PART-II Integral Calculus

ART-II Integral alculus

158 omprehensive Applied Mathematics UK EU UK UK EU UK UK UK...(1) (ii) Obtain a reduction formula for EU. Let I n EU EU EU I n EU UK EU UK [Intrgration by parts] I n EU UK EU UK I n I n \ EU UK EU EU EU UK EU EU EU EU UK EU EU UK EU EU which is the required the formula. (iii) (a) Obtain a reduction formula for. and deduce Let I n I n UGE I n UGE.....(1) This is the required reduction formula. (b) In Eq. (1) put 4 for n.....(2)

Integral alculus - II 159 Again in (1) for n put 2. From (2) we have (iv) (a). Obtain a reduction formula for Let I n E I n I n I n I n E E E EUGE E EUGE E E and deduce E ] EUGE _ E E. E E E...(1) This is the required reduction formula. (b) ut in (1), then ut n 3 in (1), then E E E...(2) E E E E NI UK Substituting this value of E in (2), we have (v) (a) Obtain a reduction formula for UGE E E E NI UK UGE deduce UGE

160 omprehensive Applied Mathematics I n UGE UGE (Integrating by parts) UGE UGE UGE UGE I n UGE UGE I n UGE UGE UGE I n UGE UGE UGE UGE UGE UGE UGE. UGE UGE UGE UGE UGE UGE UGE UGE UGE (i) This is the required reduction formula. (b) ut in (i), then UGE UGE UGE (vi) Obtain a reduction formula for Let I n I n UGE NI µ EUGE and deduce EUGE EUGE EUGE EUGE. EUGE E EUGE EUGE E E I n I n EUGE E EUGE E EUGE E EUGE EUGE I n EUGE E EUGE EUGE EUGE EUGE E EUGE EUGE E EUGE EUGE

Integral alculus - II 161 For part second put n 5 EUGE E EUGE EUGE EUGE E EUGE EUGE ¼ ½ EUGE E EUGEE EUGE NI EUGE E EUGEE NI Ans. ILLUSTRATIE EXAMLES EXAMLE 1. Apply reduction formula to evaluate EU. utting We know that in it, we have EU UK EU EU UK ¼ ½ EU UK EU UK N EU UK EU UK EU UK EU UK on putting n 4 EUUK ¼ ½, on taking EU UK EU UK EU UK EU EU UK EU UK EU UK

162 omprehensive Applied Mathematics EXAMLE 2. Evaluate : θfθ. utting We know in it, we have θ θfθ θfθ. θ θ θ θfθ F θ θ θ θ F ª º θ θ NI UGE θ F NI EU ª º θ θ θ θ θ NI NI ª º EXAMLE 3. If NI, show that. We know ª º. or I n ª º or or

Integral alculus - II 163 EXAMLE 4. Evaluate : Fθ. UK θ Here UK Fθ θ EUGE EUGE F, where θ θ θ We know that EUGE EUGE E EUGE EUGE EUGE E EUGE EUGE E E Hence, Fθ EUGE UK θ EUGE θe θ E θ ª º EXAMLE 5. UGE θfθ. Let θ UGE θfθ UGE θfθ (1 θ UGE θ F θ θ θ.

164 omprehensive Applied Mathematics EXAMLE 6. UGE First obtain the reduction formula for UGE. We have UGE UGE UGE (1) In (1) for n put 5 Again in (1) for n put 3 utting the value of UGE UGE UGE (2) UGE UGE UGE UGE NI UGE UGE in (2), we have UGE UGE UGE NI UGE UGE ¹ ª ¹ º UGE UGE NI UGE NI ª º EXAMLE 7. Evaluate. ut θ, so that UGE θfθ. Alos when 0, θ or θ and when x a, θ or θ or θ UGE θfθ UGE θ UGE UGE F θfθ θ θ...(1)

Integral alculus - II 165 Now we can prove UGE UGE µ θfθ UGE θ θ F θ θ...(2) utting n 7 in (2), we have UGE θfθ UGE θ θ UGE θfθ utting n 5 in (2) UGE UGE UGE θ θ θ θ θfθ ª º UGE UGE UGE θ θ θ θ F θ θ UGE UGE θ θ θ θ utting in (2). UGE θ θ UGE θ Fθ ª º UGE UGE UGE θ θ θ θ θ θ NI UGE θ θ F UGE θ θ UGE θ θ? UGE θ θ? UGE θ θ? NI UGE θ θ)???? NI? NI NI

