Engg. Math. I (Unit-II)

Dr. Stish Shukl of 7 Engg. Mth. I Unit-II) Integrl Clculus iemnn Integrl) The ide. Suppose, f be continuous function defined on [, b nd we wnt to clculte the re bounded by this function with the -is from point to point b. This re is shown by the shded prt in the Figure. iemnn suggested tht this re cn be clculted by dividing this re into smll rectngles of infinitely smll width. Are bounded by y f) with -is f b Fig.. Function f on [, b To understnd this, we need the following definitions: Definition. Consider closed intervl I [, b. By prtition of I we men finite set P {,,..., n } of points from I such tht, n b nd < <... < n. The intervl I [,, I [,,..., I n [ n, n re clled the subintervls of the intervl I [, b. By,,..., n we denote the length of subintervls I, I,..., I n respectively, i.e., i i i for i,,..., n. It is obvious tht, in prtition of subintervls of equl length, s we increse the number of points in the prtition P i.e., the vlue of n), the length of ech subintervl decreses. Suppose, we divide the intervl I [, b into three subintervls i.e., n ) nd we tke the prtition P {,,, b} s shown in the following figure. f b Fig.. Division of intervl into three subintervls

Dr. Stish Shukl of 7 Define m i inf I i f) nd M i sup I i f), where i,,. Then, in Figure : A re of first rectngle bounded between nd M ) M ; A re of first rectngle bounded between nd M ) M ; A re of first rectngle bounded between nd M ) M. A M ) A m ) M m f f b b Fig.. Upper Sum Fig. 4. Lower Sum Then, the sum of ll these res is clled the Upper Sum of f over the prtition P nd it is denoted by UP, f), i.e., UP, f) M + M + M M i i. i Similrly. in Figure 4 we define the Lower Sum, denoted by LP, f) nd LP, f) m + m + m m i i. i It is cler tht, the vlue of UP, f) is some lrger thn the ect re bounded by the cure f) with the -is from to b; nd the vlue of LP, f) is some smller thn the ect re bounded by the cure f) with the -is from to b. Thus, there is n ecess of re in UP, f) nd lck of re in LP, f). It is obvious tht, s we increse the number of points in the prtition P, the rectngles becomes more nrrower nd the length of subintervls decreses. As the rectngles becomes more nrrower, the upper sum UP, f) strts to decreses nd so the ecess of re in UP, f) decreses. Similrly, s the rectngles becomes more nrrower, the lower sum LP, f) strts to increse nd the lck of re in LP, f) decreses s well.

Dr. Stish Shukl of 7 f f b b f f b b Upper Sum decreses Lower Sum increses Figure 5 Finlly, s the number of points we denote it by n ) in the prtition P tends to infinite, then the upper sum UP, f) reches to definite vlue clled the Upper iemnn Integrl nd denoted by b f). Similrly, s the number of points in the prtition P tends to infinite, then the lower sum LP, f) reches to definite vlue clled the Lower iemnn Integrl nd denoted by b f). Thus: b n b n f) lim M i i nd f) lim m i i. n i n i If b f) b f), then the function f is clled iemnn integrble over [, b nd the common vlue of Upper nd Lower iemnn Integrls is denoted by b f) nd it is

Dr. Stish Shukl 4 of 7 equl to the re bounded by the curve with the -is from to b. Theorem First Fundmentl Theorem of Integrl Clculus). Let f be n integrble function over [, for ech [, b. Then the function F defined by: F ) is differentible nd F ) f) for ll, b). ft)dt,, b) Proof. By the definition of F we hve F + h) F ) lim h h lim h lim h [ +h ft)dt h [ +h ft)dt h ft)dt. ) D F C f) Figure 6. A B + h It is obvious tht, the quntity +h ft)dt is the re ABCDF A nd s h this re tends to the vlue f) h, nd so, it follows from ) tht which proves the result. F + h) F ) lim f), i.e., F ) f) h h Definition Antiderivtive or Primitive of function). A function F is clled primitive or ntiderivtive of function f on n open intervl, b) if F ) f) for ll, b). For emple, the function sin is primitive of the function cos in every intervl. Notice tht, the function sin s well s the function sin + c, where c is ny rbitrry constnt, is lso primitive of cos. Therefore, primitive of function re not unique. Definition Second Fundmentl Theorem of Integrl Clculus). If F is n ntiderivtive primitive) of continuous function f on, b), then, ft)dt F ) F ) for ll [, b.

Dr. Stish Shukl 5 of 7 Proof. Let P ) ft)dt, then by first fundmentl theorem we hve P ) f), i.e., P ) is lso primitive of f. Since difference between ny two primitives cn only constnt, we hve F ) P ) k constnt). Putting we get F ) P ) k, i.e., k F ) since P ) ft)dt ). Therefore, we hve: which proves the result. F ) k + P ) F ) + ft)dt Length of Curves ectifiction) The finding of the length of line is quite simple tsk in geometry. But, in cse of n irregulr curve given by n eqution y f) it is not such n esy tsk. For this purpose, we use the concept of clculus. Suppose, we hve to find the length of rc of the curve given by y f), from point A to the point B, i.e., the rcab. B Q+δ,y+δy) δs δy A yf) P,y) δ b Figure 7. Suppose P, y) be ny point on rcab, nd Q + δ, y + δy) be ny point on this rc in the vicinity of point P. Suppose rcp Q δs. Now consider the tringle P Q. Then it is obvious tht: P Q) P ) + Q), i.e., P Q) δ + δy. As the point Q tends to the point P, i.e., δ, δy, then P Q δs. Therefore, δs δ + δy. Since δ, δy, the smll quntities δs, δ, δy now reduce into the infinitely smll quntities ds,, respectively. Thus, we obtin ds +, or, ds + ). Now, the whole length of the rc AB cn be obtined by integrting ds from to b, i.e., b b ) ArcAB) ds +. ) Different forms of the formul for rectifiction. The formul ) is useful when the integrl cn be performed esily with respect to. The following forms cn be used s per the convenience nd requirement.

