Biyani's Think Tank
Concept based notes Maths II B.Sc I Year
Megha Sharma
Department of Science Biyani’s Group of Colleges
2
Published by : Think Tanks Biyani Group of Colleges
Concept & Copyright : Biyani Shikshan Samiti Sector-3, Vidhyadhar Nagar, Jaipur-302 023 (Rajasthan) Ph : 0141-2338371, 2338591-95 Fax : 0141-2338007 E-mail : [email protected] Website :www.gurukpo.com; www.biyanicolleges.org
ISBN:
Edition : 2013
While every effort is taken to avoid errors or omissions in this Publication, any mistake or omission that may have crept in is not intentional. It may be taken note of that neither the publisher nor the author will be responsible for any damage or loss of any kind arising to
anyone in any manner on account of such errors and omissions.
Leaser Type Setted by : Biyani College Printing Department
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Maths 3
Preface
Iam glad to present this book, especially designed to serve the needs of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach.
Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the readers for which the author shall be obliged.
I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this endeavour. They played an active role in coordinating the various stages of this endeavour and spearheaded the publishing work.
I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address.
Megha Sharma
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4
Syllabus
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Maths 5
Unit – I Series
Series : Infinite Series and convergent series:
Series :
Expression of that form in which the successive terms are always according to some definite rule. Here – nth term of series There are two types of series 1) Finite Series 2) Infinite Series
Infinite Series: If the no. of term in any series is infinite then the series is called infinite series. Eg. it’s denoted by
Convergent Series : An infinite series is said to be convergent if the sequence of its partial sum
Test for convergence of series : 1) D’ Alembert’s Ratio test : If be a series of positive terms such that then (i) if l > 1, will be convergent (ii) ifl < l, will be divergent (iii) if l = 1 may either convergent or divergent
Cauchy nth root test : If be a series of positive terms such that = (a real number) then (i) if <1, will be convergent (ii) if, >1 will be divergent (iii) if =1, may either converge or diverge
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Raabe’s tests : If be a series of positive terms such that
then (i) if > 1, will be convergent (ii) if <1, will be divergent (iii) if =1, may either converge or diverge
D’e Morgan and Bertrand’s Test : If be a series of positive terms such that
then (i) if >1, will be convergent (ii) if <1, will be divergent
Cauchy’s condensation test : If the series is a positive terms such that
Gauss tests : If be a series of positive terms and
Where sequence
Alternating Series : Any series of the type +… ( ) Where terms are alternatively positive and negative, is known as an alternating series and is denoted by Eg. 1 - - + ….
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Maths 7
A series whose terms are alternatively positive or negative is aid to be absolutely convergent if the series both are convergent.
Taylor’s theorem: (i) Taylor’s theorem with L1agrange’s form of Remainder Statement : if in the interval [a, a+h], a function f is defined in such a way that the differentials f’ , f” ,….., upto the order (n-1) are (i) Continuous in the interval [a, a+h], (ii) nth derivative of f exist in the interval (a, a+h) then there exist at least one number Q between O and 1, such that f(a+h) = f(a) + h f’ (a) + f” (a) + … +
(iii) Taylor’s theorem with cauchy’s from of Remainder:
Statement : if in the interval [a, a+h] a function f(x) is defined in such a way that (i) all derivatives of f (x) up to the order (n-1) are continuous in the interval [a, a+h] (ii) all derivatives of f (x) up to the order n exist in the interval (a, a+h), then there exist atleast one number between o and 1, such that.
f(a+h) = f(a) + h f’(a) + f” (a) +….
Maclaurin’s theorem : Statement : if a function f in is Such that (i) f, f` …. is continuous in [ (ii) in (iii) (0,1) such that f = f(o) + x f’(o) + + ….+ (o)+ where Rn = is called maclaurin’s remainder due to schlomitch and Roche.
Power Series If is a continuous real variable and ‘a’ any constant then the series
is Constant, is called a power series about the point x =a
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8
is called Standard power series.
Q.1 Show that the following series is divergent:
Sol. Let = and = = =
D ‘Alembert’s ratio test fail Again .n =
= again Raabe’s test also fail again
=
= -
= -
= Therefore by De’ morgan and Bertrand’s test the series is divergent.
Q.2 Discuss the convergence and absolute convergence of the following series
Sol. Given series is alternating series
Clearly and Un =
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Maths 9
therefore by Leibnitz test the above series is also convergent. again = Let us take auxiliary series Vn = Now (non zero finite quantity) Both will converge or diverge together. But = is divergent, Because p = 1 is conditionally convergent.
Q.3 Show that the following series is convergent if x < 1 and divergent if x> 1 + Sol. Un = Un+1 = Now. = .
.
.
.
Therefore if convergent. and if
Therefore at =
=
= at
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10
Q.4 Test for the convergence of the following series.
Sol. Un = =
= = (1+o) . 1.x = Therefore by Cauchy nth foot test the given series is divergent if > 1 convergent if < 1 if x =1 then Un = = let us take auxiliary series is where Vn = Now = e# 0 Therefore by comparison test will converge or diverse together. But therefore together is also divergent.