166 omprehensive Applied Mathematics Substituting this value in (1) we get NI NI? EXAMLE 8. If U n 4 0 tan n xdxshow that 7 7 of U 5. and deduce the value ¼ ½ 7 Now put n 5 7 7 7 7 (art I proved) 7 7 where 7 NI UGE NI 7 7 7 7 NI 7 NI 2.2 WALLI S FORMULA 7 7 NI < NI > 7 < NI > Ans. To evaluate UK and EU, Ν. By reduction formula of UK, we have

168 omprehensive Applied Mathematics roceeding in the same manner, it can be shown that the formulae (A) and (B) are true for EU for odd and even values of n respectively. The formulae (A) and (B) are called Wallis formulae. SOLED EXAMLES EXAMLE 1. Evaluate the following integrals: (i) (iii) (v) (ii) UK EU RFR EU (i) UK As we know that (iv) s s s UK when n is an even number. UK θ F θ UK s s" s s Ans. (ii) UK when n is an odd number. s s s s s s Ans. (iii) UK s s " ¹ ª ¹ º EU θfθ Let θ Fθ when θ 0 θ /4 t /2 EU s s ª º Ans. (iv) UK θ Fθ Let θ/2 t dθ/2 dt dθ/2 2dt dθ 2dt

Integral alculus - II 169 when θ 0 t 0 θ t /2 UK s s ª º Ans. (v) EU Let EU s ª º Ans. EXAMLE 2. Evaluate: d. θ, UGE θfθ we have d UGE θfθ θ EU θfθ by Wallis formula. EXAMLE 3. Let UK θ UKθ EUθFθ 0 UK θ θ0 θ θ UK UK θ UK θ UK EU θ θ Fθ UK θ EUθ UK θfθ UKθEUθFθ

170 omprehensive Applied Mathematics s s s s s ª º EXAMLE 4. rove that Here the power is even UK UK. ] _ s s " Ans. " ] " _ 2n 2 "? n 2 n. 2 Hence, proved. EXAMLE 5. /2 m n 0 sin xcos x dx Γ O Γ ª º ª º Γ O ª º. O UK EU O UK EU EU O O UK UK EU EU UK O O O UK EU UK O EU O O O UK EU UK O UK EU O O breaking UK O into two parts O O O UK EU UK O EU EU O UK EU UK O EU O O O UK EU O

Integral alculus - II 171 Transporting the last integer on the right to left, or UK O EU or O UK EU O UK EU O ¼ ½ UK O EU UK O EU O O UK O EU UK O EU O O O O ¹ O ª º UK EU UK O EU O O UK EU UK EU O...(1) O O O O Now the following cases arise: ase I. Let m and n be both even. Now from (1) O O UK EU UK EU O UK UK...(2) O O O UK EU In this way by repeated application of (1), ultimately the power of cos x be zero and we shall have UK O EU " O O O O " O O O O s UK O...(3) O O " O O s ª º

172 omprehensive Applied Mathematics "?O O "? O O O " O OO " µ µ Oµ Oµ " " ¹ ¹ ª ºª º Oµ Oµ Oµ µ µ " Γ O ª º ª º Γ O ª º Γ ase II. Let m be odd and n be even If m is odd we have UK O O O ". O O Substituting this value in (3) case above, we get UK O EU O O " " O O O O O ª º "?O O "? O O " O OO " O O " " µ µ µ µ µ µ ¹ ¹ ª º ª º O O " " µ µ µ µ Γ O ª º ª º Γ O ª º Γ Hence, proved. ase III. Let m be even and n be odd. utting for x this case becomes the same as case II and here the above result holds good in this case also. ase I. Let m and n both be odd. In this case by repeated application of (1) the power of cos ultimately diminishes to one.

174 omprehensive Applied Mathematics O O " " O µ O µ µ " " µ µ µ µ µ µ ¹ ¹ ª º ª º Γ O ª º ª º Γ O ª º Γ Hence, for all positive integer values of m and n, we have O Γ ª º ª º Γ O ª º Γ SOLED EXAMLES EXAMLE 1. Evaluate UK EU. UK EU Γ Γ Γ Γ Γ Γ?? Ans. EXAMLE 2. Evaluate UK EU. UK EU Γ Γ ª º ª º Γ ª º ΓΓ Γ Ans.

