The volume V of the solid of revolution obtained by revolving the region of Fig about the x axis is given by. (disk formula)

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1 CHAPTER Applicatios of Itegratio II: Volume A solid of revolutio is obtaied by revolvig a regio i a plae about a lie that does ot itersect the regio. The lie about which the rotatio takes place is called the ais of revolutio. Let f be a cotiuous fuctio such that f() for a b. Cosider the regio uder the graph of f, above the ais, ad betwee a ad b. (See Fig. -.) If is revolved about the ais, the resultig solid is a solid of revolutio. The geeratig regios for some familiar solids are show i Fig. -. Fig. - Disk Formula The volume V of the solid of revolutio obtaied by revolvig the regio of Fig. - about the ais is give by b V π ( f ( )) d π y d (disk formula) a b a Fig. -! " #%$ & ' '( )% " *, - *. ' / ),.

2 CHAPTER Applicatios of Itegratio II: Volume 5 See P roblem 9 for a sketch of the proof of this formula. Similarly, whe the ais of rotatio is the y ais ad the regio that is revolved lies betwee the y ais ad a curve g( y) ad betwee y c ad y d (see Fig. -), the the volume V of the resultig solid of revolutio is give by the formula V π ( g( y)) dy π dy (disk formula) c d b a Fig. - E X A M P L E. : Cosider the solid of revolutio obtaied by revolvig about the ais the regio i the first q uadrat bouded by the parabola y 8 ad the lie. (See Fig. -.) B y the disk formula, the volume is V π y d π d π 8 ( ) π ( 6 ) 6 π Fig. - E X A M P L E. : Cosider the solid of revolutio obtaied by revolvig about the y ais the regio bouded by the parabola y ad the lies ad y 6. (See Fig. -5.) To fid its volume, we use the versio of the disk formula i which we itegrate alog the y ais. Thus, 6 6 y π V dy dy y π π π 8 8 ( 56 ) π 6

3 6 CHAPTER Applicatios of Itegratio II: Volume Fig. -5 W ash e r M e th od Assume that g() f () for a b. Cosider the regio betwee a ad b ad lyig betwee y g() ad y f (). (See Fig. -6.) The the volume V of the solid of revolutio obtaied by revolvig this regio about the ais is give by the formula b V π ( f ( )) ( g( )) d (washer formula) a [ ] Fig. -6 The justificatio is clear. The desired volume is the differece of two volumes, the volumes π ( f ( )) d a of the solid of revolutio geerated by revolvig about the ais the regio uder y f () ad the volume b π ( g( )) d of the solid of revolutio geerated by revolvig about the ais the regio uder y g(). a A similar formula V π ( f ( y)) ( g( y)) dy (washer formula) c d [ ] holds whe the regio lies betwee the two curves f ( y) ad g( y) ad betwee y c ad y d, ad it is revolved about the y ais. (It is assumed that g( y) f ( y) for c y d.) The word washer is used because each thi vertical strip of the regio beig revolved produces a solid that resembles a plumbig part called a washer (a small cylidrical disk with a hole i the middle). b

4 CHAPTER Applicatios of Itegratio II: Volume 7 Fig. -7 E X A M P L E. : Cosider the solid of revolutio obtaied by revolvig about the ais the regio bouded by the curves, y,, ad y 6. (The same regio as i Fig. -5.) H ere the upper curve is y 6 ad the lower curve is y. H ece, by the washer formula, ( ) V π [ ] d π 6 [ 56 6 ] d π ( ) ( 56 5 ) π 5 5 8π 5 C y li d ric al S h e ll M e th od Cosider the solid of revolutio obtaied by revolvig about the y ais the regio i the first q uadrat betwee the ais ad the curve y f (), ad lyig betwee a ad b. (See Fig. -7.) The the volume of the solid is give by b V π f ( ) d π y d (cylidrical shell formula) a See P roblem for the justificatio of this formula. A similar formula holds whe the roles of ad y are reversed, that is, the regio i the first q uadrat betwee the y ais ad the curve f ( y), ad lyig betwee y c ad y d, is revolved about the ais V π yf ( y) dy π y dy c d b a c d E X A M P L E. : R evolve about the y ais the regio above the ais ad below y, ad betwee ad 5. B y the cylidrical shell formula, the resultig solid has volume 5 π y d π ( ) d π d π ( )] 65π 5 N ote that the volume could also have bee computed by the washer formula, but the calculatio would have bee somewhat more complicated. 5 5 Diffe re c e of S h e lls Formula Assume that g() f () o a iterval [a, b] with a. Let be the regio i the first q uadrat betwee the curves y f () ad y g() ad betwee a ad b. The the volume of the solid of revolutio obtaied by revolvig about the y ais is give by b V π ( f ( ) g( )) d (differece of shells formula) a

5 8 CHAPTER Applicatios of Itegratio II: Volume This obviously follows from the cylidrical shells formula because the req uired volume is the differece of two volumes obtaied by the cylidrical shells formula. N ote that a similar formula holds whe the roles of ad y are reversed. E X A M P L E.5 : Cosider the regio i the first q uadrat bouded above by y, below by y, ad lyig betwee ad. W he revolved about the y ais, this regio geerates a solid of revolutio whose volume, accordig to the differece of shells formula, is 5 ) π π ( ) d π ( ) d π 5 ( ( ) π 5 C ross-s e c tio Formula (S lic i g Formula) Assume that a solid lies etirely betwee the plae perpedicular to the ais at a ad the plae perpedicular to the ais at b. For each such that a b, assume that the plae perpedicular to the ais at that value of itersects the solid i a regio of area A(). (See Fig. -8.) The the volume V of the solid is give by For justificatio, see P roblem. b V A( ) d (cross-sectio formula) a Fig. -8 E X A M P L E.6 : Assume that half of a salami of legth h is such that a cross-sectio perpedicular to the ais of the salami at a distace from the ed O is a circle of radius. (See Fig. -9.) H ece, the area A() of the cross-sectio is π( ) π. So, the cross-sectio formula yields h h π h V A d d ( ) π π h Fig. -9 This formula is also called the slicig formula because each cross-sectioal area A() is obtaied by slicig through the solid.

6 CHAPTER Applicatios of Itegratio II: Volume 9 S O L V E D P R O B L E M S. Fid the volume of a coe that has height h ad whose base has radius r. The coe is geerated by revolvig about the ais the regio betwee the lie y r h ad the ais, betwee ad h. (See Fig. -(a).) B y the disk formula, the volume of the coe is h h π y d π r π π h d r r h ( ) h h ( h ) πr h. Fid the volume of the cylider of height h ad radius r. The cylider is geerated by revolvig about the ais the regio betwee the lie y r ad the ais, betwee ad h. (See Fig. -(b).) B y the disk formula, the volume of the cylider is h h V π y d r d r π π r h π h.. Fid the volume of a sphere of radius r. The sphere is geerated by revolvig about the ais the regio betwee the semicircle y r ad the ais, betwee r ad r. (See Fig. -(c).) B y the symmetry with respect to the y ais, we ca use the part of the give regio betwee ad r ad the double the result. H ece, by the disk formula, the volume of the sphere is ( ) ( ) r r r V y d r d π π ( ) π ( r ) π r r π r πr. Let be the regio betwee the ais, the curve y, ad the lie. (See Fig. -.) (a) Fid the volume of the solid obtaied by revolvig about the ais. (b) Fid the volume of the solid obtaied by revolvig about the y ais. Fig. - (a) The disk formula yields the volume V π y d d d π π 6 π 7 ( ) 8π 7 7