Q.5 Test for the convergence and absolute convergence of the following series: +…… Case - I = when P > O Sol. Given series is alternating series and = 1 If p> 1, series will be convergent If o For free study notes log on: www.gurukpo.com Maths 11 ______oscillates finitely Case - III = When p < o Let p = m, where m > o then ∑un = ∑ = ∑ = 1 - + here Un = or when n simultaneous odd and even. and therefore given series is oscillating series between - Q.6 Test for the convergence of the following series. Sol. For the given series if we neglect first term of given series then convergence of given series not affected therefore we Let Then Un = and Un +1 = Now = = therefore if ∑un is convergent and if x > 1 then < 1 is divergent. If x = 1 then D’Almbert’s ratio test fail. at x = 1 n = ( ) = test given series is convergent. Therefore if < ! the given series will be convergent and the given series will be divergent. Q.7 Find Lagrange’s and Cauchy’s remainder after n terms in the expansion of following functions. For free study notes log on: www.gurukpo.com 12 (i) log (1+x) Sol. Let f ; possesses derivatives of every = ! = Q.8 If f(x) is continuous in [a,b] and possesses finite derivatives for x = C (a,b), then prove that Sol. Since the function is differentiable at x=c, hence f’ (x) and f’ (x) exist in the neighborhood (c-h, c+h) of x =c Now applying second mean value theorem for the intervals (c-h, c) and (c,c+h), we have f f(c-h)=f(c) – hf’(c) + O< <1 F(c+h) = f(c) + h’(c) + f” (c+ ) ------(2) O< <1 Adding (1) and (2), we obtain F”(c- h) +f(c+h) = 2f(c) + - f(c-h)+ f(c+h) – 2f(c)= [f”(c- [ f”(c- ] Q.9 Test the convergence of following series: +….. Sol. Let first term of this series 0 = then f(n) = = For free study notes log on: www.gurukpo.com Maths 13 Suppose f = then Vn = nd Vn+1 = now = a>1 is convergent therefore by cauchy’s consensation test the given series is convergent. Q.10 Examine the convergence of the following series: Sol. For the given series if we leave first term of given series then convergence of given series not affected therefore we suppose then Un = . and now = . = therefore by D’ Alembert ratio test will be convergent if <1 and will be divergent if < 1 or at 1 D’Alembert Ratio test fail at = = = therefore by Raabe’s test is convergent therefore if <1 the given series is convergent if the given series is divergent Q.11 Prove that the following function cannot be expanded in maclaurin’s series (i) (i) (ii) Sol. (i) Let f(x) = f’ (x) = (- ) For free study notes log on: www.gurukpo.com 14 f” (x) = , … clearly at x =,0 the function and its derivative are not finite. Therefore Maclaurin’s expansion of not valid. (ii) Let f(x) = logx f’(x) = f”(x) = - Clearly at x = 0 the function and its derivative are not finite. Hence it is not possible to expand log x by Maclaurin’s expansion. Q.12 Show that the following series is convergent if x < and divergent if x > Sol. Here un = and Un+1 = = . = . . = = Therefore by D’Alembert’s ratio test when X< then the ∑Un is convergent. then the un is divergent When x = = e . log = loge+nlog (1 ) – (n+1) log - n log = + = 1-n + = + n log = - [ For free study notes log on: www.gurukpo.com Maths 15 By logarithmic ratio test the given series is divergent in this case. Therefore the given series is convergent when x< and divergent when x > Q.13 Show that the series + is Conditionally convergent Sol. Given series is Alternating series Let Un = Un+1 = Un-Un+1 = - > 0, and = = 0 By Leibnitz test the given series Un is convergent. Again ∑ = ∑ = ∑ Which is divergent because its Form of series and here p<1 Therefore this series is conditionally convergent. Q.14 Expansion of Suppose f(x) = Thus, in interval (o,x), f, f1 ….. exist finitely putting x = 0 in (1) ⇒ in the expansion of power series we have to prove that when n then Rn Now we consider lagrange’s form of remainder is maclaurin’s expansion Rn = ( x) O < <1 Here Rn = sin ( x + n ) = < < 1 < =0 For free study notes log on: www.gurukpo.com 16 Thus sin x satisfy both the properties of Maclaurin series. Q.15 Expansion of Ans. Here f (x) = cos x = )n - (1) Thus every order of derivative for the function f(x) = (osx exists, From (i), = Now for expansion of power series we have to prove that when n then Rn . For this we consider Lagrange’s form of remainder in Maclaurins expansion is Rn = ( cos ( = < [ Satisfy both the properties of Maclaurin series. + + … Q.17 Expansion of log (1+x) Sol. Let f(x) = log (1+x) If 1+x >0 i.e. x >-1, then log (1+x) Passesses derivatives of every order = , and x>-1 - (1) From (1) (o) = = Now for power series expansion we have to show that when n , then Rn For this lagrange’s from of remainder in Maclaurin’s expansion is = = 0 < = Case –I O< x < 1 and 0< <1 0 < < 1 + < 1 For free study notes log on: www.gurukpo.com Maths 17 Thus Case 2 : -1 < x <0 In this case it is not necessary < 1 Therefore taking cauchy’s form of remainder is maclaurin’s expansion = 0< <1 = = O < < 1 - 1 < - < O and - 1 < x < O < 1 - 1 < - < x O < 1 – < 1 + X - 0 < < 1 Also <1 and – 1 < x < O = 0 So that = From equation (2) and (3) Therefore for power series expansion, the function f(x), satisfies both the conditions of maclaurin’s series. Thus log (1+x) = x - - + … Where -1 < x < 1 Q.18 Test the convergence of the series (x>0) Sol. = +1 = = = Therefore when > 1 or x < 1 Convergent. Also when < 1 or For free study notes log on: www.gurukpo.com 18 Again if then = = Let be the auxiliary series then by comparison test and converge or diverge simultaneously. But = convergent, because P = 2>1 therefore is also converge i.e. for x = < 1 and divergent for x > 1 For free study notes log on: www.gurukpo.com Maths 19 Unit 2 Derivative of length of arc andCurvature Derivative of length of arc – (a) when equation of curve y = f (x) = + (d) when equation of curve x = f (y) = + Curvature (various formula’s) (a) Polar Formula : when curve Equation r=f( ) = (b) When equation of curve in parameter form - = (c) Polar Formula : When curve equation = f(r) = (d) = here is the angle Which the tangent makes with the positive direction of x – axis. (e) = (f) = (g) = (h) = Length of perpendicular from pole on the tangent = Pedal equation of curve whose For free study notes log on: www.gurukpo.com 20 (A) Equation is given in Cartesian form P = - y (b) Equation is given in polar form : = Formula for Radius of curvature (A) = (B) = 6. Cartesian formula for Radius of curvature = 7. Radius of curvature for parametric