Integral alculus - II 175 EXAMLE 3. Evaluate UK. UK UK EU Γ Γ ª º ª º Γ ª º ΓΓ Γ EXAMLE 4. Evaluate EU. ut θ, so that Fθ. Also when x 0, θ 0 and when EU, θ. EU θfθ UK EU θfθ Γ Γ Γ Γ Γ. Ans. EXAMLE 5. Evaluate UK φeu φfφ. ut φ, so that Fφ. UK φeu φfφ UK EU. UK EU EU UK EU ΓΓ Γ Γ Γ I Ans. EXAMLE 6(a) Evaluate. ut UK θ, so that EU θfθ. Also when, θ and when, UK θ or UK θ or θ

Integral alculus - II 177 UK θeu θfθ Γ Γ Γ Ans. EXAMLE 7(c) Show that. ut UK θ, so that EUθFθ Also when, θ and when, UK θ or θ UK θeu θfθ UK θ UK θ EUθFθ Γ? Γ? Γ Ans. UNSOLED ROBLEMS 1. Apply a reduction formula to integrate the following: (i) UK θ (ii) E (iii) (iv) EU 2. rove that UGE NI?. 3. If φ. Show that φ φ. 4. Evaluate (i) 5. Evaluate. 6. Evaluate 7. rove that UK θeu θfθ (ii) UK. UK (ii). ] _

178 omprehensive Applied Mathematics 8. (i) Evaluate UK θfθ (ii). 9. rove that 10. rove that EU.. UK EU 1. (i) UK EU ANSWERS (ii) (iii) (iv) E E NI UK NI EU EU UK EU UK EU UK UK 4. (i) (ii) 5. 6. (i) (ii) 8. (i) (ii)

180 omprehensive Applied Mathematics roperty III: D E H H D H, where < E < D E roof: The R.H.S. (? (? E D E (E (? (D (E? (D ( L.H.S. This result can be generalised as follows: D H where < E < E E < E < D roof: Here the R.H.S. Hence, the result. E E E H H E H. E E E (E ( (E (E??. (D ( L.H.S. D E H H (E (E? (D (E? roperty I: H H. roof: Let so that for the limits, when x a, t 0 and when x 0, t a. H H H By rop. II H, By rop. I. Hence, proved. roperty : H H, if f (x) is an even function; 0; if f (x) is an odd function. roof: Since a < 0 < a, H By rop. III. If x t, dx dt, we have H H H, H H...(1)

182 omprehensive Applied Mathematics ase 2. H H, then (4) gives H H H H. From the above properties, we get the following important results: (i) H EU H EU, ' HEU is an even function. (ii) H UK HUK, by rop. I. SOLED EXAMLES EXAMLE 1. Evaluate UK. UK UK. Ans. EXAMLE 2. Evaluate UK. UK EU Let UK UK EU...(i) UK (Applying rop I.) UK EU EU EU UK Adding Eqs. (i) and (ii), we have UK EU EU UK...(ii)

Definite Integrals 183 UK UK EU Ans. EXAMLE 3. rove that. UK Let UK...(i) ( (Applying rop. I) UK UK Adding Eqs. (i) and (ii), we have...(ii) UK UK UK UK UK UK EU UGE UGE UGE? 2 I? Hence, proved. EXAMLE 4 Evaluate: UK. Here UK UK, by roperty I. s ' UK UK? Ans.

Definite Integrals 185 UK UK EU UGE UGE? UGE?? I UGE Ans. UK EXAMLE 6. Evaluate. EU UK Let EU #RRN[KI H H Then UK EU UK EU UK UK EU EU or UK EU

186 omprehensive Applied Mathematics UK EU Let EU UK UK When x 0, t 1 and when x, t 1.? ] _ EXAMLE 7. Evaluate. EU D UK Let EU D UK #RRN[KI H H Then EU UK D ( EU D UK EU D UK EU D UK EU D UK EU D UK

188 omprehensive Applied Mathematics EXAMLE 9. NI NI θ F θ? NI F θ R NI NI Evaluate NI UK. Let NI UK...(1) NI UK NI EU, by rop. I....(2) Adding Eqs. (1) and (2), we have 2 I ]NI UK NI EU _ NI UK EU NI UK EU NI _ NI UK NI NI UK NI where, NI UK NI, by rop. I NI UK NI, by rop. II or NI. I NI NI.

