such that the S i cover S, or equivalently S

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1 MATH 55 Triple Integrls Fll Definition Given solid in spce, prtition of consists of finite set of solis = { 1,, n } such tht the i cover, or equivlently n i. Furthermore, for ech i, intersects i or i. i=1 Finlly i j for i j is either empty or surfce. A mesh of is number m such tht ech i is contined in bll of rdius m (the center of this bll cn chnge for ech i ). Next pick point p i = (x i, y i, z i ) in ech i. Then the iemnn sum ssocited to the prtition, the points p i nd the function f(x, y, z) is n (, {p i }, f) = f(x i, y i, z i ) volume( i ) provided this limit exists. i=1 lim (, {p i}, f) mesh A bsic result due to iemnn is the following If f is continuous on nd if is closed nd bounded, then Other bsic results which follow from the definition. If f nd g stisfy f g on nd if f(x, y, z) dv nd f(x, y, z) dv g(x, y, z) dv The integrls re equl if nd only if the functions re equl. A corollry of this result is tht if m f(x, y, z) M on then m volume() f(x, y, z) dv M volume() provided f(x, y, z) dv exists. If f nd g re defined on nd if f(x, y, z) + g(x, y, z) dv = If f is defined on, if c is constnt, nd if f(x, y, z) dv nd c c f(x, y, z) dv + f(x, y, z) dv exists. g(x, y, z) dv exist, then g(x, y, z) dv exist, then f(x, y, z) dv exists, then f(x, y, z) dv g(x, y, z) dv If = 1 nd if 1 is contined in surfce, then f(x, y, z) dv + f(x, y, z) dv 1 1

2 provided f(x, y, z) dv nd 1 f(x, y, z) dv exist. Midpoint ule: If is prtitioned into boxes x < x 1 < < x n, y < y 1 < < y m nd z < z 1 < < z r nd then n m r f(x, y, z) dv f( x i, ȳ j, z k ) x i x i 1 y j y j 1 z j z j 1 i=1 where x i = x i + x i 1, ȳ j = y j + y j 1 j=1 k=1 nd z k = z k + z k 1. Averge Vlue By definition, the verge vlue of function f on solid is 1 f(x, y, z) dv volume()

3 . Iterted integrls An iterted (triple) integrl is n expression of the form b t(x) b(x,y) b(x) (x,y) f(x, y, z) dz dy dx There will be other vritions. ix of them cn be obtined by permuting the order of the vribles. Others come from using different coordinte systems. It is lso useful in setting up triple integrls s iterted integrls to let be the region defined by b t(x) b(x) dy dx nd observe tht b t(x) b(x,y) b(x) (x,y) f(x, y, z) dz dy dx = ( ) b(x,y) (x,y) The outer double integrl is n ordinry double integrl so if you hve 3 ( ) b(x,y) you know how to get the corresponding iterted integrl. The new mteril is to work out wht b(x,y) (x,y) f(x, y, z) dz is nd you cn certinly guess the correct nswer. Here (x, y) nd b(x, y) re functions of x nd y where usully (x, y) b(x, y) for (x, y) but the inequlity is not necessry to work out the nswer so don t bother checking it t this point. All you do is do first yer clculus definite integrl treting x n y s constnts. In more detil, by the Fundmentl Theorem of Clculus, you need to find F (x, y, z) such tht F (x, y, z) = f(x, y, z) z nd then b(x,y) f(x, y, z) dz = F ( x, y, b(x, y) ) F ( x, y, (x, y) ) Notice s promised tht (x,y) b(x,y) (x,y) is just n ordinry double integrl. (x,y) f(x, y, z) dz is function of x nd y so the outer double integrl

