MULTIPLE INTEGRALS. A double integral of a positive function is a volume, which is the limit of sums of volumes of rectangular columns.

Transcription

1 5 MULTIPL INTGALS A double integrl of positive function is volume, which is the limit of sums of volumes of rectngulr columns. In this chpter we etend the ide of definite integrl to double nd triple integrls of functions of two or three vribles. These ides re then used to compute volumes, msses, nd centroids of more generl regions thn we were ble to consider in Chpters 6 nd 8. We lso use double integrls to clculte probbilities when two rndom vribles re involved. We will see tht polr coordintes re useful in computing double integrls over some tpes of regions. In similr w, we will introduce two new coordinte sstems in three-dimensionl spce clindricl coordintes nd sphericl coordintes tht gretl simplif the computtion of triple integrls over certin commonl occurring solid regions. 95

2 5. OUBL INTGALS OV CTANGLS In much the sme w tht our ttempt to solve the re problem led to the definition of definite integrl, we now seek to find the volume of solid nd in the process we rrive t the definition of double integrl. VIW OF TH FINIT INTGAL First let s recll the bsic fcts concerning definite integrls of functions of single vrible. If f is defined for b, we strt b dividing the intervl, b into n subintervls i, i of equl width b n nd we choose smple points * i in these subintervls. Then we form the iemnn sum nd tke the limit of such sums s n l to obtin the definite integrl of f from to b: n f * i i b f d lim n l n f * i i In the specil cse where f, the iemnn sum cn be interpreted s the sum of the res of the pproimting rectngles in Figure, nd b f d represents the re under the curve f from to b. Î f(* i ) i- i n- b FIGU * * * * * i n VOLUMS AN OUBL INTGALS =f(, ) In similr mnner we consider function f of two vribles defined on closed rectngle, b c, d, b, c d b c d nd we first suppose tht f,. The grph of f is surfce with eqution f,. Let S be the solid tht lies bove nd under the grph of f, tht is, S,, 3 f,,, FIGU (See Figure.) Our gol is to find the volume of S. The first step is to divide the rectngle into subrectngles. We ccomplish this b dividing the intervl, b into m subintervls i, i of equl width b m nd dividing c, d into n subintervls j, j of equl width d cn. B drw- 95

3 95 CHAPT 5 MULTIPL INTGALS ing lines prllel to the coordinte es through the endpoints of these subintervls, s in Figure 3, we form the subrectngles ij i, i j, j, i i, j j ech with re A. d ij ( i, j ) Î j j- (* ij, * ij ) c ( *, * ) FIGU 3 ividing into subrectngles i- Î i b If we choose smple point ij *, ij * in ech ij, then we cn pproimte the prt of S tht lies bove ech ij b thin rectngulr bo (or column ) with bse ij nd height f ij *, ij * s shown in Figure 4. (Compre with Figure.) The volume of this bo is the height of the bo times the re of the bse rectngle: f ij *, ij * A If we follow this procedure for ll the rectngles nd dd the volumes of the corresponding boes, we get n pproimtion to the totl volume of S: 3 V m n i j f ij *, ij * A (See Figure 5.) This double sum mens tht for ech subrectngle we evlute f t the chosen point nd multipl b the re of the subrectngle, nd then we dd the results. c f( * ij, * ij ) d b ij FIGU 4 FIGU 5

4 SCTION 5. OUBL INTGALS OV CTANGLS 953 N The mening of the double limit in qution 4 is tht we cn mke the double sum s close s we like to the number V [for n choice of ij *, ij * in ij ] b tking m nd n sufficientl lrge. Our intuition tells us tht the pproimtion given in (3) becomes better s m nd n become lrger nd so we would epect tht 4 V lim m, n l m n i j f ij *, ij * A We use the epression in qution 4 to define the volume of the solid S tht lies under the grph of f nd bove the rectngle. (It cn be shown tht this definition is consistent with our formul for volume in Section 6..) Limits of the tpe tht pper in qution 4 occur frequentl, not just in finding volumes but in vriet of other situtions s well s we will see in Section 5.5 even when f is not positive function. So we mke the following definition. N Notice the similrit between efinition 5 nd the definition of single integrl in qution. 5 FINITION The double integrl of f over the rectngle is if this limit eists. f, da lim m, n l m n i j f ij *, ij * A N Although we hve defined the double integrl b dividing into equl-sied subrectngles, we could hve used subrectngles ij of unequl sie. But then we would hve to ensure tht ll of their dimensions pproch in the limiting process. The precise mening of the limit in efinition 5 is tht for ever number there is n integer N such tht f, da m n i j f ij *, ij * A for ll integers m nd n greter thn N nd for n choice of smple points ij *, ij * in ij. A function f is clled integrble if the limit in efinition 5 eists. It is shown in courses on dvnced clculus tht ll continuous functions re integrble. In fct, the double integrl of f eists provided tht f is not too discontinuous. In prticulr, if f is bounded [tht is, there is constnt M such tht f, M for ll, in ], nd f is continuous there, ecept on finite number of smooth curves, then f is integrble over. The smple point ij *, ij * cn be chosen to be n point in the subrectngle ij, but if we choose it to be the upper right-hnd corner of ij [nmel i, j, see Figure 3], then the epression for the double integrl looks simpler: 6 f, da lim m, n l m n i j f i, j A B compring efinitions 4 nd 5, we see tht volume cn be written s double integrl: If f,, then the volume V of the solid tht lies bove the rectngle nd below the surfce f, is V f, da

5 954 CHAPT 5 MULTIPL INTGALS The sum in efinition 5, m n i j f ij *, ij * A is clled double iemnn sum nd is used s n pproimtion to the vlue of the double integrl. [Notice how similr it is to the iemnn sum in () for function of single vrible.] If f hppens to be positive function, then the double iemnn sum represents the sum of volumes of columns, s in Figure 5, nd is n pproimtion to the volume under the grph of f nd bove the rectngle. (, ) (, ) (, ) (, ) FIGU 6 6 =6- - V XAMPL stimte the volume of the solid tht lies bove the squre,, nd below the elliptic prboloid 6. ivide into four equl squres nd choose the smple point to be the upper right corner of ech squre. Sketch the solid nd the pproimting rectngulr boes. ij SOLUTION The squres re shown in Figure 6. The prboloid is the grph of f, 6 nd the re of ech squre is. Approimting the volume b the iemnn sum with m n, we hve V i j f i, j A f, A f, A f, A f, A This is the volume of the pproimting rectngulr boes shown in Figure 7. M We get better pproimtions to the volume in mple if we increse the number of squres. Figure 8 shows how the columns strt to look more like the ctul solid nd the corresponding pproimtions become more ccurte when we use 6, 64, nd 56 squres. In the net section we will be ble to show tht the ect volume is 48. FIGU 7 FIGU 8 The iemnn sum pproimtions to the volume under =6- - become more ccurte s m nd n increse. () m=n=4, VÅ4.5 (b) m=n=8, VÅ (c) m=n=6, VÅ V XAMPL If,,, evlute the integrl s da

6 SCTION 5. OUBL INTGALS OV CTANGLS 955 S FIGU 9 (,, ) (,, ) (,, ) SOLUTION It would be ver difficult to evlute this integrl directl from efinition 5 but, becuse s, we cn compute the integrl b interpreting it s volume. If s, then nd, so the given double integrl represents the volume of the solid S tht lies below the circulr clinder nd bove the rectngle. (See Figure 9.) The volume of S is the re of semicircle with rdius times the length of the clinder. Thus s da 4 M TH MIPOINT UL The methods tht we used for pproimting single integrls (the Midpoint ule, the Trpeoidl ule, Simpson s ule) ll hve counterprts for double integrls. Here we consider onl the Midpoint ule for double integrls. This mens tht we use double iemnn sum to pproimte the double integrl, where the smple point ij *, ij * in ij is chosen to be the center i, j of ij. In other words, i is the midpoint of i, i nd j is the midpoint of j, j. MIPOINT UL FO OUBL INTGALS f, da m n i j f i, j A where i is the midpoint of i, i nd is the midpoint of j, j. j 3 FIGU (, ) V XAMPL 3 Use the Midpoint ule with to estimte the vlue of the integrl m n 3 da, where,,. SOLUTION In using the Midpoint ule with m n, we evlute f, 3 t the centers of the four subrectngles shown in Figure. So,, 5 3, nd. The re of ech subrectngle is A Thus 3 da i j f, A f, A f, A f, A f (, 5 4) A f (, 7 4) A f ( 3, 5 4) A f ( 3, 7 4) A ( 67 6) ( 39 6 ) ( 5 6) ( 3 6 ) 95 8 f i, j A.875 Thus we hve 3 da.875 M NOT In the net section we will develop n efficient method for computing double integrls nd then we will see tht the ect vlue of the double integrl in mple 3 is. (emember tht the interprettion of double integrl s volume is vlid onl when the integrnd f is positive function. The integrnd in mple 3 is not positive function, so its integrl is not volume. In mples nd 3 in Section 5. we will discuss how to interpret integrls of functions tht re not lws positive in terms of volumes.) If we keep dividing ech subrectngle in Figure into four smller ones with

7 956 CHAPT 5 MULTIPL INTGALS Number of subrectngles Midpoint ule pproimtions similr shpe, we get the Midpoint ule pproimtions displed in the chrt in the mrgin. Notice how these pproimtions pproch the ect vlue of the double integrl, AVAG VALU ecll from Section 6.5 tht the verge vlue of function n intervl, b is f ve b b f d f of one vrible defined on In similr fshion we define the verge vlue of function on rectngle to be where A is the re of. If f,, the eqution f ve A A f ve f, da f, da f of two vribles defined FIGU ss tht the bo with bse nd height f ve hs the sme volume s the solid tht lies under the grph of f. [If f, describes mountinous region nd ou chop off the tops of the mountins t height f ve, then ou cn use them to fill in the vlles so tht the region becomes completel flt. See Figure.] XAMPL 4 The contour mp in Figure shows the snowfll, in inches, tht fell on the stte of Colordo on ecember nd, 6. (The stte is in the shpe of rectngle tht mesures 388 mi west to est nd 76 mi south to north.) Use the contour mp to estimte the verge snowfll for the entire stte of Colordo on those ds FIGU

8 SCTION 5. OUBL INTGALS OV CTANGLS 957 SOLUTION Let s plce the origin t the southwest corner of the stte. Then 388, 76, nd f, is the snowfll, in inches, t loction miles to the est nd miles to the north of the origin. If is the rectngle tht represents Colordo, then the verge snowfll for the stte on ecember ws where A To estimte the vlue of this double integrl, let s use the Midpoint ule with m n 4. In other words, we divide into 6 subrectngles of equl sie, s in Figure 3. The re of ech subrectngle is 76 f ve A f, da A mi FIGU Using the contour mp to estimte the vlue of f t the center of ech subrectngle, we get f, da 4 4 i j f i, j A A Therefore f ve On ecember, 6, Colordo received n verge of pproimtel 3 inches of snow. M

9 958 CHAPT 5 MULTIPL INTGALS POPTIS OF OUBL INTGALS We list here three properties of double integrls tht cn be proved in the sme mnner s in Section 5.. We ssume tht ll of the integrls eist. Properties 7 nd 8 re referred to s the linerit of the integrl. N ouble integrls behve this w becuse the double sums tht define them behve this w. 7 8 f, t, da cf, da c f, da f, da t, da where c is constnt If f, t, for ll, in, then 9 f, da t, da 5. XCISS. () stimte the volume of the solid tht lies below the surfce nd bove the rectngle, 6, 4 Use iemnn sum with m 3, n, nd tke the smple point to be the upper right corner of ech squre. (b) Use the Midpoint ule to estimte the volume of the solid in prt (). (b) stimte the double integrl with m n 4 b choosing the smple points to be the points frthest from the origin If, 3,, use iemnn sum with m 4, n to estimte the vlue of da. Tke the smple points to be the upper left corners of the squres. 3. () Use iemnn sum with m n to estimte the vlue of sin da, where,,. Tke the smple points to be lower left corners. (b) Use the Midpoint ule to estimte the integrl in prt (). 4. () stimte the volume of the solid tht lies below the surfce nd bove the rectngle,, 4. Use iemnn sum with m n nd choose the smple points to be lower right corners. (b) Use the Midpoint ule to estimte the volume in prt (). 5. A tble of vlues is given for function f, defined on, 3, 4. () stimte f, da using the Midpoint ule with m n A -ft-b-3-ft swimming pool is filled with wter. The depth is mesured t 5-ft intervls, strting t one corner of the pool, nd the vlues re recorded in the tble. stimte the volume of wter in the pool. 7. Let V be the volume of the solid tht lies under the grph of f, s5 nd bove the rectngle given b 4, 6. We use the lines 3 nd 4 to

10 SCTION 5. ITAT INTGALS 959 divide into subrectngles. Let L nd U be the iemnn sums computed using lower left corners nd upper right corners, respectivel. Without clculting the numbers V, L, nd U, rrnge them in incresing order nd eplin our resoning. 8. The figure shows level curves of function f in the squre,,. Use the Midpoint ule with m n to estimte f, da. How could ou improve our estimte? A contour mp is shown for function f on the squre, 4, 4. () Use the Midpoint ule with m n to estimte the vlue of f, da. (b) stimte the verge vlue of f. 4. The contour mp shows the temperture, in degrees Fhrenheit, t 4: PM on Februr 6, 7, in Colordo. (The stte mesures 388 mi est to west nd 76 mi north to south.) Use the Midpoint ule with m n 4 to estimte the verge temperture in Colordo t tht time vlute the double integrl b first identifing it s the volume of solid.. 3 da,,, 6. 5 da,, 5, da,,, 4. The integrl s9 da, where, 4,, represents the volume of solid. Sketch the solid. 5. Use progrmmble clcultor or computer (or the sum commnd on CAS) to estimte where,,. Use the Midpoint ule with the following numbers of squres of equl sie:, 4, 6, 64, 56, nd epet ercise 5 for the integrl. 7. If f is constnt function, f, k, nd, b c, d, show tht k da kb d c. 8. Use the result of ercise 7 to show tht where [, 4] [ 4, ]. s e da sin( s ) da sin cos da 3 5. ITAT INTGALS ecll tht it is usull difficult to evlute single integrls directl from the definition of n integrl, but the Fundmentl Theorem of Clculus provides much esier method. The evlution of double integrls from first principles is even more difficult, but in this sec-

11 96 CHAPT 5 MULTIPL INTGALS tion we see how to epress double integrl s n iterted integrl, which cn then be evluted b clculting two single integrls. Suppose tht f is function of two vribles tht is integrble on the rectngle, b c, d. We use the nottion d f, d to men tht is held fied nd c f, is integrted with respect to from c to d. This procedure is clled prtil integrtion with respect to. (Notice its similrit to prtil differentition.) Now f, d is number tht depends on the vlue of, so it defines function of : d c A d f, d If we now integrte the function A with respect to from to b, we get c b A d b d c f, d d The integrl on the right side of qution is clled n iterted integrl. Usull the brckets re omitted. Thus b d c f, d d b d c f, d d mens tht we first integrte with respect to from c to d nd then with respect to from to b. Similrl, the iterted integrl 3 d c b f, d d d c b f, d d mens tht we first integrte with respect to (holding fied) from to b nd then we integrte the resulting function of with respect to from c to d. Notice tht in both qutions nd 3 we work from the inside out. XAMPL vlute the iterted integrls. () 3 d d (b) 3 d d SOLUTION () egrding s constnt, we obtin d 3 Thus the function A in the preceding discussion is given b A 3 in this emple. We now integrte this function of from to 3: 3 d d d 3 d d 3 7

12 SCTION 5. ITAT INTGALS 96 (b) Here we first integrte with respect to : 3 d d 3 d d d 9 d 9 7 M Notice tht in mple we obtined the sme nswer whether we integrted with respect to or first. In generl, it turns out (see Theorem 4) tht the two iterted integrls in qutions nd 3 re lws equl; tht is, the order of integrtion does not mtter. (This is similr to Clirut s Theorem on the equlit of the mied prtil derivtives.) The following theorem gives prcticl method for evluting double integrl b epressing it s n iterted integrl (in either order). N Theorem 4 is nmed fter the Itlin mthemticin Guido Fubini ( ), who proved ver generl version of this theorem in 97. But the version for continuous functions ws known to the French mthemticin Augustin- Louis Cuch lmost centur erlier. b FIGU A() TC Visul 5. illustrtes Fubini s Theorem b showing n nimtion of Figures nd. FIGU c d 4 FUBINI S THOM If f is continuous on the rectngle, b, c d, then More generll, this is true if we ssume tht f is bounded on, f is discontinuous onl on finite number of smooth curves, nd the iterted integrls eist. The proof of Fubini s Theorem is too difficult to include in this book, but we cn t lest give n intuitive indiction of wh it is true for the cse where f,. ecll tht if f is positive, then we cn interpret the double integrl f, da s the volume V of the solid S tht lies bove nd under the surfce f,. But we hve nother formul tht we used for volume in Chpter 6, nmel, where A is the re of cross-section of S in the plne through perpendiculr to the -is. From Figure ou cn see tht A is the re under the curve C whose eqution is f,, where is held constnt nd c d. Therefore nd we hve f, da b f, da V b A d b A similr rgument, using cross-sections perpendiculr to the -is s in Figure, shows tht f, da d c b f, d d d c f, d d d V b A d A d f, d c c d c b f, d d f, d d

13 96 CHAPT 5 MULTIPL INTGALS N Notice the negtive nswer in mple ; nothing is wrong with tht. The function f in tht emple is not positive function, so its integrl doesn t represent volume. From Figure 3 we see tht f is lws negtive on, so the vlue of the integrl is the negtive of the volume tht lies bove the grph of f nd below. V XAMPL vlute the double integrl 3 da, where,,. (Compre with mple 3 in Section 5..) SOLUTION Fubini s Theorem gives 3 da 7 d 7 SOLUTION Agin ppling Fubini s Theorem, but this time integrting with respect to first, we hve d 3 d d [ 3 ] _4 _8 =-3 _ da 3 d d 3 d FIGU 3 6 d 3 ] M V XAMPL 3 vlute sin da, where,,. SOLUTION If we first integrte with respect to, we get sin da sin d d [cos] d cos cos d sin sin ] SOLUTION If we reverse the order of integrtion, we get N For function f tht tkes on both positive nd negtive vlues, f, da is difference of volumes: V V, where V is the volume bove nd below the grph of f nd V is the volume below nd bove the grph. The fct tht the integrl in mple 3 is mens tht these two volumes V nd V re equl. (See Figure 4.) sin da To evlute the inner integrl, we use integrtion b prts with u du d sin d d dv sin d v cos _ FIGU 4 = sin() 3 nd so cos sin d cos cos [sin] sin cos d

14 SCTION 5. ITAT INTGALS 963 If we now integrte the first term b prts with u nd dv cos d, we get du d, v sin, nd cos d sin sin d Therefore cos sin d sin N In mple, Solutions nd re equll strightforwrd, but in mple 3 the first solution is much esier thn the second one. Therefore, when we evlute double integrls, it is wise to choose the order of integrtion tht gives simpler integrls. nd so sin d d sin sin sin M V XAMPL 4 Find the volume of the solid S tht is bounded b the elliptic prboloid 6, the plnes nd, nd the three coordinte plnes SOLUTION We first observe tht S is the solid tht lies under the surfce 6 nd bove the squre,,. (See Figure 5.) This solid ws considered in mple in Section 5., but we re now in position to evlute the double integrl using Fubini s Theorem. Therefore V 6 da [6 3 3 ] d 6 d d FIGU 5 ( ) d [ ] 48 M In the specil cse where f, cn be fctored s the product of function of onl nd function of onl, the double integrl of f cn be written in prticulrl simple form. To be specific, suppose tht f, th nd, b c, d. Then Fubini s Theorem gives In the inner integrl, is constnt, so h is constnt nd we cn write d c b b f, da d c b th d d d h b t d d b t d d h d c th d d d since t d is constnt. Therefore, in this cse, the double integrl of f cn be written s the product of two single integrls: c b th d d c 5 t h da b t d d h d c where, b c, d

15 964 CHAPT 5 MULTIPL INTGALS XAMPL 5 If,,, then, b qution 5, sin cos da sin d cos d [cos ] [sin ] M N The function f, sin cos in mple 5 is positive on, so the integrl represents the volume of the solid tht lies bove nd below the grph of f shown in Figure 6. FIGU 6 5. XCISS Find nd 5 f, d f, d.. f, 3. f, e 8., da,, 3 4 Clculte the iterted integrl sin d d 6. 4 d d d d d d d d 6 5 cos d d e d d e 3 d d 9. sin da,., da. e da,., da, 6, 3,,,,,,. u v 5 du dv. 3. r sin d dr 4. s d d ss t ds dt 3 4 Sketch the solid whose volume is given b the iterted integrl d d 5 Clculte the double integrl da, 6. cos da, 7., 3,,,,,, 3 3 da 4. d d 5. Find the volume of the solid tht lies under the plne 3 nd bove the rectngle,, Find the volume of the solid tht lies under the hperbolic prboloid 4 nd bove the squre,,.

16 SCTION 5.3 OUBL INTGALS OV GNAL GIONS Find the volume of the solid enclosed b the surfce e sin nd the plnes,,, nd. 9. Find the volume of the solid enclosed b the surfce sec nd the plnes,,,, nd Find the volume of the solid in the first octnt bounded b the clinder 6 nd the plne Find the volume of the solid enclosed b the prboloid nd the plnes,,,, nd 4. ; 3. Grph the solid tht lies between the surfce nd the plne nd is bounded b the plnes,,, nd 4. Then find its volume. CAS 7. Find the volume of the solid ling under the elliptic prboloid 4 9 nd bove the rectngle,,. 33. Use computer lgebr sstem to find the ect vlue of the integrl 5 3 e da, where,,. Then use the CAS to drw the solid whose volume is given b the integrl. CAS CAS 34. Grph the solid tht lies between the surfces e cos nd for,. Use computer lgebr sstem to pproimte the volume of this solid correct to four deciml plces Find the verge vlue of f over the given rectngle. 35. f,, hs vertices,,, 5,, 5,, 36. f, e s e, 37. Use our CAS to compute the iterted integrls d d 3 o the nswers contrdict Fubini s Theorem? plin wht is hppening. 38. () In wht w re the theorems of Fubini nd Clirut similr? (b) If f, is continuous on, b c, d nd t,, 4, nd c f s, t dt ds d d 3 for b, c d, show tht t t f,. 5.3 OUBL INTGALS OV GNAL GIONS For single integrls, the region over which we integrte is lws n intervl. But for double integrls, we wnt to be ble to integrte function f not just over rectngles but lso over regions of more generl shpe, such s the one illustrted in Figure. We suppose tht is bounded region, which mens tht cn be enclosed in rectngulr region s in Figure. Then we define new function F with domin b F, f, if if, is in, is in but not in FIGU FIGU