7 5 CHAPTER Applicatios of Itegratio II: Volume (b) (First solutio) The cylidrical shells formula yields the volume V y d d d 5 π π ( ) π π 5 ( ) (Secod solutio) Itegratig alog the y ais ad usig the washer formula yields the volume V π 8 y dy y dy y π 8 π ( ) [ ] y 5 6π 5 ( ) 8 ( 5 ) π ( ) 5 6π 5 5. Fid the volume of the solid obtaied by revolvig about the y ais the regio i the first q uadrat iside the circle y r, ad betwee y a ad y r (where < a < r). See Fig. -. (The solid is a polar cap of a sphere of radius r.) Fig. - Itegratig alog the y ais, the disk formula yields the volume ( ) r r r V π dy π r y dy ( r y y ) a π a π ( ) r ( r a a ) π a ( r r a a ) 6. Fid the volume of the solid obtaied by revolvig about the y ais the regio i the first q uadrat bouded above by the parabola y ad below by the parabola y. (See Fig. -.) Fig. -

8 CHAPTER Applicatios of Itegratio II: Volume 5 The curves itersect at (,). B y the differece of cylidrical shells formula, the volume is V π (( ) ) d π ( ) d π ( ) π ( ) π 7. Cosider the regio bouded by the parabola y ad the lies ad y 6. (See Fig. -5.) Fid the volume of the solid obtaied by revolvig about the lie y. To solve this problem, we reduce it to the case of a revolutio about the ais. R aise the regio vertically upward through a distace of uits. This chages ito a regio * that is bouded below by the parabola y, o the left by the y ais, ad above by the lie y 8. (See Fig. -.) The the origial solid of revolutio has the same volume as the solid of revolutio obtaied by revolvig R* aroud the ais. The latter volume is obtaied by the washer formula: V π ( 8 ( ) ) d π ( ) d π ( 5 5 ) π 5 π 5 8 ( 5 ) As i P roblem 7, cosider the regio bouded by the parabola y ad the lies ad y 6. (See Fig. -5.) Fid the volume of the solid obtaied by revolvig about the lie. Fig. - To solve this problem, we reduce it to the case of a revolutio about the y ais. M ove the regio to the right through a distace of uit. This chages ito a regio * that is bouded o the right by the parabola y ( ), above by y 6, ad o the left by. (See Fig. -.) The desired volume is the same as that obtaied whe we revolve * about the y ais. The latter volume is got by the differece of cylidrical shells formula: V 6 d 6 π ( ( ) ) π ( 8 ) d 8 π ( 6 8 ) d π ( 8 ) 8 π ( ) 8 ( )

9 5 CHAPTER Applicatios of Itegratio II: Volume Fig. - b 9. J ustify the disk formula: V π ( f ( )) d. a D ivide the iterval [a, b] ito eq ual subitervals, each of legth b a. (See Fig. -5.) Cosider the volume V i obtaied by revolvig the regio i above the ith subiterval about the ais. If m i ad M i are the absolute miimum ad absolute maimum of f o the ith subiterval, the V i lies betwee the volume of a cylider of radius m i ad height ad the volume of a cylider of radius M i V i ad height. Thus, πmi Vi π M i ad, therefore, mi M π i. (W e have assumed that the volume of a cylider of radius r ad height h is πr h.) H ece, by the itermediate value theorem for the cotiuous fuctio ( f ()) * V, there eists i i the ith subiterval such that i f π ( ( * )) ad, therefore, i * V π f ( ). Thus, i ( i ) i ( i ) i i * V V π f ( ) Lettig, we obtai the disk formula. Fig. -5

10 CHAPTER Applicatios of Itegratio II: Volume 5 b. J ustify the cylidrical shells formula: V π f ( ) d. a D ivide [a, b] ito eq ual subitervals, each of legth. (See Fig. -6.) Let i be the regio above the * ith subiterval. Let i be the midpoit i i of the ith iterval. The solid obtaied by revolvig the regio t about the y ais is approimately the solid obtaied by revolvig the rectagle with base ad height * * yi f ( i ). The latter solid is a cylidrical shell, that is, it lies betwee the cyliders obtaied by revolvig the * rectagles with the same height f ( i ) ad with bases [, i ] ad [, i ]. H ece, it has volume * * * π f ( ) π f ( ) π f ( )( ) i i i i i i i * * * * * π f ( i )( i i )( i i ) π f ( i )( i )( ) π i f ( i )( ) Thus, the total V is approimated by * * π f ( ) which approaches π f ( ) d as. i. J ustify the cross-sectio formula: V A( ) d. b a D ivide [a, b] ito eq ual subitervals [ i, i ], ad choose a poit * i i [ i, i ]. If is large, is small ad the piece of the solid betwee i ad i, will be close to a (ocircular) disk of thickess ad base area * A( i ). (See Fig. -7.) This disk has volume A( * i ). So V is approimated by * A( i ), which approaches b i A( ) d as. a b a Fig. -6 Fig. -7

11 5 CHAPTER Applicatios of Itegratio II: Volume. A solid has a circular base of radius uits. Fid the volume of the solid if every plae sectio perpedicular to a particular fied diameter is a eq uilateral triagle. Take the circle as i Fig. -8, with the fied diameter o the ais. The eq uatio of the circle is y 6. The cross-sectio ABC of the solid is a eq uilateral triagle of side y ad area A() y (6 ). The, by the cross-sectio formula, 56 V ( 6 ) d ( 6 ). A solid has a base i the form of a ellipse with major ais ad mior ais 8. Fid its volume if every sectio perpedicular to the major ais is a isosceles triagle with altitude 6. Take the ellipse as i Fig. -9, with eq uatio y. The sectio ABC is a isosceles triagle of base 5 6 y, altitude 6, ad area A( ) 6y H ece, ( ) 5 5 V 5 5 d 6π 5 Fig. -8 (N ote that 5 d is the area of the upper half of the circle y 5 ad, therefore, is eq ual to 5π/.) 5 Fig. -9 S U P P L E M E N T A R Y P R O B L E M S. Cosider the regio bouded by the parabola y 8 ad the lie. (See Fig. -.) (a) Fid the volume of the solid geerated by revolvig about the y ais. (b) Fid the volume of the solid geerated by revolvig about the lie. As. (a) 8 5 π ; (b) 56π 5

12 CHAPTER Applicatios of Itegratio II: Volume Fid the volume of the solid geerated by revolvig the regio betwee the ais ad the parabola y about the lie y 6. As. 8π 5 6. Fid the volume of the torus (doughut) geerated by revolvig the circle ( a) y b about the y ais, where < b < a. As. π ab 7. Cosider the regio bouded by y 6 ad y. Fid the volume of the solid geerated by revolvig about: (a) the ais; (b) the lie. As. (a) 79 π ; (b) 56π 5 I P roblems 8 6, fid the volume geerated whe the give regio is revolved about the give lie. U se the disk formula. 8. The regio bouded by y, y,, 5, about the ais. Αs. 5 π 9. The regio bouded by y 6, y, 8, about the ais. As. 56π. The regio bouded by y,, y 6, about y 6. (See Fig. -5.) As. 96π 5. The regio bouded by y, y,, about the ais. Αs. π. The regio bouded by y, y,, about. As. 6π 5. The regio withi the curve y (l ), about the ais. As. π 5. The regio withi the ellipse 9y 6, about the ais. As. 6 π 5. The regio withi the ellipse 9y 6, about the y ais. Αs. π

13 5 6 CHAPTER Applicatios of Itegratio II: Volume 6. The regio withi the parabola 9 y ad betwee y 7 ad the y ais, about the y ais. As. 96π 5 I P roblems 7, fid the volume of the solid geerated by revolvig the give regio about the give lie. U se the washer formula. 7. The regio bouded by y, y,, 5, about the y ais. Αs. 6 5 π 8. The regio bouded by y 6, y, 8, about the y ais. As. 8 π 9. The regio bouded by y,, y 8, about. As. π 5. The regio bouded by y, y, about the ais. As. π. The regio bouded by y, y, about y 6. As. 6π. The regio bouded by 9 y, y 7, about. Αs. 5π 5 I P roblems 7, fid the volume of the solid geerated by revolvig the give regio about the give lie. U se the cylidrical shells formula.. The regio bouded by y, y,, 5, about 6. Αs. 7 5 π. The regio bouded by y, y,, about y 8. Αs. π 7 5. The regio bouded by y, y, about 5. As. 6π 6. The regio bouded by y 5 6 ad y, about the y ais. As. 5π 6