Definite Integrals 189 Note: This is an important result and may be applied directly in many problems. It at once follows that NI UK NI EU NI. EXAMLE 10. NI EU θ Fθ. Let I NI EU θ Fθ NI EU θ F θ θ NI F NI EU θ Fθ θ NI θ NI EU Fθ θ Let x θ 2 x dθ 2 dx when θ 0, x 0; when θ, θ NI NI EU Fθ...(1). NI EU θ F θ NI EU NI NI Now from (1) we have I NI NI NI NI EXAMLE 11. θ UK θ EU θ F θ. Let UK EU....(1) θ θ θfθ θuk θeu θ F θ ' H H

Definite Integrals 191 EUGE NI NI NI NI UK EU NI EU UK UK EU NI EU UK UK UK NI UK UK NI NI NI NI NI

192 omprehensive Applied Mathematics EXAMLE 13. Show that NI. Let NI NI NI E NI NI NI NI. EXAMLE 14. Show that UK NI 0. Let UK NI UK NI UK NI E UK NI UK NI

Definite Integrals 193 UK NI UK NI EXAMLE 15. Show that EU >?. Let...(1) EU Also, EU or Adding Eqs. (1) and (2), we have EU EU...(2) &KXKFKIWOGTTF FGOKTD[ EU UGE UGE UGE...(3) Let when x 0, t 0 when, Now from (3) we have UGE

194 omprehensive Applied Mathematics. EXAMLE 16. Evaluate NI. Let Let θ when x 0, θ 0 when, θ NI UGE θfθ NI θ E θ? F θ UK θ EU θ NI Fθ EU θ UK θ NI Fθ UK θ EU θ NI UK θ EU θ?fθ

196 omprehensive Applied Mathematics ( Trigonometrically) EXAMLE 19. NI Evaluate. Let When NI NI NI Ans.

Definite Integrals 197 1. rove that (i) UNSOLED ROBLEMS (ii) NI 2. Show that E 3. Show that UK UK 4. Evaluate EU 5. rove that 6. rove that NI UK NI UK EU 7. Evaluate UGE 8. rove that NI UK NIEU NI

2 B.2 WORKING RULE Summation of Series 199 Thus, we have the rule: to write down the corresponding integral when the general term of the series is H T. Replace by dx, T by x, N by and insert the values of T for the first and last terms or the limits of such values as the lower and upper limits respectively in the integral. EXAMLE 1.(A) SOLED ROBLEMS Find the limit of the sum (a) (b) (a) Here T r T T Hence, the series NKO T T T Lower limit of the integral NKO T is taking the first term of T NKO T T Upper limit of the intergal NKO T Ans.

Summation of Series 201 Hence, N T T NI? NI G G EXAMLE 3. Evaluate N Let the limit be A, then NI # N NI NI NI G T N NI T T T N NI T T NI (from trigonometry) EXAMLE 4. # G Ans..

202 omprehensive Applied Mathematics Here the series T r T T 6 T T T T T NKO T T T L. Limit of the integral T NKO T T U. Limit of the intergal NKO T Then the intergal is Let t If x 0, t 1; x 1, t 2 NI? NI NI? NI Ans. EXAMLE 5. ] _. Here the series T 6T NKO T NKO T T NKO T T

204 omprehensive Applied Mathematics From Eq. (A)? 0 1 ( 1) UK Ans. EXAMLE 7. Evaluate NKO Let # NKO Taking log both the side NI # NKO NI NKO NI NI NI NKO NI T T L. Limit of integral is 0 U. Limit of integral is 1. NI NI NI # ] NI _ Integrating by parts NI # NI NI ]NI _ NI NI # NI NI G NI G G ]' NIG _ # G Ans.

Summation of Series 205 EXAMLE 8. Find the limit of when n tends to be infinite. NKO # NKO Taking log on both the sides NI # NKO NI NI NI NI T NI # NKO NI T L. Limit of integral T NKO T T U. Limit of integral NKO T NI # NI NI NI NI # A e 1 G Ans. Find the limit of the sum when : (i) UNSOLED ROBLEMS

206 omprehensive Applied Mathematics (ii) (iii) (iv) ] _ T (v) Evaluate NKO T T (vi) Evaluate NKO (vii) Evaluate NKO ANSWERS (i) NI G (ii) (iii) (iv) NIG (v) NI (vi) G (vii) G