17 966 CHAPT 5 MULTIPL INTGALS If F is integrble over, then we define the double integrl of f over b grph of f f, da F, da where F is given b qution FIGU 3 FIGU 4 grph of F efinition mkes sense becuse is rectngle nd so F, da hs been previousl defined in Section 5.. The procedure tht we hve used is resonble becuse the vlues of F, re when, lies outside nd so the contribute nothing to the integrl. This mens tht it doesn t mtter wht rectngle we use s long s it contins. In the cse where f,, we cn still interpret f, da s the volume of the solid tht lies bove nd under the surfce f, (the grph of f ). You cn see tht this is resonble b compring the grphs of f nd F in Figures 3 nd 4 nd remembering tht F, da is the volume under the grph of F. Figure 4 lso shows tht F is likel to hve discontinuities t the boundr points of. Nonetheless, if f is continuous on nd the boundr curve of is well behved (in sense outside the scope of this book), then it cn be shown tht F, da eists nd therefore f, da eists. In prticulr, this is the cse for the following tpes of regions. A plne region is sid to be of tpe I if it lies between the grphs of two continuous functions of, tht is,, b, t t where t nd t re continuous on, b. Some emples of tpe I regions re shown in Figure 5. =g () =g () =g () =g () =g () =g () b b b FIGU 5 Some tpe I regions =g () d c =g () b FIGU 6 In order to evlute f, da when is region of tpe I, we choose rectngle, b c, d tht contins, s in Figure 6, nd we let F be the function given b qution ; tht is, F grees with f on nd F is outside. Then, b Fubini s Theorem, Observe tht F, if t or t becuse, then lies outside. Therefore d c f, da F, da b F, d t F, d t f, d t d c t F, d d

18 SCTION 5.3 OUBL INTGALS OV GNAL GIONS 967 becuse F, f, when t t. Thus we hve the following formul tht enbles us to evlute the double integrl s n iterted integrl. 3 If f is continuous on tpe I region such tht, b, t t d c d =h () =h () FIGU 7 Some tpe II regions =h () c =h () (_, ) =+ (, ) then The integrl on the right side of (3) is n iterted integrl tht is similr to the ones we considered in the preceding section, ecept tht in the inner integrl we regrd s being constnt not onl in f, but lso in the limits of integrtion, t nd t. We lso consider plne regions of tpe II, which cn be epressed s 4 where h nd h re continuous. Two such regions re illustrted in Figure 7. Using the sme methods tht were used in estblishing (3), we cn show tht 5, c d, h h f, da b f, da d t t where is tpe II region given b qution 4. V XAMPL vlute da, where is the region bounded b the prbols nd. SOLUTION The prbols intersect when, tht is,, so. We note tht the region, sketched in Figure 8, is tpe I region but not tpe II region nd we cn write c f, d d h f, d d h,, _ FIGU 8 = Since the lower boundr is nd the upper boundr is, qution 3 gives da d d [ ] d d d M

19 968 CHAPT 5 MULTIPL INTGALS = (, 4) = NOT When we set up double integrl s in mple, it is essentil to drw digrm. Often it is helpful to drw verticl rrow s in Figure 8. Then the limits of integrtion for the inner integrl cn be red from the digrm s follows: The rrow strts t the lower boundr t, which gives the lower limit in the integrl, nd the rrow ends t the upper boundr t, which gives the upper limit of integrtion. For tpe II region the rrow is drwn horiontll from the left boundr to the right boundr. FIGU 9 s tpe I region 4 (, 4) XAMPL Find the volume of the solid tht lies under the prboloid nd bove the region in the -plne bounded b the line nd the prbol. SOLUTION From Figure 9 we see tht is tpe I region nd,, Therefore the volume under nd bove is = =œ V da d d 3 3 d d FIGU s tpe II region N Figure shows the solid whose volume is clculted in mple. It lies bove the -plne, below the prboloid, nd between the plne nd the prbolic clinder. = = d SOLUTION From Figure we see tht cn lso be written s tpe II region: Therefore nother epression for V is V 4, 4, s da 4 3 s 3 s d d d d = ] 6 35 M FIGU V XAMPL 3 vlute da, where is the region bounded b the line nd the prbol 6. SOLUTION The region is shown in Figure. Agin is both tpe I nd tpe II, but the description of s tpe I region is more complicted becuse the lower boundr consists of two prts. Therefore we prefer to epress s tpe II region: (, ) 4, 3

20 SCTION 5.3 OUBL INTGALS OV GNAL GIONS 969 =œ +6 =- (5, 4) = -3 =+ (5, 4) _3 =_œ +6 (_, _) (_, _) _ FIGU () s tpe I region (b) s tpe II region Then (5) gives da 4 4 d d 3 3 d 4 [ ( 3) ] d d (,, ) = T ++= (,, ),, If we hd epressed s tpe I region using Figure (), then we would hve obtined da 3 s6 s6 d d 5 s6 d d but this would hve involved more work thn the other method. M FIGU 3 += (or =-/) =/, XAMPL 4 Find the volume of the tetrhedron bounded b the plnes,,, nd. SOLUTION In question such s this, it s wise to drw two digrms: one of the threedimensionl solid nd nother of the plne region over which it lies. Figure 3 shows the tetrhedron T bounded b the coordinte plnes,, the verticl plne, nd the plne. Since the plne intersects the -plne (whose eqution is ) in the line, we see tht T lies bove the tringulr region in the -plne bounded b the lines,, nd. (See Figure 4.) The plne cn be written s, so the required volume lies under the grph of the function nd bove FIGU 4,,

21 97 CHAPT 5 MULTIPL INTGALS Therefore V da [ ] d d d d d M V XAMPL 5 vlute the iterted integrl. sin d d = = SOLUTION If we tr to evlute the integrl s it stnds, we re fced with the tsk of first evluting sin d. But it s impossible to do so in finite terms since sin d is not n elementr function. (See the end of Section 7.5.) So we must chnge the order of integrtion. This is ccomplished b first epressing the given iterted integrl s double integrl. Using (3) bckwrd, we hve sin d d sin da where,, FIGU 5 s tpe I region We sketch this region in Figure 5. Then from Figure 6 we see tht n lterntive description of is,, This enbles us to use (5) to epress the double integrl s n iterted integrl in the reverse order: = = sin d d sin da FIGU 6 s tpe II region sin d cos ] cos sin d d [ sin ] d M POPTIS OF OUBL INTGALS We ssume tht ll of the following integrls eist. The first three properties of double integrls over region follow immeditel from efinition nd Properties 7, 8, nd 9 in Section f, t, da cf, da c f, da f, da t, da

22 SCTION 5.3 OUBL INTGALS OV GNAL GIONS 97 If f, t, for ll, in, then 8 f, da t, da The net propert of double integrls is similr to the propert of single integrls given b the eqution b f d. c f d b f d c If, where nd don t overlp ecept perhps on their boundries (see Figure 7), then 9 f, da f, da f, da FIGU 7 Propert 9 cn be used to evlute double integrls over regions tht re neither tpe I nor tpe II but cn be epressed s union of regions of tpe I or tpe II. Figure 8 illustrtes this procedure. (See ercises 5 nd 5.) FIGU 8 () is neither tpe I nor tpe II. (b) =, is tpe I, is tpe II. = The net propert of integrls ss tht if we integrte the constnt function f, over region, we get the re of : FIGU 9 Clinder with bse nd height da A Figure 9 illustrtes wh qution is true: A solid clinder whose bse is nd whose height is hs volume A A, but we know tht we cn lso write its volume s da. Finll, we cn combine Properties 7, 8, nd to prove the following propert. (See ercise 57.) If m f, M for ll, in, then ma f, da MA

23 97 CHAPT 5 MULTIPL INTGALS XAMPL 6 Use Propert to estimte the integrl e sin cos da, where is the disk with center the origin nd rdius. SOLUTION Since sin nd cos, we hve sin cos nd therefore e e sin cos e e Thus, using m e e, M e, nd A in Propert, we obtin 4 e e sin cos da 4e M 5.3 XCISS 6 vlute the iterted integrl.. d d d d 5. 4 s cos e sin dr d 6. d d d d v s v du dv 7. is bounded b the circle with center the origin nd rdius da, 8. da, is the tringulr region with vertices,,,, nd, vlute the double integrl s da, da, is bounded b s nd 5. 3 da, 6. da, 5 da, da,,, sin 3 da, e da,,,,,, e, ln, 4,,, cos da, is bounded b,, is the tringulr region with vertices (, ), (, ), 3, da, is enclosed b nd s 9 8 Find the volume of the given solid. 9. Under the plne nd bove the region bounded b nd 4. Under the surfce nd bove the region bounded b nd 3. Under the surfce nd bove the tringle with vertices,, 4,, nd,. nclosed b the prboloid 3 nd the plnes,,, 3. Bounded b the coordinte plnes nd the plne Bounded b the plnes,,, nd 5. nclosed b the clinders, nd the plnes, 4 6. Bounded b the clinder 4 nd the plnes,, in the first octnt 7. Bounded b the clinder nd the plnes,, in the first octnt 8. Bounded b the clinders r nd r ; 9. Use grphing clcultor or computer to estimte the -coordintes of the points of intersection of the curves 4 nd 3. If is the region bounded b these curves, estimte da.

24 SCTION 5.3 OUBL INTGALS OV GNAL GIONS 973 ; 3. Find the pproimte volume of the solid in the first octnt tht is bounded b the plnes,, nd nd the clinder cos. (Use grphing device to estimte the points of intersection.) CAS 3 3 Find the volume of the solid b subtrcting two volumes. 3. The solid enclosed b the prbolic clinders, nd the plnes, 3. The solid enclosed b the prbolic clinder nd the plnes 3, Sketch the solid whose volume is given b the iterted integrl. 33. d d Use computer lgebr sstem to find the ect volume of the solid. 35. Under the surfce 3 4 nd bove the region bounded b the curves 3 nd for 36. Between the prboloids nd 8 nd inside the clinder 37. nclosed b nd 38. nclosed b nd Sketch the region of integrtion nd chnge the order of integrtion. 39. f, d d f, d d vlute the integrl b reversing the order of integrtion d d s s 3 s9 s9 ln f, d d 3 e d d 3 rcsin cos s cos d d 5. 8 e 4 d d 3 s 3 s9 4 s s rctn f, d d cos d d e d d d d 4 f, d d 4 f, d d CAS 5 5 press s union of regions of tpe I or tpe II nd evlute the integrl. 5. _ Use Propert to estimte the vlue of the integrl. Q 53. e da, Q is the qurter-circle with center the origin nd rdius in the first qudrnt T 54. sin 4 da, T is the tringle enclosed b the lines,, nd Find the verge vlue of f over region. 55. f,, is the tringle with vertices,,,, nd, f, sin, is enclosed b the curves,, nd 57. Prove Propert. 58. In evluting double integrl over region, sum of iterted integrls ws obtined s follows: da _ f, da (, ) =(+)@ f, d d 3 3 f, d d Sketch the region nd epress the double integrl s n iterted integrl with reversed order of integrtion. 59. vlute tn 3 4 da, where,. [Hint: ploit the fct tht is smmetric with respect to both es.] 6. Use smmetr to evlute 3 4 da, where is the region bounded b the squre with vertices 5, nd, Compute s da, where is the disk, b first identifing the integrl s the volume of solid. 6. Grph the solid bounded b the plne nd the prboloid 4 nd find its ect volume. (Use our CAS to do the grphing, to find the equtions of the boundr curves of the region of integrtion, nd to evlute the double integrl.) _ da _ =-Á

25 974 CHAPT 5 MULTIPL INTGALS 5.4 OUBL INTGALS IN POLA COOINATS Suppose tht we wnt to evlute double integrl f, da, where is one of the regions shown in Figure. In either cse the description of in terms of rectngulr coordintes is rther complicted but is esil described using polr coordintes. + = + =4 + = FIGU () =s(r, ) r, πd (b) =s(r, ) r, πd P(r, )=P(, ) ecll from Figure tht the polr coordintes r, of point re relted to the rectngulr coordintes, b the equtions r r r cos r sin O FIGU (See Section.3.) The regions in Figure re specil cses of polr rectngle r, r b, which is shown in Figure 3. In order to compute the double integrl f, da, where is polr rectngle, we divide the intervl, b into m subintervls r i, r i of equl width r b m nd we divide the intervl, into n subintervls j, j of equl width. Then the circles r r i nd the rs j divide the polr rectngle into the smll polr rectngles shown in Figure 4. n = j = j- = r=b ij (r i *, j*) Î r= =å r=r i O å O r=r i- FIGU 3 Polr rectngle FIGU 4 ividing into polr subrectngles

26 SCTION 5.4 OUBL INTGALS IN POLA COOINATS 975 The center of the polr subrectngle ij r, r i r r i, j hs polr coordintes j r i * r i r i j* j j We compute the re of using the fct tht the re of sector of circle with rdius nd centrl ngle is r ij r. Subtrcting the res of two such sectors, ech of which hs centrl ngle, we find tht the re of is j j Although we hve defined the double integrl f, da in terms of ordinr rectngles, it cn be shown tht, for continuous functions f, we lws obtin the sme nswer using polr rectngles. The rectngulr coordintes of the center of ij re r i * cos j*, r i * sin j*, so tpicl iemnn sum is ij A i r i r i r i r i r i r i r i r i r i * r m n i j f r i * cos j*, r i * sin j* A i m n i j f r i * cos j*, r i * sin j* r i * r If we write tr, rfr cos, r sin, then the iemnn sum in qution cn be written s m n i j which is iemnn sum for the double integrl tr i *, j* r Therefore we hve b tr, dr d f, da lim m, n l m lim m, n l m n i j n i j f r i * cos j*, r i * sin j* A i tr i *, j* r b f r cos, r sin r dr d b tr, dr d CHANG TO POLA COOINATS IN A OUBL INTGAL If f is continuous on polr rectngle given b r b,, where, then f, da b f r cos, r sin r dr d

27 976 CHAPT 5 MULTIPL INTGALS The formul in () ss tht we convert from rectngulr to polr coordintes in double integrl b writing r cos nd r sin, using the pproprite limits of integrtion for r nd, nd replcing da b r dr d. Be creful not to forget the dditionl fctor r on the right side of Formul. A clssicl method for remembering this is shown in Figure 5, where the infinitesiml polr rectngle cn be thought of s n ordinr rectngle with dimensions r d nd dr nd therefore hs re da r dr d. r d r d da dr FIGU 5 O N Here we use the trigonometric identit sin cos See Section 7. for dvice on integrting trigonometric functions. XAMPL vlute 3 4 da, where is the region in the upper hlf-plne bounded b the circles nd 4. SOLUTION The region cn be described s It is the hlf-ring shown in Figure (b), nd in polr coordintes it is given b r,. Therefore, b Formul, 3 4 da,, 4 [r 3 cos r 4 sin ] r r d 7 cos 5 sin d [7 cos 5 cos ] d 7sin 5 3r cos 4r sin r dr d 3r cos 4r 3 sin dr d 5 4 sin 5 M (,, ) FIGU 6 V XAMPL Find the volume of the solid bounded b the plne nd the prboloid. SOLUTION If we put in the eqution of the prboloid, we get. This mens tht the plne intersects the prboloid in the circle, so the solid lies under the prboloid nd bove the circulr disk given b [see Figures 6 nd ()]. In polr coordintes is given b r,. Since r, the volume is V da d r r 3 dr r r r dr d r 4 4

28 SCTION 5.4 OUBL INTGALS IN POLA COOINATS 977 If we hd used rectngulr coordintes insted of polr coordintes, then we would hve obtined V da s s d d which is not es to evlute becuse it involves finding 3 d. M O = å =å r=h ( ) r=h ( ) FIGU 7 =s(r, ) å, h ( ) r h ( )d Wht we hve done so fr cn be etended to the more complicted tpe of region shown in Figure 7. It s similr to the tpe II rectngulr regions considered in Section 5.3. In fct, b combining Formul in this section with Formul 5.3.5, we obtin the following formul. 3 then If f is continuous on polr region of the form r, f, da, h r h h f r cos, r sin r dr d h In prticulr, tking f,, h, nd h h in this formul, we see tht the re of the region bounded b,, nd r h is A da h r h r dr d d h d nd this grees with Formul.4.3. = π 4 V XAMPL 3 Use double integrl to find the re enclosed b one loop of the fourleved rose r cos. SOLUTION From the sketch of the curve in Figure 8, we see tht loop is given b the region {r, 4 4, r cos } So the re is FIGU 8 =_ π 4 A 4 da cos 4 r dr d 4 [ r cos ] d cos d 4 4 cos 4 d 4[ 4 4 sin 4] M

29 978 CHAPT 5 MULTIPL INTGALS FIGU 9 (-)@+ = (or r= cos ) V XAMPL 4 Find the volume of the solid tht lies under the prboloid, bove the -plne, nd inside the clinder. SOLUTION The solid lies bove the disk whose boundr circle hs eqution or, fter completing the squre, (See Figures 9 nd.) In polr coordintes we hve r nd r cos, so the boundr circle becomes r r cos, or r cos. Thus the disk is given b nd, b Formul 3, we hve V r, da 4 cos4 d 8 cos, r cos r r dr d cos 4 d 8 r 4 4 cos cos d d [ cos cos 4] d FIGU [ 3 sin 8 sin 4] 3 3 M 5.4 XCISS 4 A region is shown. ecide whether to use polr coordintes or rectngulr coordintes nd write f, da s n iterted integrl, where f is n rbitrr continuous function on... 4 =- 4 _ _ 5 6 Sketch the region whose re is given b the integrl nd evlute the integrl r dr d vlute the given integrl b chnging to polr coordintes. 7. da, where is the disk with center the origin nd rdius 3 8. da, where is the region tht lies to the left of the -is between the circles nd 4 9. cos da, where is the region tht lies bove the -is within the circle 9. s4 da, where, 4, e 4 cos r dr d. da, where is the region bounded b the semicircle s4 nd the -is. e da, where is the region in the first qudrnt enclosed b the circle 5

30 SCTION 5.4 OUBL INTGALS IN POLA COOINATS rctn da, where, 4, 4. da, where is the region in the first qudrnt tht lies between the circles 4 nd 5 8 Use double integrl to find the re of the region. 5. One loop of the rose r cos 3 6. The region enclosed b the curve r 4 3 cos 7. The region within both of the circles r cos nd r sin 8. The region inside the crdioid r cos nd outside the circle r 3 cos 9 7 Use polr coordintes to find the volume of the given solid. 9. Under the cone s nd bove the disk. Below the prboloid 8 nd bove the -plne. nclosed b the hperboloid nd the plne. Inside the sphere 6 nd outside the clinder 4 3. A sphere of rdius 4. Bounded b the prboloid nd the plne 7 in the first octnt 5. Above the cone s nd below the sphere 6. Bounded b the prboloids 3 3 nd 4 7. Inside both the clinder 4 nd the ellipsoid () A clindricl drill with rdius is used to bore hole through the center of sphere of rdius r. Find the volume of the ring-shped solid tht remins. (b) press the volume in prt () in terms of the height h of the ring. Notice tht the volume depends onl on h, not on or. r 9 3 vlute the iterted integrl b converting to polr coordintes. 3 3 s9 9. sin d d 3. s r 3. d d 3. r s d d s 4 s d d 33. A swimming pool is circulr with 4-ft dimeter. The depth is constnt long est-west lines nd increses linerl from ft t the south end to 7 ft t the north end. Find the volume of wter in the pool. 34. An griculturl sprinkler distributes wter in circulr pttern of rdius ft. It supplies wter to depth of e r feet per hour t distnce of r feet from the sprinkler. () If, wht is the totl mount of wter supplied per hour to the region inside the circle of rdius centered t the sprinkler? (b) etermine n epression for the verge mount of wter per hour per squre foot supplied to the region inside the circle of rdius. 35. Use polr coordintes to combine the sum s s d d s into one double integrl. Then evlute the double integrl. 36. () We define the improper integrl (over the entire plne I lim where is the disk with rdius nd center the origin. Show tht e da (b) An equivlent definition of the improper integrl in prt () is e da lim e da l S where S is the squre with vertices,. Use this to show tht (c) educe tht e da l e da e d e d e d s (d) B mking the chnge of vrible t s, show tht e d s d d s s4 d d e d d (This is fundmentl result for probbilit nd sttistics.) 37. Use the result of ercise 36 prt (c) to evlute the following integrls. () e d (b) s e d

31 98 CHAPT 5 MULTIPL INTGALS 5.5 APPLICATIONS OF OUBL INTGALS We hve lred seen one ppliction of double integrls: computing volumes. Another geometric ppliction is finding res of surfces nd this will be done in Section 6.6. In this section we eplore phsicl pplictions such s computing mss, electric chrge, center of mss, nd moment of inerti. We will see tht these phsicl ides re lso importnt when pplied to probbilit densit functions of two rndom vribles. NSITY AN MASS (, ) In Section 8.3 we were ble to use single integrls to compute moments nd the center of mss of thin plte or lmin with constnt densit. But now, equipped with the double integrl, we cn consider lmin with vrible densit. Suppose the lmin occupies region of the -plne nd its densit (in units of mss per unit re) t point, in is given b,, where is continuous function on. This mens tht FIGU FIGU (* ij, * ij ) ij, lim m A where m nd A re the mss nd re of smll rectngle tht contins, nd the limit is tken s the dimensions of the rectngle pproch. (See Figure.) To find the totl mss m of the lmin, we divide rectngle contining into subrectngles ij of equl sie (s in Figure ) nd consider, to be outside. If we choose point ij *, ij * in ij, then the mss of the prt of the lmin tht occupies ij is pproimtel ij *, ij * A, where A is the re of ij. If we dd ll such msses, we get n pproimtion to the totl mss: m k l i j ij *, ij * A If we now increse the number of subrectngles, we obtin the totl mss m of the lmin s the limiting vlue of the pproimtions: m lim k, l l k l i j ij *, ij * A, da Phsicists lso consider other tpes of densit tht cn be treted in the sme mnner. For emple, if n electric chrge is distributed over region nd the chrge densit (in units of chrge per unit re) is given b, t point, in, then the totl chrge Q is given b Q, da XAMPL Chrge is distributed over the tringulr region in Figure 3 so tht the chrge densit t, is,, mesured in coulombs per squre meter (Cm ). Find the totl chrge.

32 SCTION 5.5 APPLICATIONS OF OUBL INTGALS 98 = (, ) =- FIGU 3 SOLUTION From qution nd Figure 3 we hve Q 5 4, da d d d 3 d Thus the totl chrge is C. M d MOMNTS AN CNTS OF MASS In Section 8.3 we found the center of mss of lmin with constnt densit; here we consider lmin with vrible densit. Suppose the lmin occupies region nd hs densit function,. ecll from Chpter 8 tht we defined the moment of prticle bout n is s the product of its mss nd its directed distnce from the is. We divide into smll rectngles s in Figure. Then the mss of ij is pproimtel *, ij * ij A, so we cn pproimte the moment of with respect to the -is b ij ij *, ij * A ij * If we now dd these quntities nd tke the limit s the number of subrectngles becomes lrge, we obtin the moment of the entire lmin bout the -is: 3 M lim m, nl m i n j ij * ij *, ij * A, da Similrl, the moment bout the -is is (, ) 4 M lim m, nl m i n j ij * ij *, ij * A, da FIGU 4 As before, we define the center of mss, so tht m M nd m M. The phsicl significnce is tht the lmin behves s if its entire mss is concentrted t its center of mss. Thus the lmin blnces horiontll when supported t its center of mss (see Figure 4). 5 The coordintes, of the center of mss of lmin occuping the region nd hving densit function, re M m m, da M m, da m where the mss m is given b m, da

33 98 CHAPT 5 MULTIPL INTGALS (, ) =- 3, 8 6 (, ) FIGU 5 V XAMPL Find the mss nd center of mss of tringulr lmin with vertices,,,, nd, if the densit function is, 3. SOLUTION The tringle is shown in Figure 5. (Note tht the eqution of the upper boundr is.) The mss of the lmin is m Then the formuls in (5) give m, da 3, da d d 4 d 4 3 d 3 d d d 3 3 d m, da d d d d The center of mss is t the point ( 3 8,. M 6) V XAMPL 3 The densit t n point on semicirculr lmin is proportionl to the distnce from the center of the circle. Find the center of mss of the lmin., 3 π + =@ SOLUTION Let s plce the lmin s the upper hlf of the circle. (See Figure 6.) Then the distnce from point, to the center of the circle (the origin) is s. Therefore the densit function is, Ks _ FIGU 6 where K is some constnt. Both the densit function nd the shpe of the lmin suggest tht we convert to polr coordintes. Then s r nd the region is given b r,. Thus the mss of the lmin is m, da K r 3 K d r dr Ks da 3 K 3 3 Kr r dr d Both the lmin nd the densit function re smmetric with respect to the -is, so the