14 CHAPTER Applicatios of Itegratio II: Volume The regio bouded by 9 y, y 7,, about y. As. 69π I P roblems 8, fid the volume geerated by revolvig the give regio about the give lie. U se ay appropriate method. 8. The regio bouded by y e, y,,, about the y ais. Αs. π( e ) 9. The regio bouded by y, y, about. Αs. 7 π. The regio bouded by y, y,,, about the y ais. Αs. π. The regio bouded by y, y, about the ais. Αs. π. The regio bouded by y, y ( ), about the ais. Αs. 7π 5. Fid the volume of the frustum of a coe whose lower base is of radius R, upper base is of radius r, ad altitude is h. As. πh( r rr R ). A solid has a circular base of radius uits. Fid the volume of the solid if every plae perpedicular to a fied diameter (the ais of Fig. -8) is: (a) a semicircle; (b) a sq uare; (c) a isosceles right triagle with the hypoteuse i the plae of the base. As. (a) 8 π ; (b) ; (c) A solid has a base i the form of a ellipse with major ais ad mior ais 8. Fid its volume if every sectio perpedicular to the major ais is a isosceles right triagle with oe leg i the plae of the base. As The base of a solid is the first-q uadrat regio bouded by the lie 5y ad the coordiate aes. Fid its volume if every plae sectio perpedicular to the ais is a semicircle. As. π 7. The base of a solid is the circle y 6, ad every plae sectio perpedicular to the ais is a rectagle whose height is twice the distace of the plae of the sectio from the origi. Fid its volume. Αs. π

15 5 8 CHAPTER Applicatios of Itegratio II: Volume 8. The sectio of a certai solid cut by ay plae perpedicular to the ais is a circle with the eds of a diameter lyig o the parabolas y ad y. Fid its volume. As. 656π 8 9. The sectio of a certai solid cut by ay plae perpedicular to the ais is a sq uare with the eds of a diagoal lyig o the parabolas y ad y. Fid its volume. As A hole of radius uit is bored through a sphere of radius uits, the ais of the hole beig a diameter of the sphere. Fid the volume of the remaiig part of the sphere. As. 6π

16 CHAPTER Techiques of Itegratio I: Itegratio by P arts If u ad v are fuctios, the product rule yields D ( uv ) uv vu which ca be rewritte i terms of atiderivatives as follows: uv uv d vu d N ow, uv d ca be writte as u dv, ad vu d ca be writte as v du. Thus, uv u dv v du therefore, The purpose of itegratio by parts is to replace a difficult itegratio v du. ad, u dv uv v du (itegratio by parts) u dv by a easy itegratio E X A M P L E. : Fid l d. I order to use the itegratio by parts formula, we must divide the itegrad l d ito two parts u ad dv so that we ca easily fid v by a itegratio ad also easily fid v du. I this eample, let u l ad dv d. The we ca set v ad ote that du d. So, the itegratio by parts formula yields: ( ) l d u dv uv v du (l )( ) l d l C ( l ) C Itegratio by parts ca be made easier to apply by settig up a rectagle such as the followig oe for E ample. u l dv d du d v d, where, after the itegratio o the right, the variable v is replaced by the correspodig fuctio of. I fact, by ( ) v ( ). H ece, u dv uv d. Similarly, v du vu d. uv d u dv the Chai R ule, D u dv D u dv D v u v : ; B C A A A B C A C A D?E8F < GHI!J9 K L%M N : O O(5 6 P%7 K Q : G R,B S QÏ T 5 O : Ï U< G 9 GV6 9G 9 PR6 VW R,G T 5 9

17 6 CHAPTER Techiques of Itegratio I: Itegratio by Parts I the first row, we place u ad dv. I the secod row, we place the results of computig du ad v. The desired result of the itegratio parts formula uv v du ca be obtaied by first multiplyig the upper-left corer u by the lower-right corer v, ad the subtractig the itegral of the product v du of the two etries v ad du i the secod row. E X A M P L E. : Fid e d. Let u ad dv e d. W e ca picture this i the bo below. u dv e d du d v e The, e d uv v du e e d e e C e ( ) C E X A M P L E. : Fid e cos d. Let u e ad dv cos d. The we get the bo u e dv cos d du e d v si So, e cos d uv v du e si e si d () N ow we have the problem of fidig e si d, which seems to be just as hard as the origial itegral e cos d. H owever, let us try to fid e si d by aother itegratio by parts. This time, let u e ad dv si d. u e dv si d du e d v cos The, e si d e cos e cos d Substitutig i formula () above, we get: e cos e cos d e cos d e si e cos e cos d e si e cos e cos d ( ) Addig e cos d to both sides yields e cos d e si e cos. So, e cos d ( e si e cos ) W e must add a arbitrary costat: e cos d ( e si e cos ) C N otice that this eample req uired a iterated applicatio of itegratio by parts.

18 CHAPTER Techiques of Itegratio I: Itegratio by Parts 6 S O L V E D P R O B L E M S. Fid e d. Let u ad dv v e dw e e w w.) e d. N ote that v ca be evaluated by usig the substitutio w. (W e get u dv e d du d v e H ece, e d e e d e e C e ( ) C. Fid l( ) d. Let u l ( ) ad dv d. u l( ) dv d du d v So, l( ) d l( ) d ( ) l( ) d l( ) ta C (l( ) ) ta C. Fid l d. Let u l ad dv d. u l dv d du d v So, l d l d l C (l ) C. Fid si d. W e have three choices: (a) u si, dv d; (b) u si, dv d; (c) u, dv si d. (a) Let u si, dv d. The du (si cos ) d, v, ad si d si (si cos ) d The resultig itegral is ot as simple as the origial, ad this choice is discarded.

19 6 CHAPTER Techiques of Itegratio I: Itegratio by Parts (b) Let u si, dv d. The du cos d, v, ad si d si cos d The resultig itegral is ot as simple as the origial, ad this choice too is discarded. (c) Let u, dv si d. The du d, v cos, ad si d cos cos d cos si C 5. Fid l d. Let u l, dv d. The du d, v, ad d d d l l l l 9 C 6. Fid si d. Let u si, dv d. u si dv d du d v So, si d si si ( ) / ( ) d d / si ( ( ) ) C (by Quick Formula I) / si ( ) C si C 7. Fid ta d. Let u ta, dv d. u ta dv d du d v So, ta ta d ta d d ta l( ) C (by Quick Formula II) 8. Fid sec d. Let u sec, dv sec d. u sec dv sec d du sec ta d v ta

20 CHAPTER Techiques of Itegratio I: Itegratio by Parts 6 Thus, sec d sec ta sec ta d sec ta sec (sec ) d sec ta sec d sec d sec ta sec d l sec ta The, sec d sec ta l sec ta H ece, sec d (sec ta l sec ta ) C 9. Fid si d. Let u, dv si d. Thus, du d ad v cos. The si d cos cos d cos cos d N ow apply itegratio by parts to cos d, with u ad dv cos d, gettig cos si si d si cos H ece, si d cos ( si cos ) C. Fid e d. Let u, dv e d. The du d, v e, ad e d e e d For the resultig itegral, let u ad dv e d. The du d, v e d e e e d e e, ad ( ) e For the resultig itegral, let u ad dv e d. The du d, v e d e e e e d e d e, ad ( ) 8 e e e e C. D erive the followig reductio formula for si m d. si Let u si m ad dv si d. m m d si cos m m m si m m u si dv si d m du ( m )si d v cos d

