Geometrical Applications of Integration

Engineering Mthemtics through Applictions Geometricl Applictions of Integrtion. INTRODUCTION In generl, we consider the integrtion s the inverse of differentition. In the epression of the sum, f (), f is considered continuous on nd we find tht limit of S s pproches to zero is the numer fd () F () F (), where F is ny nti-derivtive of f. We pply this contention in finding the re etween the -is nd the curve y f(),. We etend the ppliction to compute distnces, volumes nd volumes of revolution, length of curves, re of surfce of revolution, verge vlue of function, centre of mss, centroid, etc.. AREA OF BOUNDED REGIONS (QUADRATURE) I. Ares of Crtesin Curves The re ounded y the curve y f(), X-is nd the y B ordintes, is yd, when f() is continuous single vlued nd finite function of, nd y does not chnge sign in the intervl [, ]. (, y) P P Q Q If AB is the curve y f() etween the ordintes A LA( ) nd MB( ) with condition tht y is strictly incresing (or strictly decresing) function of in the intervl [, ]. y y+ δy Let P(, y) nd Q( + δ, y + δy) e two neighouring O L N N points on this continously incresing curve y f () nd δ M NP, N Q e their respective ordintes. Here clerly the ALNP i.e. A depends on the position of Fig.. the point P(, y) whose sciss is nd re PNN Q δa lies etween the res of the δa rectngles PN nd NQ, i.e. δa lies etween re yδ nd (y + δy) δ or δ (y + δy). lies etween y nd da On tking the limits s Q P, i.e. δ (mening therey δy ), y d Integrting oth sides etween the limits to, we get y f( ) ( + δ, y + δy) -is

Geometricl Applictions of Integrtion Are ALMB A yd Y. y M B However, if nd y re interchnged in the ove formul, we f( y) see tht the re ounded y the curve f(y), Y-is nd the sciss L y A y, y is dy. X O Fig.. Oservtions: (i) The re ounded y the curve y f(), the two ordintes t A nd B nd the X-is is often clled the re under the curve AB nd the process of clculting the re ounded y the curve is clled qudrture. (ii) An re whose oundry is descried in nti-clockwise direction is considered positive otherwise negtive. Or in other words, for the portion of the curve (under Y considertion) ove X-is for which y is positive, re enclosed is positive, wheres for the portion of the curve (under considertion) elow X-is for which y is negtive, re is negtive. But in cse of re with negtive sign, we men numericl vlue of L M X the re. O (iii) If in the intervl, c, the curve y f() is ove the -is nd in the intervl c, the curve y f() is elow the -is then we write the A re yd c yd+ yd y f( ) B c or in otherwords, if -chnges sign from to, y f() chnges sign t some intervl point c (sy), then the re for from to c nd Y c to, re clculted sepertely nd then their numericl vlue re dded (see Fig..4) Similrly, the result cn e etended if y chnges sign t more A thn one intermedite point in the intervl [, ]. (iv) Are of the region ounded etween two continuous curves f() nd g() on [, ] nd the verticl lines, is given y A [ f( ) g( )] d, where f() g() in [, ]. (see Fig..5) O L In the region under considertion, representtive rectngle is shown with height: f( i ) g( i ), width: nd P( i g( i )); Q( i f( i )) If f() g() in [, c] nd f() g() in [c, ], then we write the re s A c [() ()] + [() ()] c s shown in Fig..6. Y f () f () Y Q +ve re Fig.. c M X N ve re Fig..4 g () B f( i ) PP g ( i ) g () O i X c Fig..5 Fig..6 X

Engineering Mthemtics through Applictions Emple : (i) Find the re ounded y the prol y 4 nd its ltus-rctum. (ii) Show tht the re cut off from prol y ny doule ordinte is two-third of the corresponding rectngle otined y the doule ordintes nd its distnce from the verte. Solution (i): For the prol y 4, let the doule ordinte PP' e c Since the curve is symmetricl out X-is, therefore, for prt of the curve ove X-is, y e tken s positive, i.e. y 4 Thus, the re ounded y the prol with its doule ordinte PMP (i.e. Ltus rctum) Are P APP Are PAMP (ii) Agin for P(, y), c c c 8 yd 4 d 4 c y MP 4 4c c so tht PP' MP 4 c Now the re of the rectngle formed y the doule ordinte (PMP ) nd its distnce from the verte A (i.e. AM) PP' AM ( ) 4c c 4c Now the re cut of from the prol y doule ordintes, i.e. re P APP ( ) 8 c 4c re of the rectngle formed y PP' nd AM. Hence, the re formed y the prol nd its ltus rctum is two third of the re of the rectngle formed y the doule ordintes with its distnce from the verte, A. Note: Vice vers, the re of the rectngle is Emple : Find the re etween the curve y (y ) nd its symptotes. Solution: The given curve y (y ) is symmetricl out oth the is nd t the origin, y ± s the tngents. Further, ± re the two symptotes prllel to Y-is. The curve no where intersects with the is ecept t (, ). Whence the curve does not enclose re with its sciss or ordintes (see Fig..8). Due to its symmetry out oth the is, the whole re etween the curve nd its symptotes is yd 4 4 d (Putting sin) times the re ounded y the prol with ltus rctum. B(, ) Y B A C Y Fig..8 O (4, 4 ) Fig..7 y y P M c P A (, ) y 4 X

Geometricl Applictions of Integrtion sin cosd 4 d cos 4 sin 4 Emple : Find the re of the curve y ( ). Solution: Without going into geometricl detils of the curve, the re enclosed y it in the first qudrnt is Y A yd d (i) The curve intersects X-is t,. O X X (ii) Ais of X is the tngent t the origin. (, ) (, ) A(, ) dy dy (iii) t, t d d, Put sin (sin ) A 4 6 sin cos d sin 4sincosd 6 6 4 ( p )( p ) ( q )( q ) using sin pcosqd ( p+ q)( p+ q ) Hence the totl re is. Emple 4: For the curve y ( ) ( + ) [NIT Kurukshetr, 4] (i) Find the re of the loop Y (ii) Are of the portion ounded y the curve nd its symptote. Solution: The curve pssing through the origin is symmetricl out X-is nd hs s its symptote. It intersects the is of X t (, ) s shown in Fig... For the re of the loop, vries from to. Further, the loop is symmetricl out X-is + + yd d d (on rtionliztion) Are of the loop d + d ( ) ( ) d + d X (, ) A O Y Fig.. Fig..9 y y X

4 Engineering Mthemtics through Applictions Alterntely: Limits for ( )( ) d + d d d ( ) d+ sin + sin ( ) + sin ( ) ( sin ( ) ) ( ) +, where sin ( ) + [ 4 + ] ( 4) numericlly 4 4 + Put sin, d cos d A d,, ;, ( + sin ) A sin cosd ( sin ) + sin + sin sin cosd sin + sin sincos ( + sin ) [sin+ sin ] d d sin sin d 4 sin + ( cos ) cos+ ( 4) (4 )numericlly () Are etween the curve nd its symptote + yd d This integrnd is sme s in the cse () simply limits re to. Are ( ) + sin ( + 4) Emple 5: Show tht the re of the loop of the curve y ( + ) ( ) is equl to the re etween the curve nd its symptote. [KUK, ; NIT Kurukshetr, ]

Geometricl Applictions of Integrtion 5 Solution: The curve y ( + ) ( ) is symmetricl out X-is nd psses through the origin with y ± s the tngent. The line or is the symptote prllel to Y-is. Further the curve intersects X-is t (, ), (, ) mking one loop in the positive direction of X-is s shown in the geometry. / O / (, ) B A (, ) Here y + Clerly the re of the loop is the re which the curve encloses etween the ordintes, nd the X-is, viz. (, ) Fig.. + ( ) A (Are ABO) yd d ( ) Implying Tke ( ) cos so tht d sind nd the limits, ;, A cos + ( cos ) sin cos d ( + cos ) d 4 ( cos ) sin ( cos ) 4 ( cos )( cos ) 4 ( cos cos ) + d d sin 4 (cos cos ) d 4 sin + + 4 Further, the re etween the curve nd its symptote is pproimtely, A yd + ( ) Sme sustitution ( ) cos, d, ( ) nd limits:, ;, sin A 4 + sin which is numericlly the sme s the re of the loop of the curve.