Dr. Stish Shukl 6 of 7 A) Crtesin form: ArcAB) d yc ds d B) Prmetric form: ArcAB) t t tt ds t r θ C) Polr form: ArcAB) θ θθ ds θ r r ArcAB) ds + r ) dr. rr dr c ) +. ) + dt r + ) dr, or; Qr+δr,θ+δθ) ) dt. dt P P r,θ) P θ, Q θ + δθ, P P r, Q r + δr, P P rδθ, δs δr + rδθ). Figure 8. Emple. Find the length of the rc of the prbol 4y mesured from the verte to one etremity of the ltus-rectum. Sol: Since 4y we hve y 4 nd. S L L, ) Prbol The required length of the rc is the ArcL. Now ) + + ) ) +.

Dr. Stish Shukl 7 of 7 Therefore, ) ArcL + ) + [ ) + + ) ln + ) ) + [ 8 + ln + ) 8 [ ln ) [ + ln + ) 8 [ + ln + ). ln ) Thus, the required length, ArcL [ + ln + ). Emple. Find the length of rc of the semi-cubicl prbol y from the verte to the ordinte 5. Sol: B5,5 ) 5 A5,) Semi-cubicl prbol Here, y, therefore,. The required length of the rc is the ArcB. Now ) ) + + 4 + 9. Therefore, ArcB 5 7 5 7. ) + [4 + 9) / 5 9 [7 / / 5 4 + 9

Dr. Stish Shukl 8 of 7 Thus, the required length, ArcB 5 7. Emple. Find the whole length of: ) Crdioid: r + cos θ); b) Cycloid: θ sin θ), y cos θ); c) Astroid: cos t, y sin t or, / + y / / ); d) Circle: cos θ, y sin θ or, + y ); Sol: ) Here, r + cos θ), so dr sin θ polr form). Therefore, ) dr r + + cos θ) + sin θ) cos θ/). Now, the whole length of the Crdioid L r + π 4 [ sinθ/) π 8. ) dr cos θ/) π θπ θ A,) Aπ,) ) Crdioid b) Cycloid B,) B,) A,) A,) A,) A,) B, ) B, ) c) Astroid b) Circle b) Here, θ sin θ), y cos θ) prmetric form), so cos θ), sin θ.

Dr. Stish Shukl 9 of 7 Therefore, ) + ) [ cos θ) + sin θ) sinθ/). Now, the whole length of the Cycloid is the ArcA. Note tht, t point, θ nd t point A, θ π. Therefore, π ) ) π A + sinθ/) [ cosθ/) π 8. c) Here, cos t, y sin t prmetric form), so, Therefore, ) + dt ) dt dt cos t sin t, dt sin t cos t. [ cos t sin t + sin t cos t ) sin t cos t. Now, the whole length of the Astroid is the L ArcABA B A. Note tht, t point A, t nd t point B, θ π/. Therefore, π/ ) ) π/ L 4 + dt 4 sin t cos tdt dt dt 6 6. π/ sint)dt [ cost) π/ d) Try yourself. Home Work Assignment) Q.) Find the length of the rc of the prbol y 4 ) cut off by the y-is. B,) A,) Ans: ArcBA: [ + ln + ) Q.) By finding the length of of the curve show tht the curve θ sin θ), y cos θ) is divided in the rtio : t θ π. Q.) Find the length of the curve y lnsec ) from to π/. Ans: ln + ). Q.4) Find the length of the rc of the prbol y 4 cut off by the line y 8. Ans: 5/6 + ln ).

Dr. Stish Shukl of 7 Q.5) Find the whole length of the loop of the curve y ) Ans: 4/. Hint: y ), so, A ) ) nd so, + +) + ). Q.6) Find the length of the curve y ) cut off by the line 4. B4, 7 / ) A/, ) 4 Ans. 7.85 C Volume of Solid Generted by the evolution of Curves The ide. Suppose, n rc of the curve y f) from to b is revolved bout - is. Then solid shpe is thus generted nd we hve to find the volume of this solid. For this, we cut verticlly this solid into lrge number sy n ) of thin discs ech of thickness δ. Consider such disc P P QQ shown in the figure below. Then, the volume of the solid cn be obtined by dding the volume of ll such discs. y f) Q Q y b P P Now we obtined the volume of disc P P QQ. Let the coordinte of point Q is, y), then the volume of the disc P P QQ will be δv πy δ. As n the smll quntities δ nd δv reduce into the infinitely smll quntities nd dv respectively. Therefore, the volume of disc P P QQ is dv πy. Thus, the volume of the solid generted is: V b dv b πy. emrk. A) If the rc of curve from y c to y d) is revolved bout the -is, then the volume of generted solid is given by:

Dr. Stish Shukl of 7 y d V d yc dv d c π. y f) Q Q PP y c B) If the rc of curve from point A to B) is revolved bout the line L, then the volume of generted solid is given by: y f) V B A dv B A π ) I I ). I I C) In polr form the volume of revolution bout the initil line is: V πr sin θ. θθ Emple 4. Find the volume generted by revolving the re in the first qudrnt bounded by the prbol y 8 nd its ltus rectum bout the -is. Sol. Here y 8 nd the required volume is generted by the revolution of ArcL bout -i. Therefore, chnges from to 4. Thus, the required volume: V πy 8π 6π. y 8 θ L, 4) L

Dr. Stish Shukl of 7 Emple 5. Find the volume generted by revolving the ellipse + y, bout the -is. b Sol. Here y b ). The volume generted by revolving the ellipse bout -is will be equl to by the revolution of ArcA BA bout the -i. Therefore, chnges from to. Thus, the required volume: V πy π b ). A,) B,b) A,) π b ) 4π b. B, b) Emple 6. The curve y + ) ) revolves bout the is of. Find the volume generted by the loop. Sol. Here y ). The volume generted by + revolving the loop AA bout -is will be equl to the volume by the revolution of ArcAA bout the -i. chnges from to. Thus, the required volume: V π [ πy π ) + + 4 4 + 4 + π [8 ln. A A, ) Emple 7. Find the volume of the solid generted by the revolution of the curve: ) / y ) / cos t, y b sin t, or +, bout the -is. b

Dr. Stish Shukl of 7 Sol. Here cos t, y b sin t. The volume generted by revolving the ellipse bout -is will be equl to the twise of the volume generted by the revolution of ArcBA bout the -is. Therefore, chnges from to. Thus, the required volume: B, b) V πy πb sin t) d cos t) tπ/ π/ 6πb π/ 6πb π! 9 7 5 π π b 5. sin 7 t cos tdt 6πb Γ4)Γ ) Γ ) A, ) A, ) B, b) Emple 8. Find the volume of the solid generted by the revolution of the crdioid: r + cos θ) bout the initil line from θ to θ π. Sol. Here r + cos θ). Given tht the vlue of θ chnges from to π. Thus, the required volume: V π π θ π θ Put + cos θ t we obtin: V π πr sin θ π + cos θ) sin θ. t dt 8π. Home Work Assignment) θ π θ Q. Find the volume of the spindle-shped solid generted by revolving the stroid / + y / / bout -is. Ans. π 5 ) Q. Find the volume of sphere of rdius. Hint: revolve the circle + y bout the -is.) Q. Find the volume of the solid generted by the revolution of prbol y 4 bout the -is from to h. Ans: πh ) Q.4 Prove tht the volume of right circulr cone of height h nd bse of rdius r is πr h. Hint: It is generted by the revolution of the line y h r r ) bout -is).

Dr. Stish Shukl 4 of 7 Surfce of evolution The ide. Suppose, n rc of the curve y f) from to b is revolved bout -is. Then solid shpe is thus generted nd we hve to find the surfce re of this solid. For this, we cut verticlly the surfce of solid into lrge number sy n ) of thin rings ech of thickness ds. Consider such ring P P QQ shown in the figure below. Then, the surfce re of the solid cn be obtined by dding the surfce re of ll such rings. y f) Q Q y b P P ds Now we obtined the surfce re of the ring P P QQ. Let the coordinte of point Q is, y), then the surfce re of the ring P P QQ will be δs πyδs, where δs is the length of the rc QQ. As n the smll quntities δs nd δs reduce into the infinitely smll quntities ds nd ds respectively. Therefore, the surfce re of the ring P P QQ is ds πyds. We know tht the length of the rc QQ ds ) + ). Therefore, the surfce re of the smll ring: ds πyds πy ) + ) πy + ). Now the surfce re of the whole revolution cn be obtined by integrting ds from to b, i.e., the required surfce: b b ) S ds πy +. emrk. A) If the curve is revolved bout the -is, then the surfce of revolution: d d ) S ds π +. yc c B) If the curve is revolving bout the -is or the initil line polr form), then the surfce of revolution: θ θ ) dr S ds πr sin θ r +. θθ θ C) If the curve is revolving bout the -is or the line θ π/ polr form), then the surfce of revolution: θ θ ) dr S ds πr cos θ r +. θθ θ

Dr. Stish Shukl 5 of 7 Emple 9. Find the surfce re of the solid generted by the revolution of the rc of the prbol y 4 bounded by its ltus rectum bout the is of. Sol. Here y 4, so, y. Therefore, + ) +. ds L,) The vlue of vries from ) to L ). Thus, the required re: ) S πy + π 4 + 8π [. 4π + L Emple. Find the surfce re of the solid generted by the revolution of stroid / + y / / bout the is of. Sol. The prmetric eqution of stroid is sin t, y cos t. Therefore, ) + dt ) sin t cos t. dt B, ) Now, the surfce re of the revolution of stroid is equl to the twice the re of revolution of the AcrBA) bout -is. n this rc, the vlue of t vries from Bt ) to At π/). Thus, the required re: S 4π π/ t π/ ) πy + dt π Γ 5 )Γ) Γ 7 ) ) dt dt π/ sin t sin t cos t) dt π sin 4 t cos t dt ds A, ) π. 5 Emple. Find the surfce re of the solid generted by the revolution of ellipse +4y bout the is of.

Dr. Stish Shukl 6 of 7 Sol. The eqution of ellipse is + 4y, i.e., 4y. Therefore, + ) 4y + 6y 4 4y. Now, the surfce re of the revolution of ellipse is equl to the twice the re of revolution of the AcrBA) bout -is. n this rc, the vlue of vries from B ) to A ). Thus, the required re: S 4π π [ ) + dt 4 4y π 4 + 4 sin πy y [ π + π. 4 ) Emple. Find the surfce re of the solid generted by the revolution of crdioid r 5 + cos θ) bout the initil line. Sol. The eqution of crdioid is r 5 + cos θ), i.e., dr 5 sin θ. Therefore, ) dr r + 5 + cos θ cosθ/). Now, the surfce re of the revolution of crdioid is equl to the re of revolution of the AcrA) bout -is. n this rc, the vlue of θ vries from Aθ ) to θ π). Thus, the required re: S π π 4π πy θ π 4π 8π 6π. π π π/ r + ) dr r sin θ cosθ/) 5 + cos θ) sin θ cosθ/) sinθ/) cosθ/) sin t cos t dt θ π ds θ A, )