208 omprehensive Applied Mathematics Integrate with respect to x between limits a and b, 2.1.2 D H Area of DBA 4GSWKTGF TG H D Area in olar oordinate D F5 D F5 < 5 > If f (q) be continuous for every value of q in the domain (a, b) then the value of area R bounded by curve r f (q) and radius vector q a to q b is T F R. RB Area of sector OA is A B S ( r dr, q dq) R r (, q) D dq b a O Initial line q 0 Fig. 2.2 Area of sector O (A d A) A Þ da Area of sector OS < Area of sector O < Area of sector OR A Þ r r dq < da < (r dr) (r dr) dq Þ Þ r2 < E # ER onsidering NKO 3m2 T NKO ERm F# FR T Þ F# B Þ [A] b [A] a B 4GSWKTGF TG T ET # [Dividing by dq] E T T ET ª ER º E # ER T F R RB R RB T FR T FR T

2.1.3 Rectification Application of Integral 209 The process of determining the length of an arc of a curve between two given points is called rectification. 2.1.3.1 For artesian equation of the form y f(x) If the length of an arc of a curve measured from a fixed point A (for which x a) to another point (x, y) on the curve is S, then by Differential alculus. F5 S F[ µ F[ µ Hence, if the abscissae of points A and B are a and b, the length of the arc AB is given by S D F[ µ 2.1.3.2 For artesian equations of the form x f(y) In this case by Differential alculus, we have F5 F[ µ Hence, if the ordinates of any two points and D on the curve are c and d respectively, the length of the arc D is equal to F[ µ F dy E F[ 2.1.3.3 For the curve given in parametric form x f(t), y y(t) In this case, we know that, F5 µ F[ µ Hence, if the values of t for any two points A and B on the curve are t 1 and t 2 respectively, the length of the arc AB is given by S µ F[ µ 2.1.3.4 For polar equations of the form r f(q) In this case, we know that F5 FR T FT µ R F

Application of Integral 211 and S R B R FT µ T ¹ FR ª ¹ º F R ]_ H R ] H a_? R FR B F(q) (iii) This intrinsic equation is obtained by eliminating q and f between the relations (i), (ii) and (iii). 2.1.4.3 To find the intrinsic equation from the parametric equations Let x f(t), y f (t) be the parametric equations of a curve, Hence, tan y F[ F[ H a Ga (i) S F[ µ µ ¹ ª ¹ º 2.2 OLUMES OF SOLIDS OF REOLUTION A solid of revolution is the solid formed by the revolution of an area about a line (called the axis of revolution). It is assumed that the axis of revolution does not pass through the area. Theorem The volume of the solid generated by the revolution, about the axis of x, of the area bounded by the curve y f(x), the coordinates x a, x b and the axis of x, is equal to D [ roof. Let the equation of the curve AB be y f (x) and let the abscissae of A and B be a and b respectively. Take any point p(x, y) on the curve and let M be its ordinate. hoose another point (x dx, y dy) on the curve close to and let N be its ordinate. Fig. 2.5

Application of Integral 213 (3) If the equation of the curve, in polar coordinates, is r f (q) and the curve rotates about the initial line (i.e., q 0), then the volume generated is D [ [ F B F R R F TUK TEU F B R F R R R where x r cos q, y r sin q and q a when x a and q b where x b. 2.3 SURFAE AREA OF SOLIDS Theorem. The surface area of the solid generated by the revolution about the axis of x, the area bounded by the curve y f (x), the x-axis and the ordinates x a, goes to x b is D [FU roof. Let AB be the arc of the curve y f(x) lying between the ordinates x a and x b. hoose two points (x, y) and (x dx, y dy) on the curve close to each other and let M and N be their ordinates. Let, ds length of elementary arc and ds surface area of the solid formed by the revolution of the arc about x-axis. Then it can be safely assumed that the area of curved surface of the solid generated by the revolution of the area MN about the x-axis lies in magnitude between the areas of the curved surfaces of the two right circular cylinders, one of which has the radius M (y) and the other N( y dy). Thus 2py ds < ds < 2p(y dy)ds E5 or 2py < < 2p(y dy) (1) EU In the limit when, ds 0 and dy 0 it follows from (1) that F5 FU 2py Integrating the limits x a $ x b, we get D [FU D Þ [S] F5 b a (alue of S when x b) (alue of S when x a) Area of the surface generated by the revolution of the arc AB about the x-axis. D Hence, required surface area 5 [FU O Y a A ( x, y) R M N D Fig. 2.6 b ( x dx,y dy ) S B y f( x) X

