Vector Calculus - Math

1 CHAPTER 18 Vector CalculusIn this chapter we develop the fundamental theorem of the Calculus in two and three dimensions. Thisbegins with a slight reinterpretation of that theorem. Consider the endpointsa;bof the interval[a;b]fromatobas the boundary of that interval. Then the fundamental theorem, in this form:( )f(b) f(a)=Zbad fdx(x)dx;relates the values of a function at the boundary with the values of its derivative in the interior. Statedthis way, the fundamental theorems of the Vector Calculus (Green s, Stokes and Gauss theorems) arehigher dimensional versions of the same idea. However, in higher dimensions, things are far morecomplex: regions in the plane have curves as boundaries, andfor regions in space, the boundary is asurface, and surfaces in space have curves as boundaries.

2 This requires a reinterpretation of the termf(b) f(a), as asignedsum of the values offon the boundary, the sign being determined by the sideon which the interval lies (it is to the right ofaand to the left ofb). This leads to the understanding thatin higher dimensions both sides will be integrals; for example, for a regionRin the plane withCas itsboundary, the termf(b) f(a)becomes an integral over the curveC. And in three dimensions, we willhave two versions of the fundamental theorem, one relating integrals over a region with integrals overthe bounding surface, and another relating integrals over surfaces with integrals over the bounding curve(and with the relation involving some form of differentiation).

3 We will not give derivations, or even intuitive arguments for the proofs of these theorems. Firstof all, the idea of the proof is to reduce the theorem to the one-variable fundamental theorem; in thisprocess, the notational complexity is constantly threatening to get out of hand. The proofs then becomemasterful displays of technical control, and provide little insight. The insight comes from the physicalinterpretation of these theorems (indeed, so also did the first proofs), particularly in terms of fluid example, Gauss theorem simply says that, for a fluid in flow we can measure the rate of change ofthe amount of fluid in a given region in two ways: directly overthe region, or instead, by measuring therate of passage through the 18 Vector Vector FieldsAvector fieldis an association of a Vector to each pointXof a regionR:( )F(x;y;z)=P(x;y;z)I+Q(x;y;z)J+R(x;y;z)K: For example, the Vector field( )X(x;y;z)=xI+yJ+zKis the field of vectors pointing outward from the origin, whose length is equal to the distance from theorigin.

4 The fieldU=(1=r)X(wherer(x;y;z)=(x2+y2+z2)1= 2) is the unit Vector field with the (Gravitation). According to Newton s Law of gravitation, two bodies attract each otherwith a force proportional to the product of the masses, and inversely proportional to the square of thedistance between them. Suppose one body, of massMis situated at the origin. Then another body ofmassm, situated at the pointXexperiences the gravitational force due toM:( )F= GMmr2U;whereGis Newton s universal constant of gravitation, andUis the unit Vector pointing the directionofX. If we want to concentrate on the effect of the massMon bodies in its vicinity, we introduce thegravitational fieldofM:( )G(X)= GMr2U= GMr3X:SinceF=mA, a body of massmatXaccelerates toward the origin with accelerationG(X).

5 Definition the region R is filled with a fluid which is in motion. Wecan describe themotion by following the individual particles. LetX(X0;t)be the position at time t of the particle whichwas atX0at time t=0. Thevelocity fieldof the motion is the velocity of the particle at positionXattime t, represented byV(X;t).This is a time-dependent Vector field in the region R. We say that the flowissteadyif its velocity field is independent of studying a fluid in motion, we are not interested in the history of particular particles, but in thefluid as a whole. Thus, it is the velocity field of the fluid that is the object of study, rather than theequations of motion. It can be shown that the velocity field completely determines the a fluid is flowing on the plane radially away from the origin.