curves : = 8. Formula for radius of curvature when x and y both are functions of S. 9. Curvature of circle at point p = = = = which is Constant for any circle 10. Radius of curvature for polar equations: = For free study notes log on: www.gurukpo.com Maths 21 11. Formula for Radius of curvature for tangential polar equations = p + 12. Length of chord of curvature parallel to X – axis is - 2 sin here is the angle which the tangent makes with the positive direction of x – axis. 13. Length of chord of curvature parallel to Y – axis is - 2 cos here is the angle which the tangent makes with the positive direction of x – axis. 14. Euler’s theorem on Homogenous Function - x + y = n f Q.1 Show that the pedal equation of the ellipse = 1+ e cos s = Sol. Given equation of curve = 1 + ecos take logarithm on both side log l - log r = log (1+ecos ) take differential with respect to , = = = + = = (1+ ) = Or - (1) For free study notes log on: www.gurukpo.com 22 Because = e cos 2 Eliminate from equation (1) and (2) Or = = [ + -1] Q.2 Show that for the radius of curvature at a point (a , a ) on the curve + = is sin2 Sol. Here = a co , y = a si = - 3a co sin and = - 3a cos and = 3a (- co + 2 cos According to formula of radius of curvature – = = = = = = Q.3 If u = then prove + Sol. Here u = (given) Take logarithm on both side log u = log (1-2xy+ ) - (1) Differentiating with respect to x = = = = = = y For free study notes log on: www.gurukpo.com Maths 23 = y = y (2) Again partial differentiating with respect to y of equation - (1) = = = = (x-y) Or = – And = – ( ) = = = = = } = = – (3) Adding equation (2) and (3) {1- ) + + = 0 Q.4 If u = + y = sin2u Sol. The given function is not homogenous function therefore we can write the function. f = tan u = = = (y/x) (suppose) Here f is homogenous function and x and y is homogenous function of degree 2. Therefore from Euler’s theorem – + u = 2f Or x + y = 2 tan u Or x u + y u = 2 tan u x + y = 2 x x + y = sin 2u Q.5 If u = then prove that = (1+3xyz+ ) Sol. Given u = For free study notes log on: www.gurukpo.com 24 = - (1) = = ( ) = x ( = x [y. + ] = [ + ] - (2) Hence = = = = ( ) = ] = ] Q.6 Show that in the parabola = 4ax, the radius of curvature at any point p is 2 , where s is the focus of the parabola. Sol. Suppose p is any point on probably = 4ax and its co-orinates are (a then parametric form of parabola is y = 2at Now = x’ = 2at, = and = y’ = 2a, = Now from formula of Radius of curvature = = We know from the co-ordinate geometry SP = a+x SP = a+a = a (1+ ) Therefore = 2a = 7 If V = f (x-y, y-z, z-x) then prove the + + = 0 Sol. Suppose y –z = t, z-x = x-y = then v = f ( , ) were , is function of x, y, z now = 0 , =1, = -1 For free study notes log on: www.gurukpo.com Maths 25 = -1, = 0, = 1 = 1, = -1, = 0 = + + = - + - (1) and = - + - (2) and = + - (1) by adding equation (1), (2), (3) + + = 0 Q.8 If u = (x/y) + then Prove x + = 0 Sol. Given u = + Differentiating partially with respect to x, we get = + Or = - - ---1 Again differentiating u partially with respect to y, we get = + Or y = - ---2 Adding (1) and (2), we get x + y = 0 Q.9 If - 3a =0, then prove that + = 0 Sol. let f ( - 3a , then = , = 3 = -6a, = 6 , = 0 Now we know that = = - = - For free study notes log on: www.gurukpo.com 26 = (putting the value of from the given relation) = + = 0 Q.10 Find the pedal equation of the ellipse + = 1 Sol. Equation of the curve is + = 1 The equation to the tangent at any point (a cos , b sin ) on the curve is + sin =1 Or + ay sin = ab Or b cos + ay sin - ab =0 Now length of the perpendicular from origin to the tangent is: P = Or = - (1) Again = = = (1- (1- ) Or + - ( + ) + - [from 1] Or = Or = - Which is the required pedal equation. Q.11 For the curve y = a log , Prove that the chord of curvature parallel to y – axis is of constant length Sol. Equation of the given curve is Y = a log sec Differentiating it with respect to x, = = . = tan - (1) = tan (1) tan = tan or = For free study notes log on: www.gurukpo.com Maths 27 (1) again with respect to x, we get = = , the radius of curvature of any point (x,y) of the curve is given by , = = a sec The chord of curvature parallel to the axis of y = 2 cos = 2 a sec cos [ ] = 2 a constant Q.12 If u = log ( ) Then prove that + + = Sol. Given u = log ( = (1) = (2) = (2) Adding (1), (2) and (3) + = 3 = 3 = Q.13 If u (x,y) be a homogeneous function of x, y of degree n, then prove that – + 2xy + = n (n-1) u Sol. Since u is a homogenous function of degree n, by Euler’s theorem, we have +y =nu - (i) Differentiating (1) partially with respect to x, + 1 + y = n Or x y = (n-1) - (ii) Again differentiating (1) partially With respect to y + 1 + y = n Or x + y = (n-1) - (iii) Now multiply (2) by x and multiply (3) by y and then adding, we get For free study notes log on: www.gurukpo.com 28 + xy + (n-1) + (n-1) n.u Or + 2xy + = n (n-1) u. Q.15 If and be the radii of curvature at the extremities of any chord r= a (1+cos which passes through the pole the prove that 9 ( = Sol. r = a (1+cos ) = -a sin Now tan = r = = - cot = tan Therefore = and so = 2a therefore cos = Therefore the pedal equation of the curve will be = Differentiating with respect to r = Therefore = r = r = Hence various as the square root of the radius vector Again r =a (1+cos r =2a Therefore = = Now since the extremities of any chord through the pole have vectorial angle and , therefore value of , shall be obtained if we replace by Therefore Hence ; For free study notes log on: www.gurukpo.com Maths 29 9 Q.16 If = 3axy then find Sol. Let f = - 3axy then = – , = = – , = = – Now we know that Q.17 if u = (y/x) then prove that = 0 Sol. u = x (y/x) (given) Then we can write u = i.e. u is a homogenous function of order one. Therefore fro Euler’s theorem. + = 1.u = u - (1) Differentiating (1) partially with respect to x + + y = - Or + y = - (2) Again differentiating (1) partially with respect to x + + = Or + = - (3) Multiply equation (2) by x and multiply equation (3) by y and after that by adding we can write + + = 0 For free study notes log on: www.gurukpo.com 30 Q.18 Show that for the curve S = = S Sol. Here S = C - (1) Differentiating equation (1) with respect to S 1 = C. - . Or S = c sec - (2) Differentiating equation (2) with respect to = c sec tan = S = S = = = = Q.19 Find the pedal equation of the r = a (1-cos Sol. r = a (1-cos ) [given curve] Now = = + = = + = = + ``[ -1+ cos = 1- = = = = Therefore required equation will be For free study notes log on: www.gurukpo.com Maths 31 UNIT -3 Maxima, Minima and Asymptotes Maximum A function f ( ) is said to have a maximum value at exists a positive number such that f( ) < f (a) Minimum A function f( ) is said to have a minimum value at x=a, if there exists a positive number such that f( ) > f(a) (a- x Extremum: A function f(x) is said to have an extreme value at x=a if it has either a maximum or a minimum at this point. Also the point itself is called on extreme point. Condition for maximum and minimum : if f ( ) can be expanded in the neighbourhood of =c by Taylor’s theorem, then (i) f( ) is maximum at =c, if f’(c) =0 and f” (C) is negative (ii) f( ) is minimum at =c, if f’ (c) =0 and f” (c) is positive. Maxima and minima of a function of two variables: Definition : Let f(x,y) be a function of two independent variables ,y and (a,b) be a point of its domain then Maximum: f(a,b) is said to be a maximum value of f(x,y) if for all sufficiently small positive or negative values of h and k f(a,b) > f(a+h,b+k) further if (a,b) is said to be an extreme value of f(x,y) if f(a,b) is either a maximum or minimum of it. Asymptotes: Definition: A straight line at a finite distance from the origin is said to be an asymptote of a curve if it is the limiting position of a tangent line to the curve at the point of contact tends to infinity. Determination of asymptotes parallel to co-ordinate axes. (i) Asymptotes parallel to - axis : are obtained by equating to zero the coefficient of the highest power of in the equation of the curve provided it is not merely a constant. (ii) Asymptotes parallel to y axis : are obtained by equating to zero the co-efficient of the highest power of y in the equation of the curve, provided it is not merely a constant. Multiple point: (iii) Definition: - A point on a curve, through which more than one branches of a curve pass, it called a multiple point. Kinds of double point: For free study notes log on: www.gurukpo.com 32 A point through which two branches of the curve pass is called a double point. Kind node, cups. Q.1 Find the envelope of the family of straight lines = 1 Where ab = [product of intercepts is constant] Sol. We have + = 1 - (1) ab = - (2) Differentiation and then comparison [Equation (1) and (2) with respect to t considering a and b both the function of t and c is constant] We can write Or = = [from (1) and (2)] Or = , = i.e. a = 2x, b = 2y Therefore, the required equation of the envelope will be (2x) (2y) = Or xy = Which is the equation of hyperbola Q.2 Find the maximum value of Sol. For maximum and minimum value of u = - (1) and 0= sin x si(x+2y) = 0 - (2) from (1) and (2), = 0 = y or tan x = tan y = - tan (x+y) = x = y = 0 or x = y = Function will be stationary at points (o,o) and ( , Now r = = 2 sin y cos (2x+y) s = = sin (2x+2y) t = = at point , r = - S = - t= - and r t - = For free study notes log on: www.gurukpo.com Maths 33 At ( ), r < 0, and r t - > 0 Therefore at this point the value of u is maximum and maximum value = sin sin = 3 At (0,0), u =0 which is a minimum value. Q.3 Find the envelope of the straight lines where is a parameter. Sol. The given family of lines may be written as = - (1) (Dividing throughout by sin cos ) Differentiating (1) partially with respect to , we have Or tan = So that sin = and cos substituting these values of sin and cos in (1), we shall have the required equation of the envelope as + or = or = which is the envelope of given family of straight lines and is the equation of asteroid. Q.4 Find all the asymptotes of the following curve. = Sol. The given equation is of third degree and can be re-arranged as - +2 = 0 and y = m in, third, = = = = 0, we get -1 and ½S .e. m =- shall be obtained from the linear equation. c (m) + (m) =0 which in this case is For free study notes log on: www.gurukpo.com 34 c (m) + (m) =0 or c = - = - = -1, m =-1/2 Hence asymptote corresponding to m = - ½ will be y +2y +2=0 For repeated value of m =1, c is given by (m)+ c (m) + Or + c[4(1+m)] –(1+9m) = 0 Therefore c = + 2 when m =- 1 and thus the two parallel asymptotes and the third asymptote are and Q.5 Show that the minimum value of the following function is 3 Sol. Here u = xy + = For maximum and minimum values ------(1) = 0 = ------(2) From (1) and (2), x=y =a Now S = = 1 t = = at x = y =a, r =2, S=1, t=2 r t - = 4-1 = 3>0 and r>0 Therefore value of u at (a, a) is minimum and minimum value of u = Q.6 Show that the asymptotes of the cubic. Cut the curve again in three points which lie on the straight line x-y+1=0 Sol. The given equation of the curve is - 2 + xy (2x-y) + y (x-y) +1=0 Or ( + (xy- +1=0 - (1) Putting x =1, y= m in the third and second degree terms, we can obtain = 1 -2 +2m - and Now =0 1- 2 (1+2m) (1-m) (1+m) =0 m = - 1, -1 C will be obtained from For free study notes log on: www.gurukpo.com Maths 35 C = - = - when m = -1/2, c = ½ m = -1, c=-1 Thus the three asymptotes are X+2y – 1=0 x-y = 0, x+y+1 =0 The combined equation of the three asymptotes will be (x+2y-1) (x-y) (x+y+1) =0 Or - (2) Subtracting Eg (2) from the equation of the curve (Eg(1), we get x-y+1=3 Which is a straight line on which the point of intersection must lie and the number of points of intersection = 3 (3-2) = 2. Q.7 Find the envelopes of the circles down as diameter upon the radii vector of the cardioids r = a (1+cos = cosn Sol. Let (p ( , be any point on the curve r = ) Therefore l = a (1+cos - (1) and let Q(r, ) be any point on the circle, where OP is diameter of the circle. Then the equation of circle will be r = l cos ( -(2) which by equation (1) will be written in the form r = a (1+cos . Cos ( - ----(3) We have to find the envelope of the family given by (3) and for this will be eliminated between (3) and its differential with respect to . Taking log of both sides in (3) log r= log a + log (1+cos ) + log [cos ( )] Differentiating partially with respect to we have 0= - + Q(r, Or tan P(l,r) r - Substituting this value of in (3) , we have = a = 2a cos2 cos = 2a cos3 Which is the required equation. Q.8 Trace the cissoids y2 (2a-x) = x3 Sol Equation of the given curve y2 (2a-x) =x3 ------(1) For free study notes log on: www.gurukpo.com 36 1 Symmetry :- The curve has only even power of y so it is symmetrical about the axix of x . 2 Origin: - There is no constant term in the given curve equation so the curve passes through the origin (o,o). Tangent at the origin shall be obtained by equation lowest degree term to zero. Therefore tangent at the origin are y2=o y=o, y=o which represent two coincident sins. Hence we may expect a cusp at the origin. 3 Asymptotes: - Put the coefficient of maximum power of y2 equating to zero, 2a-x=o x=2a parallel to y-axis there is Asymptotes. 4 Points: - Cure cut the axis at the origin. 5 Region :- From eq(1) y2 = if we assume x negative then y become imaginary . the same way if we take x>2a then also y become imaginary. Therefore the curve lies between y axis and asymptotes x-2a. 6 Increasing and decreasing:- we observe that x increasing from o to 2a the same way y increase from o to therefore the given function is increasing . Y x=2a X1 O X Y1 The shape of the given curve , Q.9 Trace the following curve r = a+b cos Sol. symmetry :- The curve is symmetrical about the initial line . (2) Pole :- when = cos-1 (-a/b), then r =o, therefore the curve passes through the pole . Now cos-1 (-a/b) will have two values between =O and =2 , one which will be second quadrant and Second value will be third quadrant. Let we assume have two value 1 and 2 for which r=0 therefore two tangent on pole coil be = , = 2 and we can expect a node at the origin. (3) Asymptote- There is no asymptote (4) Points:- 0 0 0 0 0 0 0 Q 0 60 90 120 1 180 2 270 360 R a+b a+b/2 a a-b/2 0 -(b-a) O a a+b when Q value increase from 1200-1800 then r become Negative from positive . Therefore Q has one value between 1200-1800 for which r=o . For free study notes log on: www.gurukpo.com Maths 37 IN this way has one value between 1800-2700 for which r=0 therefore between = 1 and = 2 there will be a loop . Q.9 Find all the asymptotes of the curve Sol. The equation is of fourth degree. putting x=1 and y=m, we get 4 (m) = (1-m2)2 3(m) = 0 2(m) = -4m2 1(m) = m Soluing 4(m) = 0, we get m= 1,1,-1, and -1 The value of C for repeated value of m will be given by 4’’(m) + c 3’(m) + 2(m) = 0 Substitution of the value produce (-4+12m2) + c.0-4m2=0 i.e. C2(3m2-1)- 2m2=0 Therefore C2=1, when m=1 i.e. c= 1 and also c2=1, where m=-1 i.e. c= 1 Hence the four asymptotes are y=x 1, y=-x 1 Q.10 Prove that the minimum radius vector of curve is of length (a+b) Sol. Transforming the given equation into polar form , we can write or r2 = a2 sec2 +b2 cosec2 ------(1) Let R = r2 = a2 Sec2 +b2cosec2 for a maximum or minimum r2 (or R) sec2 tan -2b2cosec2 cot =0 a2tan4 -b2=0 tan2 = Now = 2a2 (sec4 +2sec2 tan2 ) + 2b2 (cosec4 +2cosec2 cot2 ---2 Substituting Value of tan2 in (2) +2b2 =Positive quantity At tan2 = , R or r2 is minimum Hence minimum value of r2 =a2 = a(a+b)+b(b+a) r2=(a+b)2 minimum value of r is (a+b) Q.11 Find the points where the value of u=x3+y3-3axy is maximum or minimum . Sol. U=x3+y3-3axy for maximum and minimum value of u, , than by u (x1y) ------(1) For free study notes log on: www.gurukpo.com 38 ------(2) Substituting value of y from (1) and (2) x4-a3 x=0 x(x-a)(x2+ax+a2)=0 x=0, x=a, other roots will be imaginary . Substituting x=0 in eqn. (1) we get y =0 and substituting x=a in equation (1) we get y=a Hence (0,0) and (a,a) are stationary parts. Now r = = 6 s = = -39 t= r = = 6 At (0,0) : r =0, rt-s2=-9a2<0 No extreme value at (0,0) and at (a,a): r = 6a=t rt –s2 = 27a2>0 Therefore at (a,a) , u will have maximum value if a For free study notes log on: www.gurukpo.com Maths 39 Q.13 Find the envelope of y=x tan - the parameter being . Sol. We can write we can write the given equtin quadratic in tan , i.e. tan2 ------(2) We know that equation of envelope of family of curve given by A 2+B +c=0 B2-4AC=0 is Therefore, envelope of eqn (2) is Or x2= - Q.14 Assuming that the evolute of a curve is the emvelope of its normals find the evolute of the ellipse Sol. The equation of the normal at any point (a cos , b sin ) of the given Ellipse is a sec tan + by cosec cot = 0 = a2-b2 ------(1) Differentiating (1) Partially w.r.t we get so that tan3 = i.e tan = Hence cosec = sec = Putting these value in (1), the required equation of the evolute of ellipse is = a2-b2 Or Or Or Q.15 Show that the maximum rectangle inscribe in a circle is a square . Sol. Let O be the Centre of given circle and readius the r, and let ABCD be a rectangle inscribed in a circle. We know that each angle of rectangle is 900, therefore its diagonals will be the diameters of circle, i.e. AC=2r Let CAB= Sides of rectangle AB=CD=2rcos BC=AD=2rsin Area of rectangle S = 4r2 sin cos S=2r2 sin2 For free study notes log on: www.gurukpo.com 40 For maximum or minimum value if S 4r2 cos2 =0 2 =900 or =450 Now For =450, Hence for =450, area of rectangle is maximum and AB=2r Maximum inscribed rectangle is square. Q.16 Find the envelope of the family of lines where the parameters a and b are connected by the following relation an+bn=cn Sol. Equation of given straight line ------(1)and the relation given by an+bn=cn ------(2) here two parameter a and b and both of paramder connected by a relation. Let us assume a and b are the function of t (arbitrary parameter) differentiation eqn (1) and (2) with respect to t . Or ------(3) And an-1 ------(4) Compare the value of in above equation , we can write ------(5) From eq (1), (2), and (5) = = Therefore an+1=cnx and bn+1=cny ------(6) Substitute the value of a and b in equation (2) (cnx) n/n+1)+(cny)n/n+1)=cn Or xn/(n+1)+yn/(n+1)=cn/(n+1) which is the equation of envelope. Q.17 Find the asymptotes of the following curve Sol. Here coefficient of maximum power of x and y is constant therefore to the axix . Now 3(m) = 1+m3 3 (m) = 0 m3+1=0 m=-1 and other two roots will be imaginary. And 2(m) = -3am 3(m) = 3m2 C = = = = therefore the usymptotes are y= -x-a or x+y+a=0 For free study notes log on: www.gurukpo.com Maths 41 Q.18 Prove that the curve y= lies above the asymptote for large positive Values of x . Sol. Equation of given curve y = Therefore asymptote of this curve will be y=x Now difference between curve and order of Asymptotes will be which sign depend upon 2a2 But if 2a2 always have positive value then curve lies above the asymptote . Q.19 Trace the curve y2 (a2+x2) = x2(a2-x2). Sol. (1) The curve is symmetrical about both the coordinate axes as in the equation both x and y occurs in even power. (2) The origin (0,0) lies on the curve. (3) By substituting y=0, we get x=0, x= a i.e the x-axis cuts the curve at x=a and x =-a Now due to symmetry about the x-axis the curve consists two loops between x=0, x=-a and x=0, x=a (4) The curve has no real asymptotes . (5) To find the tangent at arigin , writing the given eqn. as ----- Y2(a2+x2) = x2(a2-x2) or