Integrals of Functions of Several Variables

Transcription

1 Itegrals of Fuctios of Several Variables We ofte resort to itegratios i order to deterie the exact value I of soe quatity which we are uable to evaluate by perforig a fiite uber of additio or ultiplicatio operatios. For exaple, we have to evaluate a itegral to fid the area of the regio betwee the graph of fx) = x + 3 ad the iterval 0,. The itegral is x + 3 ) dx 1) 0 We caot deterie this area by calculatig the areas of a fiite uber of rectagles or triagles because the regio is ot etirely eclosed by straight lie segets. I geeral, a itegratio proble ivolves the followig basic ites: 1. Soe quatity I whose exact value we ust deterie, but we caot calculate it by perforig a fiite uber of additios/ultiplicatios.. Soe set S that has a legth, if it is a subset of the real lie R, a area if it is a subset of the plae, or a volue if it is a subset of 3-diecsioal space. A geeral ter for the legth or the area or the volue of a set S is a easure of S. It should be possible to divide S ito a fiite uber of "saller segets" S 1, S,..., S, to be called "eleets of S", which also have easures. I the itegral 1), the set S is the iterval 0,, with legth. We ca subdivide it ito saller subitervals, e.g. equal subitervals of legth l = each. The sybol l is proouced "delta l" which you should thik of as a "sall legth". 3. Soe fuctio f defied o S such that the product of the easure of a eleet S i ad the value of f at soe poit x i S i gives a reasoable estiate of the cotributio of the eleet S i to the exact value I we wish to deterie. I the itegral 1), the fuctio is fx) = x + 3. If we divide S = 0, ito saller subitervals of legth l = each ad pick a poit x i i a eleet S i = i, i+1) the the product fx i ) l gives a approxiate value of the area eclosed by the seget S i ad the graph of f. Area betwee seget S i ad the graph of f. Approxiatig area Evaluatig the itegral boils dow to doig the followig: Slice S ito a fiite uber of saller eleets S i. Multiply the easure of each eleet S i by the value of f at soe poit x i i the eleet to get a estiate of the cotributio of S i to the value of I. Add up all these cotributios to get a estiate of the value of I. Deterie the liit of all such sus as the easures of the eleets shrik to zero. If the liit it exists, it is by defiitio, the exact value of I ad it is called the itegral of f over S. 1

2 We pla to deote a subset of R, the plae by), R ad a subset of R 3, 3-diesioal space), by B or V. Exaple 1 Let R be the rectagle with vertices at 0,, 0), 5,, 0), 5, 4, 0), 0, 4, 0) ad f be the fuctio of two variables with forula fx, y) = xy + 5. We wish to deterie the volue of the solid, show below), eclosed by the graph of f ad the rectagle. 0,,0) 0,4,0) 5,,0) 5,4,0) We kow how to calculate the volue of ay rectagular box, siply ultiply its legth, width ad height), but this is ot oe of the, therefore we have to "itegrate". More precisely, we have to partitio the rectagle R ito saller rectagles, use the saller rectagles to calculate approxiate values of the required volue the deterie the liit of the approxiatios. The easiest way to partitio R ito saller rectagles is to divide the itervals 0, 5 ad, 4 ito saller subitervals. I the figure below, we divided 0, 5 ito two equal subitervals usig the poits x 0 = 0, x 1 =.5, x = 5 ad, 4 ito 3 equal subitervals usig poits y 0 =, y 1 = 0, y =, y 3 = 4. They partitio R ito 6 saller rectagles R ij = {x, y) : x i x x i+1 ad y j y y j+1, i = 0 or 1 ad j = 0, 1, or. R is partitioed ito saller rectagles.

3 Each oe of these saller rectagles deteries a sall portio of the give solid, show below). The saller solids geerated by the saller rectagles Sice we kow how to calculate volues of rectagular boxes, we approxiate each of the above sall solids with a box that has the sae base ad a suitable height. To siplify the coputatio, we chose to approxiate the sall solid above the rectagle R ij by the box with base R ij ad height fx i, y j ). I geeral, ay poit s i, t j ) i R ij ay be used to get the height of a approxiatig box.) The box approxiatig the solid above R 11 is shaded i the figure below. It has volue.5 fx 1, y 1 ) =.5 f.5, 0) = 15 The total volue of the 6 approxiatig boxes is.5 f0, ) + f0, 0) + f0, ) + f.5, ) + f.5, 0) + f.5, ) = = Therefore Volue of solid I geeral, we ay divide the iterval 0, 5 ito saller subitervals 0, 5, 5, 10,..., 5i, 5i+1),..., 5 1), 5 of equal legth x = 5, to siplify coputatios), ad divide, 4 ito saller subitervals, + 6, + 6, + 1,..., + 6j, + 6j+1),..., + 6 1), 6 of 3

4 equal legth y = 6, the for the rectagles { R ij = x, y) : 5i x 5i+1) ad + 6j 6j+1) y +, i = 0,..., 1 ad j = 0,..., 1 They are of the ad each oe has area A ij = x y. Deote by P ij the sall solid eclosed by the graph of f ad the rectagle R ij. Sice we kow how to calculate volues of boxes, we approxiate it with a box that has the sae base R ij ad height f, + ). Its volue is f ) 5i, + Aij = + 6j ) = + ) The su of these volues is 1 1 f ), + Rij = 1 1 Therefore the required volue is approxiately equal to 1 By defiitio, = =. Secodly, {{ 1 = i) 1 ter s 300 1) = 1 I case you are puzzled, we ade use of the idetity i = 1 1 i=1 j=1 = = i) 1 j 900 ) ) 1) 1) Therefore the volue of the solid is approxiately equal to + ) = ) 300 = ). 1). Fially, = ) 1 1 ) ) ) 1 1 ) The liit of this su as the widths x i ad legths y j of the rectagles R ij shrik to 0 is This ust be the volue of the solid ) + 5 1) 1) = 85 We take this opportuity to itroduce soe otatios ad teriology. The su 1 1 f ), + Rij = is called a Riea su of f deteried by the rectagles { R ij = x, y) : 5i x 5i+1) ad + 6j 6j+1) y , i = 0,..., 1 ad j = 0,..., 1 4

5 The liit of these Riea sus as all the x i s ad y j s shrik to 0 is called the Riea itegral of f over the set R ad is deoted by fx, y)da. 1 Why two itegral sigs? Because the itegral is the liit of a double suatio R 1 f ), + Aij. Ad why da? Because we partitioed R ito eleets, i this case rectagles), with areas A ij. Riea Itegral of a Fuctio of two variable We ow geeralize the above costructio. To this ed, let fx, y) be a give fuctio of two variables ad R be a set i the plae which we ay assue to be a rectagle. If it is ot a rectagle, siply eclose it i a suitable rectagle R ad defie f to have value zero outside R. Divide the rectagle ito saller rectagles R ij which we ay assue, for siplicity, to have the sae legth x ad the sae width y, hece the sae area A ij = x y. Let θ i, α j ) be a poit i the rectagle R ij. The su i=1 j=1 f θ i, α j ) A ij. ) is called a Riea su of f. The liit of these Riea sus as x ad y ted to 0, assuig the liit exists), is called the Riea itegral of f over R ad it is deoted by fx, y)da. R It ay be viewed as the volue of the solid eclosed by the graph of f ad the set R. As explaied i Exaple 1, we use two itegral sigs because we deterie the liit of a double suatio. For this reaso, it is ofte called a double itegral. The sybol da serves to poit out that the set R was divided ito eleets R ij with area A ij. 5