6 Engineering Mthemtics through Applictions Alterntively, y cn lso e epressed s then 4 ( + ) y + + nd if ( + ) 4sin cos sin 6 6 yd (4 sin ) 8 sincos d 8 (4 sin cos cos ) d 6 6 8 (sin cos ) cos d 8 sin cos d cos4 + cos 6 6 ( ) 8 d 4 cos4+ cos d sin 4 sin sin 4 sin + 4 + 4 4 6 6 4 s t, limit vlue is zero. 4 + 4 Emple 6: Find the re included etween the circle + y nd the prol y. [NIT Kurukshetr, 5] Solution: The circle + y is symmetricl out X-is, nd psses through the origin with centre s (, ), rdius, Y-is ( ) s the tngent t the origin. Further, + y or ( ) + y mens it intersects X-is t ± or,. In cse of prol, y, it intersects the circle + y t (, ), (, ). Hence the re outside the prol nd inside the circle is covered under limits nd. (see geometry). Hence the desired re Are OBCO ( ) ( ) y of the circle y of the prol d d d I II y-is (, ) (, ) Fig.. I d ( ) d cos sind O C B (, ) y A (, ) -is

Geometricl Applictions of Integrtion 7 sin d 4 II d 4 4 A ( 8) 4 6 4 (Tking cos) Emple 7: Find the re included etween the prol 4y nd the curve y 8 + 4. Solution: The curve, 4y represents n upwrd prol symmetricl out Y-is with the origin s its verte. Y 8 The curve y which is symmetricl + 4 4y out Y-is, does not pss through B (, ) the origin. Further, y, i.e. is of X is 8 n symptote to it. It meets the Y-is (, ) y C + 4 t (, ). (, ) A To find the points of intersection, equting the two vlues of y, i.e. 8 4 + 4 O Fig.. X or 4 + 4 4 or ( + 8 )( 4 ) Rejecting ( + 8 ) (which gives imginry vlues of ), we get 4, i.e. ±, nd y. Thus, these two curves intersects t (, ) nd (, ). 8 The required re OABC OBC d + 4 4 8 tn 4 4 ( 8 ) 4 ( ) Emple 8: Find the re etween the curve + y y nd its symptote + y +. [KUK, ; NIT Kurukshetr, 7]

8 Engineering Mthemtics through Applictions Solution: Clerly from the Fig..4, which hs lredy een eplined in detil under trcing of curves, the product of the slops of the line of symmetricl, y nd the eqution of symptote, + y + is, i.e. they re t right ngle to ech other. Whence if the es of references re turned through n ngle of 45, the new X-is coincides with the symmetricl line y with chnged eqution of curve hving symptote prllel to the new Y-is insted the olique Y symptote + y +. After trnsformtion, new nd y re X cos45 sin45 ( y y ) X' X + sin 45 + cos45 ( y Y y ) nd hence the new eqution of curve ecomes ( y) ( + y) ( y) ( + y) Y + y + + y y + y ( ) ( ) ( )( ) or ( + y ) ( y ) + y y ( + ) ( ), where Clerly, the symptote of this eqution prllel to Y-is is. Hence, the re etween the curve nd its symptote is given y A yd d + ( ) d ( + )( ) Putting ( cos ) so tht d sin d nd for, ;, Y' (Tking negtive sign) O(9) B A 45 + y + ( m ) Fig..4 y ( m ) ( ) ( )( ) ( cos ) ( cos ) ( cos ) ( cos ) sin A d + ( cos cos ) d

Geometricl Applictions of Integrtion 9 4 (cos+ cos ) d 4 sin sin + Emple 9: Show tht re common to the two prols y 4( + ) nd y 8 4( ) is ( + ). y 4 ( + ) Y L y 4( ) y 4( + ) () i Solution: Both the prols, 4( ) ( ) re y ii symmetricl out X-is. Prol, y 4( + ), hs its verte t A(, ) nd ltus rctum s 4, wheres the prol y 4( ), hs its verte t B(, ) nd ltus rctum s 4. For intersection of these two, we get 4( + ) 4( ) or ( + ) or ( ) Hence the two prols intersects t L nd M for. Are included etween them re ALBMA re ALBNA (Are ALNA + Are LNBL) + + 4 ( ) d 4 ( ) d ( + ) ( ) 4 4 + (, ) A 8 8 8 8 + ( + ) O N B (, ) M Fig..5 X Emple : Find the re common to the circle + y 4 nd the ellipse + 4y 9. Solution: The eqution + y 4 represents circle with centre (, ) nd rdius units where the eqution + y 9 or y + represents n ellipse with semi-mjor is s units nd semi-minor is s units nd for intersection of these two, we get

Engineering Mthemtics through Applictions + 4(4 ) 9 or 7 i.e. 7 Since the ellipse nd the circle oth re symmetricl out oth the is Required common re to the circle nd the ellipse 4 re common to them in the Ist qudrnt 4[re OAPD] 4[Are OLPD + re LAPL] 7 4 (vlue y of ellipse) d + (vlue y of circle) d 7 7 + y 4 4 + 9 d 4 d B P (, ) 7 y P D (, /) 7 A(, ) A 9 + 4 9 sin + O 7 L C(, ) (, ) 4 4 + sin Fig..6 7 7 7 7 7 7 7 9 + 9sin + + 4sin () 4 4sin 7 7 7 5 7 + 9sin + 4 4sin 7 7 5 7 7 9sin 4sin + + 7 5 7 7 + + 9sin 4sin 7 + 4 9 ASSIGNMENT y. Find the re of the ellipse +. Find the re of the circle + y. Find the re of the loop of the curve y ( ) HINT : d [ ( ) ] d 4. Clculte the re of the curve y ( y) 5. Find the whole re of the curve ( + y ) ( y ) [HINT : Put cos]

Geometricl Applictions of Integrtion 6. Find the re etween the curve y ( ) nd its symptote. 7. Find the re ounded y the curve y 4 ( ) nd its symptote. 8. Compute the re ounded y the prol y + nd the stright line, nd + y. 9. Find the re enclosed etween the curve 4y nd the stright line 4y.. Find the re of the segment cut off from the prol 8y y the line y + 8.. Find the re ounded y the prol y 4 nd the line + y.. Prove tht the re common to the prols 4y nd y 6 4 is. II. Are of Curves Given in Prmetric Form (i) The re ounded y the curves f(t), ψ φ(t), the X-is nd the ordintes t the points where, t, t is given y d y dt. dt (ii) The re ounded y the curve f(t), ψ φ(t), the Y-is nd the scissc t the points where, t c, t d is given y d dy dt. c dt Emple : Find the re included etween the cycloid ( + sin), y ( cos); nd its se. Also find the re etween the curve nd X-is. } Solution: For the inverted cycloid ( + sin ), which is symmetricl out Y-is, the y ( cos ) point O (the frthest point on it) is its verte nd the line AB which is prllel to X-is is its se. For hlf of the cycloid, vries from to. Are etween the curve nd its se BCA re BOA re OCA dy d dy d ( + sin )( sin ) d ( sin+ sin ) d cos sind+ d sin ( cos ) ( cos ) d + sin [ cos+ sin ] + B( ) Y C A( ) dy d M O P L Fig..7