4 4 3. etting up iterted integrls The gol is to reduce triple integrl to n iterted integrl. As is usul in this sort of problem, the function is irrelevnt. trt with solid nd pick coordinte plne, sy the xy plne. You re looking for the region in the xy plne with the following property. A point (x, y) if nd only if there is n intervl [α, ω] such tht the verticl line through the point (x, y), sy x, y, + t,, 1 intersects in the segment x, y, + t,, 1, α t ω. In prticulr, if (x, y) /, the line x, y, + t,, 1 does not intersect the solid t ll. For ech vlue of (x, y) in, the point x, y, α is the lowest point in which intersects the line. Hence α is function of (x, y) nd we write α(x, y). imilrly, x, y, ω is the highest point in which intersects the line so ω = ω(x, y). To proceed YOU need to find formuls for α(x, y) nd ω(x, y) nd then ( ) ω(x,y) α(x,y) The region is clled the projection of the solid into the xy plne. Hence given solid you need to determine: (1) The projection of into coordinte plne (your choice). () The functions α nd ω Projections. As we discussed erlier in the semester, projections of cylinders re esy. A cylinder is curve in coordinte plne nd consists of the set of ll lines perpendiculr to the coordinte plne pssing through the curve. xmples of cylinders: Inside x + y = 9 Inside x y = 1, y = 1 nd y =

5 3.. qutions. The solid we re going to discuss hs projection where is one of the two regions in the lst subsection nd it is the prt of 3-spce bove z = x y nd below z = x + y 8. 5 Here is picture of the cse of where is the disk. The blue surfce is the grph of z = x +y 8 while the white one is z = x y. If x + y 9, 3 x 3 nd 3 y 3. Hence x + y 18 nd x + y 18 so inside the yellow disk, z = x + y 8 is negtive nd z = x y is positive so t ny point (x, y) in, the intervl [x + y 8, x y ] lies in. Hence ll of the yellow disk lies between these two grphs nd ( ) x y x +y 8 To work out n exmple to the end, suppose we wnt to find the volume. Then f(x, y, z) = 1. ( ) x y 1 dv = 1 dz da = z da = 3(16 x y ) da = 3 x +y 8 π 4 (16 r )r dr dθ = 6π 6π (18 64) = 6 64π = 384π z= x y z=x +y 8 (8r r4 4 ) 4 =

6 6 For nother exmple, consider the surfce T bove z = x +y 8 nd below z = x y. At first there seems to be no obvious region. ch grph projects into the entire xy plne so these projections re not the issue. However, the solid is the set of ll points x + y 8 x y. To understnd this inequlity, consider the equlity x + y 8 = x y, or x + y = 16. The circle divides the plne into two regions nd inside ech region either x + y 8 x y or x + y 8 x y by continuity. At the origin (, ), x +y 8 < x y so the disk stisfies x +y 8 x y. A point in the outside region is (5, ) nd therefore x +y 8 x y in the entire outside region. Hence the disk x + y 16 is the set of ll (x, y) such tht x + y 8 x y. Then T x +y 16 ( ) x y x +y 8 We will evlute the triple integrl 4. Type? solids f(x, y, z) dv where is the tetrhedron below the plne x + y + 3z = 1 nd in the first octnt. We will do this in three wys. The definition in the book of Type? solids re simply those for which the projection into coordinte plne cn be worked out. In prticulr, we will see tht the solid is Type 1, Type nd Type 3. 1 x y If we project into the xy plne we need to find ll x, y such tht z =. 3 The line x + y = 1 divides the first qudrnt into two pieces. The point (, ) lies in the tringle nd z > there so the tringle is the projection. Hence the projection into the xy plne is the tringle in the first qudrnt x + y = 1, denoted T 1. 1 x 3z If we project into the xz plne we need to find ll x, z such tht y =. The line x + 3z = 1 divides the first qudrnt into two pieces. The point (, ) lies in the tringle nd y > there so, the tringle is the projection. Hence the projection into the xz plne is the tringle in the first qudrnt x + 3z = 1, denoted T. If we project into the yz plne we need to find ll y, z such tht x = 1 y 3z. The line y + 3z = 1 divides the first qudrnt into two pieces. The point (, ) lies in the tringle nd x > there so, the tringle is the projection. Hence the projection into the yz plne is the tringle in the first qudrnt y + 3z = 1, denoted T 3. Then T 1 = = ( 1 x y 3 ( 1 x 3z T ( 1 y 3z T 3 f(x, y, z) dz f(x, y, z) dy ) ) da da ) f(x, y, z) dx da