Dr. Stish Shukl 7 of 7 Emple. Find the surfce re of the solid generted by the revolution of circle r cos θ bout the initil line. dr Sol. The eqution of circle is r cos θ, i.e., sin θ. Therefore, r + ) dr 4 cos θ + 4 sin θ. Now, the surfce re of the revolution of crdioid is equl to the re of revolution of the AcrA) bout -is. n this rc, the vlue of θ vries from Aθ ) to θ π/). Thus, the required re: S π π/ πy θ π/ 8π [ sin θ 4π. r + ) dr r sin θ 4π π/ π/ cos θ sin θ Home Work Assignment) θ π θ π/ Q.) Find the re of the surfce generted by the revolution of the cycloid C, ) θ t sin t), y cos t) bout -is. Ans : 64π Q.) Find the surfce re of right circulr cylinder of rdius r nd hight h. Hint: right circulr cylinder is generted by the revolution of line y r bout the -is, from to h). Q.) Find the surfce re of cone of hight h nd rdius r. Hint: cone is generted by the revolution of line r h y) bout the -is, from y to y h). h Integrl Clculus Multiple Integrls) S P i,y i,z i) z i B A C D P i,y i) δy δ ) b)

Dr. Stish Shukl 8 of 7 The Ide Double Integrl) I) Double integrl s volume. Suppose, z f, y) represents surfce S, s shown in the figure ). Suppose, the region, i.e., the rectngle ABCD be the region of integrtion. We divide this region into lrge number of smll rectngles sy n ) of res δ δy, δ δy,..., n δ n δy n. Let P i, y i ) be point in the i th re i. Let P i, y i, z i ) be point on the surfce S, so tht, z i f i, y i ) nd P i, y i ) is its projection of point P on the region. Then, the volume of the rectngulr solid of hight z i, i.e., f i, y i ) nd the bse re i δ i δy i will be δv i z i δ i δy i f i, y i )δ i δy i. Similrly, we clculte the ech volume v, v,..., v n, nd clculte the sum of volumes of ll such rectngulr solids thus obtined, i.e., the sum n v i f i, y i )δ i δy i. It is cler tht this sum of volumes i i is not ectly the volume bounded by the surfce S with region. Now, when n, ech smll vlue i.e., δ) trnsform into the infinitely smll quntity i.e., d). In this cse, the vlue of the sum of volumes is clled the double integrl of the function f over the region, nd it is denoted by f, y), i.e., It is cler tht the quntity f, y) lim n surfce S i.e., z f, y)) with the region. n f i, y i )δ i δy i. i f, y) represents the ect volume bounded by the II) Double integrl s re. Suppose be given region in -plne nd we hve to find the re of this region. We consider the function z f, y). Now it is obvious tht the double integrl of function z f, y) over this region will be equl to the volume of the lmin L shown in the figure c). Since the hight of lmin is, its volume, i.e., the double integrl of function z f, y) over the region will be equl to the re of region Therefore: Are of region f, y). L c) III) Double integrl s mss of lmin. Agin, suppose the lmin L hs uniform surfce density mss per unit re) ρ ρ, y). Then the mss of the infinitely smll re will be ρ, y). Therefore, the mss of the whole lmin M ρ, y).

Dr. Stish Shukl 9 of 7 Similrly, for different menings of the function f, y) the double integrl of this function over region cn be describe in different wys. Actully, the significnce of double integrl is directly relted to the mening of function f, y). d b Solving double integrl nd limits of nd y If the double integrl is of the type f, y), then we first solve the inner integrl, i.e., d b c f, y) d c [ b f, y). Note tht, when we integrte with respect to vrible, then y must be treted s constnt nd vise vers. In cse, when the limit of integrtion is not constnt, then the order of integrtion is decided by the vrible present in the limit nd we perform the first integrl with respect to tht vrible, which is not present in the limits of inner integrl. For emple b in the double integrl f, y), we first integrte with respect to y since y is not present in the limits of inner integrl), i.e., b f, y) b f, y). Finding the limits when the region of integrtion is given. c Q Q f) f) g) P ) b We understnd this with n emple. Suppose we hve to clculte the integrl f, y), where is the given region of integrtion bounded by the lines, b nd the curves y f) nd y g) s shown in figure. Since we re obtining the limits from region, we cn integrte with respect to ny vrible first. Suppose, we integrte first with respect to y then is treted s constnt nd we move the elementry re from the bottom to the top nd prllel to y is or long to line prllel to y is) in such wy tht it lwys lies inside the region of integrtion. Thus, strip P Q is formed which is prllel to y is if you integrte first with respect to, then strip prllel to y is formed). The lower end P decides the lower limit of y, nd since the lower end P is situted on the curve y g), the lower limit of y is y g). Similrly, the upper end P decides the upper limit of y, nd since the upper end P is situted on the curve y f), the upper limit of y is y f). g) b) P b

Dr. Stish Shukl of 7 Now, we integrte with respect to therefore y is treted s constnt nd now the strip P Q will move from left to right nd long line prllel to is in such wy tht the strip lwys remins inside the region nd its lower end P lwys lie on the lower limit curve g) nd the upper end Q lwys lie on the upper limit curve f). The strip moves from to b to cover the whole region, nd so, the limits of re from to b. Thus, Emple 4. Evlute: Sol. + y ) f, y) + y ). b f) g) f, y). + y ) [ + y. [ + y Emple 5. Evlute: + y ). Sol. + y ) + y ) [ 5/ + / 5. [ y + y Emple 6. Evlute: + + + y. Sol. + + + y π 4 + + + y [ + y + tn + π [ ln + ) + + 4 π 4 ln + ). Emple 7. Evlute: y, where is the positive qudrnt of the circle +y.