214 omprehensive Applied Mathematics Remarks. The above result may be easily modified to give the area of the surface of solid of revolution when the axis of revolution is other than the axis of x or when the equation of the curve is given in other systems of coordinates. (1) If the equation of the curve is given in polar form, i.e. r f(q), the surface area R FU [ FR R FR R R FT µ T UKR T FR R where, q 1 and q 2 are the vectorial angles of points A and B. (2) When the equation of the curve is in parametric form, i.e. x f (t), y f(t), the surface area is FU [ F µ F[ µ [ 2.4 URE TRAING urve tracing is important in order to find the limits of integration for various curves under area/volume calculation. 2.4.1 Tracing in artesian oordinates 1. Symmetric (a) If all the powers of Y are even, the curve will be symmetrical about X axis. Example. y 2 4ax is symmetric about X-axis. Fig. 2.7 (b) If all the powers of X are even, the curve is symmetrical about Y-axis. Example. x 2 4ay is symmetric about y-axis Fig. 2.8 (c) If all powers of both X and Y are even, then the curve is symmetric about both the axes.

216 omprehensive Applied Mathematics 4. Find F[ and evaluate at what points it becomes zero or infinite. At these points, tangents are parallel to the axis of x and axis of y respectively. 5. Find out points on the curves whose presence can be easily detected. 6. Assymptotes. For finding direction in which the curve extends to, find assymptotes parallel to coordinate axes and also other linear assymptotes by equating coefficient of highest degree term to zero. If it is a constant then there is No assymptotes to parallel axis. 7. If possible, solve the equation for x or y whichever is more convenient to find out the range of curve. If for values of x in certain range y is found imaginary then curve will not exist in that range. 8. Sometimes artesian curves can be traced easily by converting to polar form x r cos q, y r sin q Example. x 3 y 3 EURUKR 3axy becomes r. EU R UK R 2.5 TRAING IN OLAR OORDINATES 1. Symmetry (a) The curve is symmetric about the initial line if equation remains unchanged on changing q to q Example. r a(1 cos q) Fig. 2.12 Þ r a(1 cos ( q) (b) The curve is symmetrical about the pole if the equation remains unchanged on changing x to r. (c) The curve is symmetrical about q if the equation remains unchanged on changing q to p q. (d) The curve is symmetrical about q if the equation remains unchanged on changing q to q. 2. If r is zero for any real value of q, then the pole lies on the curve. 3. Make a table of corresponding values of r and q which satisfy equation of curve. If the curve is periodic like sin q or cos q, then values from o to 2p are to be considered only. EXAMLE 1. SOLED EXAMLES Trace the curve y 2 (a x) x 2 (a x). Symmetry. Since all powers of y are even the curve is symmetrical about x-axis Origin. Since x, y 0 satisfies the curve, origin lies on the curve.

Application of Integral 217 Equation of Tangent: Equating lowest degree terms to zero ay 2 xy 2 ax 2 x 3 a(y 2 x 2 ) xy 2 x 3 0 y 2 x 2 0 y x So, two tangents are real and distinct. Then origin is said to be NODE. Intercepts: y 0 Þ x 2 (a x) 0 Þ x 0, a oints where curve cuts x-axis (0, a) and (0, 0) Fig. 2.13 Assymptote. Equating coefficient of highest degree term of y to zero. x a 0 x a (Eq. of assymptotes parallel to y-axis) While the coefficient of highest degree term of x is constant No assymptotes parallel to x-axis. \ [ p \ [? [ d EXAMLE 2. Trace a curve y 2 (2a x) x 3. Symmetry. Since all powers of y arc even the curve is symmetrical about x-axis. Origin. Since x, y 0 satisfies the curve, origin lies on curve. Tangents. Equating lowest degree term to zero: 2ay 2 0 y 2 0 y 0» ¼ IG GSU y 0½ Since tangents are coincident, origin is cusp. y 0 (tangent) (0, 0) y Fig. 2.14 x2a (assymptote) x

