18.02 Multivariable Calculus (Spring 2009): Lecture 17Double and iterated integrals in the planeMarch 13Reading Material: From Simmons: 20.1 and 20.2. From Course Notes: I.Last time: Non-indep e ndent variables. Partial differential equations.Today: Double and iterated integrals in the plane.2 Integrals of functions of one variableConsider a function y = f(x) such that x is in [a, b]. Assume also in this case that f(x) ≥ 0 on [a, b].Then we know that the area A of the sub graph is given byArea A =Zbaf(x) dxWe also know how to evaluate this integral using an anti-derivative F (x) (F0(x) = f (x)). In fact bythe fundamental theorem of calculusA =Zbaf(x) dx = F (b) − F (a)A couple of questions are natural at this point:Question: Why areas are computed through integrals? What does an integral really mean?Take the interval [a, b] and subdivide it into smaller intervals by introducing a partition℘ = {a = x0, x1, ..., xi, ..., xn= b}1CallIi= [xi, xi+1] and ∆xi= xi+1− xi.For each interval Iiwe pick a point x∗iin it and we measure the highthi= f (x∗i).Then if Riis the rectangle of base Iiand hight hiit is clear that the area of the sub graph A is approximated by the sum of the areas of the rectanglesRi’s:A ≈ An=n−1Xi=1f(x∗i)∆xiIf we consider partitions that contain more and more points (e.i finer ones), then we haveA = limn→∞An= limn→∞n−1Xi=1f(x∗i)∆xi:=Zbaf(x) dx,and the last term is a purely symbolic definition.3 Iterated IntegralsConsider now a function of two variables z = f (x, y) defined on a rectangle [a, b] × [c, d]:2Then the volume V of the sub graph of f(x, y) above the rectangle R isV =Z ZRf(x, y) dA :=ZbaZdcf(x, y) dy| {z }inner integraldx| {z }outer integral,this is a double integral defined as iterated integrals.We will go back to the method of evaluation later, now we discuss why one can compute volumeswith double integrals. As before we consider a partition ℘ for [a, b] and we also c onsider a partition˜℘ for [c, d]:˜℘ = {c = y0, y1, ..., yj, ..., ym= d},and we define the intervals Jj= [yj, yj+1]. We then define the rectanglesRi,j= Ii× Ij,of area∆Aij= ∆xi∆yj,we pick a point P ∗i,jin Ri,jand we define the hight hi,j= f (P ∗i,j). Now it is easy to see that thevolume V is approximated by the sum of the volumes of the parallelepipeds defined by Ri,j× hi,jand henceV ≈ Vn,m=n−1Xi=1m−1Xj=1f(P ∗i,j)∆xi∆yj.3Now as we make the partitions finer and finer, we haveV = limn→∞limm→∞Vn,m= limn→∞limm→∞n−1Xi=1m−1Xj=1f(P ∗i,j)∆xi∆yj:=Z ZRf(x, y) dA.4 Method of evaluationGiven a function z = f(x, y) for (x, y) in R = [a, b] × [c, d] we compute the double intervalV =Z ZRf(x, y) dAby using the method of iterated integrals:1. We fix one of the variable, say x in [a, b], that we call outer variable and perform the innerintegralA(x) =Zdcf(x, y) dy.This function A(x) represents the area of the intersection of the solid given by the sub graphof f with the plane parallel to the yz-plane and through the point (x, 0, 0):2. Now to get the volume V of the solid given by the sub graph we need to integrate all thes eareas A(x) along [a, b], hence we perform the outer integralV =ZbaA(x) dx =Zba"Zdcf(x, y) dy#dx. (4.1)Remark. When the domain of integration is a rectangle the choice of the outer or inner variable istotally arbitrary. In fact above we could have picked y as the outer variable and replaced (4.1) byV =Zdc"Zbaf(x, y) dx#dy. (4.2)In this case we say that we changed the order of integration.4Exercise 1. Consider the function f (x, y) = 20 − x2− y2. Compute the volume of the solid underthe graph of f and above the rectangle R = [1, 2] × [1, 3].Solution:5 Double integrals with non rectangular basisWe start with the following problem:Exercise 2. Consider again the function f(x, y ) = 20 − x2− y2as above. Compute the volume ofthe solid below the graph of f and above the triangle of vertices (1, 1), (2, 1) and (2, 3).5Solution: The procedure is similar to the one described above with the iterated integrals. Theproblem is that it is more tricky to find the limits of integration:1. Fix the outer variable, with respect to this variable the outer integral is done on the full range.2. Set up the inner integral where now the range depends on the outer variable (see picture below)In our resolution we fix x as the outer variable and its range is [1, 2], thenInner Integral:Outer Integral:6Study Guide 1. Answer the following questions:• I explained what is the geometric picture behind formula (4.1). Coud you explain in a similarway the geometric meaning of (4.2)?• If you compare (4.1) and (4.2) you see that changing the order of integration doesn’t changeanything for the limits of integration for the respective variables. Is this also true for
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