Engineering Mthemtics through Applictions ( ) +. Are etween the curve nd X-is re OAL d yd y d d ( cos ) (+ cos ) d / d d sin 4 sin ( p )( p ) using sind pp ( ). Hence 8 Emple : Show tht the re of the hypocycloid cos t, y sin t is deduce the re of the steriod cos t, y sin t. Solution: The given curve cos t, y sin t (hypocycloid) shown in the figure,.8 meets -is t t nd t nd to y-is t t nd t, nd is symmetricl out oth the is. Y t / (, ) t (, ) t (, ) X (, ) The required re Note: Are in cse of steriod ecomes. 8 d 4yd 4 y dt dt / Fig..8 4 sin t.cos t. sintdt / 4t t dt sin cos (4 ) (4 ) ( ) 6 4 8 ( )( ) ( )( ) using sin p cos q d p p q q ( p+ q)( p+ q ) + y which is prticulr cse of hypocycloid, when nd re equl

Geometricl Applictions of Integrtion t t t t e + e e e Emple For ny rel t,, y is point of the y. Show tht the re ounded y this prol nd the lines joining its centre to the points corresponding to t' nd t' is t'. Solution: Let P(t') nd Q( t') e two points on the hyperol y (Fig..9). Then the re ounded y the hyperol nd the two lines OP nd OQ is shown y the shded portion. The required re is the difference of the re of the OPQ nd re PAQBP. Now e + e e e OPQ PB OB e e 4 B t' t' t t t t d e e e e Are PAQBP yd y dt dt A dt t' t' t' t' t' t' Are of ( ) t' t t t' t t e e ( e e ) dt t + + O Y + y AB Pt () Qt ( ) OB e t' e t' + PB y e t' e t' Fig..9 X ( e e 4 t ') 4 t' t' POQAP e e e e t t' 4 4 t' t' t' t' The required re ( ) ( 4 ) ASSIGNMENT sin t,. Show tht the re of the loop of the curve y sint } is 4.. Show tht the re ounded y the cissiod sin t, sin t y cost nd its symptote is 4. t + t. Find the whole re of the curve t y + t (It is the prmetric form of the circle) } 4. Find the re included etween the cycloid t ( sin t) y ( cos t ) nd its se.

4 Engineering Mthemtics through Applictions ( sin ), 5. Find the re included etween the inverted cycloid y ( + cos ) } nd its se. 6. Prove tht the whole re etween the four infinite rnches of the trctri t cost + log tn, y sint is. III. Are for Curves Given in Polr Form, r f(). Are ounded y the curve r f() nd rdii vectors α, β is B β, α rd where r f() is continuous nd single vlued. S R Let AB e the curve r f(), nd OA, OB e the rdii vectors Pr (, ) for α nd β respectively Fig.. A Let P(r, ) e ny generl point on the curve such tht Q(r + δr, + δ) s its neighouring point. initil Let the re OAP (which is function of ) e A so tht sectoril is re OPQ is δa. Fig.. Evidently re OPQ lies etween the sectors OPS nd ORQ, i.e. δa lies etween δ r nd Using Are of circulr sector (rdius) ( r r ), circulr mesure of +δ δ the ngle δa lies etween δ r nd ( +δ ) r r da Proceeding to limits s δ or δr, we get r d Integrting oth sides from α to β, we get [A] β α (vlue of A for β) (vlue of A for α ) r d α β A (sectoril re OAB ) r d α β Hence the re OAB r d α Note : In the ove result, we hve supported tht r is n incresing function of in the intervl [α, β]. The sme formul is vlid even if the rdius vector r decreses s vries from α to β. However, the sme formul is not necessrily vlid if r f(), tkes oth positive nd negtive vlues in the intervl [α, β]. If there re finite numer of points of Mimum nd Minimum rdii vectors in the intervl [α, β] sy t,, n, then we divide the sectoril re OAB into (n + ) sectors with the corresponding limits s follows n β AAre OAB rd + rd + + rd + rd α n n Note : In cse of re ounded y two polr curves r f() nd r ψ() nd the rdii vectors β α nd β is (r r )d α where r is the rdius vector for outer curve nd r is the rdius vector for inner curve. β 6 β r α Q( r + δr, + δ)

Geometricl Applictions of Integrtion 5 Emple 4: Find the re of the loop of the curve r cos nd hence find the totl re of the curve. Solution: As we know tht the curve r cos n or r sinn hve equl loops if n is odd nd n equl loops if n is even. In our prolem, r cos, n is even mens the curve hs 4 equl loops. Further to find the limits of integrtion for loop, we generlly put r nd find two consecutive vlues of. Thus, there r implies cos i.e. ± or ± 4 i.e. for the first loop of the curve vries from to s shown 4 4 in Fig... /4 5 /4 O Fig.. /4 7 /4 /4 Are of one loop of the curve /4 /4 rd cos d /4 /4 /4 cos d As cos is n even function of nd for n even function fd () fd () Putting t, we get Are / dt cos t 8 The totl re of the curve is 4 times the re of the single loop, i.e. 4 or 8 Emple 5: Find the re outside the circle r cos nd inside the crdiod r ( + cos). [NIT Kurukshetr, 8] Solution: For the circle, r cos, r. r Further, the circle r cos is symmetricl out the initil is. Otherwise lso in crtisn coordintes, it ecomes r r rcos or ( + y ) ( ) + ( y ), i.e. circle with centre (, ) nd rdius. For crdiod, r ( + cos);, r ;, r ;, r

6 Engineering Mthemtics through Applictions Further, the crdiod lso is symmetricl out the initil line. For intersection, the geometry is s shown in Fig.. (III) / B Y-is / (, /) D, O X-is A(, ) O (, ) A(, ) O d A (, ) (I) (II) (III) Fig.. Aove the initil is, the crdiod is trced from to, wheres in cse of circle, goes from to. So the re outside the circle inside the crdiod, i.e. Are OABDO. [Are of the crdiod ove X-is Are enclosed y circle ove X-is] / rd rd / ( + cos) d 4 cos d / cos d 4 cos d / 4 4 cos d cos d I II / 4 / 4 cos tdt cos d In I, putting t 4 4 using / p ( p )( p ) cos d pp ( ) Emple 6: Show tht the re included etween the crdiod r ( + cos) nd r ( cos) is ( 8). [KUK, ]

