Spring 2005 Math 152Overview: Vector & Multiple IntegralsThu, 03/Marc2005, Art BelmonteSummaryLooking ahead to Calc 3, we’ll briefly examine vector integrals,multiple integrals, and vector multple integrals. While the limit-of-a-Riemann-sum approach is mentioned, in practice these typesof integrals are evaluated in a mechanistic fashion. They are quiteeasy to do with a TI-89 calculator or in MATLAB. This is whatwe’ll usually do, especially in Calc 3.Review of Single Integrals from Calc 1Definition Let f be a function defined on I = [a, b]. Split[a, b]intonsubintervals whose endpoints constitute a partitionP : a = x0< x1< x2< ···< xn−1< xn= b.(Often the xiare equally spaced and we have a regular partition.)Let x∗i∈xi−i, xi be in the ith subinterval and 1xi= xi− xi−1be the length of this subinterval. The norm of P is defined bykPk= max1 xi. Now let the number of subintervals n increaseindefinitely while the norm of P shrinks to 0. The definite integralof f from a to b is defined byZbaf (x) dx = limkPk→0nXi=1fx∗i1xi,provided this limit of a Riemann sum exists. When this occurs,f is said to be integrable on [a, b].Antiderivatives The scalar function F(x) is an antiderivative orindefinite integral of the scalar function f (x) on an interval I ifand only if F0(x) = f (x) for all x ∈ I. A vector function R(t) isan antiderivative or indefinite integral of the vector function r(t)on an interval J if and only if we have R0(t) = r(t) for all t ∈ J.Fundamental Theorem of Calculus (FTC), Part 2 Computinga definite integral as a limit of a Riemann sum is quite tedious.But computing it with the FTC is easy. Let f be a continuousfunction on [a, b]. ThenZbaf (x) dx = F(x)ba= F(b) − F(a),where F is an antiderivative of f .Vector IntegalsLet r(t) = [ f (t), g(t)], a ≤ t ≤ b, be a continuous vectorfunction and P be a partition of [a, b]. The definite integral of rfrom a to b is defined byZbar(t) dt = limkPk→0nXi=1rt∗i1ti= limkPk→0nXi=1ft∗i, gt∗i 1ti= limkPk→0nXi=1ft∗i1ti, gt∗i1ti = limkPk→0"nXi=1ft∗i1ti,nXi=1gt∗i1ti#="limkPk→0nXi=1ft∗i1ti, limkPk→0nXi=1gt∗i1ti#="Zbaf (t) dt,Zbag(t) dt#.In other words,Zba[ f (t), g(t)] dt =Zbar(t) dt ="Zbaf(t) dt,Zbag(t) dt#.We simply map the operation of integration onto the componentsof the vector function. Moreover, we may use the FTC to evaluatethe component integrals.Double IntegralsDefinition Let f (x, y) be a continuous function over arectangular regionR ={(x, y): a ≤ x ≤ b, c ≤ y ≤ d}in the xy-plane. We define the double integral of f by “slicingand dicing,” as it were. (Think of mincing that onion with yourGinsu knife...) That is, split [a, b]intomsubintervals and [c, d]into n subintervals. The normkPkof the resulting partition P isthe length of the longest diagonal among the subrectangles of thepartition. We then form a double Riemann sum and take the limitaskPkshrinks to 0. If this limit exists, we obtain the doubleintegral of f over R.ZZRf(x,y)dA = limkPk→0mXi=1nXj=1fx∗i, y∗j1xi1yjIterated integrals Computing the exact value of a doubleintegral by taking a limit of a double Riemann sum is very difficulteven when it is possible.A practical way to evaluate multiple integrals is via interatedintegration, where we compute single integrals in succession.1In other words, we repeatedly compute antiderivatives and applythe Fundamental Theorem of Calculus, working from inside-outuntil we are finished.This mechanistic approach has been fully automated, as in the smi(stepwise [multiple] integration) commands on the TI-89 and inMATLAB. Accordingly, this reduces the problem of computingmultiple integrals to simply setting them up. That said, you oughtto try some problems purely by hand to get some feeling for thework that you are being spared.Fubini’s Theorem If f is continuous on a rectangular regionR ={(x, y): a ≤ x ≤ b, c ≤ y ≤ d},thenZZRf(x,y)dA =ZbaZdcf ( x, y) dydx =ZdcZbaf(x, y) dxdy.Nonrectangular regions• A Type I region has the formD ={(x, y): a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}.It is a region in the xy-plane bounded on the left by thevertical line x = a, on the right by the vertical line x = b,below by the curve y = g1(x), and above by the curvey = g2(x). The double integral of the function f (x, y) overD may be computed asZbaZg2(x)g1(x)f (x, y) dydx.• A Type II region has the formD ={(x, y): c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}.It is a region in the xy-plane bounded on the below by thehorizontal line y = c, above by the horizontal line y = d,onthe left by the curve x = h1(y), and on the right by the curvex = h2(y). The double integral of the function f (x, y) overD may be computed asZdcZh2(y)h1(y)f (x, y) dxdy.• NOTES: The aforementioned “curves” may be lines. Also,vertical or horizontal boundaries may actually collapse to asingle point. Finally, some regions are of both types. Hereare some samples of regions.Type 1 regiona x = by = g2(x) y = g1(x)Type 2 regioncdx = h1(y) x = h2(y)Vector Double IntegralsLet r(x, y) = [ f (x, y), g(x, y)] be a continuous function definedon a region D of Type 1 and/or Type 2. The vector double integralof r over D is just what you’d think.ZZDr(x,y)dA =ZZD[f(x,y), g(x, y)] dA=ZZDf(x,y)dA,ZZDg(x, y)dAJust map the integration operation onto the components of thevector function. Moreover, we may use Fubini’s Theorem ongeneral regions to evaluate the component integrals.GeneralizationsLet’s extend the aforementioned concepts.• If our vector function has three components, thenZba[ f (t), g(t), h(t)] dt ="Zbaf(t) dt,Zbag(t) dt,Zbah(t) dt#.• If f ( x, y, z) is continuous on the rectangular boxR ={(x, y, z): a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}then the triple integral of f over R is defined byZZZRf(x,y,z)dV =ZsrZdcZbaf ( x , y,z) dx dydz= limkPk→0lXi=1mXj=1nXk=1fx∗i, y∗j, z∗k1xi1yj1zk.Similarly, Fubini’s Theorem may be extended to allowcomputation of a triple integral via iterated integration.Moreover, triple integrals over more general regions E maybe realized.ZZZEf(x,y,z)dV =ZbaZg2(x )g1(x )Zh2(x , y)h1(x , y)f ( x , y,z) dzdydx• Let r(x, y, z) = [ f (x, y, z), g(x, y, z), h(x, y, z)]beacontinuous function defined on a suitable region E in 3-Dxyz-space. The vector triple integralRRREr(x, y, z) dV of rover E isZZZE[f(x, y,z), g(x, y, z), h(x, y, z)] dV=ZZZEf(x, y, z) dV,ZZZEg(x, y,z)dV,ZZZEh(x,
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