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. 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.

2 For reasonably nice functions this doesindeed 15 Multiple Integrationcy1y2y3da x1x2x3x4x5b x yFigure rectangular subdivision of [a, b] [c, d].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].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.

3 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. 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.

4 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. 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.)

5 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. 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.

6 (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. 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;in some cases, one direction might be much easier than the other, so it s usually worthconsidering the two different we will be interested in a region that is not simply a rectangle.

7 Let scompute the volume under the surfacex+ 2y2above the region described by 0 x 1and 0 y x2, shown in figure principle there is nothing more difficult about this problem. If we imagine the three-dimensional region under the surface and above the parabolic region as an oddly shapedloaf of bread, we can still slice it up, approximate the volume of each slice, and add Volume and Average Height3910101 Figure parabolic region of up. For example, if we slice perpendicular to thexaxis atxi, the thickness of aslice will be xand the area of the slice will beZx2i0xi+ we add these up and take the limit as xgoes to 0, we get the double integralZ10Zx20x+ 2y2dy dx=Z10xy+23y3 x20dx=Z10x3+23x6dx=x44+221x7 10=14+221= could just as well slice the solid perpendicular to theyaxis, in which case we getZ10Z1 yx+ 2y2dx dy=Z10x22+ 2y2x 1 ydy=Z1012+ 2y2 y2 2y2 y dy=y2+23y3 y24 47y7/2 10=12+23 14 47= is the average height of the surface over this region? Asbefore, it is the volumedivided by the area of the base, but now we need to use Integration to compute the area392 Chapter 15 Multiple Integrationof the base, since it is not a simple rectangle.

8 The area isZ10x2dx=13,so the average height is 29 the volume under the surfacez=p1 x2and above thetriangle formed byy=x,x= 1, and s consider the two possible ways to set this up:Z10Zx0p1 x2dy dxorZ10Z1yp1 x2dx appears easier? In the first, the inner integral is easy, because we need an anti-derivative with respect toy, and the entire integrandp1 x2is constant with respect toy. Of course, the outer integral may be more difficult. In the second, the inner integralis mildly unpleasant a trig substitution. So let s try the first one, since the first step iseasy, and see where that leaves x2dy dx=Z10yp1 x2 x0dx=Z10xp1 is quite easy, since the substitutionu= 1 x2works:Zxp1 x2dx= 12Z u du= 13u3/2= 13(1 x2)3 x2dx= 13(1 x2)3/2 10= is a good example of how the order of Integration can affect the complexity of theproblem. In this case it is possible to do the other order, butit is a bit messier. Insome cases one order may lead to a very difficult or impossible integral; it s usually worthconsidering both possibilities before going very Volume and Average Height393 Exercises 20 401 +x dy dx.

9 1 1 20x+y dy dx. 21 y0xy dx dy. 10 yy2/2dx dy. 21 x1x2y2dy dx. 10 x20yexdy dx. /20 x20xcosy dy dx. /20 cos 0r2(cos r)dr d . : 10 1 y x3+ 1dx dy. : 10 1y2ysin(x2)dx dy. : 10 1x2x 1 +y2dy dx : 10 y02 1 x2dx dy : 10 33yex2dx dy 1 1 1 x20x2 y dy dx. 2/20 1 2x2 1 2x2x dy dx. x2dAover the region in the first quadrant bounded by the hyperbolaxy= 16and the linesy=x,y= 0, andx= 8. the volume belowz= 1 yabove the region 1 x 1, 0 y 1 x2. the volume bounded byz=x2+y2andz= 4. the volume in the first octant bounded byy2= 4 xandy= 2z. the volume in the first octant bounded byy2= 4x, 2x+y= 4,z=y, andy= 0. 394 Chapter 15 Multiple the volume in the first octant bounded byx+y+z= 9, 2x+ 3y= 18, andx+ 3y= 9. the volume in the first octant bounded byx2+y2=a2andz=x+y. the volume bounded by 4x2+y2= 4zandz= 2. the volume bounded byz=x2+y2andz=y. the volume under the surfacez=xyabove the triangle with vertices (1,1,0), (4,1,0),(1,2,0).

10 The volume enclosed byy=x2,y= 4,z=x2,z= 0. swimming pool is circular with a 40 meter diameter. The depth is constant along east-westlines and increases linearly from 2 meters at the south end to 7 meters at the north the volume of the pool. the average value off(x, y) =ey x+eyon the rectangle with vertices (0,0), (4,0),(4,1) and (0,1). shows a temperature map of Colorado. Use the data to estimate the averagetemperature in the state using 4, 16 and 25 subdivisions. Give both an upper and lowerestimate. Why do we like Colorado for this problem? What other state(s)might we like?Figure cylinders of radius 1 intersect at right angles at the origin, as shown in figure the volume contained inside all three cylinders. that iff(x, y) is integrable and ifg(x, y) = xa ybf(s, t)dt dsthengxy=gyx=f(x, y). the order of Integration on each of the following integralsa. 90 9 y0f(x, y)dx dyb. 21 lnx0f(x, y)dy Double Integrals in Cylindrical Coordinates395 Figure of three cylinders.