34 SCTION 5.5 APPLICATIONS OF OUBL INTGALS 983 center of mss must lie on the -is, tht is,. The -coordinte is given b m, da 3 K 3 r sin r r dr d 3 sin d 3 r 3 dr 3 3 [cos ] r 4 4 N Compre the loction of the center of mss in mple 3 with mple 4 in Section 8.3, where we found tht the center of mss of lmin with the sme shpe but uniform densit is locted t the point, Therefore the center of mss is locted t the point, 3. M MOMNT OF INTIA The moment of inerti (lso clled the second moment) of prticle of mss m bout n is is defined to be mr, where r is the distnce from the prticle to the is. We etend this concept to lmin with densit function, nd occuping region b proceeding s we did for ordinr moments. We divide into smll rectngles, pproimte the moment of inerti of ech subrectngle bout the -is, nd tke the limit of the sum s the number of subrectngles becomes lrge. The result is the moment of inerti of the lmin bout the -is: 6 I lim n m, n l m ij * ij *, ij * A, da i j Similrl, the moment of inerti bout the -is is 7 I lim n m, n l m ij * ij *, ij * A, da i j It is lso of interest to consider the moment of inerti bout the origin, lso clled the polr moment of inerti: 8 I lim m, nl m n i j [ ij * ij * ] ij *, ij * A, da Note tht I I I. V XAMPL 4 Find the moments of inerti I, I, nd I of homogeneous disk with densit,, center the origin, nd rdius. SOLUTION The boundr of is the circle nd in polr coordintes is

35 984 CHAPT 5 MULTIPL INTGALS described b, r. Let s compute first: I da d Insted of computing I nd I directl, we use the fcts tht I I I nd I I ( from the smmetr of the problem). Thus I r 3 dr r 4 4 r r dr d 4 I I I 4 4 M In mple 4 notice tht the mss of the disk is m densit re so the moment of inerti of the disk bout the origin (like wheel bout its le) cn be written s 4 I m Thus if we increse the mss or the rdius of the disk, we thereb increse the moment of inerti. In generl, the moment of inerti pls much the sme role in rottionl motion tht mss pls in liner motion. The moment of inerti of wheel is wht mkes it difficult to strt or stop the rottion of the wheel, just s the mss of cr is wht mkes it difficult to strt or stop the motion of the cr. The rdius of grtion of lmin bout n is is the number such tht 9 m I where m is the mss of the lmin nd I is the moment of inerti bout the given is. qution 9 ss tht if the mss of the lmin were concentrted t distnce from the is, then the moment of inerti of this point mss would be the sme s the moment of inerti of the lmin. In prticulr, the rdius of grtion with respect to the -is nd the rdius of grtion with respect to the -is re given b the equtions m I m I Thus, is the point t which the mss of the lmin cn be concentrted without chnging the moments of inerti with respect to the coordinte es. (Note the nlog with the center of mss.) V XAMPL 5 Find the rdius of grtion bout the -is of the disk in mple 4. SOLUTION As noted, the mss of the disk is m, so from qutions we hve I m Therefore the rdius of grtion bout the -is is, which is hlf the rdius of the disk. M

36 SCTION 5.5 APPLICATIONS OF OUBL INTGALS 985 POBABILITY In Section 8.5 we considered the probbilit densit function f of continuous rndom vrible X. This mens tht f for ll, f d, nd the probbilit tht X lies between nd b is found b integrting f from to b: Now we consider pir of continuous rndom vribles X nd Y, such s the lifetimes of two components of mchine or the height nd weight of n dult femle chosen t rndom. The joint densit function of X nd Y is function f of two vribles such tht the probbilit tht X, Y lies in region is In prticulr, if the region is rectngle, the probbilit tht X lies between nd b nd Y lies between c nd d is (See Figure 7.) P X b b f d PX, Y P X b, c Y d b f, da d c f, d d =f(, ) c FIGU 7 The probbilit tht X lies between nd b nd Y lies between c nd d is the volume tht lies bove the rectngle =[, b][c, d] nd below the grph of the joint densit function. b d Becuse probbilities ren t negtive nd re mesured on scle from to, the joint densit function hs the following properties: f, f, da As in ercise 36 in Section 5.4, the double integrl over is n improper integrl defined s the limit of double integrls over epnding circles or squres nd we cn write f, da f, d d

37 986 CHAPT 5 MULTIPL INTGALS XAMPL 6 If the joint densit function for X nd Y is given b f, C if, otherwise find the vlue of the constnt C. Then find PX 7, Y. SOLUTION We find the vlue of C b ensuring tht the double integrl of f is equl to. Becuse f, outside the rectngle,,, we hve f, d d C d d C [ ] d C d 5C Therefore 5C nd so C 5. Now we cn compute the probbilit tht X is t most 7 nd Y is t lest : PX 7, Y 7 f, d d [ ] d d d 8 96 d M Suppose X is rndom vrible with probbilit densit function f nd Y is rndom vrible with densit function f. Then X nd Y re clled independent rndom vribles if their joint densit function is the product of their individul densit functions: f, f f In Section 8.5 we modeled witing times b using eponentil densit functions f t e t if t if t where is the men witing time. In the net emple we consider sitution with two independent witing times. XAMPL 7 The mnger of movie theter determines tht the verge time moviegoers wit in line to bu ticket for this week s film is minutes nd the verge time the wit to bu popcorn is 5 minutes. Assuming tht the witing times re independent, find the probbilit tht moviegoer wits totl of less thn minutes before tking his or her set. SOLUTION Assuming tht both the witing time X for the ticket purchse nd the witing time Y in the refreshment line re modeled b eponentil probbilit densit functions, we cn write the individul densit functions s f e if if f 5e 5 if if

38 SCTION 5.5 APPLICATIONS OF OUBL INTGALS 987 Since X nd Y re independent, the joint densit function is the product: f, f f 5e e 5 if, otherwise += We re sked for the probbilit tht X Y : where is the tringulr region shown in Figure 8. Thus PX Y PX Y PX, Y f, da 5e e 5 d d FIGU 8 5 [e 5e 5 ] d e e 5 d e e 4 e d e 4 e.7476 This mens tht bout 75% of the moviegoers wit less thn minutes before tking their sets. M XPCT VALUS ecll from Section 8.5 tht if X is rndom vrible with probbilit densit function then its men is f d f, Now if X nd Y re rndom vribles with joint densit function f, we define the X-men nd Y-men, lso clled the epected vlues of X nd Y, to be Notice how closel the epressions for nd in () resemble the moments M nd M of lmin with densit function in qutions 3 nd 4. In fct, we cn think of probbilit s being like continuousl distributed mss. We clculte probbilit the w we clculte mss b integrting densit function. And becuse the totl probbilit mss is, the epressions for nd in (5) show tht we cn think of the epected vlues of X nd Y, nd, s the coordintes of the center of mss of the probbilit distribution. In the net emple we del with norml distributions. As in Section 8.5, single rndom vrible is normll distributed if its probbilit densit function is of the form f, da f s where is the men nd is the stndrd devition. f, da e

39 988 CHAPT 5 MULTIPL INTGALS XAMPL 8 A fctor produces (clindricll shped) roller berings tht re sold s hving dimeter 4. cm nd length 6. cm. In fct, the dimeters X re normll distributed with men 4. cm nd stndrd devition. cm while the lengths Y re normll distributed with men 6. cm nd stndrd devition. cm. Assuming tht X nd Y re independent, write the joint densit function nd grph it. Find the probbilit tht bering rndoml chosen from the production line hs either length or dimeter tht differs from the men b more thn. cm. SOLUTION We re given tht X nd Y re normll distributed with 4., 6., nd.. So the individul densit functions for X nd Y re FIGU 9 Grph of the bivrite norml joint densit function in mple f.s Since X nd Y re independent, the joint densit function is the product: f, f f 5 e 4. e 54 6 A grph of this function is shown in Figure 9. Let s first clculte the probbilit tht both X nd Y differ from their mens b less thn. cm. Using clcultor or computer to estimte the integrl, we hve. f.s e 4. e 6. e 6. P3.98 X 4., 5.98 Y f, d d 5.9 Then the probbilit tht either X or Y differs from its men b more thn. cm is pproimtel e 54 6 d d M 5.5 XCISS. lectric chrge is distributed over the rectngle 3, so tht the chrge densit t, is, (mesured in coulombs per squre meter). Find the totl chrge on the rectngle.. lectric chrge is distributed over the disk 4 so tht the chrge densit t, is, (mesured in coulombs per squre meter). Find the totl chrge on the disk. 3 Find the mss nd center of mss of the lmin tht occupies the region nd hs the given densit function. 3.,, ;, 4.,, b;, c 5. is the tringulr region with vertices,,,,, 3;, 6. is the tringulr region enclosed b the lines,, nd 6;, 7. is bounded b e,,, nd ;, 8. is bounded b s,, nd ;, 9., sinl, L;,. is bounded b the prbols nd ;, s

40 SCTION 5.5 APPLICATIONS OF OUBL INTGALS 989 CAS CAS. A lmin occupies the prt of the disk in the first qudrnt. Find its center of mss if the densit t n point is proportionl to its distnce from the -is.. Find the center of mss of the lmin in ercise if the densit t n point is proportionl to the squre of its distnce from the origin. 3. The boundr of lmin consists of the semicircles s nd s4 together with the portions of the -is tht join them. Find the center of mss of the lmin if the densit t n point is proportionl to its distnce from the origin. 4. Find the center of mss of the lmin in ercise 3 if the densit t n point is inversel proportionl to its distnce from the origin. 5. Find the center of mss of lmin in the shpe of n isosceles right tringle with equl sides of length if the densit t n point is proportionl to the squre of the distnce from the verte opposite the hpotenuse. 6. A lmin occupies the region inside the circle but outside the circle. Find the center of mss if the densit t n point is inversel proportionl to its distnce from the origin. 7. Find the moments of inerti,, for the lmin of ercise Find the moments of inerti,, for the lmin of ercise. 9. Find the moments of inerti,, for the lmin of ercise 5.. Consider squre fn blde with sides of length nd the lower left corner plced t the origin. If the densit of the blde is,., is it more difficult to rotte the blde bout the -is or the -is? Use computer lgebr sstem to find the mss, center of mss, nd moments of inerti of the lmin tht occupies the region nd hs the given densit function.., sin, ;. is enclosed b the crdioid r cos ;, s 3 6 A lmin with constnt densit, occupies the given region. Find the moments of inerti I nd I nd the rdii of grtion nd. 3. The rectngle b, h 4. The tringle with vertices,, b,, nd, h 5. The prt of the disk in the first qudrnt 6. The region under the curve sin from to I I I I I I I I I, CAS 7. The joint densit function for pir of rndom vribles X nd Y is C if, f, otherwise () Find the vlue of the constnt C. (b) Find PX, Y. (c) Find PX Y. 8. () Verif tht f, 4 is joint densit function. (b) If X nd Y re rndom vribles whose joint densit function is the function f in prt (), find (i) P(X (ii) P(X, Y ) ) (c) Find the epected vlues of X nd Y. 9. Suppose X nd Y re rndom vribles with joint densit function f,.e.5. () Verif tht f is indeed joint densit function. (b) Find the following probbilities. (i) PY (ii) PX, Y 4 (c) Find the epected vlues of X nd Y. 3. () A lmp hs two bulbs of tpe with n verge lifetime of hours. Assuming tht we cn model the probbilit of filure of these bulbs b n eponentil densit function with men, find the probbilit tht both of the lmp s bulbs fil within hours. (b) Another lmp hs just one bulb of the sme tpe s in prt (). If one bulb burns out nd is replced b bulb of the sme tpe, find the probbilit tht the two bulbs fil within totl of hours. 3. Suppose tht X nd Y re independent rndom vribles, where X is normll distributed with men 45 nd stndrd devition.5 nd Y is normll distributed with men nd stndrd devition.. () Find P4 X 5, Y 5. (b) Find P4X 45 Y. 3. Xvier nd Yolnd both hve clsses tht end t noon nd the gree to meet ever d fter clss. The rrive t the coffee shop independentl. Xvier s rrivl time is X nd Yolnd s rrivl time is Y, where X nd Y re mesured in minutes fter noon. The individul densit functions re f e if if if, otherwise if, otherwise f 5 if otherwise (Xvier rrives sometime fter noon nd is more likel to rrive promptl thn lte. Yolnd lws rrives b : PM nd is more likel to rrive lte thn promptl.) After Yolnd rrives, she ll wit for up to hlf n hour for Xvier, but he won t wit for her. Find the probbilit tht the meet.