6 In this case theorigin is called asource; if the fluid were flowing toward the origin, we call it asink. The equation ofmotion is given by( )X(X0;t)=f(t)X0for some scalar functionfwithf(0)=1:Let s look at the casef(t)=eat. We find the velocity field as follows. First, the velocity of the particleoriginally atX0is( ) tX(X0;t)=ddt(eat)X0=aeatX0 Fields283 But this isaX, so the velocity field isV(X)=aX, and the flow is steady. However, if, sayf(t)=1+tso thatX(X0;t)=(1+t)X0,we have( ) tX(X0;t)=X0=(1+t) 1X;so the flow is terminology may seem confusing: in the first case, the particle s speed is increasing exponen-tially, while in the second case the particle s speed is constant. But, if we look at a particular pointXinspace, then in the first case, the fluid is always moving with the same velocity through that point, whilein the second case, the fluid slows down at that point over a fluid is rotating on the plane about the origin in thecounterclockwise directionat constant angular velocity.

7 From the description, this is a steady flow; let s find its velocity a pointX, particles move throughXalong the circle of radiusjXjat angular velocity .Thus thevelocity of the fluid atXis of magnitude jXjand in the direction tangent to to the circle throughX, soV(X)= X?.Definition differentiable function w=f(x;y;z)has associated to it itsgradient field( ) w= f xI+ f yJ+ f zK:The surfaces f(x;y;z)=const. are orthogonal to the Vector field ( ), and are called theequipoten-tials, and the function f , apotentialfor the , the flow associated to a gradient field is easily visualized as being in the direction perpendicularto these equipotential surfaces. A natural question is: when is a Vector fieldFthe gradient of a function;that is, when does a Vector field have a potential function?

8 Ifthe Vector field with the componentsF=PI+QJ+RKis a gradient, so looks like ( ), then, because of the equality of mixed derivatives,we must have( ) P y= Q x; P z= R x; Q z= R y:If these conditions are satisfied, then we can try to find the potential function by integrating one variableat a (2xy+x)I+x2 yJ. IsFa gradient field? If so, find the potential , we check that the condition ( ) is satisfied:( ) P y= y(2xy+x)=2x Q x= y(x2 y)=2x:So, we have a chance of finding a functionfsuch that f=F. To findfwe have to solve the equations( ) f x=2xy+x; f y=x2 y:We can find a function satisfying the first equation by integrating with respect tox; so we tryf(x;y)=x2y x2=2. Now we see if thisfsatisfies the second equation:( ) f y=x2;Chapter 18 Vector Calculus284which unfortunately is notx2 y.

9 However, since the derivative with respect toxof any function ofyiszero, we could also have tried( )f(x;y)=x2y+x2=2+ (y)for some yet-to-be-determined (y). Now, we have, instead of ( ),( ) f y=x2+ 0(y);setting that equal toQgives the equation 0(y)= y, so we can take (y)= y2=2. We conclude thatour solution is( )f(x;y)=x2y+x22 y22+C;for any constantC. The reason that the terms involvingxdisappear in equation ( ) is precisely thatthe condition P= y= Q= xis satisfied; if it were not, this procedure would break down at this procedure in three dimensions is the same, but longer. Suppose we are given thevector fieldF=(y2z+1)I+(2xyz+z)J+(xy2+y+1)K, and we are told that it is the differential of afunctionf.

10 We are told that there is a potential function, we need not verify conditions ( ). We startwith( ) f x=y2z+1:Integrating both sides with respect tox, (thinking ofyandzas constants), we obtain( )f(x;y;z)=xy2z+x+ (y;z)where is an unknown function ofyandzalone. Now, differentiating this equation, since f= y=2xyz+z, we obtain( )2xyz+z=2xyz+ y;or( ) y=z:Now we do the same, integrating both sides with respect toy:( ) (y;z)=yz+ (z);for some unknown function (z). Thus ( ) now becomes( )f(x;y;z)=xy2z+x+yz+ (y;z):Differentiating now with respect toz:( )xy2+y+1=xy2+y+ Fields285so = z=1, and thus (z)=z+C. Putting this back in ( ), we have found( )f(x;y;z)=xy2z+x+yz+z+C:The reason that the variablexdisappeared from ( ) andxandyfrom ( ) is precisely because ofthe conditions ( ); if they did not hold there would be nosuch functionf, and we could not havesolved equations ( ) and ( ).