Dr. Stish Shukl of 7 Sol. The region of integrtion is the shded prt AB in the figure. We integrte first with respect to y. Then we consider strip P Q prllel to -is lying inside the region AB. The lower end P of strip P Q is situted on the -is, therefore the lower limit of y is y the eqution of -is). The upper end Q is situted on the circle + y, therefore the upper limit of y is y. Now, to complete the region of integrtion, this strip moves from i.e., the - is) to the point i.e., the point A), nd so, the limits of re from to. Therefore: y Emple 8. Evlute: 4 8. [ y y +y [ y ) + y), where is bounded by + y b. B,) Q P A,) Sol. The region of integrtion is the re of ellipse s shown in the figure. We integrte first with respect to y. Then we consider strip P Q prllel to -is lying inside ellipse. The lower end P of strip P Q is situted on the the ellipse + y b below the -is, therefore the lower limit of y is y b. The upper end Q is gin situted on the sme ellipse but this time on the bove of -is, therefore the upper limit of y is y b. Now, to complete the region of integrtion, this strip moves from i.e., the point A ) to the point i.e., the point A), nd so, the limits of re from to. Therefore: + y) 4 b b + y) A, ) b b b + y ) [ y + y [ b + b ) / [ b + b ) /. B,b) B, b) Q P + y + y ) b A,)

Dr. Stish Shukl of 7 Putting sin θ, the bove eqution reduces into the following form: π/ ) + y) 4b sin θ cos θ + b cos4 θ 4b { Γ /) Γ /) + b Γ) πb 4 + b ). } Γ 5/) Γ /) Γ) Emple 9. Evlute: r dr, where is the re of the circle r cos θ. Sol. The region of integrtion is the shded circle s shown in the figure. We integrte first with respect to r. Then we consider strip P long the rdius vector r lying inside the region. The lower end of strip is situted on the pole, therefore the lower limit of r is r. The upper end P is situted on the circle r cos θ therefor the upper limit of r is r cos θ. Now, to complete the region of integrtion, this strip rottes from θ π/ to θ π/, nd so, the limits of θ re from θ π/ to θ π/. Therefore: r dr π/ cos θ 4 9. π/ π/ r dr cos θ θπ/ θ π/ π/ [ cos θ π/ Γ)Γ/) Γ5/) dr A,) r dr P r cos θ Emple. Evlute: r sin θdr, where is the region bounded by r +cos θ) bove the initil line. Sol. The region of integrtion is the shded crdioid s shown in the figure. We integrte first with respect to r. Then we consider strip P long the rdius vector r lying inside the region. The lower end of strip is situted on the pole, therefore the lower limit of r is r. The upper end P is situted on the crdioid r + cos θ) therefor the upper limit of r is r + cos θ). Now, to complete the region of integrtion, this strip rottes from θ to θ π, nd so, the limits of θ re from θ to θ π. Therefore: r sin θdr π +cos θ) π [ r +cos θ) r sin θdr sin θ θ π π dr [ +cos θ) rdr sin θ π [ + cos θ) sin θ. θ P θ

Dr. Stish Shukl of 7 Substitute + cos θ t we obtin sin θ dt nd now the new limits of t re from t to t. Therefore r sin θdr t dt [ t 4. Emple. Evlute e +y ) by trnsforming it into polr coordintes. Sol. Q y P θ π/ dr P θ ) Crtesin coordintes b) Polr coordintes Here, in the limits of integrtion, nd y both vries from to, therefore the region of integrtion is the first qudrnt s shown in the figure ). Now, to chnge the integrl into polr coordintes, we note tht the elementry re in polr coordintes is rdr. Also, since r cos θ, y r sin θ, we hve + y r. For the new limits of r nd θ we see the figure b). We first integrte with respect to r. Then we consider strip P long the rdius vector r nd its lower end is situted on pole nd the upper end in on, therefore, the limits of r re from r to r. To complete the region of integrtion i.e., the first qudrnt) this strip rottes from θ to θ π/, which re the limits of θ. Therefore: π/ π/ [ e +y ) e r rdr e r rdr [ π/ π 4. e r π/ Emple. Evlute by chnging the order of integrtion: Q y Sol. e y y. y P y P Q y ) b)

Dr. Stish Shukl 4 of 7 To chnge the order, we first find the region of integrtion. Since in the given integrl the first integrtion is performed with respect to y, therefore we consider strip prllel to -is whose lower end is situted on the line y the lower limit of y) nd the upper end on y the upper limit of y). This strip moves from the lower limit of, i.e., from the -is) to the upper limit of, i.e., in such wy tht its lower end P lwys lie on the line y nd the upper end on y. Therefore, we get the tringulr region which is shown in the figure ). Now to chnge the order, we hve to integrte first with respect to, nd so, we consider strip P Q prllel to -is within the region of integrtion. Its lower end P is situted on the -is, i.e., the line nd the upper end is on the line y. Therefore, the new limits of re from to y. Now, to complete the region, this strip moves from point, i.e., from y to the line y. Therefore, the new limits of y re from y to y. Thus, e y y y e y y y e y y e y. Emple. Evlute by chnging the order of integrtion: Sol. B, ) B, ) y. Q y P Q y + y A, ) C, ) + y A, ) P ) P Q b) To chnge the order, we first find the region of integrtion. Since in the given integrl the first integrtion is performed with respect to y, therefore we consider strip P Q prllel to -is whose lower end is situted on the curve y the lower limit of y) nd the upper end on the line y the upper limit of y) or + y. This strip moves from the lower limit of, i.e., from the -is) to the upper limit of, i.e., the point A, )) in such wy tht its lower end P lwys lie on the curve y nd the upper end on the line + y. Therefore, we get the shded region AB which is shown in the figure ). Now to chnge the order, we hve to integrte first with respect to. For this we divide the whole region into two prts, i) the region AC ii) the region CABC. i) We consider strip P Q prllel to -is within the region AC. Its lower end P is situted on the -is, i.e., the line nd the upper end is on the curve y. Therefore, the new limits of in the region AC re from to y. Now, to