Geometricl Applictions of Integrtion 7 Solution: Both the curves r ( + cos) nd r ( cos) re symmetricl out the initil is nd they intersects t ± s ( + cos) ( cos) implies cos, i.e. ± From the geometry it is cler tht re included etween the two curves is times the re OPBQO ove the initil is for to. Further, this re OPBQO is tken up s the rekup of the re OPBO for to nd re BQOB for to Fig.. Whence the re included etween the two curves, / + rd rd / I II (r ( cos) nd r ( + cos)) / ( + cos cos ) d+ ( + cos + cos ) d / / cos cos cos d cos + + + + d / / sin sin sin sin + + + + 4 4 / + ( 8) 4 Emple 7: Prove tht the re enclosed y one loop of the curve + y y is three times the re enclosed y the curve r cos. Solution: The curve + y y is symmetricl out the line y nd insect this line y t points (, ) nd,. For detils see the Fig..6 under trcing. It trnsforms } this curve to polr coordintes, rcos, y r sin resulting in, r (cos + sin ) r sin cos sin cos r sin + cos or / B r ( + cos ) Q P C O A (, ) r ( cos )

8 Engineering Mthemtics through Applictions Clerly the loop is ounded for r, i.e. sincos or,. / / sin cos The required re of the loop rd 9 d (sin + cos ) / 9 tn sec d (+ tn ) Put tn t, so tht tn sec d dt nd limits re t to 9 dt ( + t) ( + t) + t For the curve r cos, one of the loop is ounded in etween the rdii vectors to, s r gives ±/4. 4 For detils, see the Fig..66 under trcing. Are of the loop /4 /4 /4 sin rd d cos 4 Whence the re of the loop of the curve + y y is three times the re of the one of the loops of r cos. Emple 8: Find the rtio of the two prts into which the prol r( + cos) divides the re of the crdiod r ( + cos). Solution: The curve r ( + cos) is stndrd crdiod with vlues, r 4, point A(4, ), r, point B,, r, point O (, ), r, point C, Likewise, for prol r, + cos, r, pointa (, ), r, k point B,, r, k point C,

Geometricl Applictions of Integrtion 9 Clerly the two curves intersects t ±, otherwise lso, r ( + cos ) ( + cos) + cos implying cos, i.e. r + cos y / ± B(, /) r ( + cos ) E (, ) O D A y F C(, /) / Fig..4 The whole re of the crdiod rd 4 ( cos ) d + 4 (+ cos+ cos ) d + cos + + + + 4 4 cos d 4 sin sin 6 () Are of the unshded region etween the two curves [Are OABO re ODBO] / / rd rd, where r ( + cos), crdiod nd r, the prol + cos / / 4 (+ cos ) d d ( + cos ) / / cos 4 + cos+ d cos / / sin 4 sin + + sec sec d 4 4

Engineering Mthemtics through Applictions / 4 tn + + sec d 4 4, Put tn t so tht sec d dt 4 + + ( t ) dt 4 4 t (9 6) 4 + t + + 4 () The re of the shded region The whole re of the crdiod the re of the unshded region of the crdiod (9+ 6) (9 6) 6 () Hence the required rtio viz. the rtio of shded region to tht to unshded region 9 6 9+ 6 units (4) ASSIGNMENT. Find the whole re of the crdiod r ( + cos ). Find the whole re of the curve r + cos. Find the re of one loop of the curve r sin. 4. Find the re common to the circle r nd the crdiod r ( + cos). 5. Find the re common to the circle r. nd r cos. [NIT Kurukshetr, ] sin 6. Show tht the re etween the cissiod r nd its symptote is. cos 4 HINT : Here s increse from to, r increses from to nd s increses from to, r increses from to. Asymptote is rcos i.e. 5 7. Show tht the re etween the curve r (sec + cos ) nd its symptote is 4. ± HINT : Here we see,, r ;, r ± i.e. s increses from to, r increses from to nd s increses from to, r increses to. r Otherwise lso, in crtisn form, the given curve is r ( ) + or y r 8. Find the re lying etween the crdiod r ( + cos) nd its doule tngent.

Geometricl Applictions of Integrtion HINT : Here nd for ll position P( r, ),. If φ ψ p +φ+ ψ then. i.e. tngent t P will e perpendiculr to the initil line.. LENGTH OF CURVES (RECTIFICATION) The process of finding the length of n rc of curve etween the two points on it is clled rectifiction. I. Length of Curves in Crtisn Coordinte System nd its Intrinsic Eqution Length of the rc of the curve y f(), etween two points nd (s two scisse) is given y dy + d d where in y nd dy re continuous nd single vlued d functions in the internl [, ] nd the integrnd either positive (or negtive) throughout the intervl. Let AB e the curve y f() etween the two point A( ) nd B( ) s the two scisse points nd CA nd DB their respective ordintes. Let P(, y) e ny point on the curve with MP s perpendiculr on -is. If s denotes the length of the rc AP mesured from fied point A to the vrile point P, then s clerly is function of nd ds dy + d d y A s Fig..5 P(, y) O C( ) M D ( ) y B [ ] dy S (vlue of s for ) (vlue of s for ) (Arc AB O) + d d Hence the rc length AB + yd ds dy Oservtions: As ± + d d, ut we hve ssumed tht s increses with increse in (tking + sign of the rdicl). Negtive sign mens, s decreses with increse in, throughout. If ds or the integrnd chnges sign t some intermedite vlue c, then we divide the intervl to into d ds two prts one from to c nd the nother c to nd, vlue of is tken positive(or negtive) throughout in d ech su intervl ccordingly.

Engineering Mthemtics through Applictions If the eqution of the curve is off the form F(y), then the length of the rc etween points d dy y c nd y d is + dy c d y tking s n eplicit function of y(mens F(y) contining only terms of y in right hnd side). Intrinsic Eqution: A reltion etween the vrile s (the rc) nd ψ(the tngent t vrile point P(, y) mkes with the X-is) is clled intrinsic eqution of curve. For prcticl dy dy purposes, eliminte etween eqution of S + d nd tnψ d d. Emple 9: Show tht the whole length of the curve ( ) 8 y is. [KUK, ] Solution: The curve ( ) 8 y is symmetricl out oth the is nd y ± re the tngents t the origin. Further, it intersects -is t (, ); (, ) nd, y ± ( + )( ). Mens the curve wholly lies etween nd, nd the origin is node. so tht Here dy d dy ( ) + + d 8 ( ) ( ) ( ) ( ) + ( ) + nd hence the length of the curve, 9 + 4 8 ( ) 4 4 dy s 4 + d d + d d sin sin + + sin (, ) B O Y Fig..6 (, ) A X

Geometricl Applictions of Integrtion Emple : Find the prmeter of the loop y ( ). Solution: y dy ( ), on differentition, implies 6y d or y 6y so tht 6 y + ( ) + y + 6 y (6 y) y ( ) + ( ) ( ) [ + 4 9 ] ( ) [6 4 + 9 ] ( ) (4 ) (4 ) ( ) ( ) y (, ) y Fig..7 So the desired perimeter of the loop, s + y d ( ) + d, t ; Let ( ) t so tht d dt nd the limits re, t } s t+ dt t + t t dt (4 ) d ( ) t t 4 + + 4 e Emple : Determine the length of the curve y log from to. e + Solution: The given eqution of the curve e y log log( e ) log( e + ) e + dy e e e d e e + e Required length of the curve, dy S + d d