a2y2 + x2y2 = a2x2-x4 Equating the lowest degree term to zero , we get a2(y2-x2) = 0 or y=±x Therefore and y=x are tangents at the origin . Here origin in a double point and as the two tangents at (0,0) are real and distinct hence origin is a node. (6) Now , for find the nature of points (a,0) and (-a,o) through which the curve passes . for this shifting the origin to (a,0) and putting in the given equation of the curve , new equation to the curve becomes Y2[a2+(x+a)2] = (x+a)2 [ a2-(x+a)2] i.e. Y2 [x2+2a +2a2] = (x2+2ax+a2) [-2ax-x2] To find the tangent at (a,0) equating the lowest degree terms to zero in the above equation . i.e. 2a3x=0 or x=0 i.e. x-a =0 or x=a is tangent at (a,0) similarly x=-a is tangent at the other point (- a,0) (7) Now y2>0 for a2-x2 > o i.e. for x2 For free study notes log on: www.gurukpo.com 42 = , =6x-30 = 6x-30 Now for extreme value of f(x1y), =0 3( After solve these we obtain the following pairs of values of x and y which are the stationary points of the function f(x,y) (4,0), (6,0), (5,1) and (5,-1) for (4,0), A = -6, B=0, c=-6 AC-B2=36 So that AC-B2>0 and A<0 f(x,y) is maximum at (4,0) and maximum value is 112 . For (6,0), A=6, B=0, c=6 AC-B2 =36 so that AC-B2>0 and A>0 f(x,y) is minimum at (6,0) and minimum value is 108 For (5,1) A=0, B=6, C=0 AC-B2= -36<0 f(5,1) is not an extreme value. For (5,1) A=0, B=-6 C=0 AC-B2= -36<0 f(5,1) is not an extreme value. Hene the given function is maximum at (4,0) . minimum at (6,0) and neither maximum or minimum at (5,1) and (5,-1) . For free study notes log on: www.gurukpo.com Maths 43 Unit – IV Integrals, Length, Volumes and Surface Point :- Area of curves given by Cartesian equations. The area bounded by the Cartesian curve y=f(x), the axis of x and the ordinates x=a and x=b is given by Area = f(x) dx= y dx . Note:- I f the given curve is symmetrical about any one side of the axis , then the area of the curve on any one side of the axis of symmetry be calculated and then it should be multiplied by 2 . Area of curves given by polar equatin . The area enclosed by the curve r=f and the radio vectors = , = ( < ) is equal to r2 d Area of closed curve . The area of a closed curve given by x=f1(t) y=f2(t) is equal to to Approximate value of definite Integral or Areas :- 1. Simpson’s rule :-An Approximate value of f(x) dx is Here n is any positive integer. n = and yr = f 2. Prismoidal Rule :- The approximate area bounded by the curve y=f(x), x-axis and ordinates x=a and x=b is given by f(x)dx= (b-a) Where = f(a), = f(b) and Y = f 3. Trapezoidal Rule:- The approximate area bounded by the curve y=f(x), x-axis, and ordinate x=a and x=b is given by f(x) dx = Where yr = f{a+(r-1)h} and h= LENGTH’S OF CURVES :- (i) Cartesian form :-Equation of the curve x=f(y) then = If the qrdinates of any two points A and B be a and b respectively. the length of the are AB is given by (ii) Parametric form :- The equation of the curve is in parametric form is given by X=f(t) y = (t) then For free study notes log on: www.gurukpo.com 44 The length of the arc included between two points A and B whose parametric values are t1 and t2 is given by arc AB = (iii) Polar Form :- The equation of the curve in polar form r=f( ) then The length of the arc included between two points A and B whose vectorial angles and is given by arc AB = Volume of a solid of revolution: - If the area bounded by the curve y=f(x), x-axis and the ordinates at points x=a, x=b, revolves about x-axis , the volume of the solid generated is given by Being assumed that f(x) is finite , single valued and continuous function in the domain (a,b) and does not cross the x-axis of revolution . Note I If the curve given by an equation in polar co-ordinate r=f( ) and the curve revolves about the initial line then volume generated = = Where and are the values of corresponding to the extremities c and D of the curve respectively . Now x=r cos y=r sin volume = Note II If the curve is given by the parametric equations x= (t) , y= (t) Then volume generated = = Where m and n are the values of t corresponding to the point at x=a and x=b, x- axis is the axis of rotation . NoteIII If the axix of revolution is not x-axis, but a line parallel to x-axis , say y=c, the volume will be = Surface of a solid of revolution :- If the arc of the curve y=f(x) intercepted between the points whose abscissa are a,b revolves about x-axis, the curved surface of the solid generated is given by where s is the length of the arc of the curve measured from x=a to any arbitrary point (x,y) on the curve , it is being assumed that f(x) is finite, single valued and continuous in the domain (a,b) 1 If the equation of the curve is in polar form r=f( ) and revolves about the initial line , then the surface area of the solid generated is = For free study notes log on: www.gurukpo.com Maths 45 2 If the equation of the curve is in parametric form x= (t), y= then surfaced generated = = Q.1 Find the whole length of the asteroid Sol: The equation of the given curve is ------(i) Differentiating both sides with respect to x, Or ------(ii) Now since the curve is symmetrical about both the co-ordinate exes The required length = = = = = = Q.2 Find the perimeter of the cardioid r=a (1+cos ). Also show that the upper half arc of the cardioids r=a(1+cos ) is bisected by the line = /3 Sol. Suppose at upper arc of curve there is a point p such that = = = =4 [1-0] =4 ------(iii) For free study notes log on: www.gurukpo.com 46 Perimeter of Cardioid = 2 upper half arc = 2 4a=8a Again length of Arc AB = = = = = 4 [ -0]=2a ------(iv) therefore upper half arc of cardioids is dissected by the line = /3 Q.3 Find the volume of the solid generated by revolving the ellipse about the x-axis . Sol. Since Ellipse is symmetrical about the x-axis therefore volume of Ellipse area ABA ‘B’ A is same to volume enclosed by Area ABA’A Again Ellipse is also symmetriol with respect to y therefore volume enclosed by Area ABA’A will be double to the volume enclosed by area ABo A. Area AB A is Bounded by Arc AB, x-Axis and x=0 and x=a .X Volume =2 B (O,b) = p(x,y) = A’ A A = X(-9,0) = B’ ( O,-b) = Y Q.4 Find the area common to circles r = a and r=2a cos Sol. Centre of circle r= is at origin and radius is and circle r = 2a cos which radius is and its passes from origin and its centre is situated at initial line. Suppose Both of circle cut to each other at point p and Q. therefore common area of both circle is OEPAQO which is clearly