Dr. Stish Shukl 5 of 7 complete the region AC, this strip moves from point, i.e., from y to the line CA, i.e., the line y. Therefore, the new limits of y in this region re from y to y. ii) Agin, we consider strip P Q prllel to -is within the region CABC. Its lower end P is situted on the -is, i.e., the line nd the upper end is on the line + y. Therefore, the new limits of in the region CABC re from to y. Now, to complete the region CABC, this strip moves from the line CA, i.e., from y to the point B, i.e., the y. Therefore, the new limits of y in this region re from y to y. Thus, y AC 8. y + y + CABC y y y) y y + y y Emple 4. Evlute by chnging the order of integrtion: + y. Sol. +y B, ) Q A,) P y +y B, ) P Q C,) A,) P Q y ) b) To chnge the order, we first find the region of integrtion. Since in the given integrl the first integrtion is performed with respect to y, therefore we consider strip P Q prllel to -is whose lower end is situted on the line y the lower limit of y) nd the upper end on the circle y the upper limit of y) or + y. This strip moves from the lower limit of, i.e., from the point ) to the upper limit of, i.e., the point A, )) in such wy tht its lower end P lwys lie on the line y nd the upper end on the circle + y. Therefore, we get the shded region AB which is shown in the figure ). Now to chnge the order, we hve to integrte first with respect to. For this we divide the whole region into two prts, i) the region AC ii) the region CABC.

Dr. Stish Shukl 6 of 7 i) We consider strip P Q prllel to -is within the region AC. Its lower end P is situted on the -is, i.e., the line nd the upper end is on the line y. Therefore, the new limits of in the region AC re from to y. Now, to complete the region AC, this strip moves from point, i.e., from y to the line CA, i.e., the line y. Therefore, the new limits of y in this region re from y to y. ii) Agin, we consider strip P Q prllel to -is within the region CABC. Its lower end P is situted on the -is, i.e., the line nd the upper end is on the circle + y. Therefore, the new limits of in the region CABC re from to y. Now, to complete the region CABC, this strip moves from the line CA, i.e., from y to the point B, i.e., the y. Therefore, the new limits of y in this region re from y to y. Thus, + y AC y. + y + + y + CABC + y y + y Home Work Assignment) Q.) Evlute ) y ). π Ans: 4 Q.) Evlute Q.) Evlute y e y/. Ans: +y. Ans: π 4 Q.4) Evlute y over the region, where + y in the positive qudrnt.ans: 4 Q.5) Evlute Ans: 56 Q.6) Evlute e y + y) over the region bonded by the curves y nd y. e lny). Hint: First chnge the order) Ans: e Q.7) Evlute by chnging the order of integrtion Q.8) Evlute by chnging the order of integrtion 4 y + y). y. Ans: 4

Dr. Stish Shukl 7 of 7 Are by Double Integrl Emple 5. Find the re between the curves y 4 nd 4y. Sol. The required re is the re P AQ. Consider the elementry re in the region P AQ. We first integrte with respect to y. Then, this elementry re moves long the y is in the region P AQ nd form the strip P Q. The lower end P of strip is situted on the prbol 4y nd the upper end Q on the prbol y 4. Therefore, the limits of y re from y 4 to y. Now, to complete the region, this strip moves long the -is from point, i.e., from to the point A, i.e., to 4, therefore, the limits of re from to 4. Thus, the required re: 4 [ 4 AreP AQ 4 6. 4 4 ) 4 y 4 Q 4 P 4y [y 4 A4, 4) Emple 6. Find the whole re the ellipse + y b. Sol. The required re is the shded re nd is equl to 4 reab. Consider the elementry re in the region AB. We first integrte with respect to y. Then, this elementry re moves long the y is in the region AB nd form the strip P Q. The lower end P of strip is situted on the -is, i.e., y nd the upper end Q on the rc AB of ellipse + y b Therefore, the limits of y re from y to y b. Now, to complete the region, this strip moves long the -is from the line B, i.e., from to the point A, i.e., to, therefore, the limits of re from to. Thus, the required re: 4 4 πb. b b 4 ) 4 b [ b 4 ) [ B, b) [y b Q P + sin ) A, )

Dr. Stish Shukl 8 of 7 Emple 7. Find the whole re the stroid / + y / /. Sol. The required re is the shded re nd is equl to 4 reab. Consider the elementry re in the region AB. We first integrte with respect to y. Then, this elementry re moves long the y is in the region AB nd form the strip P Q. The lower end P of strip is situted on the -is, i.e., y nd the upper end Q on the rc AB of the stroid / + y / / Therefore, the limits of y re from y to y / /) /. Now, to complete the region, this strip moves long the -is from the line B, i.e., from to the point A, i.e., to, therefore, the limits of re from to. Thus, the required re: B, b) Q P A, ) 4 4 / / ) / 4 [y / / ) / 4 / / ) / / /) /. Putting sin θ, we obtin sin θ cos θ nd the new limits of θ re from θ to θ π/. Therefore, the required re is π/ sin θ cos 4 θ π π! Emple 8. Find the whole re the crdioid r + cos θ). π 8. Sol. The required re is the shded re nd is equl to reac. Consider the elementry re rdr in the region AC. We first integrte with respect to r. Then, this elementry re moves long the rdil vector r in the region AC nd form the strip P. The lower end of strip is situted on the pole, i.e., r nd the upper end P on the rc AC of the crdioid r + cos θ) Therefore, the limits of r re from r to r + cos θ). Now, to complete the region, this strip rottes from the line, i.e., from θ to the point, i.e., to θ π, therefore, the limits of θ re from θ to θ π. Thus, the required re: π π [ r +cos θ) +cos θ) θ π [ π +cos θ) rdr rdr π π + cos θ) θ 4 cos 4 C B dr ). P A θ