4 Engineering Mthemtics through Applictions 4e ( e ) + 4e + d ( e ) ( e ) d ( e + ) ( e + ) d d ( e ) ( e ) ( e + e ) d log( e e ), ( ) e e log( e e ) log( e e ) log e e e e + log( e e ) log + e e f' ( ) using d log f ( ) f ( ) II. Arc Length of Curves in Prmetric Form: If the prmetric form of curve is given y φ(t), y ψ(t), t where φ(t) nd ψ(t) re continuously differentile function on [, ] then the rc length of the curve is given y d dy s ( ) ( ') dt φ + ψ + dt dt dt Note: For intrinsic eqution of the curves, eliminte t etween s ( ') ( ') dy y' + y dt nd tnψ d '. y Emple : Rectify the ellipse cos nd y sin, or +. Solution: The ellipse cos, y sin, y i.e. + is symmetricl out oth the es. On differentiting with respect to, Y d sin / d B dy cos d A The perimeter of the ellipse O / d dy 4 Arc AB 4 + d d / 4 sin + cos d / 4 sin + ( e )cos d (Since for n ellipse ( e ), where e is the ecentricity of the ellipse) Fig..8

Geometricl Applictions of Integrtion 5 / e d 4 (sin + cos ) cos / 4 ( e cos ) d / 4 + ( e cos ) + ( e cos ) ( e cos ) + + nn ( ) By Binomil theorem, ( ) n (, ) C B Y (, ) D(, ) Fig..9 d n + + + + / e cos 4 4 6 6 4 e cos e cos d 4 4 6 / p ( p )( p ) using cos d /, for p even pp ( ) 5 4 4 4 4 6 6 4 4 6 e e e 4 6 e 5 e e 4 4 6 5 Emple : Find the perimeter (full length) of the hypocycloid cos, y sin, i.e. y + [KUK, ] Solution: The given eqution of the curve is y + i.e. y if >, y y is negtive, i.e. or y is negtive. Thus, the curve does not lie eyond ±. Similrly, the curve does not lie eyond y ±. Hence the shpe of the curve is s shown A (, ) X Thereore, whole length of the curve 4 the length of the rc in Ist qudrnt 4 Arc AB Clerly from rc AB, vries from to.

6 Engineering Mthemtics through Applictions On differentiting the given curve with respect to, we get y dy + d or y dy d Thus, Therefore, y dy + + + d ( ) ( ) the required length 4 4 dy + d + d d, ; Let sin, so tht d sin cosd nd for, / ( sin ) Required length 4 + sin cosd sin / sin cos sin cos + d Agin tking sin + cos t, we get { sin cos + cos ( sin)}d t dt i.e. ( ) sincos d t dt t i.e. sincosd dt ( ) Also, for, t ;, t Thus with ove sustrctions, the required integrl ecomes, S ( tdt t t dt ) ( ) t 4 ( ) 4 + + ( ) ( ) ( + ) ( ) Alterntively: In prmetric form cos, y sin, so tht

Geometricl Applictions of Integrtion 7 d dy + 4 + 4 d d 9 cos sin 9 sin cos 9sin cos ( cos + sin ) Thus, the required length in the first qudrnt is / d dy s + d d Putting nd for, t ;, t / cos sin ( cos sin ) + d cos + sin t so tht ( ) cos sind tdt tdt s t tdt t ( )( ) + + ( )( + ) ( ) + + ( + ) Hence the full length of the curve ( + + ) 4 ( + ) t Emple 4: Find the length of the trctri cost + log tn, y sint fied point (, ) on the curve. Also find the intrinsic eqution of the curve. Solution: From the given eqution, t sec d + + cos t sint sint dy t nd cost dt tn sint sint dt dy dy cost dt tn t d d cos t dt sint At the point (, ), dy t nd (tn t) t/. dt/ Therefore, y-is is the tngent to the curve t the point (, ). from the ()

8 Engineering Mthemtics through Applictions Whence Now tn ψ cot t tn t dy d nd, therefore ψ t. t t 4 d dy cos t dt cos t dt / / S + + dt dt sin t t / cos cot t t+ dt t t / () cottdt log(sin t) / log(sint) () Therefore, the intrinsic eqution, S log sin ψ log(cos ψ) (4) (using ()) Emple 5: Find length of one full rc of the cycloid (t + sint), y ( cost) nd show tht the intrinsic eqution of the cycloid is s + ρ 6. t ( + sin t) Solution: The eqution of cycloid y ( cos t )} implies Step: I Mesuring S from the point where t d ( + cos t), dt dy sint dt t d dy t t S + dt ( + cos t ) + sin t dt ( + cos t ) dt dt dt t t sin t t t t cos cos t dt dt 4sin Here for full rnch, vries from to nd to, for two hlves. Length of one full rnch is () 4sin 8 dy t t sin cos dy sint Step II tn t ψ dt tn d d + t t cos dt t implying ψ ( cos ) t Step III Eliminting t y putting s Ψ t ds From (), s 4sin 4sinψ or ρ 4cosψ dψ () ()

Geometricl Applictions of Integrtion 9 s + ρ (4sin Ψ) + (4 cos Ψ) 6 III Length of Polr Curves The rc length of curve r f() etween the points α nd β is given y β β dr r + d r + r d α d α dr where is continuous nd single vlued in [α, β]. d When the eqution is of the form f(r), the rc length etween two rdii vectors r nd r is given y where d dr r r d r + r dr + dr dr r r r is continuous nd single vlued in [α, β]. β Note: For intrinsic eqution of the curve, eliminte nd φ etween r f(), r s r + r d nd tnφ α r (where ψ + φ) / Emple 6: Find the whole length of the crdiod r ( + cos ). Also show tht the upper hlf is isected y. [NIT Kurukshetr, 4, 5] O B P / A Solution: The crdiod r ( + cos) is symmetricl out the initil is nd for the upper hlf, goes from to. Here Fig.. dr sin d Length of the whole rc dr r + d (+ cos ) + sin d d ( cos ) cos + d d sin 4 8( ) 8 Thus, the length of the upper hlf of the curve is 4. Here length of the rc AP(for vrying from to /) is hlf the length of the upper hlf of the crdiod. / ( + cos ) d, Emple 7: For the curve r e cotα, prove tht s r constnt, s eing mesured from the origin.

Engineering Mthemtics through Applictions dr Solution: For given curve r e cotα cotα, we get e cot α r cotα d So the length of the curve r f() e cotα etween two points is given y β dr s r + d r + r cot αd rcosec αd ( rcosec α) α d (s eing mesured from the origin) or s cosec α, constnt numer (sy λ) r Hence the proof. Emple 8: Find the length of the rc of the equingulr spirl r e cotα etween the two points for which rdii vectors re r nd r. dr Solution: Given r e cotα cot α so tht e cotα rcotα d Here length of the curve etween two points for which rdii vectors re r nd r is given s: r r d r s r dr + + dr r dr r r cotα r + tn αdr r r secαdr r ( r r)secα Emple 9: Find the length of the rc of the hyperolic spirl (Reciprocl Spirl) r from the point r to r. Solution: Rewrite the given eqution s r, so tht d dr r The required length from r to r is given y d + r 4 S + r dr + r dr dr dr r r Put + r t so tht rdr tdt nd for r, t ; r, t 5 5 5 t Implying S dt dt + t t t+ t log t + 5

Geometricl Applictions of Integrtion ( ) 5 5 + log log 5 + + ( ) 5 + 5 + log + 5 ( 5 + ) ( + ( 5 + ) ( + ) ( ) + 5 5 log ( ) 5 + log Emple : Find the whole length of lemniscte r cos. Solution: The curve r cos is symmetricl out the initil line nd mkes two loops in totl. One loop is enclosed etween rdius vectors /4 to /4, elow nd ove the initil line. Full length /4 /4 /4 r + r d 4 r + r d Here r cos rr ( sin) or /4 4 sin 4 r + d r r /4 4 4 /4 cos + sin 4 4 d cos cos sin r Tke t so tht d dt nd for t, ; t, 4 / / dt Implying, length l 4 dt cost cost Agin, tke cost cos φ so tht sint dt cos φ sin φ dφ, (Limits remins unchnged) φ φ l / cos sin φ cosφ sint d / sin φ 4 dφ (sin t) / sinφ / sinφ 4 dφ 4 4 dφ ( cos t) ( cos φ) / sinφ 4 dφ ( cos φ ) ( + cos φ)

