# Chapter Twelve. Integration. We now turn our attention to the idea of an integral in dimensions higher than one. Consider a real-valued function f : D

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1 Chapter Twelve Integraton 12.1 Introducton We now turn our attenton to the dea of an ntegral n dmensons hgher than one. Consder a real-valued functon f : R, where the doman s a nce closed subset of Eucldean n-space R n. We shall begn by seeng what we mean by the ntegral of f over the set ; then later we shall see just what such an abstract thng mght be good for n real lfe. Mrs. Turner taught us all about the case n = 1. As t was n etendng the defnton of a dervatve to hgher dmensons, our defnton of the ntegral n hgher dmensons wll nclude the defnton for dmenson 1 we learned n grammar school as always, there wll be nothng to unlearn. Let us agan hark back to our youth and revew what we know about the ntegral of f : R n case s a nce connected pece of the real lne R. Frst, n ths contet, the only nce closed peces of R are the closed ntervals; we thus have s a set [ a, b ], where b > a. Recall that we defned a partton P of the nterval to be smply a fnte subset { 0, 1, K, n } of [ a, b ] wth a = 0 < 1 < 2 < K < n = b. The mesh of a partton s ma{ 1 : = 12K,, n}. We then defned a Remann sum S( P) for ths partton to be a sum n S( P) = f ( ) = 1, where = 1 s smply the length of the subnterval [ 1, ] and s any pont n ths subnterval. (Thus there s not just one Remann sum for a partton P; the sum obvously also depends on the choces of the ponts. Ths s not reflected n the notaton.) Now, f there s a number L such that we can make all Remann sums as close as we lke to L by choosng the mesh of the partton suffcently small, then f s sad to be 12.1

2 ntegrable over the nterval, and the number L s called the ntegral of f over [a, b]. Ths number L s almost always denoted b a f ( ) d. More formally, we say that L s the ntegral of f over [ a, b ] f for every ε > 0, there s a δ so that S( P) L < ε for every partton P havng mesh < δ. You no doubt remember from your frst encounter wth ths ntegral that t ntally seemed lke an mpossble thng to compute n any reasonable stuaton, but then some verson of the Fundamental Theorem of Calculus came to the rescue Two mensons Let us begn our study of hgher dmensonal ntegrals wth the two dmensonal case. As we have seen so often n the past, n etendng calculus deas to hgher dmensons, most of the ectement occurs n takng the step from one dmenson to two dmensons seldom s the step from 97 to 98 dmensons very nterestng. We shall thus begn by lookng at the ntegral of f : R for the case n whch s a nce closed subset of the plane. Complcatons appear at once. On the real lne, nce closed sets are smply closed ntervals; n the plane, nce closed sets are consderably more nterestng: 12.2

3 A moment's reflecton convnces us that the doman can, even n just two dmensons, be consderably more complcated than t s n one dmenson. Frst, capture nsde a rectangle wth sdes parallel to the coordnate aes; and then dvde ths rectangle nto subrectangles by parttonng each of ts sdes: Now, label the subrectangles that meet, say wth subscrpts = 12,, K, n. The largest area of all such rectangles s called the mesh of the subdvson. In each such rectangle, choose a pont (, y ) n. A Remann sum S now looks lke n S = f (, y ) A = 1, where A s the area of the rectangle from whch (, y ) s chosen. Now f there s a number L such that we can get as close to L as we lke by choosng the mesh of the subdvson suffcently small, then f s sad to be ntegrable over, and the number L s the ntegral of f over. The number L s usually wrtten wth two snake sgns: f (, y ) da. Such ntegrals over two dmensonal domans are frequently referred to as double ntegrals. 12.