Application of Integral 219 for y to be real «EXAMLE 5. Trace the curve x(x 2 y 2 ) a(x 2 y 2 ) Symmetry. Since all powers of y are even the curve is symmetrical about x-axis. Origin. Since x, y 0 satisfies curve, origin lies on it. Tangent. x 2 y 2 0 y x Since tangents are real and distinct, so origin is called node. Intercepts y 0 x 3 ax 2 x 2 (a x) 0 x a, 0 oints on x-axis (0, 0) (a, 0). Assymptotes. For assymptotes parallel to y-axis, equating coefficient of highest power of y, i.e. y 2 (x a) x a 0 x a Range y x for y to be real «Fig. 2.17

Application of Integral 221 The latus rectum is the line x a. Therefore, the units of integration are from x 0 to x a because at x 0 at, x a y 2 4ax Y p \ The parabola is symmetrical about x-axis Required area of the curve [ Required area Ans. ª º O q S(,0) a Fig. 2.20 X EXAMLE 7. To find the volume of the region bounded by parabola y 2 4ax and cut by the latus rectum and revoluate about x-axis. Required volume [ ª ¹ º 2ap[a 2 ] 2pa 3. Ans. EXAMLE 8. To find length of an arc of the parabola y 2 4ax cuts off by its latus rectum. To find length of arc of the parabola, we need to calculate F[, then S µ F[ Hence, S µ F[ Fig. 2.21 2a Let x at 2 dx 2at ] UK _ Þ ] UK _ 5 NI? EXAMLE 9. Find the surface area of solid generated by revolution of arc of parabola y 2 4ax bounded by its latus rectum. Formula: Surface area of the solid which is revoluvate about x-axis is [ FU \ y 2 F[ 4ax Þ [

222 omprehensive Applied Mathematics A A A FU s [ s s ª ¹ º s? s? EXAMLE 10. Find the area of the real generated when parabola y 2 4ax all off by the latus rectum about the tangent. Required surface area FU s µ µ NI µ NI ª ¹ º Fig. 2.22 NI NI ¹ ¹ ª ª º º NI NI ª º ª º NI? Astroid EXAMLE 11. Intrinsic equation of the astroid x a cos 3 t, y a sin 3 t taking (a, 0) as the fixed point.

224 omprehensive Applied Mathematics µ µ EU EU (1 cos 2t) (1 cos 2t) (1 cos 2t)dt (1 cos 2t) (1 cos 2 2t)dt (1 cos 2t) sin 2 2t dt UK UK EU EU µ UK ª ¹ º EU UK ª º Area of astroid is. EXAMLE 13. Find length of the perimeter of the astroid Length of a parametric curve L µ F[ µ We compute the length of the perimeter of astroid L 4 EU UK UK EU L EU UK UK EU UK EU UK EU

UK EU UK EU? Application of Integral 225 3a(cos p cos 0) 3a( 1 1) 3a 2 L 6a Length of the perimeter of astroid is 6a. EXAMLE 14. Find the surface area of the solid of the curve x 2/3 y 2/3 a 2/3 which is revoluveted about the x-axis thus generated. Surface area S S S µ F[ µ [ UK EU UK UK EU UK EU UK UK EU UK UK EU UK EU UK ª ¹ º. Surface area obtained by rotating the astroid about the x-axis is pa2. EXAMLE 15. Find the volume of the solid thus generated by revolving the curve x 2/3 y 2/3 a 2/3 about x-axis. If we revolve a curve about the x-axis we have For the astroid D [ p(a sin 3 t) 2 ( 3a cos 2 t sin t)dt UK EU UK EU EU UK (by using property)

226 omprehensive Applied Mathematics Let u cos t EU EU UK (1 u2 ) 3 u 2 ( du) (1 3u2 3u 4 u 6 ) u 2 du (by using integrating property) (u2 3u 4 3u 6 u 8 ) du W W W ¹ ª ¹ º ª º µ ¹ ª º µ ¹ ª º ª º. 2.7 YLOID There are four types of cycloid: (i) x a(t sin t) (ii) x a(t sin t) y a(1 cos t) y a(1 cos t) t 0 y t p 0 x t p (i) (ii)

228 omprehensive Applied Mathematics B t 0 Y A 2a dy x 2a t p O t p x Fig. 2.25 learly curve passes through the origin and x > 0 and x < ap y is positive and increases as x increases. EXAMLE 16. (1 cos t). Find the whole length of an arc of the curve cycloid x a(t sin t), y a µ µ FU F[ As from the curve whole length of the arc of the curve. S S µ F[ µ EU S EU S EU, Let x Þ dt 2dx EXAMLE 17. S UK? S 8a[1 0] t 0 x 0 t p n p/2 S 8a Find the intrinsic equation of the cycloid x a(t sin t) y a (1 cos t) F[ F[ a(1 cos t) UK S F[ UK EU EU F5 µ F[ µ EU UK Ans.

