INTEGRATION OF FUNCTIONS OF SEVERAL VARIABLES. Contents 1. Integration 1 2. Double integrals 3 3. Iterated integrals and Fubini s Theorem 4

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1 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES Contents. Integrtion 2. Double integrls. Iterted integrls nd Fubini s Theorem 4. Integrtion Now tht the quik review of differentil lulus of severl vribles is finished, let s strt with the new mteril in this lss. We re interested in developing theory of integrtion for funtions of severl vribles. Let us begin with the se of funtions f(x, y) of two rel vribles. We wnt to define n integrl for this funtion. How should we proeed? Notie tht there is no obvious ndidte for n ntiderivtive, so the strtegy of trying to define n integrl s n ntiderivtive is not going to work. We should therefore emulte the definition of definite integrl for funtions of single vrible. How is definite integrl f(x) dx defined? ell tht the definition of this integrl is s the vlue of limit lim f(x i ) x i P where the x i re prtition of [, b], with lrgest x i equl to P, nd x i number between x i nd x i+. Eh of these sums is lled iemnn sum, nd n be thought of s n pproximtion to the re under the grph of y = f(x) on the intervl [, b]. This limit is just fny wy of sying tht we re interested in the limit of better nd better pproximtions to this re, whih we obtin by mking the retngles thinner nd thinner. Presumbly, our intuition tells us tht this should mke the error in pproximtion smller nd smller. Wht is the orret nlogue of the notion of iemnn sum for funtions of two vribles? Suh sum should estimte the volume under the grph of z = f(x, y) on some region. The losest nlogue to n intervl [, b] in might be retngle in 2, perhps of the form [, b] [, d]. We wnt sum whih estimtes the volume of the region under z = f(x, y) over the retngle. Just like how we use retngles to pproximte the re under y = f(x), we n use retngulr prisms to pproximte the re under z = f(x, y). In prtiulr, over retngle [x i, x i+ ] [y i, y i+ ], we n use retngulr prism with tht retngle s bse, nd height given by f(x i, yi ) s n pproximtion to the volume over this smller retngle. This retngulr prism hs volume

2 2 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES f(x i, y i ) x i y i. ( x i = x i+ x i, nd similrly for y i.) Therefore, the sum of these retngulr prisms, whih is pproximting the volume under z = f(x, y) over the retngle, is f(x i, y i ) x i y i where the summtion is over ll smll retngles mking up the lrge retngle. This sum is lled iemnn sum for z = f(x, y) over the retngle. If we mke these smll retngles smller nd smller, then the limit of these sums (if the limit exists) will be lled the definite integrl of f(x, y) over nd will be denoted f(x, y) da. The two integrl signs remind us tht we re integrting over two-dimensionl region; nmely, the retngle whih is written underneth the two integrl signs. The differentil da tells us tht we re integrting with respet to n re element. It is mostly to be thought of s nottion further reminding us tht we re integrting over two-dimensionl region. Of ourse, in prtie, we never use this definition of definite integrl to tully evlute definite integrls like f(x) dx. We sometimes see how to evlute these integrls diretly using the limit definition for ertin speil funtions, like f(x) = x 2, but these lultions re long nd tedious. Insted, we use the fundmentl theorem of Clulus, whih tells us tht to evlute definite integrl we need only find n ntiderivtive for f(x), nd then evlute it t, b. The iemnn sum definition is still useful from omputtionl point of view sometimes. For instne, there re mny funtions whih re impossible to integrte extly in ny sort of useful wy, but whih pper frequently in rel life (suh s the norml distribution, perhps the most importnt funtion in probbility nd sttistis). Nevertheless, we my wnt to obtin numeril estimte for these integrls. Sine iemnn sums re ment to be pproximtions of definite integrls, nd the omputtion of prtiulr iemnn sum is finite lultion, we n use iemnn sums to pproximte integrls whih we my not be ble to do not wnt to evlute extly. In generl, the more terms we tke in iemnn sum, the more urte our estimte will be, lthough this is not lwys true. Exmple. Consider the funtion f(x, y) = x 2 +y 2 over the retngle = [, 4] [, 4]. Use iemnn sum to estimte the integrl of f(x, y) over, where the iemnn sum splits up into four ongruent squres. Use the bottom-left points in squre to obtin one estimte, nd the midpoints to obtin nother.

