside the domain, then, Stokes’ formula (B .81) implies /contintegraldisplay Cv·dx=/integraldisplay/integraldisplay S(∇×v)·ndS= 0. Since this happens for every†C⊂Ω, then the path-independence condition (B.80) is satisfied, and hence v=∇ϕadmits a potential. Example B.34. The Newtonian gravitational force field v(x) =x /ba∇dblx/ba∇dbl3=(x,y,z)T (x2+y2+z2)3/2 is well defined on Ω = R3\{0}, and is divergence-free: div v≡0. Nevertheless, this vector field does not admit a vector potential. Indeed, on the sphere Sa={/ba∇dblx/ba∇dbl=a}of radius a, the unit normal vector at a point x∈Saisn=x//ba∇dblx/ba∇dbl. Therefore, /integraldisplay/integraldisplay Sav·ndS=/integraldisplay/integraldisplay Sax /ba∇dblx/ba∇dbl3·x /ba∇dblx/ba∇dbldS=/integraldisplay/integraldisplay Sa1 /ba∇dblx/ba∇dbl2dS=1 a2/integraldisplay/integraldisplay SadS= 4π, sinceSahas surface area 4 πa2. Note that this result is independent of the radius of the sphere. If v=∇×w, this would contradict Proposition B.32. The problem is, ofcourse, thatthe domainΩ isnot 2–connecte d, andso Theorem B.28 does not apply. However, it would apply to the vector field von any 2–connected sub- domain, for example the domain /tildewideΩ =R3\ {x=y= 0,z≤0}obtained by omitting the negativez-axis. Exercise asks you to construct a vector potential in this case. We further note that vis curl free: ∇ ×v≡0. Since the domain of definition Ω is 1–connected, Theorem B.28 tells us that vadmits a scalar potential — the Newtonian gravitational potential. Indeed, ∇/parenleftbig /ba∇dblx/ba∇dbl−1/parenrightbig =v, as the reader can check. The Divergence Theorem The last of the three fundamental integral theorems is the Divergence Theorem , also known as Gauss’ Theorem . This result relates a surface flux integral over a closed sur face to a volume integral over the domain it bounds. Theorem B.35. LetΩ⊂R3be a bounded domain whose boundary ∂Ωconsists of one or more piecewise smooth simple closed surfaces. Let ndenote the unit outward normal to the boundary of Ω. Letvbe a smooth vector field defined on Ωand continuous up to its boundary. Then /integraldisplay/integraldisplay ∂Ωv·ndS=/integraldisplay/integraldisplay/integraldisplay Ω∇·vdxdydz. (B.83) †It suffices to know this for unknotted curves to conclude it for arbit rary closed curves. 12/11/12 1262 c/ci∇cleco√y∇t2012 Peter J. Olver In terms of the alternative notation (B.47) for surface inte grals, the divergence for- mula (B.83) can be rewritten in the form /integraldisplay/integraldisplay Su dydz+vdzdx+wdxdy=/integraldisplay/integraldisplay/integraldisplay Ω/parenleftbigg∂u ∂x+∂v ∂y+∂w ∂z/parenrightbigg dxdydz. (B.84) Example B.36. Let us compute the surface integral /integraldisplay/integraldisplay Sxydzdx +zdxdy of the vector field v= (0,xy,z)Tover the sphere S={/ba∇dblx/ba∇dbl= 1}of radius 1. A direct evaluationineithergraphicalorsphericalcoordinatesis notsopleasant. Butthedivergence formula (B.84) immediately gives /integraldisplay/integraldisplay Sxydzdx +zdxdy=/integraldisplay/integraldisplay/integraldisplay Ω/parenleftbigg∂(xy) ∂y+∂z ∂z/parenrightbigg dxdydz =/integraldisplay/integraldisplay/integraldisplay Ω(x+1)dxdydz =/integraldisplay/integraldisplay/integraldisplay Ωxdxdydz +/integraldisplay/integraldisplay/integraldisplay Ωdxdydz =4 3π, where Ω = {/ba∇dblx/ba∇dbl<1}is the unit ball with boundary ∂Ω =S. The final two integrals are, respectively, the xcoordinate of the center of mass of the sphere multiplied by i ts volume, which is clearly 0, plus the volume of the spherical ball. Example B.37. Suppose v(t,x) is the velocity vector field of a time-dependent fluid flow. Let ρ(t,x) represent the density of the fluid at time tand position x. Then the surface flux integral/integraldisplay/integraldisplay S(ρv)·ndSrepresents the mass flux of fluid through the surface S⊂R3. In particular, if S=∂Ω represents a closed surface bounding a domain Ω, then, by the Divergence Theorem B.35, /integraldisplay/integraldisplay ∂Ω(ρv)·ndS=/integraldisplay/integraldisplay/integraldisplay Ω∇·(ρv)dxdydz represents the net mass flux out of the domain Ω at time t. On the other hand, this must equal the rate of change of mass in the domain, namely −∂ ∂t/integraldisplay/integraldisplay/integraldisplay Ωρdxdydz =−/integraldisplay/integraldisplay/integraldisplay Ω∂ρ ∂tdxdydz, the minus sign coming from the fact that we are measuring net m ass loss due to outflow. Equating these two, we discover that /integraldisplay/integraldisplay/integraldisplay Ω/parenleftbigg∂ρ ∂t+∇·(ρv)/parenrightbigg dxdydz = 0 foreverydomain occupied by the fluid. Since the domainis arbitrary, t his can only happen if the integrand vanishes, and hence ∂ρ ∂t+∇·(ρv) = 0. (B.85) 12/11/12 1263 c/ci∇cleco√y∇t2012 Peter J. Olver The latter is the basic continuity equation of fluid mechanics, which takes the form of a conservation law. For a steady state fluid flow, the left hand side of the divergen ce formula (B.83) measures the fluid flux through the boundary of the domain ∂Ω, while the left hand side integrates the divergence over the domain Ω. As a consequenc e, the divergence must represent the net local change in fluid volume at a point under the flow. In particular, if ∇v= 0, then there is no net flux, and the fluid flow is incompressibl e. The Divergence Theorem B.35 is also consistent with Theorem B.28. Let vis a divergence-free vector field, ∇ ·v= 0, defined on a 2–connected domain Ω ⊂R3. Every simple closed surface S⊂Ω bounds a subdomain, so S=∂D, withD⊂Ω also contained inside the domain of definition of v. Then, by the divergence formula (B.83), /integraldisplay/integraldisplay Sv·ndS=/integraldisplay/integraldisplay/integraldisplay Ω∇·vdxdydz = 0. Therefore, by Theorem B.28, v=∇×wadmits a vector potential. Remark: The proof of all three of the fundamental integral theorems , can, in fact, be reduced to the Fundamental Theorem of (one-variable) Calcu lus. They are, in fact, all special cases of the general Stokes’ Theorem, which forms th e foundation of the profound theory of integration on manifolds, [ 2,23,70]. Stokes’ Theorem has deep and beautiful connections with topology — and is of fundamental importanc e in modern mathematics and physics. However, the full ramifications lie beyond the s cope of this introductory text. 12/11/12 1264 c/ci∇cleco√y∇t2012 Peter J. Olver