21 6 CHAPTER Techiques of Itegratio I: Itegratio by Parts The si m cos si m d ( m ) si m cos d m cos si m ( m ) si ( si ) d m m m cos si ( m ) si d ( m ) si d H ece, m si m d cos si m ( m ) si m d ad divisio by m yields the req uired formula.. Apply the reductio formula of P roblem to fid si d. W he m, we get si si cos d si si cos d d si cos C si cos C. Apply the reductio formula of P roblem to fid si d. W he m, we get si si cos d si d si cos cos C cos ( si ) C S U P P L E M E N T A R Y P R O B L E M S I P roblems, use itegratio by parts to verify the specified formulas.. cos d si cos C 5. sec d ta l sec C 9 6. cos d cos C 7. ta d ( )ta C 8. e d e ( 9 ) C 9. si d cos si 6 cos 6 si C

22 CHAPTER Techiques of Itegratio I: Itegratio by Parts 6 5. si ( ) d si ( ) C. l d l C π. Show that si d π for ay positive iteger.. P rove the followig reductio formula: sec d ta sec sec d.. Apply P roblem to fid sec d. As. ta (sec ) C 5. P rove the reductio formula: d d ( a ) ( a ) ( a ) 6. Apply P roblem 5 to fid d. ( a ) As. ( ta a a a ) C 7. P rove l d ( ) [( )l )] C for. 8. P rove the reductio formula: e d a e a a a ead. 9. U se P roblem 8 ad E ample to show that: e d e ( ) C.

23 CHAPTER Techiques of Itegratio II: Trigoometric Itegrads ad Trigoometric Substitutios T rig o ome tric I te g ra d s k. Let us cosider itegrals of the form si cos d, where k ad are oegative itegers. T yp e. At least oe of si ad cos occurs to a odd p ower: The a substitutio for the other fuctio works. E X A M P L E. : si cos d. Let u cos. The du si d. H ece, 7 E X A M P L E. : si cos d. Let u si. The du cos d, ad si cos d si cos si d ( cos )cos si d ( u ) u du ( u u ) du 5 u u C 5 cos cos C si cos d si cos cos d 6 u ( u ) du u ( u u u ) du ( u u u u ) du u 7 u u u C 9 si si si si C E X A M P L E. : si 5 d. Let u cos. The du si d ad : ; B C A A A B C A C A D?E8F < GHI!J9 K L%M N : O O(5 6 P%7 K Q : G R,B S QÏ T 5 O : Ï U< G 9 GV6 9G 9 PR6 VW R,G T

24 CHAPTER Techiques of Itegratio II si d si si d ( cos ) si d ( u ) du ( u u ) du 5 ( u u 5 u ) C cos cos cos C 5 5 T yp e. Both p owers of si ad cos are eve: This always ivolves a more tedious computatio, usig the idetities cos cos cos ad si E X A M P L E. : cos si d (cos )(si ) d cos cos ( )( ) d cos cos cos ( )( ) ( ( cos cos 8 ) (cos )( cos cos )) d ( cos cos cos cos cos ) d 8 ( cos cos cos ) d 8 ( ) d cos d cos d cos d 8 8 si cos ( d d (cos )( si ) ) ( ) si si d u d si si si si ( 8 ) C si si ( 8 6 ) C ( ) si si C d cos u [ lettig u si ] k. Let us cosider itegrals of the form ta sec d, where k ad are oegative itegers. R ecall that sec ta. T yp e. is eve: Substitute u ta. E X A M P L E.5 : ta sec d Let u ta, du sec d. So, ta sec d ta ( ta )sec d u ( u ) du ( u u ) du u u C ta ta C

25 6 8 CHAPTER Techiques of Itegratio II T yp e. is odd ad k is odd: Substitute u sec. E X A M P L E.6 : ta sec d Let u sec, du sec ta d. So, ta sec d ta sec ta d (sec )sec ta d ( u ) du u u C sec sec C T yp e. is odd ad k is eve: This case usually req uires a tedious calculatio. E X A M P L E.7 : ta sec d (sec )sec d (sec sec ) d (sec ta l sec ta ) l sec ta C ( by P roblem 8 of Chapter ) (sec ta l sec ta ) C. Let us cosider itegrals of the form si A cos B d, si A si B d, ad cos A cos B d. W e shall eed the idetities si A cos B (si( A B) si( A B) ) si A si B (cos( A B) cos( A B) ) cos A cos B (cos( A B) cos( A B) ) E X A M P L E.8 : si 7 cos d (si( 7 ) si( 7 ) ) d (si si ) d ( cos cos ) C ( cos 5cos ) C E X A M P L E.9 : si 7si d (cos( 7 ) cos( 7 ) ) d (cos cos ) d ( si si ) C ( 5si si ) C E X A M P L E. : cos7 cos d (cos( 7 ) cos( 7 ) ) d (cos cos ) d ( si si ) C ( 5si si ) C T rig o ome tric S ub stitutio s There are three pricipal kids of trigoometric substitutios. W e shall itroduce each oe by meas of a typical eample. E X A M P L E. : Fid d. Let ta q, that is, q ta (/). The d sec θ dθ ad ta θ ta θ sec θ sec θ

26 CHAPTER Techiques of Itegratio II 6 9 B y defiitio of the iverse taget, π/ < q < π/. So, cos q > ad, therefore, sec q >. Thus, sec θ sec θ /. H ece, z d sec θ dθ ta θ( sec θ) secθ dθ cosθ dθ ta θ si θ ( (si θ) ) C C siθ (si θ) cosθ dθ N ow we must evaluate si q. Aalytic method: siθ taθ sec θ / /. G eometric method: D raw the right triagle show i Fig. -. From this triagle we see that siθ /. (N ote that it follows also for q <.) H ece, d C Fig. - This eample illustrates the followig geeral rule: S tr a te g y I. If a occurs i a itegrad, try the substitutio a ta q. E X A M P L E. : Fid d. 9 Let si q, that is, q si (/). The d cos q dq ad 9 9 9si θ si θ cos θ cos θ B y defiitio of the iverse sie, π/ < q < π/ ad, therefore, cos q >. Thus, cos θ cos θ 9 /. N ow, d 9 cos θ dθ 9 si θ ( cos θ) 9 csc θ dθ cot θ C cos θ C 9 9 si θ 9 9 / C / 9 9 C This eample illustrates the followig geeral method: S tr a te g y I I. If a occurs i a itegrad, try the substitutio a si q. E X A M P L E. : Fid d.

27 7 CHAPTER Techiques of Itegratio II Let sec q, that is, q sec (/). The d sec q dq ad sec θ sec θ ta θ ta θ B y defiitio of the iverse secat, q is i the first or third q uadrat ad, therefore, ta q >. So, ta θ ta θ /. N ow, d sec θ( secθ ta θ) dθ taθ sec θ dθ (secθ taθ l secθ ta θ C) ( by P roblem 8 of Chapter ) l C l C l K where K C l This eample illustrates the followig geeral method: S tr a te g y I I I. If a occurs i a itegrad, try the substitutio a sec q. S O L V E D P R O B L E M S I P roblems, verify the give solutios. R ecall the idetities si u ( cos u) cos u ( cos u) si si cos. si d ( cos ) d ( si ) C ( si cos ) C.. cos ( ) d ( cos 6) d ( 6 si 6) C.. si d si si d ( cos )si d si d cos ( si ) d cos cos C ( by Quick Formula I). si cos d si cos cos d si ( si )cos d si cos d si cos d 5 si 5 si C ( by Quick Formula I)

28 CHAPTER Techiques of Itegratio II 7 5. si ( )cos 5 ( ) d ( cos ( ))cos 5 ( )si( ) d 5 7 cos ( )si( ) d cos ( )si d 5 7 cos ( )( si( )) d cos ( )( si( )) d cos ( ) 8 cos ( ) C ( by Quick Formula I) 8 6 ( cos ( ) cos ( )) C 7 ( ) ( ) ( ) ( ) ( ) 6. cos d si cos d si ( ) cos d cos d si cos d si ( )( ) si cos d si si ( ) C ( by Quick Formula I) si si ( ) 7. si (si ) d d ( cos( )) d d d cos( ) cos ( ) d si( ) ( cos )) 8 d si ( ) 8 ( si ( )) C si ( ) si ( ) C 8 8. si cos si ( ) d ( cos( )) d d 8 C ( si( )) C si( ) C si ( )cos ( ) d (si ( )cos ( ))si ( ) d si ( 6)( cos( 6)) d 8 8 si ( 6) d si ( 6)cos( 6) d 8 ( cos( )) d 6 8 si ( 6)( 6cos( 6)) d ( si( )) si ( 6) C ( by Quick Formula I) si( )) si ( ) C 6. si si d (cos( ) cos( )) d (cos cos 5)) d (si 5 si 5) C si si C 5

29 7 CHAPTER Techiques of Itegratio II. si cos 5 d (si( 5) si( 5)) d 8 (si( ) si( )) d 8 ( si( ) si( )) d ( cos( ) cos( 8)) C cos( ) cos( ) C cos cos d (cos( ) cos( 6)) d ( si( ) si( 6)) C si( ) si( 6) C. cos d si ( ) d by si ( ) cos /. ( cos ) d cos 6 ( ) ( ( )) ( ) cos C cos d ( ) si ( ) cos ( ) C sice cos ( ) cos( ) d d cos d si cos ( ) ( ) ( ) si( ) si ( ) cos ( ) si( ) si ( ) C d si( ) si ( ) 9 C 5. d si d cos π ( ) d si π ( ) sice si ( ) cos π π ( ) ( ) ( ) ( ) csc π l csc π cot π d C 6. ta d ta ta d ta (sec ) d ta sec d ta d ta (sec ) d ( by Quick Formula I) ta (ta ) C ta ta C

30 CHAPTER Techiques of Itegratio II ta d ta ta d ta (sec ) d ta sec d ta d ta ta (sec ) d ta ta sec d t a ( by Quick Formula I) d ta ta l sec C ( by Quick Formula I) 8. sec ( ) d sec ( )sec ( ) d sec ( )( ta ( )) d sec ( ) d sec ( )ta ( )) d ta( ) ta ( )( sec ( )) d ta( ) ta ( ) C ( by Quick Formula I) ta( ) ta ( ) C 9. ta ( )sec ( ) ta ( )( ta ( d ))sec ( ) d 6 ) d 5 ta ( )sec ( ) d ta ( )sec ( ta ( ) ta ( ) C ta 6 6 ( ) 6 8 ta ( ). cot ( ) d cot( )(csc ( ) ) d C cot ( ) l csc( ) C. cot ( ) d cot ( )(csc ( ) ) d cot ( )csc ( ) d cot ( ) d 9 cot ( ) (csc ( ) ) d cot ( ) cot( ) C 9. 6 csc d csc ( cot ) d csc d cot csc d cot csc d cot cot 5 cot 5 C. 5 cot csc d cot csc csc cot d (csc ) csc csc cot d 6 csc csc cot d csc csc cot d csc 5 csc C

31 7 CHAPTER Techiques of Itegratio II. Fid 9 d So, let siθ. The d cosθ dθ ad 9 9 9si θ cos θ cos θ cosθ H ece, B ut 9 cos θ( cos θ)dθ d siθ cos θ siθ d θ θ si θ d θ si (cscθ si θ) dθ l cscθ cot θ cosθ C cscθ siθ ad cotθ cosθ siθ 9 / / 9 So, 9 9 d l 9 K where K C l 5. Fid d. 9 Let taθ. (See Fig. -.) The d sec θ ad 9 secθ. H ece, d sec θ dθ 9 ( ta θ)( sec θ) cscθ dθ θ θ l csc cot C l 9 K Fig Fid ( 6 9 ) / 6 d. Let siθ. (See Fig. -.) The d cosθ dθ ad 6 9 cosθ. H ece, / ( 6 9 ) ( 6cos θ)( cos θ dθ) 6 d 96 6 si θ 79 5 cot θ csc θ θ cot θ 6 d C 8 5 ( 6 9 ) ( 6 9 ) 5 C / 5 / C

32 CHAPTER Techiques of Itegratio II Fid d d ( ). Fig. - Let si q. (See Fig. -.) The d cos q dq ad cosθ. H ece, ( si θ ) d cosθ cosθ dθ ( si θ) dθ ( siθ cos θ ) dθ θ cosθ si θ C si ( ) ( ) C si ( ) ( ) C 8. Fid d d ( 7) ( ( ) 9) / /. Fig. - Let secθ. (See Fig. -5.) The d secθ taθ dθ ad 7 taθ. So, d secθ taθ dθ / ( 7) 7ta θ 8 cscθ cotθ dθ cscθ C C (from Fig. -5) Fig. -5 S U P P L E M E N T A R Y P R O B L E M S 9. cos d si C. si d 6 cos cos C

33 7 6 CHAPTER Techiques of Itegratio II. si d 8 8 si 6 si 8 C. cos d 8 si si C si d 7 cos 5 cos cos cos C 6 5. cos d 6 si si si C si cos d si 5 si 7 si C 5 6. si cos d 5 cos cos C 7. si cos d 8 cos 6 cos C 8. si cos d 8 ( si 8 si 8 C 9. si cos d cos cos 6 C. cos cos d si si 5 C. si 5si d 8 si si 6 C... cos d si si si C / cos / 8/ cot si 5 5 d C cos csc csc si d C 5. (cos si ) d (si cos )( si ) C 6. ta d ta l cos C 7. ta sec d 9 sec sec C / / / 8. ta sec d ta 5 ta 9 C ta sec d 7 ta 5 ta C

34 CHAPTER Techiques of Itegratio II cot d cot l si C 6 5. cot csc d cot 6 cot C 5 5. cot csc d 5 csc csc C 5. csc d cot 6 cot C sec d ta ta ta ( ) C cot si csc csc d C 5 6. ta sec d sec C 5 7. d ( ) / C d 5l 5 C 5 9. d a a a C 6. d l( ) C 6. d ( a ) a / a si ( ) C 6. d l C a d a a a a l C a a d 5 ( ) ( ) / / C 6 5. d ( a ) / a a C 6 6. d C 6 7. d 6 8l 6 C 6

35 7 8 CHAPTER Techiques of Itegratio II 5/ 6 8. a d a a 5 a / ( ) ( ) C 6 9. d l( ) C 7. d ( ) / C 7. d 5 ta C ( 9 ) 8( 9 ) ( ) I P roblems 7 ad 7, first apply itegratio by parts. 7. si d ( )si C 7. cos d ( )cos C

36 CHAPTER Techiques of Itegratio III: Itegratio by P artial Fractios N( ) W e shall give a geeral method for fidig atiderivatives of the form D( ) d, where N() ad D() are polyomials. A fuctio of the form N ( ) is called a ratioal fuctio. (N() is the umerator ad D() is D( ) the deomiator.) As eamples, cosider d ad 8 d Two restrictios will be assumed, either of which limits the applicability of our method: (i) the leadig coefficiet (the coefficiet of the highest power of ) i D() is ; (ii) N() is of lower degree tha D(). A q uotiet N()/D() that satisfies (ii) is called a p rop er ratioal fuctio. Let us see that the restrictios (i) (ii) are ot essetial. E X A M P L E. : ote that Cosider the case where N ( ) D( ) is 8. H ere, our first restrictio is ot satisfied. H owever, 5 8 d 5 5 The itegral o the right side satisfies restrictios (i) ad (ii) d E X A M P L E. : Cosider the case where N ( ) D( ) is 5 7. H ere, our secod restrictio is ot satisfied. B ut we ca divide N() by D(): H ece, 5 7 d 8 7 d ad the problem is reduced to evaluatig 8 7 d, which satisfies our restrictios. A polyomial is said to be irreducible if it is ot the product of two polyomials of lower degree. Ay liear polyomial f() a b is automatically irreducible, sice polyomials of lower degree tha f() are costats ad f () is ot the product of two costats. N ow cosider ay q uadratic polyomial g() a b c. The g() is irreducible if ad oly if b ac < X Y Z [ \ ] ^ _`abc cd e f d d d e f d d ce f d g bh[i _ jkl!m\ o%p q ] r r(x Y s%z t ] j u,e v tl w X r ] l _ j \ jyy \`j \ suy yz u,j w 7 9