41 99 CHAPT 5 MULTIPL INTGALS 33. When studing the spred of n epidemic, we ssume tht the probbilit tht n infected individul will spred the disese to n uninfected individul is function of the distnce between them. Consider circulr cit of rdius mi in which the popultion is uniforml distributed. For n uninfected individul t fied point A,, ssume tht the probbilit function is given b f P dp, A where dp, A denotes the distnce between P nd A. () Suppose the eposure of person to the disese is the sum of the probbilities of ctching the disese from ll members of the popultion. Assume tht the infected people re uniforml distributed throughout the cit, with k infected individuls per squre mile. Find double integrl tht represents the eposure of person residing t A. (b) vlute the integrl for the cse in which A is the center of the cit nd for the cse in which A is locted on the edge of the cit. Where would ou prefer to live? 5.6 TIPL INTGALS B Just s we defined single integrls for functions of one vrible nd double integrls for functions of two vribles, so we cn define triple integrls for functions of three vribles. Let s first del with the simplest cse where f is defined on rectngulr bo: B,, b, c d, r s The first step is to divide B into sub-boes. We do this b dividing the intervl, b into l subintervls i, i of equl width, dividing c, d into m subintervls of width, nd dividing r, s into n subintervls of width. The plnes through the endpoints of these subintervls prllel to the coordinte plnes divide the bo B into lmn sub-boes B ijk i, i j, j k, k which re shown in Figure. ch sub-bo hs volume V. Then we form the triple iemnn sum Î B ijk Î Î l m n i j k f ijk *, * ijk, ijk * V where the smple point ijk *, ijk *, ijk * is in B ijk. B nlog with the definition of double integrl (5..5), we define the triple integrl s the limit of the triple iemnn sums in (). 3 FINITION The triple integrl of f over the bo B is FIGU if this limit eists. B f,, dv lim l, m, n l l i m n j k f ijk *, ijk *, ijk * V Agin, the triple integrl lws eists if f is continuous. We cn choose the smple point to be n point in the sub-bo, but if we choose it to be the point i, j, k we get simpler-looking epression for the triple integrl: B f,, dv lim l, m, n l l i m n j k f i, j, k V Just s for double integrls, the prcticl method for evluting triple integrls is to epress them s iterted integrls s follows.

42 SCTION 5.6 TIPL INTGALS 99 4 FUBINI S THOM FO TIPL INTGALS If f is continuous on the rectngulr bo B, b c, d r, s, then B f,, dv s r d c b f,, d d d The iterted integrl on the right side of Fubini s Theorem mens tht we integrte first with respect to (keeping nd fied), then we integrte with respect to (keeping fied), nd finll we integrte with respect to. There re five other possible orders in which we cn integrte, ll of which give the sme vlue. For instnce, if we integrte with respect to, then, nd then, we hve B f,, dv b s r d c B dv f,, d d d V XAMPL vlute the triple integrl, where B is the rectngulr bo given b B,,,, 3 SOLUTION We could use n of the si possible orders of integrtion. If we choose to integrte with respect to, then, nd then, we obtin B dv 3 3 d d d 4 d d d d d d M FIGU A tpe solid region =u (, ) =u (, ) Now we define the triple integrl over generl bounded region in threedimensionl spce ( solid) b much the sme procedure tht we used for double integrls (5.3.). We enclose in bo B of the tpe given b qution. Then we define function F so tht it grees with f on but is for points in B tht re outside. B definition, This integrl eists if f is continuous nd the boundr of is resonbl smooth. The triple integrl hs essentill the sme properties s the double integrl (Properties 6 9 in Section 5.3). We restrict our ttention to continuous functions f nd to certin simple tpes of regions. A solid region is sid to be of tpe if it lies between the grphs of two continuous functions of nd, tht is, 5 f,, dv B F,, dv,,,, u, u, where is the projection of onto the -plne s shown in Figure. Notice tht the upper boundr of the solid is the surfce with eqution u,, while the lower boundr is the surfce u,.

43 99 CHAPT 5 MULTIPL INTGALS B the sme sort of rgument tht led to Formul 5.3.3, it cn be shown tht if is tpe region given b qution 5, then 6 f,, dv u, f,, u d da, b =g () =u (, ) =u (, ) =g () The mening of the inner integrl on the right side of qution 6 is tht nd re held fied, nd therefore u, nd u, re regrded s constnts, while f,, is integrted with respect to. In prticulr, if the projection of onto the -plne is tpe I plne region (s in Figure 3), then,, b, t t, u, u, nd qution 6 becomes FIGU 3 A tpe solid region where the projection is tpe I plne region 7 f,, dv b t t u, u, f,, d d d c =u (, ) =u (, ) =h () d If, on the other hnd, is tpe II plne region (s in Figure 4), then nd qution 6 becomes 8,, c d, h h, u, u, f,, dv d c h h u, u, f,, d d d =h () FIGU 4 A tpe solid region with tpe II projection dv XAMPL vlute, where is the solid tetrhedron bounded b the four plnes,,, nd. SOLUTION When we set up triple integrl it s wise to drw two digrms: one of the solid region (see Figure 5) nd one of its projection on the -plne (see Figure 6). The lower boundr of the tetrhedron is the plne nd the upper (,, ) =-- =- (,, ) (,, ) = = FIGU 5 FIGU 6

44 SCTION 5.6 TIPL INTGALS 993 boundr is the plne (or ), so we use u, nd u, in Formul 7. Notice tht the plnes nd intersect in the line (or ) in the -plne. So the projection of is the tringulr region shown in Figure 6, nd we hve 9 This description of s tpe region enbles us to evlute the integrl s follows: dv d d d d d,,,, d d 3 d d M A solid region is of tpe if it is of the form,,,, u, u, where, this time, is the projection of onto the -plne (see Figure 7). The bck surfce is u,, the front surfce is u,, nd we hve f,, dv u, f,, u d da, =u (, ) =u (, ) FIGU 7 A tpe region Finll, tpe 3 region is of the form =u (, ) =u (, ) FIGU 8 A tpe 3 region,,,, u, u, where is the projection of onto the -plne, u, is the left surfce, nd u, is the right surfce (see Figure 8). For this tpe of region we hve f,, dv u, f,, u d da,

45 994 CHAPT 5 MULTIPL INTGALS In ech of qutions nd there m be two possible epressions for the integrl depending on whether is tpe I or tpe II plne region (nd corresponding to qutions 7 nd 8). TC Visul 5.6 illustrtes how solid regions (including the one in Figure 9) project onto coordinte plnes. s dv V XAMPL 3 vlute, where is the region bounded b the prboloid nd the plne 4. SOLUTION The solid is shown in Figure 9. If we regrd it s tpe region, then we need to consider its projection onto the -plne, which is the prbolic region in Figure. (The trce of in the plne is the prbol.) = +@ FIGU 9 egion of integrtion 4 FIGU Projection on -plne =4 = +@=4 _ From we obtin s, so the lower boundr surfce of is s nd the upper surfce is s. Therefore the description of s tpe region is nd so we obtin Although this epression is correct, it is etremel difficult to evlute. So let s insted consider s tpe 3 region. As such, its projection 3 onto the -plne is the disk 4 shown in Figure. Then the left boundr of is the prboloid nd the right boundr is the plne 4, so tking u, nd u, 4 in qution, we hve,,, 4, s s s dv s dv s 4 s Although this integrl could be written s 3 4 s d da s d d d 3 4 s da FIGU Projection on -plne The most difficult step in evluting triple integrl is setting up n epression for the region of integrtion (such s qution 9 in mple ). emember tht the limits of integrtion in the inner integrl contin t most two vribles, the limits of integrtion in the middle integrl contin t most one vrible, nd the limits of integrtion in the outer integrl must be constnts. it s esier to convert to polr coordintes in the -plne: r cos, r sin. This gives s dv s4 s4 4 s d d 3 4r 3 4 s da 4 r r r dr d 3 r d 4r r 4 dr M

46 SCTION 5.6 TIPL INTGALS 995 APPLICATIONS OF TIPL INTGALS ecll tht if f, then the single integrl f d represents the re under the curve f from to b, nd if f,, then the double integrl f, da represents the volume under the surfce f, nd bove. The corresponding interprettion of triple integrl f,, dv, where f,,, is not ver useful becuse it would be the hpervolume of four-dimensionl object nd, of course, tht is ver difficult to visulie. (emember tht is just the domin of the function f ; the grph of f lies in four-dimensionl spce.) Nonetheless, the triple integrl f,, dv cn be interpreted in different ws in different phsicl situtions, depending on the phsicl interprettions of,, nd f,,. Let s begin with the specil cse where f,, for ll points in. Then the triple integrl does represent the volume of : b V dv For emple, ou cn see this in the cse of tpe region b putting Formul 6: dv u, u, d da u, u, da f,, in nd from Section 5.3 we know this represents the volume tht lies between the surfces u, nd u,. XAMPL 4 Use triple integrl to find the volume of the tetrhedron T bounded b the plnes,,, nd. SOLUTION The tetrhedron T nd its projection on the -plne re shown in Figures nd 3. The lower boundr of T is the plne nd the upper boundr is the plne, tht is,. (,, ) = ++= += (or =- /) T (,, ),, =/, FIGU FIGU 3 Therefore we hve VT T dv d d 3 b the sme clcultion s in mple 4 in Section 5.3. d d d

47 996 CHAPT 5 MULTIPL INTGALS (Notice tht it is not necessr to use triple integrls to compute volumes. The simpl give n lterntive method for setting up the clcultion.) M All the pplictions of double integrls in Section 5.5 cn be immeditel etended to triple integrls. For emple, if the densit function of solid object tht occupies the region is,,, in units of mss per unit volume, t n given point,,, then its mss is 3 m,, dv nd its moments bout the three coordinte plnes re 4 M,, dv M,, dv M,, dv The center of mss is locted t the point,,, where 5 M m M m M m If the densit is constnt, the center of mss of the solid is clled the centroid of. The moments of inerti bout the three coordinte es re 6 I,, dv I I,, dv,, dv As in Section 5.5, the totl electric chrge on solid object occuping region nd hving chrge densit,, is Q,, dv If we hve three continuous rndom vribles X, Y, nd Z, their joint densit function is function of three vribles such tht the probbilit tht X, Y, Z lies in is In prticulr, PX, Y, Z f,, dv P X b, c Y d, r Z s b The joint densit function stisfies d c s f,, d d d r f,, f,, d d d

48 SCTION 5.6 TIPL INTGALS 997 V XAMPL 5 Find the center of mss of solid of constnt densit tht is bounded b the prbolic clinder nd the plnes,, nd. = SOLUTION The solid nd its projection onto the -plne re shown in Figure 4. The lower nd upper surfces of re the plnes nd, so we describe s tpe region:,,,, Then, if the densit is,,, the mss is m dv d d d = = d d d 4 d 4 d FIGU Becuse of the smmetr of nd bout the -plne, we cn immeditel s tht M nd therefore. The other moments re M dv d d d d d 3 3 d 3 6 d M dv d d d 3 d d 6 d 7 d d Therefore the center of mss is,, M m, M m, M m ( 5 7,, 5 4) M

49 998 CHAPT 5 MULTIPL INTGALS 5.6 XCISS. vlute the integrl in mple, integrting first with respect to, then, nd then.. vlute the integrl, where using three different orders of integrtion. 3 8 vlute the iterted integrl d d d e d d d vlute the triple integrl. 9., where., where,,,,., where is bounded b the plnes,,, nd 4 3., where is bounded b the prbolic clinder nd the plnes,, nd 4., where is bounded b the prbolic clinders nd nd the plnes nd 5., where T is the solid tetrhedron with vertices,,,,,,,,, nd,, 6., where T is the solid tetrhedron with vertices,,,,,,,,, nd,, 7., where is bounded b the prboloid 4 4 nd the plne 4 8., where is bounded b the clinder 9 nd the plnes, 3, nd in the first octnt 9 Use triple integrl to find the volume of the given solid s s dv {,,, s4,} cos 5 dv 6 dv., where lies under the plne nd bove the region in the -plne bounded b the curves s,, nd dv e dv dv T dv T dv dv dv sin d d d 3 dv,,,, cos d d d d d d e d d d The tetrhedron enclosed b the coordinte plnes nd the plne 4 CAS CAS. The solid bounded b the clinder nd the plnes, 4, nd 9. The solid enclosed b the clinder 9 nd the plnes 5 nd. The solid enclosed b the prboloid nd the plne 6 3. () press the volume of the wedge in the first octnt tht is cut from the clinder b the plnes nd s triple integrl. (b) Use either the Tble of Integrls (on eference Pges 6 ) or computer lgebr sstem to find the ect vlue of the triple integrl in prt (). 4. () In the Midpoint ule for triple integrls we use triple iemnn sum to pproimte triple integrl over bo B, where f,, is evluted t the center i, j, k of the bo B ijk. Use the Midpoint ule to estimte B s dv, where B is the cube defined b 4, 4, 4. ivide B into eight cubes of equl sie. (b) Use computer lgebr sstem to pproimte the integrl in prt () correct to the nerest integer. Compre with the nswer to prt (). 5 6 Use the Midpoint ule for triple integrls (ercise 4) to estimte the vlue of the integrl. ivide B into eight sub-boes of equl sie. 5. B, where ln dv B,, 4, 8, 4 6., where B,, 4,, 7 8 Sketch the solid whose volume is given b the iterted integrl. 7. B sin 3 dv d d d 9 3 press the integrl s n iterted integrl in si different ws, where is the solid bounded b the given surfces , 3. 9,, 3.,, 4 8. f,, dv 4 3.,,, d d d

50 SCTION 5.6 TIPL INTGALS The figure shows the region of integrtion for the integrl ewrite this integrl s n equivlent iterted integrl in the five other orders. 34. The figure shows the region of integrtion for the integrl ewrite this integrl s n equivlent iterted integrl in the five other orders Write five other iterted integrls tht re equl to the given iterted integrl s f,, d d d 37 4 Find the mss nd center of mss of the solid with the given densit function. 37. is the solid of ercise ; =- f,, d d d f,, d d d =œ =-,, 38. is bounded b the prbolic clinder nd the plnes,, nd ;,, is the cube given b,, ;,, =- f,, d d d CAS CAS 4. is the tetrhedron bounded b the plnes,,, ;,, 4 44 Assume tht the solid hs constnt densit k. 4. Find the moments of inerti for cube with side length L if one verte is locted t the origin nd three edges lie long the coordinte es. 4. Find the moments of inerti for rectngulr brick with dimensions, b, nd c nd mss M if the center of the brick is situted t the origin nd the edges re prllel to the coordinte es. 43. Find the moment of inerti bout the -is of the solid clinder, h. 44. Find the moment of inerti bout the -is of the solid cone s h Set up, but do not evlute, integrl epressions for () the mss, (b) the center of mss, nd (c) the moment of inerti bout the -is. 45. The solid of ercise ;,, s 46. The hemisphere, ;,, s 47. Let be the solid in the first octnt bounded b the clinder nd the plnes,, nd with the densit function,,. Use computer lgebr sstem to find the ect vlues of the following quntities for. () The mss (b) The center of mss (c) The moment of inerti bout the -is 48. If is the solid of ercise 8 with densit function,,, find the following quntities, correct to three deciml plces. () The mss (b) The center of mss (c) The moment of inerti bout the -is 49. The joint densit function for rndom vribles X, Y, nd Z is f,, C if,,, nd f,, otherwise. () Find the vlue of the constnt C. (b) Find PX, Y, Z. (c) Find PX Y Z. 5. Suppose X, Y, nd Z re rndom vribles with joint densit function f,, Ce.5.. if,,, nd f,, otherwise. () Find the vlue of the constnt C. (b) Find PX, Y. (c) Find PX, Y, Z.

51 CHAPT 5 MULTIPL INTGALS 5 5 The verge vlue of function f,, over solid region is defined to be where V is the volume of. For instnce, if is densit function, then is the verge densit of. ve f ve f,, dv V 5. Find the verge vlue of the function f,, over the cube with side length L tht lies in the first octnt with one verte t the origin nd edges prllel to the coordinte es. 5. Find the verge vlue of the function f,, over the region enclosed b the prboloid nd the plne. 53. Find the region for which the triple integrl is mimum. 3 dv ISCOVY POJCT VOLUMS OF HYPSPHS In this project we find formuls for the volume enclosed b hpersphere in n-dimensionl spce.. Use double integrl nd trigonometric substitution, together with Formul 64 in the Tble of Integrls, to find the re of circle with rdius r.. Use triple integrl nd trigonometric substitution to find the volume of sphere with rdius r. 3. Use qudruple integrl to find the hpervolume enclosed b the hpersphere w r in 4. (Use onl trigonometric substitution nd the reduction formuls for sin n d or cos n d.) 4. Use n n-tuple integrl to find the volume enclosed b hpersphere of rdius r in n-dimensionl spce n. [Hint: The formuls re different for n even nd n odd.] 5.7 TIPL INTGALS IN CYLINICAL COOINATS O r P(r, )=P(, ) In plne geometr the polr coordinte sstem is used to give convenient description of certin curves nd regions. (See Section.3.) Figure enbles us to recll the connection between polr nd Crtesin coordintes. If the point P hs Crtesin coordintes, nd polr coordintes r,, then, from the figure, r cos r sin r tn FIGU In three dimensions there is coordinte sstem, clled clindricl coordintes, tht is similr to polr coordintes nd gives convenient descriptions of some commonl occurring surfces nd solids. As we will see, some triple integrls re much esier to evlute in clindricl coordintes.

52 SCTION 5.7 TIPL INTGALS IN CYLINICAL COOINATS CYLINICAL COOINATS O r P(r,, ) (r,, ) FIGU The clindricl coordintes of point In the clindricl coordinte sstem, point P in three-dimensionl spce is represented b the ordered triple r,,, where r nd re polr coordintes of the projection of P onto the -plne nd is the directed distnce from the -plne to P. (See Figure.) To convert from clindricl to rectngulr coordintes, we use the equtions r cos r sin wheres to convert from rectngulr to clindricl coordintes, we use r tn XAMPL () Plot the point with clindricl coordintes, 3, nd find its rectngulr coordintes. (b) Find clindricl coordintes of the point with rectngulr coordintes 3, 3, 7. π, 3, SOLUTION () The point with clindricl coordintes, 3, is plotted in Figure 3. From qutions, its rectngulr coordintes re cos 3 FIGU 3 π 3 sin 3 s3 s3 Thus the point is (, s3, ) in rectngulr coordintes. (b) From qutions we hve r s3 3 3s (c,, ) FIGU 4 r=c, clinder (, c, ) tn Therefore one set of clindricl coordintes is (3s, 74, 7). Another is (3s, 4, 7). As with polr coordintes, there re infinitel mn choices. M Clindricl coordintes re useful in problems tht involve smmetr bout n is, nd the -is is chosen to coincide with this is of smmetr. For instnce, the is of the circulr clinder with Crtesin eqution c is the -is. In clindricl coordintes this clinder hs the ver simple eqution r c. (See Figure 4.) This is the reson for the nme clindricl coordintes. so 7 4 n

53 CHAPT 5 MULTIPL INTGALS V XAMPL escribe the surfce whose eqution in clindricl coordintes is r. SOLUTION The eqution ss tht the -vlue, or height, of ech point on the surfce is the sme s r, the distnce from the point to the -is. Becuse doesn t pper, it cn vr. So n horiontl trce in the plne k k is circle of rdius k. These trces suggest tht the surfce is cone. This prediction cn be confirmed b converting the eqution into rectngulr coordintes. From the first eqution in () we hve FIGU 5 =r, cone r We recognie the eqution (b comprison with Tble in Section.6) s being circulr cone whose is is the -is. (See Figure 5.) M =u (, ) VALUATING TIPL INTGALS WITH CYLINICAL COOINATS Suppose tht is tpe region whose projection on the -plne is convenientl described in polr coordintes (see Figure 6). In prticulr, suppose tht f is continuous nd,,,, u, u, r=h ( ) = =u (, ) =b r=h ( ) where is given in polr coordintes b r, We know from qution tht, h r h FIGU 6 3 f,, dv u, f,, u d da, But we lso know how to evlute double integrls in polr coordintes. In fct, combining qution 3 with qution 5.4.3, we obtin 4 f,, dv h h u r cos, r sin u r cos, r sin f r cos, r sin, r d dr d d r r d dr d FIGU 7 Volume element in clindricl coordintes: dv=r d dr d Formul 4 is the formul for triple integrtion in clindricl coordintes. It ss tht we convert triple integrl from rectngulr to clindricl coordintes b writing r cos, r sin, leving s it is, using the pproprite limits of integrtion for, r, nd, nd replcing dv b r d dr d. (Figure 7 shows how to remember this.) It is worthwhile to use this formul when is solid region esil described in clindricl coordintes, nd especill when the function f,, involves the epression. V XAMPL 3 A solid lies within the clinder, below the plne 4, nd bove the prboloid. (See Figure 8.) The densit t n point is proportionl to its distnce from the is of the clinder. Find the mss of.

54 SCTION 5.7 TIPL INTGALS IN CYLINICAL COOINATS 3 =4 (,, 4) SOLUTION In clindricl coordintes the clinder is r nd the prboloid is r, so we cn write r,,, r, r 4 Since the densit t,, is proportionl to the distnce from the -is, the densit function is f,, Ks Kr FIGU 8 (,, ) (,, ) =-r@ where K is the proportionlit constnt. Therefore, from Formul 5.6.3, the mss of is m Ks dv 4 Kr r d dr d r Kr 4 r dr d K d 3r r 4 dr Kr 3 r 5 5 K 5 M = =œ + XAMPL 4 vlute. s4 d d d s s4 SOLUTION This iterted integrl is triple integrl over the solid region,,, s4 s4, s nd the projection of onto the -plne is the disk 4. The lower surfce of is the cone s nd its upper surfce is the plne. (See Figure 9.) This region hs much simpler description in clindricl coordintes: r,,, r, r FIGU 9 Therefore, we hve s4 d d d dv s4 s r r r d dr d d r 3 r dr [ r 4 5 r 5 ] 6 5 M

55 4 CHAPT 5 MULTIPL INTGALS 5.7 XCISS Plot the point whose clindricl coordintes re given. Then find the rectngulr coordintes of the point.. (), 4, (b) 4, 3, 5. (),, e (b), 3, 3 4 Chnge from rectngulr to clindricl coordintes. 3. (),, 4 (b) (, s3, ) 4. () (s3,, ) (b) 4, 3, 5 6 escribe in words the surfce whose eqution is given Identif the surfce whose eqution is given r 8. 9 Write the equtions in clindricl coordintes. 9. () (b). () 3 6 (b) Sketch the solid described b the given inequlities.. r,,., 3. A clindricl shell is cm long, with inner rdius 6 cm nd outer rdius 7 cm. Write inequlities tht describe the shell in n pproprite coordinte sstem. plin how ou hve positioned the coordinte sstem with respect to the shell. ; 4. Use grphing device to drw the solid enclosed b the prboloids nd Sketch the solid whose volume is given b the integrl nd evlute the integrl. 4 4 r 5. r d d dr Use clindricl coordintes. s dv 7. vlute, where is the region tht lies inside the clinder 6 nd between the plnes 5 nd 4. 3 dv 8. vlute, where is the solid in the first octnt tht lies beneth the prboloid. e dv r r 5 r 9r r d dr d 9. vlute, where is enclosed b the prboloid, the clinder 5, nd the -plne.. vlute, where is enclosed b the plnes nd 5 nd b the clinders 4 nd 9.. Find the volume of the solid tht lies within both the clinder nd the sphere () Find the volume of the region bounded b the prboloids nd (b) Find the centroid of (the center of mss in the cse where the densit is constnt). 4. () Find the volume of the solid tht the clinder r cos cuts out of the sphere of rdius centered t the origin. ; (b) Illustrte the solid of prt () b grphing the sphere nd the clinder on the sme screen. 5. Find the mss nd center of mss of the solid S bounded b the prboloid 4 4 nd the plne if S hs constnt densit K. 6. Find the mss of bll B given b if the densit t n point is proportionl to its distnce from the -is. 7 8 vlute the integrl b chnging to clindricl coordintes dv dv. vlute, where is the solid tht lies within the clinder, bove the plne, nd below the cone 4 4. s4 s4 s s9 d d d s d d d 9. When studing the formtion of mountin rnges, geologists estimte the mount of work required to lift mountin from se level. Consider mountin tht is essentill in the shpe of right circulr cone. Suppose tht the weight densit of the mteril in the vicinit of point P is tp nd the height is hp. () Find definite integrl tht represents the totl work done in forming the mountin. (b) Assume tht Mount Fuji in Jpn is in the shpe of right circulr cone with rdius 6, ft, height,4 ft, nd 3 densit constnt lbft. How much work ws done in forming Mount Fuji if the lnd ws initill t se level? P

56 SCTION 5.8 TIPL INTGALS IN SPHICAL COOINATS 5 ISCOVY POJCT TH INTSCTION OF TH CYLINS The figure shows the solid enclosed b three circulr clinders with the sme dimeter tht intersect t right ngles. In this project we compute its volume nd determine how its shpe chnges if the clinders hve different dimeters. CAS. Sketch crefull the solid enclosed b the three clinders,, nd. Indicte the positions of the coordinte es nd lbel the fces with the equtions of the corresponding clinders.. Find the volume of the solid in Problem. 3. Use computer lgebr sstem to drw the edges of the solid. 4. Wht hppens to the solid in Problem if the rdius of the first clinder is different from? Illustrte with hnd-drwn sketch or computer grph. 5. If the first clinder is, where, set up, but do not evlute, double integrl for the volume of the solid. Wht if? 5.8 TIPL INTGALS IN SPHICAL COOINATS Another useful coordinte sstem in three dimensions is the sphericl coordinte sstem. It simplifies the evlution of triple integrls over regions bounded b spheres or cones. SPHICAL COOINATS The sphericl coordintes,, of point P in spce re shown in Figure, where is the distnce from the origin to P, is the sme ngle s in clindricl coordintes, nd is the ngle between the positive -is nd the line segment OP. Note tht OP P(,, ) O FIGU The sphericl coordintes of point

57 6 CHAPT 5 MULTIPL INTGALS The sphericl coordinte sstem is especill useful in problems where there is smmetr bout point, nd the origin is plced t this point. For emple, the sphere with center the origin nd rdius c hs the simple eqution (see Figure ); this is the reson for the nme sphericl coordintes. The grph of the eqution is verticl hlf-plne (see Figure 3), nd the eqution represents hlf-cone with the -is s its is (see Figure 4). c c c c c c <c<π/ π/<c<π FIGU =c, sphere FIGU 3 =c, hlf-plne FIGU 4 =c, hlf-cone Q The reltionship between rectngulr nd sphericl coordintes cn be seen from Figure 5. From tringles OPQ nd OPP we hve P(,, ) P(,, ) cos But r cos nd r sin, so to convert from sphericl to rectngulr coordintes, we use the equtions r sin O r sin cos sin sin cos Pª(,, ) Also, the distnce formul shows tht FIGU 5 We use this eqution in converting from rectngulr to sphericl coordintes. V XAMPL The point, 4, 3 is given in sphericl coordintes. Plot the point nd find its rectngulr coordintes. SOLUTION We plot the point in Figure 6. From qutions we hve O FIGU 6 π 4 π 3 (, π/4, π/3) sin cos sin sin sin sin cos cos 3 cos 3 sin 3 ( ) Thus the point, 4, 3 is (s3, s3, ) in rectngulr coordintes. M 4 s3 s 3 4 s3 s 3

58 SCTION 5.8 TIPL INTGALS IN SPHICAL COOINATS 7 WANING There is not universl greement on the nottion for sphericl coordintes. Most books on phsics reverse the menings of nd nd use r in plce of. V XAMPL The point (, s3, ) is given in rectngulr coordintes. Find sphericl coordintes for this point. SOLUTION From qution we hve s s 4 4 nd so qutions give cos 4 3 TC In Module 5.8 ou cn investigte fmilies of surfces in clindricl nd sphericl coordintes. 3 cos sin (Note tht becuse s3.) Therefore sphericl coordintes of the given point re 4,, 3. M VALUATING TIPL INTGALS WITH SPHICAL COOINATS i sin k Î k Î r i = i sin k Î Î i Î r i Î = i sin k Î FIGU 7 In the sphericl coordinte sstem the counterprt of rectngulr bo is sphericl wedge where,, nd d c. Although we defined triple integrls b dividing solids into smll boes, it cn be shown tht dividing solid into smll sphericl wedges lws gives the sme result. So we divide into smller sphericl wedges ijk b mens of equll spced spheres i, hlf-plnes j, nd hlf-cones. Figure 7 shows tht ijk is pproimtel rectngulr bo with dimensions, i (rc of circle with rdius i, ngle ), nd i sin k (rc of circle with rdius i sin k, ngle ). So n pproimtion to the volume of is given b In fct, it cn be shown, with the id of the Men Vlue Theorem (ercise 45), tht the volume of is given ectl b ijk,, b, ijk V ijk i i sin k, c d i sin k k where i, j, k is some point in ijk. Let ijk *, ijk *, ijk * be the rectngulr coordintes of this point. Then f,, dv lim l, m, n l l i lim l, m, n l l i m n j k V ijk m n j k f i sin f ijk *, ijk *, ijk * V ijk But this sum is iemnn sum for the function i sin k k cos j, i sin k sin j, i cos k i sin k F,, f sin cos, sin sin, cos sin Consequentl, we hve rrived t the following formul for triple integrtion in sphericl coordintes.

59 8 CHAPT 5 MULTIPL INTGALS 3 f,, dv d c b f sin cos, sin sin, cos sin d d d sin d d where is sphericl wedge given b,, b,, c d d d FIGU 8 Volume element in sphericl coordintes: sin d d d Formul 3 ss tht we convert triple integrl from rectngulr coordintes to sphericl coordintes b writing sin cos using the pproprite limits of integrtion, nd replcing dv b sin d d d. This is illustrted in Figure 8. This formul cn be etended to include more generl sphericl regions such s,, sin sin, c d, t, t, cos In this cse the formul is the sme s in (3) ecept tht the limits of integrtion for re t, nd t,. Usull, sphericl coordintes re used in triple integrls when surfces such s cones nd spheres form the boundr of the region of integrtion. V XAMPL 3 vlute B e 3 dv, where B is the unit bll: B,, SOLUTION Since the boundr of B is sphere, we use sphericl coordintes: B,,, In ddition, sphericl coordintes re pproprite becuse, Thus (3) gives B e 3 dv sin d [cos ] e 3 sin d d d d e 3 d [ 3e 3] 4 3 e M NOT It would hve been etremel wkwrd to evlute the integrl in mple 3 without sphericl coordintes. In rectngulr coordintes the iterted integrl would hve been s s s s e 3 d d d

60 SCTION 5.8 TIPL INTGALS IN SPHICAL COOINATS 9 V XAMPL 4 Use sphericl coordintes to find the volume of the solid tht lies bove the cone s nd below the sphere. (See Figure 9.) (,, ) + +@= π 4 =œ + FIGU 9 N Figure gives nother look (this time drwn b Mple) t the solid of mple 4. SOLUTION Notice tht the sphere psses through the origin nd hs center (,, ). We write the eqution of the sphere in sphericl coordintes s cos The eqution of the cone cn be written s or cos cos s sin cos sin sin sin FIGU TC Visul 5.8 shows n nimtion of Figure. This gives sin cos, or. Therefore the description of the solid in sphericl coordintes is Figure shows how is swept out if we integrte first with respect to, then, nd then. The volume of is,, V d 3 4 dv cos sin cos 3 d cos , 4, cos 3 sin 3 sin d d d cos d 8 FIGU vries from to cos while nd re constnt. vries from to π/4 while is constnt. vries from to π. M

61 CHAPT 5 MULTIPL INTGALS 5.8 XCISS Plot the point whose sphericl coordintes re given. Then find the rectngulr coordintes of the point.. (),, (b), 3, 4. () 5,, (b) 4, 34, 3 9 Set up the triple integrl of n rbitrr continuous function f,, in clindricl or sphericl coordintes over the solid shown Chnge from rectngulr to sphericl coordintes. 3. () (, s3, s3 ) (b),, 4. () (, s3, ) (b) (,, s6 ) 5 6 escribe in words the surfce whose eqution is given Identif the surfce whose eqution is given Write the eqution in sphericl coordintes. 4 Sketch the solid described b the given inequlities..,,. 3, 3., 4., 5. A solid lies bove the cone s nd below the sphere. Write description of the solid in terms of inequlities involving sphericl coordintes. 6. () Find inequlities tht describe hollow bll with dimeter 3 cm nd thickness.5 cm. plin how ou hve positioned the coordinte sstem tht ou hve chosen. (b) Suppose the bll is cut in hlf. Write inequlities tht describe one of the hlves. 7 8 Sketch the solid whose volume is given b the integrl nd evlute the integrl sin sin csc sin d d d sin d d d 3 sin sin cos 9 9. () (b) 9. () (b) 3 34 Use sphericl coordintes.. vlute, where H is the solid hemisphere 9,. 3. vlute, where lies between the spheres nd 4 in the first octnt. 4. vlute, where is enclosed b the sphere 9 in the first octnt. 5. vlute, where is bounded b the -plne nd the hemispheres s9 nd s6. 6. vlute, where lies between the spheres nd nd bove the cone. 7. Find the volume of the prt of the bll tht lies between the cones nd. 8. Find the verge distnce from point in bll of rdius to its center. 9. () Find the volume of the solid tht lies bove the cone nd below the sphere. (b) Find the centroid of the solid in prt (). 3. Find the volume of the solid tht lies within the sphere 4, bove the -plne, nd below the cone s. 3. Find the centroid of the solid in ercise Let H be solid hemisphere of rdius whose densit t n point is proportionl to its distnce from the center of the bse. () Find the mss of H. (b) Find the center of mss of H. (c) Find the moment of inerti of H bout its is. 3 B dv. vlute, where B is the bll with center the origin nd rdius 5. H 9 dv dv e s dv dv dv cos 33. () Find the centroid of solid homogeneous hemisphere of rdius. (b) Find the moment of inerti of the solid in prt () bout dimeter of its bse.