3 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES Splitting up into four ongruent squres gives the squres [, 2] [, 2], [, 2] [2, 4], [2, 4] [, 2], [2, 4] [2, 4]. The re of eh squre is 4, so using the left endpoints gives: 4 ( ) = 64. Using the midpoints, whih re (, ), (, ), (, ), (, ) gives 4 ( ) = 6. By wy of omprison, the ext nswer, whih we will shortly see how to ompute, is 52/. The reltively poor qulity of our estimtes is due to the ft tht we only used four retngles, nd the funtion f(x, y) is fr from pproximtely onstnt on eh of those retngles. And of ourse, in prtie, we would probbly not need to use iemnn sum to estimte n integrl whih we ould evlute extly. Wht we seek now is onvenient wy to evlute double integrls. Even though we should not expet fundmentl theorem of Clulus for double integrls yet, we n still mke use of the FTC for funtions of single vrible. 2. Double integrls ell tht we disussed how definite integrl for funtion of two vribles f(x, y) over retngle is defined, s the limit of iemnn sums whih represented sums of volumes of lots of smll retngulr prisms. In prtiulr, the definite integrl should represent the (signed) volume under z = f(x, y) over the retngle. Exmples. Evlute the double integrl da in terms of the re of the retngle, where is some onstnt. This double integrl is equl to the volume of retngulr prism with bse of re A() nd height, so this double integrl should equl A(). This is the twodimensionl nlogue of the formul dx = (b ). Using the interprettion of double integrl s volume, lulte the definite integrl x2 da where is the retngle [, ] [, ]. We begin by mking sketh of the grph of this funtion over the retngle. Notie tht x 2 does not depend on y, so the ross-setions of the grph of this funtion tken when we fix y will ll look identil. In prtiulr, these ross-setions re of the form z = x 2, whih is the upper hlf of irle of rdius. So the double integrl we re evluting represents the volume of region whose

4 4 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES ross-setions by plnes of the form y = C re ll hlf-diss of rdius. In other words, our region is hlf of ylinder, whose bse is irle of rdius nd whose length is 2. The volume of suh solid is given by 2 π 2 = π. The double integrl lso hs n interprettion s the verge vlue of funtion over the retngle. In the single-vrible se, the expression b f(x) dx represents the verge vlue of f(x) on the intervl [, b]; in the two-vrible se, the expression f(x, y) da A() represents the verge vlue of f(x, y) on the retngle. For exmple, if represents the boundries of ity, nd f(x, y) is the mount of rinfll (tht is, the number of inhes of rin) the ity reeived t the point (x, y), then this verge vlue represents the verge mount of rinfll the ity reeived.. Iterted integrls nd Fubini s Theorem The previous exmple (the hlf-ylinder) provides the lue to nswering the question of how to evlute double integrls. Notie tht we solved tht problem by reognizing ross-setion of the solid we wnted to lulte the volume of when we ut it with plnes of the form y = C. Suppose we wnt to integrte f(x, y) da We might pproh this problem by tking ross-setions of the solid under z = f(x, y), over the retngle, when we fix either x or y. Conretely, suppose = [, b] [, d]. Suppose we fix x, so we tke ross-setions with plnes of the form x = C. Then the pproprite ross-setion then hs re given by the formul A(x) = d f(x, y) dy where in this integrl we tret x s fixed number. For exmple, if we re interested in evluting x 2 + y 2 da where = [, 2] [, ], then A(x) = x 2 + y 2 dy = ) (yx 2 + y y= = x y= In other words, the ross-setionl re of the solid defined by z = x 2 + y 2 over the retngle [, 2] [, ] when we fix x is x We ple y =, y = in the bounds of integrtion to remind ourselves tht we re evluting the expression yx 2 + y /