37 8 CHAPTER Techiques of Itegratio III To see why this is so, assume that g() is reducible. The g() (A B)(C D). H ece, B/A ad D/C are roots of g(). The q uadratic formula b ± b ac a should yield these roots. Therefore, b ac caot be egative. Coversely, assume b ac. The the q uadratic formula yields two roots of g(). B ut, if r is a root of g(), the g() is divisible by r. H ece, g() is reducible. E X A M P L E. : (a) is irreducible, sice b ac ()() 6 <. (b) is reducible, sice b ac (l)( ) 7. W e will assume without proof the followig fairly deep property of polyomials with real coefficiets. T H E O R E M. : Ay polyomial D() with leadig coefficiet ca be epressed as a product of liear factors of the form a ad of irreducible q uadratic factors of the form b c. (R epetitio of factors is permitted.) E X A M P L E. : (a) ( ) ( )( ) (b) ( ) ( is irreducible.) (c) 9 ( )( ) ( )( )( ) (d) ( l)( ) ( is irreducible.) M e th od of P artial Frac tio s N( ) Assume that we wish to evaluate D( ) d, where N ( ) is a proper ratioal fuctio ad D() has leadig D( ) coefficiet. First, write D() as a product of liear ad irreducible q uadratic factors. O ur method will deped o this factoriz atio. W e will cosider various cases ad, i each case, we will first eplai the method by meas of a eample ad the state the geeral procedure. C ase I D() is a product of distict liear factors. E X A M P L E.5 : Fid d. I this case, D() ( )( ). W rite A B ( )( ) It is assumed that A ad B are certai costats, that we must ow evaluate. Clear the deomiators by multiplyig both sides by ( )( ): A( ) B( ) () First, substitute for i (): A() B( ) B. Thus, B. Secod, substitute for i (): A() B() A. Thus, A. H ece, ( )( ) I geeral, if a polyomial h() has r as a root, the h() must be divisible by r.

38 CHAPTER Techiques of Itegratio III 8 So, ( ) d d l l C (l l ) C l C ( ) d E X A M P L E.6 : Fid. 6 Factorig the deomiator yields ( 6) ( )( ). The itegrad is R epreset it i the followig form: ( )( ). A B C ( )( ) Clear the deomiators by multiplyig by ( )( ): A( )( ) B( ) C( ) () Let be i (): A( )() B()() C()( ) 6A. So, A 6. Let be i (): A()(5) B()(5) C()() B. So, B. Let be i (): A( 5)() B( )() C( )( 5) 5C. So, C 5. H ece, ( ) d d ( ) l l l C 6 5 G e e ral R ule for C ase I R epreset the itegrad as a sum of terms of the form A for each liear factor a of the deomiator, where A is a ukow costat. Solve for the costats. Itegratig yields a sum of terms of the form a A l a. Remark: W e assume without proof that the itegrad always has a represetatio of the req uired kid. For every particular problem, this ca be verified at the ed of the calculatio. C ase II D() is a product of liear factors, some of which occur more tha oce. E X A M P L E.7 : ( 5) d Fid. First factor the deomiator: l ( l)( l) The represet the itegrad 5 as a sum of the followig form: 5 A B C ( ) I tryig to fid liear factors of a deomiator that is a polyomial with itegral coefficiets, test each of the divisors r of the costat term to see whether it is a root of the polyomial. If it is, the r is a factor of the polyomial. I the give eample, the costat term is. B oth of its divisors, ad, tur out to be roots.

39 8 CHAPTER Techiques of Itegratio III N ote that, for the factor ( ) that occurs twice, there are terms with both ( ) ad ( ) i the deomiator. N ow clear the deomiators by multiplyig both sides by ( l)( ) : 5 A( l) B( l)( ) C( ) () Let. The 8 ()A ()()B ()C C. Thus, C. Let. The ()A ()( )B ()C A. Thus, A. To fid B, compare the coefficiets of o both sides of (). O the left it is, ad o the right it is A B. H ece, A B. Sice A, B. Thus, Therefore, B y Quick Formula I, 5 ( ) ( 5) d l l d ( ) d ( ) ( ) d ( ) So, ( 5) d l l C l C E X A M P L E.8 : Fid ( ) d ( ). R epreset the itegrad ( ) i the followig form: ( ) ( ) A B C D E ( ) ( ) Clear deomiators by multiplyig by ( ) : A ( ) B( ) C( ) D ( ) E Let. The C. So, C. Let. The 8E. So, E 8. Compare coefficiets of. The B C. Sice C, B. Compare coefficiets of. The A B C. Sice B ad C, A. Compare coefficiets of. A D. Sice A, D. So, ( ) ( ) 8 ( ) ad ( ) d ( ) l l 8 8 C l 8 8 C

40 CHAPTER Techiques of Itegratio III 8 G e e ral R ule for C ase II For each repeated liear factor ( r) that occurs k times i the deomiator, use A A A r r k k... ( ) ( r) as part of the represetatio of the itegrad. E very liear factor that occurs oly oce is hadled as i Case I. C ase III D() is a product of oe or more distict irreducible q uadratic factors ad possibly also some liear factors (that may occur more tha oce). G e e ral R ule for C ase III Liear factors are hadled as i Cases I II. For each irreducible q uadratic factor b c, place a term A B i the represetatio of the itegrad. b c ( ) d E X A M P L E.9 : Fid ( )( ). R epreset the itegrad as follows: Clear the deomiators by multiplyig by ( l)( ). ( ) A B C D E ( )( ) M ultiply out o the right: A( l)( ) (B C)( ) (D E)( ) (A B D) (B E) (A C D) (C E) A Comparig coefficiets, we get: A, C E, A C D, B E, A B D H ece, A ad, therefore, C D, B D. From the latter two eq uatios, C B. From C E ad B E, we get C B. N ow, from C B ad C B, we get C. H ece, from C B, B. The, from B D, it follows that D. Fially, from B E, E. Thus, ( ) ( )( ) H ece, ( ) d l d ( )( ) d N ow, Also, d d l( ) (by Quick Formula II) d d d d ta l( ) ta Therefore, ( ) d l l( ) ( )( ) ta C

41 8 CHAPTER Techiques of Itegratio III C ase IV D() is a product of z ero or more liear factors ad oe or more irreducible q uadratic factors. G e e ral R ule for C ase IV Liear factors are hadled as i Cases I II. For each irreducible q uadratic factor b c that occurs to the kth power, isert as part of the represetatio of the itegrad. A B A B b c ( b c) Ak Bk... ( b c) k E X A M P L E. : Fid d. ( ) Let A B C D ( ) ( ). The (A B)( ) C D A B (A C) (B D) Compare coefficiets: A, B, A C, B D. H ece, C, D. Thus, d d ( ) ( ) d ta ( ) d I the secod itegral, let ta q. The sec θ dθ d ( ) cos d sec ( si θ θ θ θ θ cos θ) ( ) ( ta ) θ taθ ta θ Thus, ( ) 5 d ta C S O L V E D P R O B L E M S. Fid d. The itegrad is a improper fractio. B y divisio, ( ) W e write A B C ad obtai ( ) A( ) B( ) C