Dr. Stish Shukl 9 of 7 Putting θ φ, we obtin dφ nd the new limits of φ re from φ to φ π/. Therefore, using gmm function, the required re is π/ π/ 8 cos 4 φdφ 8 sin φ cos 4 φdφ + Γ π. )Γ 4+ ) Γ +4 ) 8 π π Emple 9. Find the re between the curves y 4 nd y. Sol. The required re is the shded re P BQ. Consider the elementry re in the region P BQ. We first integrte with respect to y. Then, this elementry re moves long the -is in the region P BQ nd form the strip P Q. The lower end P of strip is situted on the line y nd the upper end Q on the rc QB of the prbol y 4 Therefore, the limits of y re from y to y 4. Now, to complete the region, this strip moves from the point, i.e., from to the point B, i.e., to, therefore, the limits of re from to. Thus, the required re: [ 4 9. 4 [y 4 Q.) Find the re of circle of rdius. y 4 Q ) [ Home Work Assignment) Q.) Find the re of crdioid r cos θ). P y B, ) A4, ) Q.) Find the re between the curves y nd y. A, ) Hint: The limits of y re from y to y nd those of re from to. y y Triple Integrl nd volume Triple integrl s mss of solid: Suppose, the mss per unit volume i.e., the volume density of mss) of solid is given by its mss distribution function w f, y, z), where

Dr. Stish Shukl of 7, y, z) represent the coordintes of points inside the solid. Let V represents the volume of the solid. Then, we divide the whole volume V into n smll volumes δv i δ i δy i δz i, i,,..., n. Suppose, P i, y i, z i ) be point in the smll volume δv i. If we choose the smll volume δv i sufficiently smll, then we cn ssume tht the volume density of mss in the volume δv i is constnt nd is equl to f i, y i, z i ), nd so, the mss of this i th smll volume is given by δm i f i, y i, z i )δv i f i, y i, z i )δ i δy i δz i. We clculte ll such smll msses δm i of smll volumes δv i, where i,,..., n nd then sum up them nd get the sum n n n δm i f i, y i, z i )δv i f i, y i, z i )δ i δy i δz i. i i It is cler tht this sum msses of ll smll volumes is not ectly the mss of solid. Now, when n, ech smll vlue i.e., δ) trnsform into the infinitely smll quntity i.e., d). In this cse, the vlue of the sum of msses is clled the triple integrl of the function f over the volume, nd it is denoted by f, y, z), i.e., It is cler tht the quntity Triple integrl s Volume: H V f, y, z)dz lim n G V V i n f i, y i, z i )δ i δy i δz i. i f, y, z)dz represents the ect mss of the solid. H G H G D C D C D C dz E F E F E F A B A B A ) b) c) B Suppose, we hve to find the volume of rectngulr prllelepiped cuboid) ABCDEF GH formed by the plnes, α, y, y β, z, z γ, i.e., EA α, EF β nd EH γ. Consider n infinitely smll volume, i.e., the elementry volume dv dz inside this cuboid see, figure )). We sum up such elementry volumes long the -is i.e., during the ddition the coordintes nd y remin constnt), in such wy tht the elementry volume remins inside the cuboid. Thus, the integrtion with respect to z is completed nd verticl collum P Q is formed inside the cuboid ABCDEF GH see, figure b)). The volume of this collum P Q is obviously γ. Now, we sum up this collum long the -is i.e., during the ddition the coordinte remins constnt), in such wy tht the collum remins inside the cuboid. Thus, the integrtion with respect to y is completed nd rectngulr lmin is formed inside the cuboid ABCDEF GH see, figure c)). The volume of this rectngulr lmin is obviously βγ. Finlly, we sum up this rectngulr lmin

Dr. Stish Shukl of 7 long the -is in such wy tht the rectngulr lmin remins inside the cuboid. Thus, the integrtion with respect to is completed nd the whole cuboid ABCDEF GH is formed nd we get the volume of cuboid, i.e., V αβγ. Thus: V V dv α Emple. Evlute: 4 +y y z zdz. Sol. Given integrl cn be solved s follows: 4 y +y z Emple. Evlute: e zdz 4 4 4 y β [ +y z y [ + y) [ 8 6 6 lny) e lnz)dz. Sol. Given integrl cn be solved s follows: e ln y e ln zdz e ln y e ln y e [ e γ z dz. zdz 7. ln zdz [ )e + 4 4 e ln y y [ z [ + y) 6 e [ ln y e +y [z ln z z e )e + [ )e e + ln y [y + ) ln y y + e [ ) y + y ln y ) y e y + y y + e )y 4 e 8e + ). Sphericl coordintes. The sphericl coordintes of point P in the spce re given by P r, θ, φ) see the figure below). The reltion between crtesin nd sphericl coordintes is given by: r cos φ sin θ y r sin φ sin θ z r cos θ. The elementry volume in sphericl coordintes is given by dv r sin θdrdφ see the figure below). Some times it is esy to solve the integrl by chnging the crtesin coordintes into sphericl coordintes. There is no hrd nd fst rule to decide whether it is esy to use the sphericl coordintes, we cn decide only by observtions. Although, converting to sphericl coordintes cn mke triple integrls much esier to work out when the region you re integrting over hs some sphericl symmetry.

Dr. Stish Shukl of 7 r sin θdϕ r sin θ P r,θ,ϕ) dr r ϕ θ r ϕ θ r dϕ Sphericl coordintes Elementry volume in sphericl coordintes dv r sin θdrdϕ Cylindricl coordintes. The cylindricl coordintes of point P in the spce re given by P r, φ, z) see the figure below). The reltion between crtesin nd sphericl coordintes is given by: r cos φ y r sin φ z z. The elementry volume in cylindricl coordintes is given by dv rdrdφdz see the figure below). Some times it is esy to solve the integrl by chnging the crtesin coordintes into cylindricl coordintes. There is no hrd nd fst rule to decide whether it is esy to use the cylindricl coordintes, we cn decide only by observtions. Although, converting to cylindricl coordintes cn mke triple integrls much esier to work out when the region you re integrting over hs some cylindricl symmetry. P r,ϕ,z) ϕ r z ϕ dϕ dr rdϕ dz Cylindricl coordintes P r, QP z, Qϕ Elementry volume in cylindricl coordintes dv rdrdϕdz Emple. Find the volume of the sphere + y + z. Sol. The volume of the given sphere is V 8 V p dz, where V p is the volume of the sphere in the positive octnt. Chnging the coordintes into the sphericl coordintes we get dz r sin θdrdφ nd in the volume V p, r chnges from r to, θ chnges from θ to π/ nd φ chnges from φ to π/. Thus, the required volume: V 8 π/ π/ φ θ φ r sin θdrdφ 8 π/ π/ φ θ sin θdφ 8 π/ φ dφ 4π.

Dr. Stish Shukl of 7 Emple. Find the volume of the tetrhedron bounded by the co-ordinte plnes nd the plne + y b + z c. Sol. The required volume is the volume of the tetrhedron s shown in the figure. The vlue of z chnges from z to z c y ) nd then y chnges b from y to y b ). Finlly, chnges from to. Thus, the required volume: ) b c V y b ) dz bc y b y b y ) ) [ c z [z c y b ) z c y ) b ) ) b y y b y ) bc 6. A,, ) C,, c) + y b B, b, ) Emple 4. Find the volume bounded by the cylinder + y 4 nd the plnes y + z 4 nd z. Sol. Suppose the required volume is V which is the drk shded prt in the figure. Then V dz. The limits of z re from z to z 4 y nd then nd y vries ccording to the limits of circle C : + y 4. The vrible y vries from y 4 to y 4, nd then vries from to. Thus, the required volume: V 4 4 z 8 6π. V 4 4 y dz 4 y) 4 4 4 4 y) 4 4 [ [ 4 6 4 + 4 sin ) Intersection of plne nd cylinder + y 4 Emple 5. Find the volume bounded by the y-plne, the prboloid z + y nd the cylinder + y 4.

Dr. Stish Shukl 4 of 7 The drk shded region is the required volume + y 4 The required volume Sol. Suppose the required volume is V. Then V 4 V dz. The limits of z re from z to z + y nd then nd y vries ccording to the limits of circle C : + y 4. The vrible y vries from y 4 to y 4, nd then vries from to. Thus, the required volume: V 4 +y 4 z 4 dz + y ) 4 4 [ 4 + 4 ) / + y [ y + y 4. Putting sin θ, the limits now chnged to θ to θ π/, nd cos θ. Therefore, V 6 4π. π/ sin θ cos θ + 6 π/ sin θ cos 4 θ 6 Γ/)Γ/) Γ) + 6 Γ/)Γ5/) Γ) Emple 6. Find the volume cut from the sphere +y +z by the cylinder +y. Sol.

Dr. Stish Shukl 5 of 7 r cos θ r cos θ Side view of volume the shded prt) Top view The required volume V is the times of the volume V shown in the figure by shded prt hlf of the volume is bove of the y-plne nd hlf is below). To mke esy, we will use cylindricl coordintes. The cylindricl coordintes re: r cos θ, y r sin θ, z z nd the volume element dv rdrdz. The eqution of sphere will become r + z, nd the eqution of cylinder will become r cos θ. In the shded prt, the limits of z re from z to z r nd the r nd θ vries throughout the circle r cos θ see the figure). Therefore, r vries from r to r cos θ nd θ from θ π/ to θ π/. Therefore, the required volume V dz rdrdz V V π/ π/ 4 4 4 4 cos θ π/ π/ π/ π/ r r dr sin θ ) 4 π/ π/ π/ [ sin θ cos θ ) [ sin θ + sin θ cos θ [ θ + cos θ cos θ π ). π/ π/ π/ cos θ r rdrdz [ r ) cos θ / sin θ ) Emple 7. Find the volume between the prboloid + y z, cylinder + y y nd the plne z.

Dr. Stish Shukl 6 of 7 Sol. r sin θ r sin θ Side view of volume the drk shded prt) Top view The required volume V is shown in the figure by shded prt. To mke esy, we will use cylindricl coordintes. The cylindricl coordintes re: r cos θ, y r sin θ, z z nd the volume element dv rdrdz. The eqution of prboloid will become r z, nd the eqution of cylinder will become r sin θ. In the shded prt, the limits of z re from z to z r nd the r nd θ vries throughout the circle r sin θ see the figure). Therefore, r vries from r to r sin θ nd θ from θ to θ π. Therefore, the required volume V V π dz sin θ π 4 V r dr 4 rdrdz π π [ r 4 sin θ sin θ sin 4 θ ) π/ 8 sin 4 θ cos θ ) 8 Γ5/)Γ/) Γ) 8 π π π. Home Work Assignment) r / rdrdz Q.) Evlute: Q.) Evlute: Q.) Evlute: z +z z ln +ln y + y + z)dz. Ans. e +y+z dz. Ans. e4 8 e 4 + e 8. y + z)dz, where is the region determine by, y, z + y. Ans. 9 5

Dr. Stish Shukl 7 of 7 Q.4) Evlute: y dz. Hint: From the limits of integrtion, y z it is cler tht the region of integrtion is the prt of sphere + y + z in the positive qudrnt. Chnge the crtesin coordintes into the sphericl coordintes, the new limits will be r to, φ to π/ nd θ to π/.) Ans. π 8 Q.5) Find the volume cut from the sphere + y + z by the cone + y z. A filled Ice-crem cone) Hint. Suppose the required volume is V. Then V 8 V dz where V is prt of volume in the positive octnt nd it is the shded prt in the figure. The sphericl coordintes re r cos φ sin θ, y r sin φ sin θ, z r cos θ. Eqution of sphere r nd of cone is tn θ, i.e., θ ± π 4. The limits of r re from r to r nd then φ vries from φ to π nd θ vries from θ to π 4. Thus, the required volume is π ). The prt V