4 I hope the defnton of the ntegral n case s a nce subset of R s evdent. We capture nsde a bo, and subdvde the bo nto boes, etc., etc. There wll be more of the hgher dmensonal stuff later. Let's look a bt at some geometry. For the purpose of drawng a reasonable pcture, let us suppose that f (, y) 0 everywhere on. Each term f (, y ) A s the volume of a bo wth base the rectangle A and heght f (, y ). The top of the bo thus meets the surface z = f (, y). The Remann sum s thus the total volume of all such boes. Convnce yourself that as the sze of the bases of the boes goes to 0, the boes "fll up" the sold bounded below by the -y plane, above by the surface z = f (, y), and on the sdes by the cylnder determned by the regon. The ntegral f (, y ) da s thus equal to the volume of ths sold. If f (, y) 0, then, of course, we get the negatve of the volume bounded below by the surface z above by the -y plane, etc. = f (, y), Suppose a and b are constants, and = E F, where E and F are nce domans whose nterors do not meet. The followng mportant propertes of the double ntegral should be evdent: 12.4

5 [ af (, y) + bg(, y)] da = a f (, y) da + b g(, y) da, and f (, y) da = f (, y) da + f (, y) da. E F Now, how on Earth do we ever fnd an ntegral f (, y ) da? Let's see. Agan, we shall look at a pcture, and agan we shall draw our pcture as f f (, y) 0. It should be clear what happens f ths s not the case. We assume our doman has a specal form; specfcally, we suppose t to be bounded above by the curve y the rght by = b : = h( ), below by y = g( ), on the left by = a, and on It s convenent for us to thnk of the ntegral f (, y ) da as the volume of the blob bounded below by n the -y plane and above by the surface z = f (, y). Thnk of fndng ths volume by dvdng the blob nto slces parallel to the y-as and addng up the volumes of the slces. To appromate the volumes of these slces, we use slabs: 12.5

6 We partton the nterval [a, b ]: a = < < K < < = b. In each subnterval 0 1 n 1 [ 1, ] choose a pont. Our appromatng slab has as ts base the rectangle of "wdth" = 1 and heght h( ) g( ) ; the roof s z = f (, y) h( ) of the slab s the cross secton area tmes the thckness, or [ f (, y) dy]. n g( ). The volume The sum of the volumes of the appromatng slabs s thus n h( ) S = [ f (, y) dy] = 1 g( ). The double ntegral we seek s just the "lmt" of these as we take thnner and thnner slabs; or fner and fner parttons of the nterval [a, b]. But Lo! The above sums are 12.6

7 Remann sums for the ordnary one dmensonal ntegral of the functon h( ) F ( ) = f (, y) dy, and so the double ntegral s gven by g( ) f (, y) da = F ( ) d b a = b h( ) [ f (, y) dy] d a g( ) The double ntegral s thus equal to an ntegral of an ntegral, usually called an terated ntegral. It s tradtonal to omt the brackets and wrte the terated ntegral smply as b h( ) f (, y) dyd. a g( ) Eample lnes y 2 2 Let's fnd the double ntegral [ + y ] da, where s the area enclosed by the =, = 0, and + y = 2. The frst tem of busness here s to draw a pcture of (We always need a pcture of the doman of ntegraton.): 12.7

8 It should be clear from the pcture that n the language of our dscusson, g( ) =, h( ) = 2, a = 0, and b = 1. So slce parallel to the y as: The lower end of the slce s at y = and the upper end s at y = 2. The "volume" s thus 2 y= y 2 ( 2 ) 2 ( 2 ) [ + y ] dy = y + = ( 2 ) + = 2 + y= 7, and we have such a slce for all from = 0 to = 1. Thus ( 2 ) [ + y ] da = [ ( 2 ) 7 = = = ] d Eercses 2 1. Fnd da, where s the doman bounded by the curves y = 4 2 and y =. 12.8

9 2. Fnd ( 2 y) da, where s the area n the frst quadrant enclosed by the coordnate aes and the lne 2 + y = 4.. Use double ntegraton to fnd the area of the regon enclosed by the curves y = 2 and y = Fnd the volume of the sold cut from the frst octant by the surface z = 4 y. 5. Sketch the doman of ntegraton and evaluate the terated ntegral: y y e dyd. 6. Sketch the doman of ntegraton and evaluate the terated ntegral: log8log + y 1 0 e dyd. 7. Fnd the volume of the wedge cut from the frst octant by the cylnder z = 12 y 2 and the plane + y = Suppose you have a double ntegral f (, y) da n whch the doman s bounded on the left by the curve above by y = b. = g( y), on the rght by = h( y), below by y = a, and 12.9

10 Gve an terated ntegral for the double ntegral n whch the frst ntegraton s wth respect to, and eplan what's gong on. 9. Gve a double ntegral for the area of the regon bounded by = y 2 and = 2y y and evaluate the ntegral. 2, 12.10

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