Engineering Mthemtics through Applictions / dφ / sin φ 4 dφ ( sin ) φ Now, φ sin 5 ( ) + + + +,!! where sin φ sin φ sin φ 5sin φ + + + + 4 4 6 r / sin φ 4 5 6 l + + sin φ + sin φ + dφ 4 4 6 / p ( p )( p ) using sin d for p even φ pp ( ) 5 5 + + 4 4 + + 4 6 6 4 + + 5 + + 4 4 6 Emple : Show tht the whole length of the lemicon r + cos ( > ) is equl to tht of n ellipse whose semi-is re equl to length to the mimum nd minimum rdii vectors of the lemicon. Further, prove tht the perimeter of the lemicon r + cos, if e smll, is pproimtely, +. 4 Solution: The eqution of the lemicon is r + cos ( > ). (i) The curve is symmetricl out the initil line. (ii) Further r implies cos > numericlly (s given > ) which is impossile, since lwys cos <. Whence, r is never zero nd the curve does not pss through the pole, though it goes on decresing from r ( + ) to r ( ) upto. (iii) Some of the rod vlues of r for the vlues of () re s: C / B O Fig.. A

Geometricl Applictions of Integrtion : r : + + Thus, clerly for upper hlf of the curve, vries from to. Whence the whole length of the lemicon + dr r d d l () ( + cos ) + sin d + + cosd Now, for the lemicon mimum vlue of r ( + ) nd minimum vlue of r ( ). Therefore, the prmetric eqution of the ellipse whose semi-es vectors of the mimum nd minimum rdii vectors of the lemicon r + cos re ( + )cost y ( )sint so tht d dy + ( + ) sin + ( ) cos dt dt t t d ( + )sint dt dy nd ( )cos t dt () (sin t + cos t) + (sin t + cos t) (cos t sin t) + cost () Further, the ellipse is symmetricl out oth the es, nd in the first qudrnt, t vries for to. Whole length of the ellipse, l / d dy 4 + dt Tke t so tht dt d nd when l dt 4 / cos + tdt d t, ; t, (4) + cos( ) + + cos d (5) which is clerly equl to the whole length of the lemicon + cos, ( > ) lredy otined ove in (). Further from (), llemicon + cos + d, (Putting into inomil formt)

4 Engineering Mthemtics through Applictions cos + + cos d + + +! 4 8 cos cos d + + + cos ( cos ) d + + s is smll, neglecting nd its higher powers cos sin d + + cos cos d + + cos cos d + + 4 4 + sin+ sin 4 4 + pproimtely 4 Emple : Find the intrinsic eqution of the crdiod r ( cos). Solution: Mesuring s from the pole where, Step : dr ( cos ) sin s r + d + d d ( + cos cos + sin ) d d d ( cos ) sin cos

Geometricl Applictions of Integrtion 5 4 cos () 8sin 4 dr ( cos ) Step : tnφ r tn imply φ () d sin Step : We know, ψ+φ+ ψ ψ or or () 4 6 ψ Step 4: Eliminting from () nd (), we get s 8sin 6 which is the required intrinsic eqution of the curve. Note : For finding length of crdiod, we know for upper hlf, goes from to r sin i.e. from (i), 8 8 sin 8 4 implying s 8 4 4 Emple : Find the length of the rc of the prol r + cos from its verte nd lso otin the intrinsic eqution of the curve. [NIT Kurukshetr, 7] Solution: The eqution of the given prol my e written s dr r sec nd sec tn + cos cos d () Therefore the length s of the rc from the verte A to ny point P(r, ) is given y s r + r d 4 4 sec + sec tn d sec + tn d Put tn sotht sec t d dr O(, ) P φ ψ A T / + ψ s tn + t dt ( ) tn t + t + log t+ + t T Fig..

6 Engineering Mthemtics through Applictions tn sec + log tn + sec () Now d tnφ r sec cot tn dr sec tn φ () Since ψ is the ngle mde y the tngent to the curve t P(r, ) with the tngent AT t the verte A, s shown in the figure.. Whence +ψ+φ or ψ (using ) (4) Therefore on using (4), () ecomes s [tn ψ secψ + log(tnψ + secψ)] (5) s the desired intrinsic eqution of the curve. IV. Arc Length of Curves in Pedl Form: The rc of the curve p f(r) etween the points r, r is r dr, where p r sinφ is the length of the perpendiculr from the pole on ny tngent nd φ is the ngle etween the tngent nd the rdius vector t tht point. ds dr r Note: Intrinsic eqution of the curve in pedl form is otined y eliminting r from ρ r dψ dp f'r () r r nd s dr r p Emple 4: Find the length of the rc of the prol p r for r vrying to. r p Solution: Length of the rc of the given prol p r is given y r r ( r ) + s dr dr dr r p r r r r ( ) r dr dr + ( r r) ( r r) ( r ) dr dr + ( r r) r r + 4 4 dr ( r) dr + dr ( r ) ( )

Geometricl Applictions of Integrtion 7 ( r r) + cosh r ( ) using d cosh log + + log r r + + + log log log + + ( ) + log + Emple 5: Find the intrinsic eqution of the curve whose pedl eqution is p r. dp Solution: From the eqution, p r, we get p r dr or dr p () dp r ds dr ρ r r p p dψ dp r () Step : Let s e mesured from the point where r (since for r, p re imginry entity) Step : r r r r r r p ( ) r p () s dr dr r Step : From () nd (), eliminting p, we get ds dψ ds s or s dψ ds or d s ψ On integrtion, s ψ+α, where is contnt of integrtion. (4) Step 4: Let ψ for s. Therefore from (4), α Thus, s ψ or which is the desired intrinsic eqution of the given curve. s ψ (5)

8 Engineering Mthemtics through Applictions ASSIGNMENT 4. Find the length of the rc of the semi-cuicl prol y from the verte to the ordinte 5.. Find the length of the rc of the prol y 4 (i) included etween the ordintes nd h (ii) cut of y the line y 8.. Find the prmeter of the loop of the curve y ( ) 4. Find the length of the curve y log(sec ) from to /. t t 5. Show tht the prmeter of the curve, y is r. + t + t [The given curve is the circle + y ] [MDU, ] 6. Find the length of rc of the curve e sin, y e cos from to /. 7. Find the length of one full rc of the cycloid ( + sin), y ( + cos) or ( ( sin), y ( cos)) [Jmmu Univ, ] t 8. Find the length of the loop of the curve t, y t [NIT Kurukshetr, 8] 9. Find the length of the perimeter of the curve r cos.. Find the length of spirl of rchimedes, r etween the points whose rdii vectors re r nd r.. In four cuped hypocycloid + y, show tht (i) s cos ψ, s eing mesured from the verte. 4 (ii) s (/) sin ψ when (, ) is tken s the fied point. At the verte, 4. Show tht the prol ( cos ) r + ds,. Hence show tht the rc dψ sin ψ intercepted etween the verte nd the etremity of the ltus rctum is + log( + ). Find the intrinsic eqution of the curve whose eqution is p r sin α..4 VOLUMES OF REVOLUTION When plne re mde to revolve out fied stright line lying in its own plne, genertes solid ody of revolution nd its oundry genertes surfce of revolution. The fied line out which this plne re is rotted is clled the is of revolution.