7 Notice tht this gives triple iterted integrl problems where the integrl my be hrd to do in one setup but esier in nother From iterted to triple Just s in the -dimensionl cse, given n iterted integrl it is pretty esy to work out description of the solid. trt with b b(x) ω(x,z) (x) α(x,z) f(x, y, z)dy dz dx nd describe the solid over which we re integrting. We hve projected into the xz plne nd there we hve the region bove z = (x), below z = b(x) nd x b. Over the region in the xz plne we re bove the grph y = α(x, z) nd below the grph y = ω(x, z). id nother wy = { (x, y, z) } x b, (x) z b(x), α(x, z) y ω(x, z) Of course if you re ever lucky enough to get solid described to you by three such inequlities, the setup of one iterted integrl is immedite. Wrning: You my choose, b, (x), b(x), α(x, z) nd ω(x, z) rbitrrily nd the iterted integrl mkes sense if the functions re continuous. However, the inequlities bove my not describe solid. Built into the nottion re the ssumptions tht < b; tht for ny x [, b], (x) b(x); nd for ny (x, z) with x [, b] nd (x) z b(x), α(x, z) ω(x, z). 6. A philosophiclly stisfying lternte pproch One cn lso set up n iterted integrl s follows. As usul, let be our solid nd this time, pick n xis, sy z. ince is bounded, there is number such tht the plne z = just touches the solid from underneth. There is lso b such tht the plne z = b just touches the solid from bove. For n rbitrry z between nd b, the plne t tht height intersects the solid in region (z) (the region chnges s you chnge z). Then b f(x, y, z) da dz This method is less common becuse for some reson people re not so fond of vrying regions. It cn be useful in solving the rewriting problem s in xmple 4 from the book (pge 145). Note tht even with the 7th edition there re still typos. (ee the coordintes of the regions.) The prt tht I don t like bout the exmple is tht the book just sserts the projections re wht they re without much help to the reder. But figuring out the projections is the hrd prt! Anywy, we strt with (z) 1 x y f(x, y, z) dz dy dx To switch the outer two vribles is esy (or t lest it is problem you hve lredy studied). y where is

8 8 This is esy to set up the other wy: y = 1 1 y y f(x, y, z) dz dx dy A hrder problem is to switch the two inner vribles, but here the lternte pproch to setup is useful. Let us switch x nd z in this lst integrl. 1 1 y y f(x, y, z) dz dx dy = 1 ( ) f(x, y, z) da dy (y) Here (y) is the rectngle in the xz plne where x runs between nd 1 nd z runs between nd y. ince this is rectngle, it is esy to switch the order of integrtion nd get 1 1 y 1 y 1 f(x, y, z) dz dx dy = f(x, y, z) dx dz dy y which is the nswer in the book. If you now wnt to switch z nd y in this lst integrl, this is the esy version. 1 y 1 1 f(x, y, z) dx dz dy = f(x, y, z) dx da where T is the region in the yz plne, y 1, z y. T Hence 1 y 1 f(x, y, z) dx dz dy = T 1 f(x, y, z) dx da = z f(x, y, z) dx dy dz

9 9 To switch x nd y in this lst integrl, write z where (z) is the region f(x, y, z) dx dy dz = 1 (z) f(x, y, z) da dz eversing the order on (z) gives z f(x, y, z) dx dy dz = 1 f(x, y, z) da dz = (z) 1 1 z x z f(x, y, z) dy dx dz