double to area OEPAO. Now solving eqn. of circle a =2a cos or cos = for point P = /4 = Area OEPAQO = 2[Area OEPAO] P = = 2[Area OEPO + Area OPAO r=29cos 0; A For free study notes log on: www.gurukpo.com Maths 47 Now Area OEPO = 0 =0 x = =a2 =a2 =a2 And Area OPAO = = = repuired Area = 2[Area OEPAO] = = a2( -1) Q.5 Prove that the length of the arc from the vertex to any point on the cycloid x=a( +sin ), y=a(1-cos ) is . Also prove that the wnole length of an arch of the curve is 8a . Sol. Here origin is o, for which =o, vertex of given curve and P is any point on curve . Now x=a( +sin ) Y=a(1-cos ) = a(1+cos ) Therefore length of arc op = = =a =a =2a =4a sin = = Again length of AoB =2[length of arc oB] For free study notes log on: www.gurukpo.com 48 =2 =2 =4a =4a[2sin =8a [1-0] = 8a Q.6 Show that the volume of the solid generated by the revolution of the following tractrix about its asymptotes is (2/3) a3 x=a (cost + log tan2 t) y=a sin t? Sol. x=a cost + a log tan y=a sin t Since the curve is symmetric about the both axis and its asymptotes is x-axis . required volume = 2 =2 =2 =2 a3 = Q.7 Find the surface of the solid generated by the revolution of the asteroid x2/3 + y2/3=a2/3 about the x-axis . Sol. The equation of given curve is x2/3 + y2/3=a2/3 ------(1) 1) Differentiating w.r.t.x, or ds = = = ds = -----(ii) Now since the curve is symmetrical about both the axes and for the portion of the curve in first quadrant x varies from o to a so the required surface = 2 For free study notes log on: www.gurukpo.com Maths 49 =4 [using ii] put x= a sin3 so that dx = 3a sin2 cos d Also when x=a, = /2 and when x=o, =o the required surface =4 a1/3 = = = Q.8 Find the area common to curve y2=ax and x2+y2=4ax . Sol. The given curve x2+y2=4ax -----(1) represents a circle of radius 2a and centre at (2a,o) and y2=ax-----(2) represents a parabola of latus rectum a and verter at (o,o) Both the curves are symmetrical abut axis of x and their shapes as shown in diagram . on solving (1) and (2), we get the co-ordinates of point of Intersection at x=0 and x=3a Now the common area =area Opa QO Y =2 area OPAO P =2[area OPMO + area MPAM] =2 X’ O M A C(29,0) =2 (I1+I2) say Q Now I1= Y’ = = I2= = = = =2a2 =2a2 Required area = For free study notes log on: www.gurukpo.com 50 = Q.9 Ellipse , revolves round its major axix . find the surface area of the prolate spheroid generated . Sol. The given Equation of the Ellipse can be written as y2= or y= = Now the given curve is symmetrical about both the axes . Surface area = = ====== = = =2 ab Q.10 Find the area enclosed by the cardioids r=a(1+cos ) Sol. The equation of the curve remaine unaltered if is replaced by – The curve is symmetrical about the initial curve . When =0 , r=2a = /2 , r=a = . r=0 Thus when goes on increasing , r goes on decreasing and the required area = = For free study notes log on: www.gurukpo.com Maths 51 = =4a2 =8a2 =8a2 = Q.11 Find the surface of the solid formed by revolution of the eardioid r=a(1+cos ) about the initial line . Sol. Equation of given curve r=a(1+cos ) = =2acos( /2) Since curve is symmetric about to initial line . therefore required surface = =16 =16 = Q.12 Find the area enclosed by the parabolas y2=4ax and x2=4ay Sol. y2=4ax= 64a3x-x4=0 x=0 , x=4a Area between the parabolas x2=4ay and y2=4ax = = = = = = = Q.13 Find the area bounded by the curve x=a cost , y=b sint For free study notes log on: www.gurukpo.com 52 Sol. At t=0, x=a, y=0 and at t=2 , x=a, y=0 given curve is closed curve and can be trace from t=0 to t =2 therefore differentiating given curve w.r.t.t required area = = = = = Q.14 Find the surface area of the solid generated by the revolution of the tractrin x=a cost + a log (t/2) y = a sin t about its asymptote . Sol. From the given eqn. of curve =-a sin t + ------(1) and ------(2) = [from (i) and (ii)] = a cost Because curve is symmetric about the both axes and its asymtotes is x-axis therefore required Area = 2 =2 =4 =4 a2 =4 a2 [sin t] =4 a2 Q.15 The lemiscate r2=a2 cos2 , revolves about a tangent at the pole. show that surface of the solid generated is 4 a2. Sol. The curve is symmetrical about the initial line . The equation of tangents at the pole are =± /4 let the loops revolve about the tangent oA. let P (r,o) be any point on the loop . we draw PM perpendicular to OA. SO For free study notes log on: www.gurukpo.com Maths 53 PoM = + /4 b y PM = r sin { + /4} p or PM = O x’ x Also m . A y’ = = The required surface area =2 =4 =4 a2 = – – =4 a2 Q.16 Find the volume of the spindle shaped solid generated by revolving the asteroid x2/3+y2/3 =a2/3 about the x-axis. Sol. The equation of the given curve is x2/3+y2/3=a2/3 y=(a2/3-x2/3)2/3 ------(i) Now since the curve is symmetrical abut both the co-ordinate a yes , so half the volume of the solid is generated by revolving the area bounded by the arc in the positive quadrant , the x-axis and the ordinates x=0, x=a about O, Hence the required volume =2 = 2 put x=a sin3 so that dx=3a sin2 cos d Also when x=a = /2 , when x=o, =0 volume =2 =6 a3 =6 = = For free study notes log on: www.gurukpo.com 54 Q.17 Find the whole area of the lemniscates of Bunoulli r2=a2 cos2 Sol. The curve is symmetrical about the initial line . Further r=0 when = /4 again r=0 when = /4 Thus the loop extends from =- /4 to /4 The required Area =2 =2 =2a2 =2a2 =a2 /2 = a2 1 = a2 Q.18 Find the area of the loop of the curve ay2=x2 (a-x). Sol. The curve is symmetrical about the axis of x and y=0 when x=0 and x=a i.e. the loop extends from x=0 to x=a Required area of loop =2 =2 Put x=a sin2 so that dx=2a sin cos d and limit for are O to /2 Required Area =2 =4a2 = =4a2 = Q.19 Find the area Bounded by the curve x=a cos3t ,y = a sin3 t Sol. Here at t=0, then x=a, y=0 and at t=2 then x=a , y=0 therefore the given curve is closed curve and can be traced , when 0 t 2 Now differentiating the given equation with respect to t =-3a sin t cos2t therefore required area = = = = For free study notes log on: www.gurukpo.com Maths 55 = = =6a2. = = UNIT –V Double Integrals, Triple Integral nd Drilihelts Integral Q.1 Evaluate where R is the Region of Integration given below Sol. Here the region of Integration can be considered as bounded by the curves y= y y= x’ x x=-a x=a =2 [xy is an odd function of y] y’ 2 =2 =4 putting x=-a sin on the right hand side , so that dx=a cos d , we get =4ab =4ab =4ab =4ab For free study notes log on: www.gurukpo.com 56 = Q.2 Evaluate Sol. = = = = = Q.3 Show that Sol. The