Multiple Integration - Whitman College

1 15 Multiple a surfacef(x, y); you might temporarily think of this as representing physicaltopography a hilly landscape, perhaps. What is the averageheight of the surface (oraverage altitude of the landscape) over some region?As with most such problems, we start by thinking about how we might approximatethe answer. Suppose the region is a rectangle, [a, b] [c, d]. We can divide the rectangleinto a grid,msubdivisions in one direction andnin the other, as indicated in figure pickxvaluesx0,x1,..,xm 1in each subdivision in thexdirection, and similarly intheydirection.

2 At each of the points (xi, yj) in one of the smaller rectangles in the grid,we compute the height of the surface:f(xi, yj). Now the average of these heights shouldbe (depending on the fineness of the grid) close to the averageheight of the surface:f(x0, y0) +f(x1, y0) + +f(x0, y1) +f(x1, y1) + +f(xm 1, yn 1) bothmandngo to infinity, we expect this approximation to converge to a fixedvalue, the actual average height of the surface. For reasonably nice functions this doesindeed 15 Multiple Integrationcy1y2y3da x1x2x3x4x5b x yFigure rectangular subdivision of [a, b] [c, d].

3 Using sigma notation, we can rewrite the approximation:1mnn 1Xi=0m 1Xj=0f(xj, yi) =1(b a)(d c)n 1Xi=0m 1Xj=0f(xj, yi)b amd cn=1(b a)(d c)n 1Xi=0m 1Xj=0f(xj, yi) x two parts of this product have useful meaning: (b a)(d c) is of course the area ofthe rectangle, and the double sum adds upmnterms of the formf(xj, yi) x y, which isthe height of the surface at a point times the area of one of thesmall rectangles into whichwe have divided the large rectangle. In short, each termf(xj, yi) x yis the volume of atall, thin, rectangular box, and is approximately the volume under the surface and aboveone of the small rectangles; see figure When we add allof these up, we get anapproximation to the volume under the surface and above the rectangleR= [a, b] [c, d].

4 When we take the limit asmandngo to infinity, the double sum becomes the actualvolume under the surface, which we divide by (b a)(d c) to get the average sums like this come up in many applications, so in a wayit is the most impor-tant part of this example; dividing by (b a)(d c) is a simple extra step that allows thecomputation of an average. As we did in the single variable case, we introduce a specialnotation for the limit of such a double sum:limm,n n 1Xi=0m 1Xj=0f(xj, yi) x y=ZZRf(x, y)dx dy=ZZRf(x, y)dA,thedouble integraloffover the regionR.

5 The notationdAindicates a small bit ofarea, without specifying any particular order for the variablesxandy; it is shorter Volume and Average Height387 Figure the volume under a generic than writingdx dy. The average height of the surface in this notation is1(b a)(d c)ZZRf(x, y) next question, of course, is: How do we compute these double integrals? Youmight think that we will need some two-dimensional version of the Fundamental Theoremof Calculus, but as it turns out we can get away with just the single variable version,applied back to the double sum, we can rewrite it to emphasize a particular order inwhich we want to add the terms:n 1Xi=0 m 1Xj=0f(xj, yi) x the sum in parentheses, only the value ofxjis changing;yiis temporarily constant.

6 Asmgoes to infinity, this sum has the right form to turn into an integral:limm m 1Xj=0f(xj, yi) x=Zbaf(x, yi) after we take the limit asmgoes to infinity, the sum isn 1Xi=0 Zbaf(x, yi)dx! 15 Multiple IntegrationOf course, for different values ofyithis integral has different values; in other words, it isreally a function applied toyi:G(y) =Zbaf(x, y) we substitute back into the sum we getn 1Xi=0G(yi) sum has a nice interpretation. The valueG(yi) is the area of a cross section of theregion under the surfacef(x, y), namely, wheny=yi.

7 The quantityG(yi) ycan beinterpreted as the volume of a solid with face areaG(yi) and thickness y. Think of thesurfacef(x, y) as the top of a loaf of sliced bread. Each slice has a cross-sectional area anda thickness;G(yi) ycorresponds to the volume of a single slice of bread. Adding theseup approximates the total volume of the loaf. (This is very similar to the technique weused to compute volumes in section , except that there we need the cross-sections to bein some way the same .) Figure shows this sliced loaf approximation using thesame surface as shown in figure Nicely enough, this sum looks just like the sort ofsum that turns into an integral, namely,limn n 1Xi=0G(yi) y=ZdcG(y)dy=ZdcZbaf(x, y)dx s be clear about what this means: we first will compute theinner integral, temporarilytreatingyas a constant.

8 We will do this by finding an anti-derivative with respect tox, then substitutingx=aandx=band subtracting, as usual. The result will be anexpression with noxvariable but some occurrences ofy. Then the outer integral will bean ordinary one-variable problem, withyas the shows the function sin(xy)+6/5 on [ , ] [ , ].The volume under this surface (xy) +65dx inner integral (xy) +65dx= cos(xy)y+6x5 cos( )y+cos( )y+ , this gives a function for which we can t find asimple anti-derivative. Tocomplete the problem we could use Sage or similar software toapproximate the Volume and Average Height389 Figure the volume under a surface with slices.

9 (AP)Doing this gives a volume of approximately , so the average height is addition and multiplication are commutative and associative, we can rewritethe original double sum:n 1Xi=0m 1Xj=0f(xj, yi) x y=m 1Xj=0n 1Xi=0f(xj, yi) y if we repeat the development above, the inner sum turns into an integral:limn n 1Xi=0f(xj, yi) y=Zdcf(xj, y)dy,and then the outer sum turns into an integral:limm m 1Xj=0 Zdcf(xj, y)dy! x=ZbaZdcf(x, y)dy other words, we can compute the integrals in either order,first with respect toxtheny, or vice versa.

10 Thinking of the loaf of bread, this corresponds to slicing the loaf in adirection perpendicular to the haven t really proved that the value of a double integral is equal to the value of thecorresponding two single integrals in either order of Integration , but provided the functionis reasonably nice, this is true; the result is calledFubini s 15 Multiple IntegrationEXAMPLE computeZZR1 + (x 1)2+ 4y2dA, whereR= [0,3] [0,2], intwo ,Z30Z201 + (x 1)2+ 4y2dy dx=Z30y+ (x 1)2y+43y3 20dx=Z302 + 2(x 1)2+323dx= 2x+23(x 1)3+323x 30= 6 +23 8 +323 3 (0 1 23+ 0)= the other order:Z20Z301 + (x 1)2+ 4y2dx dy=Z20x+(x 1)33+ 4y2x 30dy=Z203 +83+ 12y2+13dy= 3y+83y+ 4y3+13y 20= 6 +163+ 32 +23= this example there is no particular reason to favor one direction over the other.