5 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES 5 t the vlues y =, insted of x =,. This proedure, where we lulte n integrl by fixing one vrible, is very strongly reminisent of prtil differentition; sometimes this is lled prtil integrtion. Therefore, it is plusible tht the volume of the solid we re interested in n be obtined by integrting the ross-setionl re A(x) with respet to the remining vrible x. Tht is, we should expet f(x, y) da = A(x) dx = ( d ) f(x, y) dy dx. From now on, we will drop the prentheses whih enlose the inner integrl. We ll suh n expression for double integrl, where we tke n integrl with respet to one vrible first, nd then with respet to the remining vrible, n iterted integrl. In the exmple we were looking t, then, we should hve x 2 + y 2 da = x dx = x + 9x 2 = = 26. There is nothing speil bout tking ross-setions by fixing x. We ould just s well hve tken ross-setions by fixing y, nd obtined formul for the rosssetionl re t y: A(y) = nd then integrted this formul over y: d A(y) dy = d f(x, y) dx f(x, y) dx dy. One probbly expets tht the order of integrtion should not hnge the finl nswer. While this is not lwys true (see Problem 7 in Chpter 6.2), the following theorem tells us tht this is true in virtully every sitution we will enounter: Theorem. (Fubini s Theorem) If f(x, y) is ontinuous on retngle = [, b] [, d], then f(x, y) da = d f(x, y) dy dx = d f(x, y) dx dy More generlly, this is true whenever f(x, y) is bounded on, ontinuous on exept possibly t finite number of smooth urves, nd both iterted integrls exist. You n hek tht the hypotheses of Fubini s Theorem hold true for bsilly every exmple we will look t, nd if we enounter ny situtions where Fubini s Theorem does not hold (whih is rther unlikely) this will be expliitly noted. Notie tht one we write n iterted integrl down, the ordering of dx dy or dy dx determines the order in whih we should perform the prtil integrtions. Py very lose ttention to the ordering of the differentils, nd remember tht you begin by integrting with respet to the left-most vrible nd then work your wy to the right. (While we only hve two differentils right now, we will be performing triple integrls in the future.)

6 6 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES Exmples. Let = [, 2] [, ]. Evlute the double integrl xe y da using both possible orders of integrtion. Sine Fubini s Theorem is obviously vlid, we n write this double integrl s either of the iterted integrls xe y dy dx, The former integrl is equl to xe y y= dx = y= The ltter integrl is equl to e y x2 x=2 dy = 2 x= x(e ) dx = xe y dx dy. (e )x2 2 2 = 2(e ). 2e y dy = 2e y = 2(e ). As expeted, these two iterted integrls re equl to eh other. Sometimes it is esier to integrte with respet to one vrible first insted of the other vrible. For exmple, let = [, π] [, ], nd evlute the double integrl x os(xy) da. Whih vrible is it esier to integrte with respet to first? If we wnt to integrte with respet to x, we will need to perform n integrtion by prts. However, if we integrte with respet to y, we need only use quik u-substitution, u = xy. Then du = x dy, nd we get π π ( ) π x os(xy) da = x os(xy) dy dx = sin(xy) dx = sin x dx = os x y= y= While in priniple it does not mtter whih vrible you integrte with respet to first, in prtie it n be omputtionlly esier to integrte with respet to one vrible first insted of using the other vrible. Of ourse, there is nothing speil bout the vribles x, y. For exmple, we n evlute n iterted integrl The first integrtion gives ( (u + v) ) v= du = v= This integrl is equl to (u + ) 4 (u + ) (u + v) 2 dv du. (u + ) (u ) du. ( ) = ( )4 = = 2 π = 2.

7 INTEGATION OF FUNCTIONS OF SEVEAL VAIABLES 7 If you wnt, you n hek tht this is equl to the nswer you would hve found hd you integrted with respet to u first insted of v.