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Geometrical Applications of Integration

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Geometrical Applications of Integration 200 Engineering Mathematics through Applications 3 Geometrical Applications of Integration aaaaa 3.1 INTRODUCTION In general, we consider the integration as the inverse of differentiation. In the expression of b the sum, ∑fx()∆ x , f is considered continuous on a ≤ x ≤ b and we find that limit of S as ∆x a b approaches to zero is the number ∫fxdx()=− Fb () Fa (), where F is any anti-derivative of f. We a apply this contention in finding the area between the x-axis and the curve y = f(x), a ≤ x ≤ b. We extend the application to compute distances, volumes and volumes of revolution, length of curves, area of surface of revolution, average value of function, centre of mass, centroid, etc. 3.2 AREA OF BOUNDED REGIONS (QUADRATURE) I. Areas of Cartesian Curves y The area bounded by the curve y = f (x), X-axis and the B b ∫ ydx x) ordinates x = a, x = b is , when f(x) is continuous single y = f( a valued and finite function of x, and y does not change sign in the P´ Q δx, y +δy) interval [a, b]. (,xyP ) (x + Q´ If AB is the curve y = f(x) between the ordinates A LA(x = a) and MB(x = b) with a condition that y is strictly yy + δ y increasing (or strictly decreasing) function of x in the interval [a, b ]. Let P(x, y) and Q(x + δx, y + δy) be two neighbouring O L N N´ M x-axis points on this continously increasing curve y = f (x) and x δ x NP, N’Q be their respective ordinates. Fig. 3.1 Here clearly the ALNP i.e. A depends on the position of the point P(x, y) whose abscissa is ‘x’ and area PNN’Q = δA lies between the areas of the δ δ δ δ δ A rectangles PN’ and NQ, i.e. A lies between area y x and (y + y) x or δx lies between y and (y + δy). dA On taking the limits as Q → P, i.e. δx → 0 (meaning thereby δy → 0), = y dx Integrating both sides between the limits x = a to x = b, we get Geometrical Applications of Integration 201 b Y Area ALMB== Ab ∫ ydx a . yb = a M B However, if x and y are interchanged in the above formula, we xfy = ( ) see that the area bounded by the curve x = f(y), Y-axis and the abscissa L b ya = A ∫ y = a, y = b is xdy. X a O Fig. 3.2 Observations: (i) The area bounded by the curve y = f(x), the two ordinates at A and B and the X-axis is often called the area under the curve AB and the process of calculating the area bounded by the curve is called quadrature. (ii) An area whose boundary is described in anti-clockwise direction is considered positive otherwise negative. Or in other words, for the portion of the curve (under Y consideration) above X-axis for which y is positive, area enclosed is positive, whereas for the portion of the curve (under consideration) below X-axis for which y is negative, area is negative. L M But in case of area with negative sign, we mean numerical value of X O the area. xa = xb = (iii) If in the interval, a ≤ x ≤ c, the curve y = f(x) is above the x-axis and in the ≤ ