5 Multiple Integrls nd their Applictions 5. INTDUCTIN T DEFINITE INTEGALS AND DUBLE INTEGALS Definite Integrls The concept of definite integrl b f d () is phsicll the re under curve f(), (s), the f A -is nd the two ordintes nd b. It is defined s the limit of the sum b f( )δ + f( )δ + + f( n )δ n when n nd ech of the lengths δ, δ,, δ n Fig. 5. tends to zero. Here δ, δ,, δ n re n subdivisions into which the rnge of integrtion hs been divided nd,,, n re the vlues of ling respectivel in the Ist, nd,, nth subintervls. Double Integrls A double integrl is the counter prt of the bove definition in two dimensions. Let f(, ) be single vlued nd bounded function of two independent vribles nd defined in closed region A in plne. Let A be divided into n elementr res δa, δa,, δa n. Let ( r, r ) be n point inside the rth elementr re δa r. Consider the sum n f, δ A + f, δ A + + f, δ A f, δa ( ) ( ) n n n r r r r 55 () Then the limit of the sum (), if eists, s n nd ech sub-elementr re pproches f, da. to zero, is termed s double integrl of f(, ) over the region A nd epressed s A δa r Fig. 5. A
56 Engineering Mthemtics through Applictions (, ) (, ) Thus n A f da Lt f δa n r δar r r r () bservtions: Double integrls re of limited use if the re evluted s the limit of the sum. However, the re ver useful for phsicl problems when the re evluted b treting s successive single integrls. Further just s the definite integrl () cn be interpreted s n re, similrl the double integrls () cn be interpreted s volume (see Figs. 5. nd 5.). 5. EVALUATIN F DUBLE INTEGAL Evlution of double integrl f (, ) dd -is is discussed under following three possible cses: Ψ C D Cse I: When the region is bounded b two continuous curves ψ () nd φ () nd the two lines (ordintes) nd b. In such cse, integrtion is first performed with respect to keeping s constnt nd then the A resulting integrl is integrted within the limits B φ nd b. b is Mthemticll epressed s: Fig. 5. b Ψ f (, ) d d ( f (, ) d ) d φ Geometricll the process is shown in Fig. 5., is where integrtion is crried out from inner rectngle φ( ) Ψ (i.e., long the one edge of the verticl strip from to ) to the outer rectngle. Cse : When the region is bounded b two continuous b B D curves φ () nd Ψ () nd the two lines (bsciss) nd b. In such cse, integrtion is first performed with A C respect to. keeping s constnt nd then the is resulting integrl is integrted between the two limits Fig. 5. nd b. Mthemticll epressed s: -is b Ψ f (, ) d d f (, ) d d D C Geometricll the process is shown in Fig. 5., where integrtion is crried out from inner rectngle (i.e., long the one edge of the horizontl strip from to ) to the outer rectngle. Cse : When both pirs of limits re constnts, the region of integrtion is the rectngle ABCD (s). b S A B b -is Fig. 5.5
Multiple Integrls nd their Applictions 57 In this cse, it is immteril whether f(, ) is integrted first with respect to or, the result is unltered in both the cses (Fig. 5.5). bservtions: While clculting double integrl, in either cse, we proceed outwrds from the innermost integrtion nd this concept cn be generlized to repeted integrls with three or more vrible lso. Emple : Evlute + ( + + ) dd [Mdrs ; jsthn 5]. Solution: Clerl, here f() vries from to + nd finll (s n independent vrible) goes between to. + I d d ( + ) + + + d d, ( + ) + tn d + tn tn d + + { } d log + + + log + Emple : Evlute e + dd over the tringle bounded b the lines, nd +. Solution: Here the region of integrtion is the tringle AB s the line + intersects the es t points (, ) nd (, ). Thus, precisel the region (s) cn be epressed s:, (Fig 5.7). + I e dd d d + e + e d A (, ) (, ) (, ) B (. 7, ) C B (,. ) Fig. 5.6 (, ) D (,. 6) (, ) (, ) () (. 7, ) Fig. 5.7 (, ) A
58 Engineering Mthemtics through Applictions e e d e e e e e + + e e ( e )( e ) 6 + + 6. Emple : Evlute the integrl nd. + dd over the re between the curves Solution: We hve nd which implies i.e. either or Further, if then ; if then. Mens the two curves intersect t points (, ), (, ). The region of integrtion is doted nd cn be epressed s:,. ( + ) ( + ) dd dd + d 6 7 + + d 5 6 7 d 6 A(, ) (, ) Fig. 5.8 5 7 8 5 6 5 7 8 6 56 Emple : Evlute + dd over the re bounded b the ellipse +. b [U Tech., 5; KUK, 9] Solution: For the given ellipse +, the region of integrtion cn be considered s b
Multiple Integrls nd their Applictions 59 bounded b the curves b, b nd finll goes from to I dd d d b / b / + ( + + ) b / I ( + ) d d b / [Here d s it hs the sme integrl vlue for both limits i.e., the term, which is n odd function of, on integrtion gives zero vlue.] b / I ( + ) d d / b I + d b I b + d n putting sin, d cos d; we get / b I b ( sin cos ) + cos cosd / b b sin cos + cos d Fig. 5.9 Now using formul / sin p q cos d p+ q+ p+ q+ nd / n cos d n + n +, (in prticulr when p, q n) 5 + b dd b +
6 Engineering Mthemtics through Applictions b + b.... ( b ) b b + b + 6 6 ASSIGNMENT. Evlute d d ( )( ). Evlute dd, where A is the domin bounded b the -is, ordinte nd the curve. [M.D.U., ] + b. Evlute e dd, where is the re of the tringle,, + b ( >, b > ). [Hint: See emple ] e d d e dd.. rove tht ( + ) ( + ) 5. Show tht d d d d ( + ). ( + ) 6. Evlute + e ( ) dd [Hint: ut ( + ) t, tking s const.] 5. CHANGE F DE F INTEGATIN IN DUBLE INTEGALS The concept of chnge of order of integrtion evolved to help in hndling tpicl integrls occurring in evlution of double integrls. b Ψ When the limits of given integrl f () (, ) dd φ re clerl drwn nd the region of integrtion is demrcted, then we cn well chnge the order of integrtion be performing integrtion first with respect to s function of (long the horizontl strip from to ) nd then with respect to from c to d. Mthemticll epressed s: d Ψ I f (, ) dd. c φ Sometimes the demrcted region m hve to be split into two-to-three prts (s the cse m be) for defining new limits for ech region in the chnged order.
Multiple Integrls nd their Applictions 6 Emple 5: Evlute the integrl dd b chnging the order of integrtion. [KUK, ; NIT Kurukshetr, ] Solution: In the bove integrl, on verticl strip (s ) vries s function of nd then the strip slides between to. Here is the -is nd i.e., + is the circle. In the chnged order, the strip becomes, resting on the curve, on the circle nd finll the strip sliding between to. I d d I [ ] d ( ) I d Substitute sin, so tht d cos d nd vries from to. I sin cos d ( ) ( ) I 6 p ( p )( p ) ( q )( q ) sin cos d, onl if both p nd q re + ve even integers] ( p+ q)( p+ q ) Emple 6: Evlute dd b chnging the order of integrtion. [M.D.U. ; TU, 9] Solution: In the given integrl, over the verticl strip (s), if chnges s function of such tht lies on the curve nd lies on the curve nd finll the strip slides between to. Here the curve i.e. is prbol with impling impling ± (, ) Fig. 5. A Fig. 5.
6 Engineering Mthemtics through Applictions i.e., it psses through (, ) (, ), (, ). Likewise, the curve or is lso prbol with nd ± i.e., it psses through (, ), (, ), (, ). Clerl the two curves re bounded t (, ) nd (, ). n chnging the order of integrtion over the strip, chnges s function of such tht lies on the curve nd lies on the curve nd finll slides between to. whence I d d [ ] d d 6 6. Emple 7: Evlute ( + ) dd b chnging the order of integrtion. / Solution: In the given integrl ( + ) d d, vries long verticl strip s / function of nd finll s n independent vrible vries from to. Here / i.e. is stright line nd /, i.e. is prbol. For ; nd. Mens the stright line psses through (, ), (, ). For ; nd ±. Mens the prbol psses through (, ), (, ), (, ),. Further, the two curves nd intersect t common points (, ) nd (, ). n chnging the order of integrtion, / ( + ) dd ( ) dd / + (t ) / (, ) Fig. 5.
Multiple Integrls nd their Applictions 6 I + d. ( ) d + + 6 + d 7 5 + 7 5 + 7 5 ( 5 7) + +. 8 Emple 8: Evlute dd. [SVTU, 6] Solution: In the bove integrl, on the verticl strip (s ) vries s function of nd then the strip slides between to. Here the curve i.e., is the prbol nd the curve is the stright line. n the prbol, ; ± i.e., the prbol psses through points (, ), (, ) nd (, ). n chnging the order of integrtion, I d d ( t ) -is d d sin d (, ) (, ) (, ) Fig. 5. -is
6 Engineering Mthemtics through Applictions sin sin d d. 6 Emple 9: Chnge the order of integrtion of dd nd hence evlute the sme. [KUK, ; Cochin, 5; TU, 5; U Tech, 5; SVTU, 7] Solution: In the given integrl d d, on the verticl strip (s), vries s function of nd finll s n independent vrible, vries from to. Here the curve is prbol with B(, ) impling impling ± i.e., it psses through (, ), (, ), (, ). Likewise, the curve is stright line with i.e. it psses though (, ), (, ) nd (, ) Fig. 5. n chnging the order integrtion, the re AB is divided into two prts AC nd ABCA. In the re AC, on the strip, chnges s function of from to. Finll goes from to. Likewise in the re ABCA, over the strip p, chnges s function of from to nd finll the strip slides between to. - C A (, ) (, ) + d d d d d + d ( ) d + d + 6 + 5 I +. 6 8
Multiple Integrls nd their Applictions 65 Emple : Evlute dd b chnging order of integrtion. [KUK, ; MDU, ; JNTU, 5; NIT Kurukshetr, 8] + Soluton: Clerl over the strip, vries s function of such tht lies on the curve nd lies on the curve nd slides between ordintes nd. The curves re, stright line nd, i.e. +, circle. The common points of intersection of the two re (, ) nd (, ). n chnging the order of integrtion, the sme region NM is divided into two prts NL nd LNML with horizontl strips nd sliding between to nd to respectivel. whence I dd+ dd + + Now the ep. d + d + ( + ) + ( + ) I d d I + d + + d + ( ) ( ) M L Fig. 5.5 N (, ) + Emple : Evlute + dd b chnging the order of integrtion. + Solution: Given d d o Clerl in the region under considertion, strip is horizontl with point ling on the curve nd point ling on the curve + nd finll this strip slides between two bsciss nd s shown in Fig 5.6.
66 Engineering Mthemtics through Applictions Now, for chnging the order of integrtion, the region of integrtion under considertion is sme but this time the strip is (verticl) which is function of with etremities nd t nd nd. respectivel nd slides between Thus I d d d d d Tke sin so tht d sin cos d, Also, For, nd for, Therefore, I sin sin sin cos d ( )( ) 8 sin cos d 8 ( ) p q ( p )( p ) ( q )( q ) using sin cos d, ( p+ q)( p+ q ) p nd q both positive even integers Emple : Chnging the order of integrtion, evlute ( + ) A (, ) Fig. 5.6 d d. [MDU, ; Delhi, ; Ann, ; VTU, 5] C (, ) ' ' B (, ) + Solution: Clerl in the given form of integrl, chnges s function of (viz. f() nd s n independent vrible chnges from to. Thus, the two curves re the stright line nd the prbol, nd the common re under considertion is ABCA. For chnging the order of integrtion, we need to convert the horizontl strip to verticl strip over which chnges s function of nd it slides for vlues of to s shown in Fig. 5.7. ( ) I ( + ) dd + d A Fig. 5.7 C(, ) ' B (, )
Multiple Integrls nd their Applictions 67 ( ) ( ) + d ( ) + 8+ d 5 + 8+ + 8 + 5 5 5 8 6 + 8 +. 6 Emple : Evlute log ( + ) d d ( > ) chnging the order of integrtion. [MDU, ] Solution: ver the strip (s), chnges s function of such tht lies on the curve nd lies on the curve nd the strip slides between to Here the curves, is stright line nd i.e. it psses through (, ) nd,. + (, ) A HG B,, K J Also, i.e. + is circle with centre (, ) nd rdius. Thus, the two curves intersect t,. n chnging the order of integrtion, the sme region AB is divided into two prts with verticl strips nd sliding between to respectivel. Fig. 5.8 nd I + d d + + d d / Whence, to log log / ()
68 Engineering Mthemtics through Applictions Now, log( + ) d log( + ) d + Ist IInd Function Function n using (), log + + d + ( + ) + log d ( + ) ( + ) + log tn / I log( ) tn + + d / log tn + d / + d log () / / log d + d For first prt, let t so tht d dt nd limits re t nd t. / log dt I t + t log t +, (B prts with logt logt ) ( log ) + () 8 Agin, using (), log I tn + + d / + / log tn d ()
Multiple Integrls nd their Applictions 69 Let sin so tht d cos d nd limits, to / sin I log sin + sin tn cosd / sin / / ( log ) cos sin cos tn ( cot ) d+ d / / ( cos ) / + / log d+ sin tn tn d / / / / sin / / log + + sin d Ist IInd Fun. Fun. / / cos cos log ( ) d + / / / I ( log ) cosd, / the limits) cos is zero for both + 8 log log ( sin ) log + log + 8 n using results () nd (5), we get I I + I (5) log log log + + + + 8 8 log log 8 8 8 ( ) ( log ) log. 8 Emple : Evlute b chnging the order of integrtion. e / dd [VTU, ; U Tech., 5; SVTU, 6; KUK, 7; NIT Kurukshetr, 7]
7 Engineering Mthemtics through Applictions ( b) f / / Solution: We write e dd e dd ( ) f Here first integrtion is performed long the verticl strip with s function of nd then is bounded between to. We need to chnge, s function of nd finll the limits of. Thus the desired geometr is s follows: In this cse, the strip chnges to with s function of, nd nd finll vries from to. Therefore Integtrl / I e dd, t, ut t so tht d dt Further, for, t t/ dt I e d, t/ e d / e d e d (B prts) e e d e e ( ). ' Fig. 5.9 ' e Emple 5: Evlute the integrl dd. [NIT Jlndhr,, 5; VTU, 7] Soluton: In the given integrl, integrtion is performed first with respect to (s function of long the verticl strip s, from to ) nd then with respect to from to. n chnging the order, of integrtion integrtion is performed first long the horizontl strip '' ( s function of ) from ' to ' nd finll this strip '' slides between the limits to. (, ) Fig. 5.
Multiple Integrls nd their Applictions 7 e I d d e ( d ) e d e e e ( ) f dd. - Emple 6: Chnge the order of integrtion in the double integrl (, ) [jsthn, 6; KUK, -5] Solution: Clerl from the epressions given bove, the region of integrtion is described b line which strts from nd moving prllel to itself goes over to, nd the etremities of the moving line lie on the prts of the circle + the prbol in the first qudrnt. For chnge nd of order of integrtion, we need to consider the sme region s describe b line moving prllel to -is insted of -is. In this w, the domin of integrtion is divided into three sub-regions I, II, III to ech of which corresponds double integrl. Thus, we get f(, ) dd f (, ) dd / rt I + f (, ) dd + f(, ) dd + / rt II rt III Emple 7: Find the re bounded b the lines sin, cos nd. Ι ΙΙ (, ) ΙΙΙ (, ) ( ) + Fig. 5., sin Solution: See Fig 5.. Clerl the desired re is the doted portion where long the strip, lies on the curve sin nd lies on the curve cos nd finll the strip slides between the ordintes nd. cos Fig. 5.
7 Engineering Mthemtics through Applictions cos d d d d sin ( cos sin ) d ( sin cos) / + +. ( ) ASSIGNMENT. Chnge the order of integrtion dd +. Chnge the order integrtion in the integrl (, ) α cos. Chnge the order of integrtion in f (, ) tnα f dd dd l. Chnge the order of integrtion in fdd (, ) [TU, 8] m 5. EVALUATIN F DUBLE INTEGAL IN LA CDINATES β rψ() To evlute f ( r, ) dr d, we first integrte with respect to r between the limits α rφ() r φ() to r ψ() keeping s constnt nd then the resulting epression is integrted with respect to from α to β. Geometricl Illustrtion: Let AB nd CD be the two continuous curves r φ() nd r Ψ() bounded between the lines α nd β so tht ABDC is the required region of integrtion. Let be rdil strip of ngulr thickness δ when mkes n ngle with the initil line. f r dr rφ refers to the integrtion with rψ Here (, ) respect to r long the rdil strip nd then integrtion with respect to mens rottion of this strip from AC to CD. r φ β δ B α Fig. 5. r Ψ A D C
Multiple Integrls nd their Applictions 7 Emple 8: Evlute rsindrd over the crdiod r ( cos) bove the initil line. Solution: The region of integrtion under considertion is the crdiod r ( cos ) bove the initil line. In the crdiod r ( cos ); for, r,, r,, r As cler from the geometr long the rdil strip, r (s function of ) vries from r to r ( cos ) nd then this strip slides from to for covering the re bove the initil line. Hence r ( cos ) I rdrsind ( cos) r sind ( cos) sind ( cos), n f f d f n+ ( cos ) ( cos) [ 8 ]. 6 6 B (, ) n+ Fig. 5. / (, ) A (, /) C Emple 9: Show tht the initil line. r sin drd, where is the semi circle r cos bove / Solution: The region of integrtion is the semi-circle r cos bove the initil line. For the circle r cos, r r therwise lso, r cos r rcos + ( + ) + ( ) + ( ) (, ) (, ) (, ) Fig. 5.5 r cos
7 Engineering Mthemtics through Applictions i.e., it is the circle with centre (, ) nd rdius r Hence the desired re cos r sindrd cos rdrsind / cos r sind / cos sind 8 cos. /, using n+ n f () f () f'() d n+ Emple : Evlute rdrd + r over one loop of the lemniscte r cos. [KUK, ; MDU, 6] Solution: The lemniscte is bounded for r impling ± nd mimum vlue of r is. See Fig. 5.6, in one complete loop, r vries from to r cos nd the rdil strip slides between to. Hence the desired re A / cos r dr d / ( + r ) / cos d+ r drd / / / cos + r d / / ( r, ) / ( cos ) + d / Fig. 5.6 / / ( cos ) d
Multiple Integrls nd their Applictions 75 / cos d / sin. Emple : Evlute rdrd, over the re included between the circles r cos nd r b cos (b < ). [KUK, ] Solution: Given r cos or r rcos + ( + ) + ( ) i.e this curve represents the circle with centre (, ) nd rdius. Likewise, r b cos represents the circle with centre (b, ) nd rdius b. We need to clculte the re bounded between the two circles, where over the rdil strip, r vries from circle r b cos to r cos nd finll vries from to. Thus, the given integrl cos rdrd bcos / cos rdrd r d / bcos / ( cos ) ( bcos ) d / cos b d 8 b cos d 8( b ) ( b ). (, ) rticulr Cse: When r cos nd r cos i.e., nd b, then 5 I ( b ) ( ) units. r b cos Fig 5. 7 b r cos
76 Engineering Mthemtics through Applictions ASSIGNMENT. Evlute rsindrd over the re of the cridod r ( + cos) bove the initil line. Hint : ( + ) cos I rsindrd. Evlute rdrd, over the re included between the circles r cos nd r b cos (b > ). [Mdrs, 6] Hint : r bcos I r dr d r cos (See Fig. 5.7 with nd b interchnged). Find b double integrtion, the re ling inside the crdiod r ( + cos) nd outside the prbol r( + cos). [NIT Kurukshetr, 8] Hint : / (+ cos ) + cos rdr d 5.5 CHANGE F DE F INTEGATIN IN DUBLE INTEGAL IN LA CDINATES β rψ In the integrl f ( r, ) drd α rφ, intertion is first performed with respect to r long rdil strip nd then this trip slides between two vlues of α to β. In the chnged order, integrtion is first performed with respect to (s function of r long circulr rc ) keeping r constnt nd then integrte the resulting integrl with respect to r between two vlues r to r b (s) Mthemticll epressed s β r Ψ ( ) r b () r f r, drd I η f () ( r, ) ddr α rφ r f r / cos Emple : Chnge the order of integrtion in the integrl f r drd, Solution: Here, integrtion is first performed with respect to r (s function of ) long rdil strip (s) from r to r cos nd finll this rdil strip slides between to. Curve r cos r rcos + ( ) + i.e., it is circle with centre (, ) nd rdius. n chnging the order of integrtion, we hve to first integrte with respect to (s function of r) long r cos or (, ) (, ) Fig. 5.8 cos r (, )
Multiple Integrls nd their Applictions 77 the circulr strip (s) with pt. on the curve nd pt. on the curve cos nd finll r vries from to. cos r cos I f r drd f r d dr (, ) (, ) Emple : Sketch the region of integrtion f( r ) of integrtion. e/ / log r Solution: Double integrl f ( r, ) rdrd is identicl to r e / r log, rdrd nd chnge the order rβ f() r f(, r ) rdrd, whence rα f() r integrtion is first performed with respect to s function of r i.e., f(r) long the circulr strip (s) with point on the curve log r nd point on the curve nd finll this strip slides between between r to r e /. (See Fig. 5.9). r r The curve log implies log e or r e / Now on chnging the order, the integrtion is first performed with respect to r s function of viz. r f() long the rdil strip (s) nd finll this strip slides between to. (Fig. 5.). / / r M L / r e r log r / or / r e (, /) B Ce ( /, /) δ / r e (, ) A r Fig. 5.9 Fig. 5. / / r e I f r rdr d r (, )
78 Engineering Mthemtics through Applictions 5.6. AEA ENCLSED B LANE CUVES. Crtesin Coordintes: Consider the re bounded b the two continuous curves φ() nd Ψ() nd the two ordintes, b (Fig. 5.). Now divide this re into verticl strips ech of width δ. Let (, ) nd S( + δ, + δ) be the two neigbouring points, then the re of the elementr shded portion (i.e., smll rectngle) δδ But ll the such smll rectngles on this strip re of the sme width δ nd chnges s function of from φ() to Ψ() The re of the strip Now on dding such strips from, we get the desired re ABCD, Ψ b Ψ Ψ b Ψ δ φ δ φ φ( ) φ( ) Lt d d d dd Lt Lt Ψ d Ψ δ δ δ δ d δ δ φ φ Likewise tking horizontl strip (s) s shown, the re ABCD is given b b Ψ dd φ olr Coordintes: Let be the region enclosed b polr curve with (r, ) nd (r + δr, + δ) s two neighbouring points in it. Let be the circulr re with rdii nd equl to r nd r + δr respectivel. Here the re of the curviliner rectngle is pproimtel δr r sin δ δr r δ r δr δ. If the whole region is divided into such smll curviliner rectngles then the limit of the sum Σrδrδ tken over is the re A enclosed b the curve. i.e., A Lt rδrδ rdrd δ r δ Emple : Find b double integrtion, the re ling between the curves nd. Solution: The given curve is prbol. b (, ) C φ δ A A C δ S δ D B b Fig. 5. δ δ Fig. 5. rδ Fig. 5. B D δr Ψ
Multiple Integrls nd their Applictions 79 where in i.e., it psses through points (, ), (, ), (, ), (, ), (, ). Likewise, the curve is stright line where i.e., it psses through (, ), (, ), (, ) C B(, ) (, ) A(, ) Now for the two curves nd to intersect, or + i.e.,, which in turn implies, respectivel. D(, ) Thus, the two curves intersect t (, ) nd (, ), Clerl, the re need to be required is ABCDA. Fig. 5. A dd d ( ) 9 units. Emple 5: Find b double integrtion, the re ling between the prbol nd the line. [KUK, ] Solution: For the given curve ; i.e. it psses through the points (, ), (, ), (, ) nd (, ). Likewise, the curve psses through (, ) nd (, ), nd hence, (, ) nd (, ) re the common points. therwise lso putting into, we get,. (, ) (, ) A B(, ) Fig. 5.5 C(, ) (, )
8 Engineering Mthemtics through Applictions See Fig. 5.5, ABC is the re bounded b the two curves nd Are ABC dd d 9 ( ) d units Emple 6: Clculte the re of the region bounded b the curves nd + [JNTU, 5] Solution: The curve is prbol where, i.e., it psses through (, ), (, ), (, ). ± Likewise, for the curve +, Hence it psses through points (, ), (, ), (, ), (, ). Also for the curve ( + ), (i.e. -is) is n smptote. For the points of intersection of the two curves nd + we write + or ( + ) Then i.e. (, ) nd (, ) re the two points of intersection. (, ) A Fig. 5.6
Multiple Integrls nd their Applictions 8 The re under considertion, A d d d + + log ( + ) ( log6 log) log. Emple 7: Find b the double integrtion, the re ling inside the circle r sin nd outside the crdiod r ( cos). [KUK 5; NIT Kurukshetr 7] Soluton: The re enclosed inside the circle r sin nd the crdiod r ( cos) is shown s doted one. For the rdil strip, r vries from r ( cos) to r sin nd finll vries in between to. / For the circle r sin r r r Likewise for the crdiod r ( cos); r r r Thus, the two curves intersect t nd. sin A rdrd ( cos ) / sin r d ( cos) / sin cos cos + d (, ) / [ cos + cos ] d, since (sin cos ) cos A Fig. 5.7 r sin r ( cos )
8 Engineering Mthemtics through Applictions / sin. sin + Emple 8: Clculte the re included between the curve r (sec + cos) nd its smptote r sec. [NIT Kurukshetr, 7] Solution: As the given crve r (sec + cos) i.e., r + cos cos contins cosine terms onl nd hence it is smmetricl bout the initil is. Further, for, r nd, r goes on decresing bove nd below the initil is s pproches to nd t, r. Clerl, the required re is the doted region in which r vries long the rdil strip from r sec to r (sec + cos) nd finll strip slides between to. (sec+ cos ) A rdrd sec / ( sec+ cos) r d sec / cos + d cos ( cos ) / + d ( 5 cos ) / + d cos / sin 5 5 +. ASSIGNMENT. Show b double integrtion, the re bounded between the prbol nd 6 is. [MDU, ; NIT Kurukshetr, ]. Using double integrtion, find the re enclosed b the curves, nd. r sec r (sec + cos ) [TU, 5] Emple 9: Find b double integrtion, the re of lminiscte r cos. [Mdrs, ] Solution: As the given curve r cos contins cosine terms onl nd hence it is smmetricl bout the initil is. Fig. 5.8
Multiple Integrls nd their Applictions 8 Further the curve lies wholl inside the circle r, since the mimum vlue of cos is. Also, no portion of the curve lies between to nd the etended is. See the geometr, for one loop, the curve is bounded between to / / / B A r cos Are rdrd r Fig. 5.9 / cos r d / / sin cosd 5.7 CHANGE F VAIABLE IN DUBLE INTEGAL The concept of chnge of vrible hd evolved to fcilitte the evlution of some tpicl integrls. Cse : Generl chnge from one set of vrible (, ) to nother set of vribles (u, v). If it is desirble to chnge the vribles in double integrl (, ) f da b mking φ(u, v) nd Ψ(u, v), the epression da (the elementr re δδ in ) in terms of u nd v is given b,, da J dudv, uv, J uv, J is the Jcobin (trnsformtion coefficient) or functionl determinnt., f dd F u v J dudv uv, (, ) (, ) Cse : From Crtesin to olr Coordintes: In trnsforming to polr coordintes b mens of r cos nd r sin,, r cos sin J r, rsin rcos r da r dr d nd (, ) (, ) f dd F r rdrd
8 Engineering Mthemtics through Applictions Emple : Evlute ( + ) d d where is the prllelogrm in the plne with vertices (, ), (, ), (, ), (, ) using the trnsformtion u +, v. [KUK, ] Solution: is the region bounded b the prllelogrm ABCD in the plne which on trnsformtion becomes úv i.e., the region bounded b the rectngle S, s shown in the Figs. 5. nd 5. respectivel. V C (, ) (, ) (, ) D (, ) A (, ) B (, ) (, ) u u U S (, ) (, ) Fig. 5. Fig. 5. u + u + With v }, A (, ) trnsforms to v } i.e., (, ) B(, ) trnsforms to (, ) C(, ) trnsforms to (, ) D(, ) trnsforms to S(, ) nd (, ) J u v ( uv, ) u v Hence the given integrl u dudv ududv [ v] udu + udu u 6 I units
Multiple Integrls nd their Applictions 85 Emple : Using trnsformtion + u, uv, show tht + e dd e [TU, ]. Solution: Clerl f() represents curves nd, nd which is n independent vrible chnges from to. Thus, the re AB bounded between the two curves nd + nd the two ordintes nd is shown in Fig. 5.. B (, ) n using trnsformtion, + u u( v) uv uv + Now point (, ) implies u( v) () nd uv () A (, ) (, ) From (), either u or v or both zero. From (), we get u, v Fig. 5. Hence (, ) (, ) trnsforms to (u, v) (, ), (, ) oint A(, ), implies u( v) () nd uv () From () either u or v, If v then from () we hve u, gin if u, eqution () is inconsistent. Hence, A(, ) trnsforms to (, ), i.e. itself. From oint B(, ), we get u( v) (5) nd vu (6) B (, ) (, ) From (5), either u or v If u, eqution (6) becomes inconsistent. If v, the eqution (6) gives u. Hence (, ) trnsform to (, ). See Fig. 5.. Hence (, ) ( uv, ) + v where e dd ue dudv J u v u u e dv du u e du e e (, ) A (, ) Fig. 5. Emple : Evlute the integrl dd + b trnsforming to polr coordintes.
86 Engineering Mthemtics through Applictions Solution: Here the curves or is prbol pssing through (, ), (, ). Likewise the curve is stright line pssing through points (, ) (, ). Hence the two curves intersect t (, ), (, ). In the given form of the integrl, chnges (s function of ) from to nd finll s n independent vrible vries from to. For trnsformtion to polr coordintes, we tke r cos, r sin nd (, ) ( r, ) J r The prbol implies r sin rcos so tht r(s function of ) vries from cos r to r nd vries from to sin Therefore, on trnsformtion the integrl becomes ( cos sin ) cos / r sin r / I rdrd r cos / sin r cos / d / 6 cos ( sin ) d / sin ( sin )( sin ) / 8 d / sin / sin + sin 8 d / sin / / 8 cosec + cot cosec + d / / 8 cot cosec cosec + d / (, ) A(, ) B (, ) Fig. 5.
Multiple Integrls nd their Applictions 87 8 cot cosec + cot + / / d / / Let cot t so tht cosec d dt. Limits for, t, t 8 t dt + ( ) + t 8 + 5 8. / Emple : Evlute the integrl ( + ) / dd b chnging to polr coordintes. Solution: The bove integrl hs lred been discussed under chnge of order of integrtion in crtesin co-ordinte sstem, Emple 7. For trnsforming n point (, ) of crtesin coordinte to polr coordintes (r, ), we tke r cos, r sin nd (, ) ( r, ) J r. The prbol implies r cos cos r sin i.e., r rsin either r or Limits, for the curve, nd for the curve cos r sin BA tn tn tn B tn cos Here r (s function of ) vries from to sin nd chnges from tn to. / (, ) Fig. 5.5 A(, ) B (, )
88 Engineering Mthemtics through Applictions Therefore, the integrl, / / ( + ) cos / sin trnsforms to. I r dr d tn cos / r sin I drd cot / cos d cot (sin ) cot cot cosec + d cot I Let cot t so tht cosec d dt ( ) nd cot t t I t t dt + I 5 7 6 t t t + t dt + 5 7 I +. 8 n Emple : Evlute ( + ) dd over the positive qudrnt of +, supposing n + >. [SVTU, 7] Solution: The double integrl is to be evluted over the re enclosed b the positive qudrnt of the circle +, whose centre is (, ) nd rdius. Let r cos, r sin, so tht + r. Therefore on trnsformtion to polr co-ordintes, Circle r / r n I rcosrsin r J drd, r / n+ r dr sincos d, (, ) ( r, ) J r / r n + sincosd n + Fig. 5.6
Multiple Integrls nd their Applictions 89 n+ sincosd n + n+ / sin ( n+ ), using f f d f n+, + >. ( n ) n + Emple 5: Trnsform to crtesin coordintes nd hence evlute the r sincosdrd. [NIT Kurukshetr, 7] Solution: Clerl the region of integrtion is the re enclosed b the circle r, r between to. Here I r sin cos drd r sin r cos rdrd n using trnsformtion r cos, r sin, I dd dd d ( ) d As nd both re odd functions, therefore net vlue on integrtion of the bove integrl is zero. d. i.e. I ( ) Fig. 5.7 Circle r or + ASSIGNMENTS 5 Evlute the following integrls b chnging to polr coordintes: () + dd () + dd
9 Engineering Mthemtics through Applictions () dd () + ( e ) dd [MDU, ] 5.8 TILE INTEGAL (HSICAL SIGNIFICANCE) The triple integrl is defined in mnner entirel nlogous to the definition of the double integrl. Let F(,, z) be function of three independent vribles,, z defined t ever point in region of spce V bounded b the surfce S. Divided V into n elementr volumes δv, δv,, δv n nd let ( r, r, z r ) be n point inside the rth sub division δv r. Then, the limit of the sum n F( r, r, zr) δvr, () r if eists, s n nd δv r is clled the triple integrl of (,, z) over the region V, nd is denoted b FzdV (,, ) () In order to epress triple integrl in the integrted form, V is considered to be subdivided b plnes prllel to the three coordinte plnes. The volume V m then be considered s the sum of number of verticl columns etending from the lower surfce s, z f (, ) to the upper surfce s, z f (, ) with bse s the elementr res δa r over region in the -plnce when ll the columns in V re tken. n summing up the elementr cuboids in the sme verticl columns first nd then tking the sum for ll the columns in V, it becomes F ( r, r, zr) δzδar () r r with the pt. ( r, r, z r ) in the rth cuboid over the element δa r. When δa r nd δz tend to zero, we cn write () s z f(, ) F,, z dz z f, da Note: An ellipsoid, rectngulr prllelopiped nd tetrhedron re regulr three dimensionl regions. Z Z f(, ) Fig. 5.8 δa r Z f (, ) 5.9. EVALUATIN F TILE INTEGALS For evlution purpose, (,, ) V FzdV is epressed s the repeted integrl ()
z Multiple Integrls nd their Applictions 9 (,, ) F z dzdd z () where in the order of integrtion depends upon the limits. If the limits z nd z be the functions of (, ); nd be the functions of nd, be constnt, then bφ z f, I F(,, z) dz d d φ () z f, which shows tht the first F(,, z) is integrted with respect to z keeping nd constnt between the limits z f (, ) to z f (, ). The resultnt which is function of, is integrted with respect to keeping constnt between the limits f () to f (). Finll, the integrnd is evluted with respect to between the limits to b. Note: This order cn ccordingl be chnged depending upon the comfort of integrtion. Emple 6: Evlute + + + e zdzdd. [KUK,, 9] Solution: n integrting first with respect to z, keeping nd constnts, we get ( + ) + + z I e dd, [Here ( + ), (s), like some constnt] ( + ) + ( + ) ( + ) + e e dd ( + ) ( + e e ) dd + + e e d, (Integrting with respect to, keeping constnt) n integrting with respect to, e e e e d e e e e + 8 e e e e + + 8 8. e I e + e 8 8 Emple 7: Evlute r / sin. rdrddz [VTU, 7; NIT Kurukshetr, 7, ]
9 Engineering Mthemtics through Applictions Solution: n integrting with respect to z first keeping r nd constnts, we get / sin r I z rdrd / sin r rdrd / sin r r d, (n integrting with respect to r) / sin sin d sin sin d, / p ( p ) ( p ) sin d ;onl if p is even ()( p p ) 5 I 8 6 Emple 8: Evlute e log e log zdzdd. [MDU, 5; KUK,, 5] Solution: e log e log zdz dd [Here z f(, ) with z nd z e + e log e log z dzdd Ist IInd fun. fun. e log log z z z dz dd z e log ( log log) e e e z dd e
Multiple Integrls nd their Applictions 9 e log e log e e d d ( ) e + d d e e I function n integrting b prts, log e e + d log ( ) + + d II function e e I log + + d + e e (log e) e log + + + d + ( e e ) ( ) e e + e e e + + e e + e e + + + e e ( 8 e e + ). Emple 9: Evlute ( + + ) z + z. z dddz [JNTU, ; Cochin, 5] z Solution: Integrting first with respect to, keeping nd z constnt, + z z I + + z ddz ( ) z z + z d dz z z z dz + z + z z z dz z zdz e
9 Engineering Mthemtics through Applictions ASSIGNMENT 6 Evlute the following integrls: () b c zdddz () ( + + ) b c z dddz [VTU, ] () z z dd dz () log + log e + + zdzdd [NIT Kurukshetr, 8] 5. VLUME AS A DUBLE INTEGAL (Geometricl Interprettion of the Double Integrl) ne of the most obvious use of double integrl is the determintion of volume of solids viz. volume between two surfces. If f(, ) is continuous nd single vlued function Z defined over the region in the -plne with z f(, ) s the eqution of the surfce. Let be the closed curve which encloses. Clerl, the surfce (viz. z f(, )) is the orthogonl projection of S(viz z F(, )) in the -plne. Divided into elementr rectngles of re δδ b drwing lines prllel to the is of nd. n ech of these rectngles errect prisms hving their lengths S C prllel to the z-is. The volume of ech such prism is zδ δ. (Division of is performed with the lines i (i,,, m) nd j (j,,, n). Through ech line δ i, pss plne prllel to z-plne, nd through δ, δ ech line j, pss plnce prllel to z-plne. The rectngle ij whose re is A ij i j will be the Fig. 5.9 bse of rectngle prism of height f( ij, h ij ), whose volume is pproimtel equl to the volume between the surfce nd the -plne i, n i ; i i. Then (, ) i j f ξ η ij ij i j gives n pproimte vlue for volume V of the prism of the clinder enclosed between z f(, ) nd the -plne. The volume V is the limit of the sum of ech elementr volume z δδ. δ δ (, ) V Lt z zdd f da δ δ Note: In cllidricl co-ordintes, the eqution of the surfce becomes z f(r, ), elementr re da r dr d nd volume f( r, ) rdrd
Multiple Integrls nd their Applictions 95 roblems on Volume of Solid with the Help of Double Integrl z Emple : Find the volume of the tetrhedron bounded b the plne + + nd b c the co-ordinte plnes. [Burdwn, ] z Solution: Given, z f (, ) c + + b c b () If f(, ) is continuous nd single vlued function over the region (see Fig. 5. 5) in the plne, then z f(, ) is the eqution of the surfce. Let C be the closed curve tht is the boundr of. Using s bse, construct clinder hving elements prllel to the z-is. This clinder intersects z f(, ) in curve Γ, whose projection on the -plne is C. Z Z C( / b / ) f(, ) (,, c) c (,, ) b (, b, ) C Fig. 5.5 Fig. 5.5 The eqution of the surfce under which the region whose volume is required, m be written in the form () i.e., z c. b Hence the volume of the region da c dd b The eqution of the inter-section of the given surfce with -plne is + () b If the prisms re summed first in the -direction the will be summed from to the line b Therefore, b V c dd b b( / ) c d b
96 Engineering Mthemtics through Applictions c b + d cb + 6 bc bc + 6 6. Emple : rove tht the volume enclosed between the clinders + nd z 8 is. 5 Solution: Let V be required volume which is enclosed b the clinder + nd the prboloid z. nl hlf of the volume is shown in Fig 5.5. Z Now, it is evident from tht z is to be evluted over the circle + (with centre t (, ) nd rdius Z. Here vries from to on the circle + nd finll vries from to V [ ] z d d s z f(, ) dd d d d d d Let sin, so tht d sin cos d. Further, for,,. V sin cos sincosd 6 sin cos d (, ) (, ) Fig. 5.5 (, ) +
Multiple Integrls nd their Applictions 97 ( p )( p ) ( q )( q ) ( p+ q)( p+ q ) p q 6,, ( ) 6 8. 5 5 roblems bsed on Volume s Double Integrl in Clindricl Coordintes Emple : Find the volume bounded b the clinder + nd the hperboloid + z. Solution: In crtesin co-ordintes, the section of the given hperboloid + z in the plne (z ) is the circle +, where s t the top nd t the bottom end (long the z-is i.e., z ± ) it shres common boundr with the circle + (Fig. 5.5 nd 5.5). Here we need to clculte the volume bounded b the two bodies (i.e., the volume of shded portion of the geometr). Z + ( r ) Fig. 5.5 Fig. 5.5 + ( r ) (Best emple of this geometr is solid dmroo in concentric long hollow drum.) In clindricl polr coordintes, we see tht here r vries from r to r nd vries from to. V zdd f r r dr d (, ) r rdrd (³ + z z dr d + )
98 Engineering Mthemtics through Applictions ( r ) d d. Emple : Find the volume bounded b the clinder + nd the plnes + z nd z. [KUK, ; MDU, ; Cochin, 5; SVTU, 7] Solution: From Fig. 5.55, it is ver cler tht z is to be integrted over the circle + in the -plne. To cover the shded portion, vries from to nd vries from to. Hence the desired volume, V zdd dd ( ) d d d 8 d d (The second term vnishes s the integrnd is n odd function) 8 + sin 6. Z Fig. 5.55 ASSIGNMENT 7. Find the volume enclosed b the coordinte plnes nd the portion of the plne l + m + nz ling in the first qudrnt.. btin the volume bounded b the surfce z c nd the qudrnt of b the elliptic clinder + b [Hint: Use elliptic polr coordintes rcos, brsin ]
Multiple Integrls nd their Applictions 99 5. VLUME AS A TILE INTEGAL Divide the given solid b plnes prllel to the coordinte plne into rectngulr prllelopiped of elementr volume δδδz. Then the totl volume V is the limit of the sum of ll elementr volume i.e., V Lt δδδ z ddd δ δ δz roblems bsed on Volume s Triple Integrl in crtesin Coordinte Sstem Emple : Find the volume common to the clinders + nd + z. Solution: The sections of the clinders + nd + z re the circles + nd + z in nd z plne respectivel. Here in the picture, one-eighth prt of the required volume (covered in the st octnt) is shown. Clerl, in the common region, z vries from to i.e.,, nd nd vr on the circle +. The required volume z V 8 z ( z ) 8 dd dzdd C Z D 8 d d 8 d d 8 d (,, ) A Fig. 5.56 B 8 ( ) d 8 6 8.
Engineering Mthemtics through Applictions Emple 5: Find the volume bounded b the plne, the clinder + nd the plne + + z. Solution: Let V(,, z) be the desired volume enclosed lterll b the clinder + (in the -plne) nd on the top, b the plne + + z ( s). Clerl, the limits of z re from (on the Z -plne) to z ( ) nd nd vr on the circle + V(,, z) dzdd ( ) ( z ) dd ( ) d d d + + ( 6 ) I d n tking sin, we get d d; For, For, Thus, / / V 6 sin sin sin cosd Fig. 5.57 / / 6cos sincos d / / 6 cos d sincos d / Ist IInd Using / cos + + pp ( ) / / cos d, onl if p is even nd p p p
Multiple Integrls nd their Applictions n+ n f f f d for Ist nd IInd integrl respectivel n+ z Emple 6: Find the volume bounded b the ellipsoid + +. b c [MDU, ; KUK, ; Kottm, 5; TU, 6] Solution: Considering the smmetr, the desired volume is 8 times the volume of the ellipsoid into the positive octnt. Z The ellipsoid cuts the plne in the ellipse C + nd z. b Therefore, the required volume lies between the ellipsoid z c b nd the plne (i.e., z ) nd is bounded on the sides b the plnes nd Hence, b c b V 8 b c 8 b dzdd 8 α c α d d b dd c α α V 8 sin b + α α Using formul c 8 + sin d b α α tking b c c α α d b d b b b b bc bc. d, A Fig. 5.58 tn d + B
Engineering Mthemtics through Applictions dddz Emple 7: Evlute the integrl tken throughout the volume z of the sphere. [MDU, ] Solution: Here for the given sphere + + z, n of the three vribles,, z cn be epressed in term of the other two, s z ±. In the -plne, the projection of the sphere is the circle +. Thus, I 8 d ddz z 8 dz d d α z, α ( ) 8 sin d α z d α C Z Z 8 ( sin sin ) d d 8 d d d A d + sin Circle + Fig. 5.6 B + I. Emple 8: Evlute ( + + ) z dddz over the tetrhedron bounded b the plnes,, z nd + + z. Solution: The integrtion is over the region (shded portion) bounded b the plne,, z nd the plne + + z. The re AB, in plne is bounded b the lines +,, Hence for n pt. (, ) within this tringle, z goes from plne to plne ABC (viz. the surfce of the tetrhedron) or in other words, z chnges from z to z. Likewise in plne, s function vries from to nd finll vries from to. I whence, ( + + z) dd dz ( over) ( + + z ) dz d d
Multiple Integrls nd their Applictions z ( + ) z+ dd ( ) ( ) + + dd ( )( + + ) dd ( + ) d d ( + ) d, ( ) + + 8 d (,, ) A Z C (,, ) (,, ) Fig. 5.6 (,, ) B ASSIGNMENT 8. Find the volume of the tetrhedron bounded b co-ordinte plnes nd the plne z + +, b using triple integrtion [KUK, ] b c. Find the volume bounded b the prboloid + z, the clinder + nd the plne z. 5.. VLUMES F SLIDS F EVLUTIN AS A DUBLE INTEGAL Let (, ) be n point in region enclosing n elementr re d d round it. This elementr re on revolution bout -is form ring of volume, δv [( + δ) δ ] δ δδ () Hence the totl volume of the solid formed b revolution of this region bout -is is, V dd () Similrl, if the sme region is revolved bout -is, then the required volume becomes V d d () Fig. 5.6 δ (, )
Engineering Mthemtics through Applictions Epressions for bove volume in polr coordintes bout the initil line nd bout the pole re drd nd r sin drd respectivel. r cos Emple 9: Find b double integrtion, the volume of the solid generted b revolving the ellipse + bout -is. b B Solution: As the ellipse + is smmetricl bout the b -is, the volume generted b the left nd the right hlves overlp. Hence we shll consider the revolution of the right-hlf ABD for which -vries from to nd -vries from b to b. b C D Fig. 5.6 A b b b b V dd b b b b b d b d b b b ( b ) d b b b b. Emple 5: The re bounded b the prbol nd the stright lines nd, in the first qudrnt is revolved bout the line. Find b double integrtion the volume of the solid generted. Solution: Drw the stndrd prbol to which the stright line meets in the point (, ), Fig. 5.6. Now the dotted portion i.e., the re enclosed b prbol, the line nd is revolved bout the line. The required volume, V dd d d b (, ) 8 Fig. 5.6
Multiple Integrls nd their Applictions 5 Emple 5: Clculte b double integrtion, the volume generted b the revolution of the crdiod r ( cos) bout its is. [KUK, 7, 9] Soluton: n considering the upper hlf of the crdiod, becuse due to smmetr the lower hlf genertes the sme volume. ( cos ) sin ( cos) V r drd r sind ( cos ) sin d ( cos) 8. Emple 5: B using double integrl, show tht volume generted b revolution of Fig. 5.65 / crdiod r ( + cos) bout the initil line is Solution: The required volume (+ cos ) r sin drd ( cos ) + r sin d ( cos ) sin + d ( + cos) 8. 8. / Fig. 5.66 ASSIGNMENT 9. Find b double integrtion the volume of the solid generted b revolving the ellipse + bout the -is. b. Find the volume generted b revolving qudrnt of the circle +, bout its dimeter.. Find the volume generted b the revolution of the curve ( ), bout its smptote through four right ngles.. Find the volume of the solid obtined b the revolution of the leminiscte r cos bout the initil line. [Jmmu Univ., ]
6 Engineering Mthemtics through Applictions 5.. CHANGE F VAIABLE IN TILE INTEGAL For trnsforming elementr re or the volume from one sets of coordinte to nother, the necessr role of Jcobin or functionl determinnt comes into picture. () Triple Integrl Under Generl Trnsformtion ( z,, ) Here f(,, z) dddz F( u, v, w) J dudvdw; where J ( ) z (,, ) '( uvw,, ) ( uvw,, ) () Since in the cse of three vribles u(,, z), v(,, z), w(,, z) be continuous together with their first prtil derivtives, the Jcobin of u, v, w with respect to,, z is defined b u v w u v w u v w z z z (b) Triple Integrl in Clindricl Coordintes Here (,, ) (,, ) f z dd dz F r z J dr ddz, where J r The position of point in spce in clindricl coordintes is determined b the three numbers r,, z where r nd re polr co-ordintes of the projection of the point on the -plne nd z is the z coordinte of i.e., distnce of the point () from the -plne with the plus sign if the point () lies bove the -plne, nd minus sign if below the -plne (Fig. 5.67). Z Z (,, z) z M N r z Fig. 5.67 Fig. 5.68 In this cse, divide the given three dimensionl region ' (r,, z) into elementr volumes b coordinte surfces r r i, j, z z k (viz. hlf plne djoining z-is, circulr clinder is coincides with Z-zis, plnes perpenduculr to z-is). The r r
Multiple Integrls nd their Applictions 7 curviliner prism shown in Fig. 5. 68 is volume element of which elementr bse re is r r nd height z, so tht v r r z. Here is the ngle between nd the positive -is, r is the distnce nd z is the distnce. From the Fig. 5.6, it is evident tht r cos, r sin, z z nd so tht, cos sin z,, J rsin rcos r uvw,, () Hence, the triple integrl of the function F(r,, z) over becomes V F( r,, z) rdrddz ( r,, z) (c) Triple Integrl in Sphericl olr Coordintes Here (,, ) (,, ) V f z dddz F r φ J drddφ, where J r sin () The position of point in spce in sphericl coordintes is determined b the three vribles r,, φ where r is the distnce of the point () from the origin nd so clled rdius vector, is the ngle between the rdius vector on the -plne nd the -is to count from this is in positive sense viz. counter-clockwise. For n point in spce in sphericl coordintes, we hve r,, φ. Divide the region into elementr volumes V b coordinte surfces, r constnt (sphere), constnt (conic surfces with vertices t the origin), φ constnt (hlf plnes pssing through the Z-is ). To within infinitesiml of higher order, the volume element v m be considered prllelopiped with edges of length r, r, r sin φ. Then the volume element becomes V r sin r φ. Z φ A 9 r 9 (,,. z) z L φ r sin Fig. 5.69 Fig. 5.7 Z φ r φ L S S r z r
8 Engineering Mthemtics through Applictions For clcultion purpose, it is evident from the Fig. 5.69 tht in tringles, AL nd L, L cosφ cos(9 ). cosφ r sin cosφ, L sinφ sin. sinφ r sin sinφ, z r cos. Thus, ( z,, sincosφ sinsinφ cos ) J rcoscosφ rcossinφ rsin r sin ( r,, φ) rsinsinφ rsincosφ roblems Volume s Triple Integrl in Clindricl Co-ordintes Emple 5: Find the volume intercepted between the prboloid + z nd the clinder +. Solution: Let V be required volume of the clinder + intercepted b the prboloid + z. Trnsforming the given sstem of equtions to polrclindricl co-ordintes. rcos Let rsin sotht V (,, z) Vr (,, z) z z (, ) r B bove substitution the eqution of the prboloid becomes r r + z z nd the clinder + clinder gives r r cos r(r cos) with r nd Fig. 5.7 r cos. r Thus, it is cler from the Fig. 5.7 tht z vries from to nd r s function of vries from to cos with s limits to. Geometr clerl shows the volume covered under the +ve octnt onl, i.e. th of the full volume. cos / r z r / ( z,, ) ( r,, z) r z V V' rdzdrd, s J r / cos r / [ ] rz rdr d / cos r dr d / cos r d Z + z rboloid
Multiple Integrls nd their Applictions 9 / cos d. Emple 5: Find the volume of the region bounded b the prboloid z + nd the clinder + b. Also find the integrl in cse when nd b. Solution: n using the clindricl polr co-ordintes (r,, z) with r cos, r sin, so tht the equtions of the clinder nd tht of the prboloid re r b nd z respectivel. See Fig. 5.7, onl one-fourth of the common volume is shown. Hence in the common region, z vries from z to from to b nd to respectivel. The desired volume / b r / V rdrddz rdr dz d / b r / r z nd r nd vries on the circle r r r dr d / b / b r d / b b As prticulr cse, when, b, then V Fig. 5.7 roblmes on Volume in olr Sphericl Co-ordintes Emple 55: Find the volume common to the sphere + + z nd the cone + z Find the volume cut b the cone + z from the sphere + + z. [NIT Kurukshetr, ]
Engineering Mthemtics through Applictions Solution: For the given sphere, + + z nd the cone + z, the centre of the sphere is (,, ) nd the verte of the cone is origin. Therefore, the volume common to the two bodies is smmetricl bout the plne z, i.e. the required volume, V dddz rsincosφ In sphericl co-ordintes, we hve rsinsin φ ; J r sin z rcos Thus, + + z becomes r i.e., r nd + z becomes r sin (cos φ + sin φ) r cos i.e., sin cos i.e. /. Clerl, the volume shown in the figure (Fig. 5.7) is one-fourth, i.e. in first qudrnt onl nd, in the common region, Z + z Hence the required volume, r vries from to, vries from to, φ vries from to / / V r sindrddφ / / rdr / / 8 sin d d φ r 8 sinddφ 8 φ / / [ cos ] d 8 / dφ Alterntel: In polr-clindricl co-ordintes, intersection of the two curves + + z nd + z results in z + z or Further, z. + z r, i.e. r vries from to / Hence, V r r rdrd 9 φ Fig. 5.7 Z Fig. 5.7 + z
Multiple Integrls nd their Applictions ³ lies on the cone wheres lies on the sphere s function of (r, ) / r r r d dr / r ( r ) + since r r r r d r ( ) ( ) ( ) Emple 56: B chnging to shpericl polr co-ordinte sstem, prove tht V z dddz bc b c where V,,z Solution: Tking u,, so tht v, b z w c Now trnsformtion co-efficient, V z : + + b c z + + u + v + w b c J u v w u v w z z z u v w z d d dz b c Vz (,, ) u v w bc dudvdw V ( u, v, w) b bc c To trnsform to polr sphericl co-rodinte sstem, let u rsincos φ, v rsinsin φ, w rcos Then V (u, v, w) {(u, v, w): u + v + w, u, v, w } reduces to V (r,, φ) {r i.e., r,, φ } u v w bcdudvdw V ( u, v, w)
Engineering Mthemtics through Applictions bc r J drddφ V" ( r, φ, ) where J r sin φ V ( r,, ) bc r r dr sin φ d dφ φ Now put r sin t so tht dr cost dt nd for r, t, r, t / V" ( r, φ, ) bc cost sin tcostdt sin d d φ ( ) ( ) ( + )( ) bc sin d d φ bc sind dφ [ cos ] bc dφ 6 bc bc d d bc. 6 φ 8 φ Emple 57: B chnge of vrible in polr co-ordinte, prove tht dzdd z 8 Evlute the integrl being etended to octnt of the sphere + + z. Evlute bove integrl b chnging to polr sphericl co-ordinte sstem. Solution: Simple Evlution: Treting dz I d d s ( ) z z sin z I d d. z
Multiple Integrls nd their Applictions sin z d d, s d d ( ) d d sin, + using d + sin + 8 B chnge of vrible to polr sphericl co-ordintes, the region of integrtion V {(,, z); + + z ;, z,.} becomes I (r,, φ); r, i.e. r,, φ } where Now rsincos φ, rsinsin φ, z rcos ( z,, ) ( r,, ) J coefficient of trnsformtion r sin. φ whence V / / d ddz r sin dr ddφ z r r I dφ sin dr d r / / Let r sin t so tht dr cos t dt. Further, when r, t, r, t / φ / / sin t I d sin d costdt cost / / dφ dsin ;