We know that tan Y EU UK EU EU EU S 4 asin t/2 F[ Þ Y 5 UK: which is the required intrinsic equation of the cycloid. Application of Integral 229 EXAMLE 18. Find the area of the curve cycloid x a(t sin t), y a(1 cos t). F[ The required area F[ [ F[ A UK UK ' UK A 2a 2 UK UK ª º A EU < EU > EU A 2a 2 p 2a 2 (sin t) p 0 a2 A 2a 2 p a 2 p 3a 2 p Required area 3a 2 p. EXAMLE 19. µ UK Length of an arc of the cycloid x a(t sin t) y a(1 cos t). The origin O (for which t 0) is the vertex of the given cycloid and is any point on it. Hence, the length of arc O µ µ F[ EU R UK EU RUK REU R EU EU EU 4a sin t/2 EU [ since a(1 cos t) y Ans.

230 omprehensive Applied Mathematics EXAMLE 20. rove that the volume of the reel formed by the revolution of the cycloid x a(t sin t), y a(1 cos t) about the tangent at the vertex is p 2 a 3. olume [ EU EU EU EU UK UK? s UK EU UK UK EU Let t/2 x t 0 Þ x 0 t p Þ x p/2 2 16pa 3 UK EU 32pa 3 ª º s s 32pa 3 s s s ª º 8NWOG Ans. EXAMLE 21. Show that the surface generated by the revolution of the cycloid x a(t sin t), y a(1 cos t) about the tangent at the vertex. Surface area [FU FU [ FU µ F[ µ EU UK

232 omprehensive Applied Mathematics Equations of ardiod The curve is generally given by its polar equation: r a(1 cos q) The cardioid has artesian equation: (x 2 y 2 ax) 2 a 2 (x 2 y 2 ) and the parametric equations: X a cost (1 cos t) y a sin t(1 cos t) urve Tracing of ardiod Trace the curve r a(1 cos q) (i) Symmetry ut q ( q) So, r a[1 cos ( q)] So, when q is replaced by ( q), the function value of the curve remains unchanged. Then the curve is symmetrical about the initial line q 0. (ii) ut r 0 Þ 0 a(1 cos q) Þ 1 cos q 0 Þ cos q 1 Þ q 0 So, r is zero for some real value of q. That means the pole lies on the curve. (iii) Table of corresponding values of r and q for the curve: q 0 p/2 p r 0 a 2a Fig. 2.27

Application of Integral 233 EXAMLE 22. Find the area bounded by the cardioid r a(1 cos q) We know that area 2 T F R. ut r a(1 cos q) \ Area of cardiod R R EU F UK RFR UK q p q 0 Fig. 2.28 ut R t Þ dq dt when q p Þ t p/2 By a property of definite integral, Area of cardiod 8a 2 A pa2 Ans. EXAMLE 23. erimeter of ardiod Find the perimeter of r a(1 cos q). We know that, F5 FT µ T FR FR ª ¹ º So, S 2 2 FT µ T ¹ FR ¹ ª º F R R R EU UK F R T FT FR EU R Fig. 2.29 UK R

Application of Integral 235 R UK R FR R < EU R > 4a(1 cos q/2) : 8a sin 2 q/4 8 a sin 2 : S 8 a sin 2 Ans. EXAMLE 25. Find the volume of the solid generated by the revolution of the cardiod r a(1 cos q) revolves about the initial line. The curve starts from O when x 0, q p, reaches the extreme left point B where x, q and turns back to right and reaches A where x 2a, q 0. \ Required volume volume generated by area ÐBA volume generated by area ÐBO [ [ [ [ where y for either integral r sin q a(1 cos q)sin q. Hence, the required volume, EU UK R R FR R F R R FR R EU UK F F EU R UK R FR R F [' x r cos q and r a(1 cos q)] EU R UK R ] EU R EU R_ FR FR EU R UK R EU R F R EU R EU R EU RUK R F R UK REU RUK R F R [' EU RUK RFR EU RUK RFR] UK REU RUK R FR Fig. 2.31