Geometricl Applictions of Integrtion 9 e.g. D C C A B A C B A B () i () ii ( iii) Fig.. (i) When rectngle is rotted out one of its se, right circulr cylinder is generted. (ii) When semicircle is rotted out its ounding dimeter, sphere is generted. (iii) When right ngle tringle, is rotted out its se, right circulr cone is formed nd so on there could e uncountle numer of emples. Whence plne when revolved out fied stright line lying in its own plne, then the ody so generted y the plne re is termed s volume of revolution nd the curved surfce (i.e. the outer fce) generted y the plne re so revolved in clled surfce of revolution. I. Volume of Revolution for Crtesin Curves (i) Revolution out X-is: The volume of the solid generted y the revolution of the re ounded y the curve y f(), the -is nd the ordintes, is yd where y f() is continuous finite nd single vlued function in the intervl. B Let AB e the curve y f() nd CA nd BD Q e the two ordintes nd respectively. Let P(, y ), Q( + δ, y + δy) e two neighouring points on the curve nd LP, NQ e their respective ordintes. A C P L N D Let V denotes the volume of the solid O(, ) generted y revolution out -is of the re ACLP, which is clerly function of. Then the volume of the solid generted y the revolution of the re PLNQ is δv. Complete the rectngles PN nd LQ. Fig..4 Clerly, δv lies etween the volumes of the right circulr cylinders generted y the revolution of the rectngles PN nd LQ. i.e. δv lies etween y δ nd (y + δy) δ s PL y nd NQ (y + δy). δv δ lies etween y nd (y + δy) Since y is continuous function of, therefore, δy s δ dv Tking limits s δ, d y

4 Engineering Mthemtics through Applictions or dv yd d [ V] d (Vol. V when ) (Vol. V when ) vol. generted y the revolution of the re ACDB O Hence the volume of the solid generted y the revolution of the re ACDB out -is is yd (ii) Revolution out Y-is: The volume of the solid generted y revolution out y-is of the re ounded y the curve f(y), y-is nd scisse y, y is dy where f() is continuous finite nd single vlued function in the intervl. (iii) Revolution out ny is: Let the curve AB, y f(), revolves out ny line sy LM other thn -is or y-is nd tke one fied point on this line. If P nd Q, e two djusnt points on the curve with perpendiculr meeting the line LM in N nd N respectively, then NN d(ln) nd the volume of revolution of the elementry disc of revolution out LM is (PN) d(ln) nd hence the totl volume is given y ( PN) d( LN) with proper limits of integrtion. y A L P Q N N' y Fig..5 B M y Emple 6: The hyperol revolves out the is of X. Show tht the volume cut off from one of the two solids thus otined y plne perpendiculr to X-is, nd distnt h from the verte, is h ( + h) A' P O(, ) A (, ) M ( + h, ) Solution: Cler from the figure.6, volume of the solid cut off y plne to X-is nd t distnce h from the verte is the sme s volume otined y rotting the portion of the curve from to + h (i.e. portion of the curve to the right hnd side of Y-is). Fig..6 + h h + h The required volume yd ( ) d + h y using

Geometricl Applictions of Integrtion 4 ( + h) h + h ( h ) + h( + h) Emple 7: Otin the volume of the solid of revolution of the loop of the curve y ( ) ( + ) out Y-is. Solution: As shown in the geometry, the curve y ( + ) ( ) hs one loop lying etween. The volume of the solid of revolution of the loop out X-is is equl to ( + ) V y d d ( ) Put ( ) t so tht d dt, (, ) A y O t t dt t ( ) ( ) Fig..7 5 t+ 4t t dt t 5 4 + t t t dt logt 5 t t + t ( ) log 5 ( ) + ( ) log 5 ( ) log Emple 8: Find the volume of sphericl cp of height h cut of from sphere of rdius. Solution: Let the eqution of the circle e + y () Let the plne PA A cut the circle t distnce h from A so tht AA h nd OA h, since OA.

4 Engineering Mthemtics through Applictions The required volume of the sphericl cp (shown shded in fig.8) is the volume of the solid generted y the revolution of the Arc AP (of the circle + y ) out the X-is ounded etween the ordintes h to. y P O h A' N h A (, ) P Fig..8 V y d ( ) d h h h h h { ( )} { ( ) } h h h + h { } h h h h Cor. If the segment is cut off y plne t distnce / either from or from the centre, then h ecomes / nd V ecomes h 5 4 6 4 units, wheres the volume of 4 the sphere of rdius is (i.e. put limits to ). Whence, the volume of the segment is 5 4 5 times tht of the volume of the sphere. 4 Emple 9: A sin is formed y revolution of the curve 64y(y > ) out the Y-is. If the depth of the sin is 8 inches, how mny cuic metres of wter will it hold? Solution: The eqution of the generting curve is 64y

Geometricl Applictions of Integrtion 4 This curve is symmetricl in opposite qudrnts i.e. y rotting through 8, the geometry remins the sme. The height of the sin is given 8 inches, so tht y 8 Therefore, 64 8 i.e. 8. Hence, the portion OA of the curve with point A(8, 8) only y is considered for generting volume y revolution out Y-is. The required volume 8 8 (64 y) dy 8 6 ydy 8 dy 5 y 6 5 5 48 (8) 5 5 48 48 56 ( ) 6 cuic inches 5 5 5 Emple 4: Show tht the volume of the solid generted y the revolution of the curve N O 8 Fig..9 P A(8, 8) y y 8 ( )y, out its symptote is [NIT Kurukshetr, 7] Solution: As this prolem lies under the ctegory of revolution out ny is. Let P(, y) e ny point on the curve, shown in Fig..4. The desired volume is otined y revolution of the curve out its symptote, V ( PM) d( PN), where PM OA ON V ( ) dy y dy + y, using given curve (, ) O ( ) y (, y) P N M A (, ) Asymptote 6 ( + y ) dy Fig..4

44 implying Engineering Mthemtics through Applictions Put y tn so tht dy sec d V 6 / sec d ( + tn ) cos d / Alterntely: In PM d ( PN ), we my keep it unchnged nd simply find dpn ( ) dy ( ) d d ( ) Thus, implying V ( ) d d ( ) Put sin, d sin cosd, limits re to. implying V cos d / Emple 4: Find the volume of the solid in the form of Torus formed y the revolution of the circle + (y ) ( > ) out -is. Solution: For ny generl point P(, y) on the circle + (y ), (y ) ( ) or y ± or y ± i.e. P tkes two positions P nd P ove nd elow the origin (, ) such tht y nd y + Therefore the required volume of the Torus so formed y revolving the shded re (shown in the figure) out Y-is is given y ( ) V yd yd y y d ( y + y ) ( y y ) d ; d Since only sin 8 + sin 8 is non-zero t (, ) A B (, + ) (, y ) P O (, ) O P Fig..4 (, y )