integral on L.H.S = = = = ----(1) by interchanging x and y in the above result , the integral on R.H.S. = Hence , the required result follows from (1) and (2) . Q.4 Change the order of integration in the integral Sol. The given limits show that the region of integration is bounded by the curves y= y=2a-x, x=0 and x=a First curve is the parabola x2=ay through the origin and second the line x+y=2a . These two curves intersect at a point (a,a). Therefore the region of Integration is OPB as shown in fig. (1) For free study notes log on: www.gurukpo.com Maths 57 Now if the order of Integration is changed . i.e. if we have ot integrate first with respect to x keeping y constant and then with respect to y , we divide the region into horizontas strips as shown in fig. (ii) The strips starts from x=0 . But some of the strips end op of while others and on BP. The line of demarcation is the line PL, y=a . so the region OPB must be divided in the sub regions OPL and LPB. These are respectively bounded by the curves x=0, x= y=0 and y=a x=0, x=2a-y, y=a and y=2a Here on changing the order of integration the given integral becomes = Q.5 Evaluate Sol. = = = = = = Q.6 Evaluate over the region in the positive quadrant for which x+y<1 Sol. The region of integration is the area A bounded by the two co-ordinate axes and the straight line x+y=1, let we consider as the area bounded by the lines x=0, x=1 and y=0, y=1-x = = = = = = For free study notes log on: www.gurukpo.com 58 = Q.7 Evaluate Sol. = = = = = Q.8 Evaluate Sol. The given integral = = = = = = = = = = Q.9 Find the volume in the first octant bounded by the cylinders x2+y2=a2 and x2+z2=a2 Sol. We first divide the required volume into vertical columns The volume of such a column is dv=zdxdy Now on adding the volumes of all such columns in a slab parallel to the yz plane extending from y=0 to y= we get the volume of this slab . Finally adding the volumes of all such slabs from x=0 to x=a, we get the required volume . Therefore the required volume is given by For free study notes log on: www.gurukpo.com Maths 59 v= y Y= z=0 plane = B = X=0 A X = Y=0 X=A = = = Q.10 Find by double integration the area lying between x2+y2=a2 and x+y=a in the first quadrant. Sol. Here the two curves intersect x axis and y axis at the same points A and B respectively. So the area can be considered as lying between the curves. y=a-x, y= x=0 and x=a The required area = = = = = = = Q.11 Find the mass of a circular plan of diameter a whose densiy at any point k times its distance from a final point on the circumference . sol. If we take the fixed point on the circumference of the circle as the pole and the diameter through it as the initial line , then the eq. of the circle will be r=a cos -- -(1) Now , From the given condition , the density at any point at a distance r from the fixed point 0 is given by P=kr ----(2) The mass of the given plate given by m= The area A is bounded by r=0 , r=a cos = /2 and = /2 m= For free study notes log on: www.gurukpo.com 60 =K = = = Q.12 Find the mass of a plate in the form of a quadrant of an ellipse whose density per unit area is given by P=kxy . Sol. Here the area can be considered as lying between the curves y=0, y= x=0 and x=a The required mass = = = = = = = Q.13 Evaluate taken throughout the ellipsoid Sol. Put x2=a2u y2=b2v z=c2w so that xdx= ydy= , zdz= then the given integral = du dv dw where u+v+w = = (By Dirichlet’s integral) = = Q.14 Evaluate the integral where x,y,z are always positive but limited by the condition Sol. Put so that dx= dy= dx= For free study notes log on: www.gurukpo.com Maths 61 then the given integral = = Q.15 Evaluate the following integral by changing to polar co-ordinates. Sol. The given limits of integration show that the region of integration lies between the curves y=x, y= x=0 and x=1 Thus the region of the integration is the area lying between the line y=x and the circle x2+y2=2x=0 In polar co- ordinates , the equation of the circle is r2cos2 + r2sin2 -2r cos =0 cr r (r-2cos ) = 0 or r =2cos Hence , in polar co-ordinates, the region of integration is included between the curves : r=0 , r=2cos , = and = = = = = = = = Q.16 Find the volume cut off the sphere x2+y2+z2=a2 by the cone x2+y2=z2 Sol. Here the required volume is = where A is the area above which the volume stands. Now eliminding z between the equations x2+y2+z2=a2 and x2+y2=z2 we get 2(x2+y2)=a2 ------(2) This is the eq. of the cylinder through the curve of Intersection of the sphere and cone. whose generators are parallel to z-axis. The section of this cylinder by the xy- plane gives the area A. we now change the integral (1) in polar co-ordinates. Then the eq. (2) becomes 2(r2cos2 +r2sin2 )=a2 or 2r2=a2 or r=a/ by symmetry the volumes above the xoy plane in four octants are equal . The fig. (ii) shows only one fourth of the area A . The required volume =4 volume in the one octant For free study notes log on: www.gurukpo.com 62 =4 =4 =4 =4 = = = ) Q.17 Find the mass of plate in the form of a quadrant of an ellipse where density per unit area is given by =kxy Q.17 Evaluate Sol. Here we have to integrate first with respect to y . from the given integrand it is evident that integration with respect to y is not possible . Thus we use here the changing the order of integration . The given limits show that the region of integration is bounded by the curves y=x,y= , x=0 and x= The first curve is the line y=x through the origion and second line y= parallel to x-axis . These two line intersect at a point ( Now if the order of integration is changed we divide the region into horizontal strips as fig. (ii) . In this case stripos startys from x=0 and end on x=y . thus here the limits fro the given region arex=0, x=y, y=0 and y= hence = = = = = =1 Q.18 Find the mass of a solid in the form of the positive octant of the sphere x2+y2+z2=a2, if the density at any point is kxyz. Sol. Here the required mass is given by M= = = = For free study notes log on: www.gurukpo.com Maths 63 where A is the area bounded by curves y=0, y= , x=0 and x=a M= = = = = = = For free study notes log on: www.gurukpo.com 64 Bibliography 1. II ifferential calculus- JPH Publication 2. II ifferential calculus- Dr. V.B.L. Chaurasia , Dr suman Jain, Dr. M.C. Goyal , Dr suveen Agarwal 3. Integral Calculus- Dr. D.C. Gokhroo, S.R. Saini, S.M. Agarwal. ```````````````````````````````````````````````````````````````````````````````` For free study notes log on: www.gurukpo.com