42 CHAPTER Techiques of Itegratio III 8 5 For, B ad B. For, C. For, A B C ad A. Thus, d d d d d. Fid ( )( ). d l l C l C Let A B ( )( ). Clear the deomiators: A( ) B( ) Let. The A. Let. The B. So, B. d d ( )( ) d l l C l(( ) ) l( ) C l ( ) ( ) C. Fid d. ( )( ) Let A B C. Clear the deomiators: ( )( ) A( )( ) B( ) C( ) Let. The A. So, A. Let. The 6 6B. So, B. Let. The C. So, C. H ece, d d d ( )( ) d l l l l ( C )( ) C. Fid ( )( ) d. Let A B C D. Clear the deomiators: ( )( ) ( ) ( ) l A( ) B( )( l) C( )( ) D( ) Let. The 7 7A. So, A 7 7. Let. The D. So, D. Compare coefficiets of. The A B. Sice A 7 7, B 7. Compare coefficiets of. A C. Sice A 7 7, C 7 9. Thus, d 7 d d d ( )( ) 9 ( ) ( ) 7 l l ( ) C d

43 8 6 CHAPTER Techiques of Itegratio III 5. Fid d. ( l)( ). W e write A B C D (A B)( ) (C D)( ) (A C) (B D) (A C) (B D) ad obtai H ece A C, B D, A C, ad B D. Solvig simultaeously yields A, B, C l, D. Thus, d d d ta l( ) C 5 6. Fid 8 d. ( ) 5 W e write 8 A B C D ( ) ( ) E F ( ). The 5 8 (A B)( ) (C D)( ) E F A 5 B (A C) (B D) (A C E) (B D F) from which A, B, C, D, E, F. Thus the give itegral is eq ual to ( ) d d d d ( ) d ( ) B y Quick Formula II, d d l( ) ad by Quick Formula I, d ( ) ( ) ( ) d ( )( ) ( ) So, 5 8 d l( ) ta ( ) ( ) C S U P P L E M E N T A R Y P R O B L E M S I P roblems 7 5, evaluate the give itegrals. 7. d l 6 9 C 8. d C 5 l ( )( ) 9. d ( ) l C

44 CHAPTER Techiques of Itegratio III 8 7. d 7 6 l 5 6 C.. d l ( )( ) C 8 d l C ( ). ( ) 6 d l( ) ( ) C. d l C 5. d l ta C ( )( ) 6. d l C 7. d ( ) l( ) C 8. d ( ) l( ) ta ( ) C 9. d l C ta ( ). 8 d l ( )( ) ( ). ta C 5 5 d 5 l ta ta ( 5 )( ) d C ( ) ( ) l 5 C.. 5. d e l C (H it: Let e e e e 9 e u.) si d cos l C (H it: Let cos u.) cos ( cos ) cos ( ta θ)sec θ l ta dθ θ ta ta θ taθ C

45 CHAPTER 6 Applicatios of Itegratio III: Area of a Surface of R evolutio If a arc of a curve is revolved about a lie that does ot itersect the arc, the the resultig surface is called a surface of revolutio. B y the surface area of that surface, we mea the area of its outer surface. Let f be a cotiuous fuctio o [a, b] that is differetiable i (a, b) ad such that f () for a b. The the surface area S of the surface of revolutio geerated by revolvig the graph of f o [a, b] about the ais is give by the formula b dy b S y d f f d a d π π ( ) ( ( )) (6.) a For a justificatio of this formula, see P roblem. There is aother formula like (6.) that is obtaied whe we echage the roles of ad y. Let g be a cotiuous fuctio o [c, d ] that is differetiable o (c, d) ad such that g( y) for c y d. The the surface area S of the surface of revolutio geerated by revolvig the graph of g o [c,d] about the y ais is give by the formula: d d d S dy g y g y dy c dy π π ( ) ( ( )) (6.) c Similarly, if a curve is give by parametric eq uatios f (u), y g(u) (see Chapter 7), ad, if the arc from u u to u u is revolved about the ais, the the surface area of the resultig surface of revolutio is give by the formula u d dy S π y du (6.) u du du H ere, we have assumed that f ad g are cotiuous o [u, u ] ad differetiable o (u, u ), ad that y g(u) o [u, u ]. Aother such formula holds i the case of a revolutio aroud the y ais. S O L V E D P R O B L E M S. Fid the area S of the surface of revolutio geerated by revolvig about the ais the arc of the parabola y from to. B y implicit differetiatio, dy d 6 6 ad dy y y d y X Y Z [ \ ] ^ _`abc cd e f d d d e f d d ce f d g bh[i _ jkl!m\ o%p q ] r r(x Y s%z t ] j u,e v tl w X r ] l _ j \ jyy \`j \ suy yz u,j w

46 CHAPTER 6 Applicatios of Itegratio III B y (6.), S π y y 6 d π d y 6 π( 8( 6) )] ( ) π. Fid the area S of the surface of revolutio geerated by revolvig about the y ais the arc of y from y to y. d dy y ad d 9 dy y. So, by (6.), S 9 y dy y 9 π π y dy π 9y 8 ( ) π 8 9 ( y ) ] π 7 ( ) ( 6y ) dy. Fid the area of the surface of revolutio geerated by revolvig about the ais the arc of y l y from y to y. S d y d y dy y dy y dy y e π π π ( ) dy π. Fid the area of the surface of revolutio geerated by revolvig a loop of the curve 8a y a about the ais. (See Fig. 6-.) Fig. 6 - H ere dy d a 8a y ad d ( a ) ( a ) dy 8a ( a ) 8a ( a ) H ece a dy S π d π d a π a d a a ( ) π a a a a a a d 5. Fid the area of the surface of revolutio geerated by revolvig about the ais the ellipse y. 6 S π y 6y d π d y 6 π 6 si 8 8π 9 π

47 CHAPTER 6 Applicatios of Itegratio III 6. Fid the area of the surface of revolutio geerated by revolvig about the ais the hypocycloid a cos q, y a si q. The req uired surface is geerated by revolvig the arc from q to q π. W e have d dy a cos θ si θ, asi θ cos θ, ad d dy ( a dθ dθ dθ ) θ θ dθ 9 cos si. The π S y d dy ( d a d ) π ( π ) θ ( π ) ( si θ) cos θ siθ θ θ dθ a d a π (sq uare uits) 5 7. Fid the area of the surface of revolutio geerated by revolvig about the ais the cardioid cos q cos q, y si q si q. The req uired surface is geerated by revolvig the arc from q to q π. (See Fig. 6-.) W e have d siθ si θ, dθ dy cosθ cos θ, dθ Fig. 6 - ad ( ) d dθ dy θ θ θ θ dθ 8 ( si si cos cos ) 8( cosθ ) The π S π ( siθ si θ )( cos θ ) dθ 8 π π si θ( cos ) 6 5 ( cos ) θ dθ π θ 5 8π (sq uare uits) 5 8. Show that the surface area of a cylider of radius r ad height h is πrh. The surface is geerated by revolvig about the ais the curve y r from to h. Sice dy d, dy d. The, by (6.), h S π r d π ( r) π rh h 9. Show that the surface area of a sphere of radius r is πr. The surface area is geerated by revolvig about the ais the semicircle y r from r to r. B y symmetry, this is double the surface area from to r. Sice y r, π y dy ad therefore d dy d ad dy y d y y y r y

48 CHAPTER 6 Applicatios of Itegratio III H ece, by (6.), r S y r r π d r d r r y π π π r. (a) Show that the surface area of a coe with base of radius r ad with slat height s (see Fig. 6-) is πrs. (b) Show that the surface area of a frustum of a coe havig bases of radius r ad r ad slat height u (see Fig. 6-) is π(r r )u. (N ote that the frustum is obtaied by revolvig the right-had segmet of the slat height aroud the base of the triagle.) Fig. 6 - Fig. 6 - (a) Cut ope the coe alog a slat height ad ope it up as part of a circle of radius s (as show i Fig. 6-5). N ote that the portio of the circumferece cut off by this regio is p r (the circumferece of the base of the coe.) N ow the desired area S is the differece betwee p s (the area of the circle i Fig. 6-5) ad the area A of the circular sector with cetral agle q. This area A is θ π π θ ( s ) s. Sice the arc cut off by q is p s p r, we get θ π s πr. Thus, A s p (s r)s. H ece, S p s p (s r)s p rs sq uare uits. Fig. 6-5 (b) From the similar triagles i Fig. 6-, we get u u u ru. The r u r r r u l r u. So, u r r. N ow, by part (a), the surface area of the frustum is p r (u u) p r l u l p (r - r l )u l p r u p r u p r u p (r r )u sq uare uits.