Geometricl Applictions of Integrtion 45 y Emple 4: Show tht the volume of solid formed y revolving the ellipse + (or cos, y sin) out the line is 4. [NIT Kurukshetr, 4] Solution: For ny point P(, y) on the ellipse we get, y y or y +, ± y or y Mens P(, y) corresponds two vlues of ginst points P nd P s (Tking +ve sign) nd + from the line. Now volume of the solid formed y revolving the given ellipse out the will e two times the volume generted y revolving the upper hlf of the curve out P ( +, y ) (, ) C B(, ) y D(, ) P(, y) y A (, ) Fig..4 i.e. V dy dy dy ( ) dy ( )( ) dy + 4 dy s + 4 nd 6 6 dy 6 ydy ydy 6 y y y + sin 6 4 (As only nd term is non-zero t limit )

46 Engineering Mthemtics through Applictions Emple 4: Find the volume (generted) of the frustrum of right circulr cone whose lower se hs rdius R, upper se is rdius r nd ltitude is h. [NIT Kurukshetr, ] Solution: From the geometry of the frustrum it is cler tht if the cone AB is revolved out Y-is (ny is PN) it will generte volume giving shpe of the frustrum. So the eqution of AB cn e written with the help of two point form with coordintes of A(R, ), B(r, h) s ( y y) ( y y) ( ) ( ) or h y ( R) r R ( r R) so tht y + R h Now volume of the frustrum is the volume otined y revolving AB out Y-is, i.e. h h ( ) ( ) V PN don dy r R h y R dy y R dy h + +, h when h ( y + R) ( ) ( ) h R R + + h r R h R + R r R h (, r h) B h h r R ( r + R + rr) ( r R) N O A ( R, ) y-is O N h N Q Fig..4 AR (, ) P (, y) Br (, h) y r R (const.) h OO' h O'B r -is Alterntely: NP NN + N P OA + N P R + AN tn r R R+ ytn R + y h NP ' Since in AN'P, tn or N'P AN' tn ytn, AN' N'B r R where in the ABN', tn or tn N''A h h y h V ( PN) don ( ) ( ytn + R) dy ( ytn R) + tn h

Geometricl Applictions of Integrtion 47 h h r R r R y + R h + R tn h tn h h [ ] h( R r r + R + rh) ( r R) Note: In cse of the revolution out X-is, limits will e R to r nd the frustrum will e long horizontl line. Emple 44: A qudrnt of circle of rdius, revolves out its chord. Show tht the volume of the spindle generted is ( ) y 6 Solution: The eqution of the generting circle of rdius e + y From the geometry, if A(, ) nd B(, ) e the etremities of the rc AB (in the positive qudrnt), then the eqution of the line AB, using two point form, is given y y y y y ( ) or y ( ) or + y Fig..44 Let P(, y) e ny point on the rc AB. Drw PN on the Chord AB nd form AP. PN is perpendiculr distnce of P(, y) from the chord AB or the line + y PN + y ( ) + y ( ) + implying ( ) ( ) + + ( ) ( ) PN ( ) And ( ) ( ( ) ) ( + y) AN AP PN + y ( ) ( ) ( ) + y y y ( ) ( ) + y y ( ) y implying AN ( y) (( ) ) O B(, ) N P (, y) A (, ) + dan ( ) d d +

48 Engineering Mthemtics through Applictions As for the rc AB, vries from to, therefore the required volume, ( ) + V ( PN) d( AN) ( ) d Put sin, so tht d cos d nd limits re, /. cos+ sin ( sin )( ( cos )) cosd cos / / ( sin )( cos )(cos+ sin ) d / sin sin sin cos cos sin cos + + cos sincos cos sin d / (sin+ cos ) sincos (sin + cos ) d + (sin cos+ cos sin ) / d(cos ) d(sin ) sin d(sin ) d(sin ) d(cos ) d + + / sin cos sin sin cos + + 5 ( ) 6 II. Volume of Revolution for Prmetric Curves If f(t) nd y φ(t) re prmetric equtions of curve, then the volume of the solid generted y revolving the re out X-is is d y dt dt with proper limits of integrtion. Likewise, the volume of the solid formed in revolving the sme curve out Y-is is dy dt dt with proper limits of integrtion. Emple 45: Find the volume of the rel formed y the revolution of the cycloid ( + sin), y ( cos) out the tngent t the verte (or out the X-is). Solution: Clerly in the geometry of the given cycloid, the point O (the frthest point on it) is its verte nd the is OX is the tngent to it t the verte O. Fig..45 B( ) A( ) O L

Geometricl Applictions of Integrtion 49 Thus, the volume of the reel formed [volume generted y the revolution of re OQALO out the tngent of O viz out X-is]. Or in otherwords, the volume of the reel is the volume generted y revolution of the re formed elow y one full rnch BOA of the cycloid out X-is., where is the prmeter d The required vol. yd y d d ( cos ) (+ cos ) d sin cos d 4 6 sin cos d / 4 sin t cos tdt, when t 6 4, / ( ) ( ) using sin m tcos n tdt m n ( m+ n)( m+ n ) Emple 46: Find volume generted y revolving one rch of the cycloid ( sin), y ( cos) (i) out its se (ii) out Y-is. (NIT Kurukshetr, ) Solution: (i) The eqution of the cycloid re ( sin), y ( cos) See the geometry, for the first hlf of the cycloid in the first qudrnt, vries from to. d V y d y d d Tke ( cos ) ( cos ) d d sin t, d dt nd for ; t for, Thus t / V (sin t) dt / 6 sin t dt 5 5 6 4 A / w B C O Fig..46 /

5 Engineering Mthemtics through Applictions (ii) For volume of revolution out Y-is, see the fig..47 Here, we first otin the volume of the reel generted y the revolution of the cycloid out Y-is, i.e. it is the volume of revolution of the re ABLA out y-is, V y y dy dy d ( sin ) sind d ( sin sin sin ) + d Now integrl I I I I sin d ( cos ) ( cos ) d + + + cos sin sin d cos sin cos I d d d sin ( cos ) ( cos ) { d } sin sin sin cos 4 I d d d sin sin sin ( cos )sin sin sin cos d sin d+ ( sin )cos d cos cos n+ + n ( f( )), using f' ()(()) f d n + sin cos cos V cos+ sin+ cos + + cos+ 4444 4444444 44444444 4 4444 + + + 4 4, (All sine terms re zero for oth the limits.) 8 Now volume generted y the revolution of the re ALBCA out Y-is, dy V' d ( sin ) sind d B cos+ sin+ cos + sin + cos + cos 4 L Q P (, y) B A C Fig..47

Geometricl Applictions of Integrtion 5 4 4 + + + + + 4 4 8 Thus the desired volume of the solid generted y revolution of the cycloid out Y-is 8 8 V' V 6 III. Volumes of Revolution for Polr Curves The volume of the solid generted y the revolution of the re ounded y the curve r f() nd the rdii vector α, β: β (i) out the initil line OX ( ) r sin d α β ð (ii) out the line OY è d cos α (The proof of these formule depend on Pppus theorem). Emple 47: Find the volume of the solid generted y the revolution of the crdiod r ( + cos) out the intil lines. [KUK, ] Solution: The Crdiod r ( + cos) is symmetricl out the initil line nd for its upper hlf, vries from to (see fig..) Required volume r sin d ( + cos ) sind ( cos ) ( sin ) + d n, using f () f' () d 4 ( cos ) + 8 ( 6) 4 6 Emple 48: Find the volume of the solid otined y revolving the lemniscte r cos out the initil line. [NIT Kurukshetr, 9] y n f + () n+ /4 Solution: The curve r cos is symmetricl out the initil line nd consists of two equl loops. The required volume is therefore twice the volume generted y the revolution of the portion OA of the curve out the initil line. B O Fig..48 A