49 CHAPTER 6 Applicatios of Itegratio III 5. Sketch a derivatio of formula (6.). Assume that [a, b] is divided ito eq ual subitervals, [ k-, k ], each of legth b a. The total surface area S is the sum of the surface areas S k geerated by the arcs betwee the poits ( k-, f ( k- )) ad ( k, f ( k )), each of which is approimated by the surface area geerated by the lie segmet betwee ( k-, f ( k- )) ad ( k, f ( k )). The latter is the area of a frustum of a coe. I the otatio of Fig. 6-6, this is, by virtue of P roblem (b): f ( k f k π ( f ( k ) f ( k ) ) ( ) ) ( ) ( y) π ( ) ( y) N ow, f ( f k ) ( k ), beig the average of f ( k- ) ad f ( k ), is betwee those two values ad, by the itermediate value theorem, is eq ual to * f ( k ) for some * i ( ). Also, y ( ) ( y) k k-, k. B y the mea value, theorem y # f # ( k ) for some k i ( k-, k ). Thus, S is approimated by the sum k ( ) ( k ) ( k ) * # π f f ad it ca be show that this sum ca be made arbitrarily close to π f ( ) f ( ) d. H ece, the latter is a eq ual to S. b ( ) Fig. 6-6 S U P P L E M E N T A R Y P R O B L E M S I P roblems, fid the area of the surface of revolutio geerated by revolvig the give arc about the give ais:. y m from to ; ais As. mπ m. y from to ; ais As. p (8 8- )/9 I geeral, the followig result ca be proved: B lis s s T h e o r e m : Assume f ad g are cotiuous o [a, b]. D ivide [a, b] ito subitervals [ k-, k ] with a < < < < b, * ad let D k k k-l. I each [ k-, k ], choose k ad # * # k. The the approimatig sum f k g k kca be made arbitrarily close b k to f ( ) g( ) d by lettig ad makig the maimum legths of the subitervals approach. a ( ) ( )

50 6 CHAPTER 6 Applicatios of Itegratio III. y from to ; y ais As. ( ) π l 5. O e loop of 8y (l ); ais As. 6. y /6 l/ from to ; y ais As. π ( ) 5 l π ( ) 7. y l from to 7; y ais As. l π 8. O e loop of 9y ( ) ; y ais As. 8p /5 9. A arch of a(q si q ), y a( cos q ); ais As. 6p a /. e t cos t, y e t si t from t to t π; ais As. p (ep l)/5. F id the su rface are a of a z oe cu t from a sp he re of rad iu s r b y tw o p aralle l p lae s, e ach at a d istace a from the ce te r. As. par. F id the su rface are a of a toru s (d ou g hu t) g e e rate d b y re v olv ig the circle (y b) a ab ou t the ais. Assu me < a < b. As. π ab

51 CHAPTER 7 Parametric Represetatio of Curves Parametric Equatios If the coordiates (, y) of a poit P o a curve are give as fuctios f(u), y g(u) of a third variable or p aram eter, u, the equatios f(u) ad y g(u) are called p aram etric eq uatio s of the curve. EXAMPLE 7.: (a) cos θ, y si θ are parametric equatios, with parameter θ, of the parabola y, sice y cos θ si θ. (b) t, y t is aother parametric represetatio, with parameter t, of the same curve. It should be oted that the first set of parametric equatios represets oly a portio of the parabola (F ig. 7 -(a)), whereas the secod represets the etire curve (F ig. 7 -(b)). q y q p p q p q p O (a) q Fig. 7- EXAMPLE 7. : (a) The equatios r cos θ, y r si θ represet the circle of radius r with ceter at the origi, sice y r cos θ r si θ r (cos θ si θ) r. The parameter θ ca be thought of as the agle from the positive ais to the segmet from the origi to the poit P o the circle (Fig. 7-). (b) The equatios a r cos θ, y b r si θ represets the circle of radius r with ceter at (a, b), sice ( a) (y b) r cos θ r si θ r (cos θ si θ) r. { } ~ ƒ ˆ ˆ ˆ Š ~Œ Ž! % ({ %},ˆ š { œ ƒ œ, š 7

52 8 CHAPTER 7 Parametric Represetatio of Curves Fig. 7- A ssume that a curve is specified by meas of a pair of parametric equatios f (u) ad y g(u). The the first ad secod derivatives dy d ad d y are give by the followig formulas. d (7.) F irst D eriv ativ e dy d dy du d du This follows from the C hai R ule formula dy dy du d d du. (7. ) S ecod D eriv ativ e d y d d dy du d d du This follows from the C hai R ule formula d dy d y d du d d du. Arc Leg th for a Parametric C urv e If a curve is give by parametric equatios f (t), y g(t), the the legth of the arc of the curve betwee the poits correspodig to parameter values t ad t is L t t d dy dt dt dt This formula ca be derived by a argumet similar to that for the arc legth formula (9.). S O LV ED PR O B LEMS. Fid dy d ad d y if t si t, y cos t. d d dt cos t ad dy dt si t. B y (7.), dy d si t. The cos t d dy t t t t dt d ( cos ) (cos ) (si ) (si ) ( cos t) cos t (cos t si t) cos t ( cos t) ( cos t) cos t

53 CHAPTER 7 Parametric Represetatio of Curves 9 H ece, by (7.), d y d cos t ( cos t) ( cos t). Fid dy d ad d y if e d t cos t, y e t si t. d t dy e (cos t si t) ad t dt e (cos t si t). B y (7.), dy dt d cos t si t cos t si t. The, d dy t t t t t dt d (cos si ) (cos si )( si cos t ) (cos t si t) (cos t si t) (cos t si t) (cos t si t) (cos t si t ) (cos t si t) (cos t si t) S o, by (7.), d y t e (cos t si t) t d (cos t si t) e (cos t si t). Fid a equatio of the taget lie to the curve t, y t at the poit where t. t d dy ad. B y (7.), dy t. S o, the slope of the taget lie whe t is dt t dt t / d t 7. W he t, ad y A equatio of the taget lie is y ( ).. The positio of a particle that is movig alog a curve is give at time t by the parametric equatios cos t, y si t, where ad y are measured i feet ad t i secods. (S ee Fig. 7-.) N ote that 9 ( ) ( y ), so that the curve is a ellipse. Fid: (a) the time rate of chage of whe t π/; (b) the time rate of chage of y whe t 5π/; (c) the time rate of chage of the agle of icliatio θ of the taget lie whe t π/. dy dy dy si t ad cos t. The taθ cot t. dt dt d (a) W he t π, d dt ft/sec (b) W he t 5 π dy, dt ( ) ft/ sec (c) θ ta ( cot t ). S o, d θ dt 9 csc cot t t 6 csc. t t 9 cot t t O y t t p q t (, ) 5 p p Fig. 7-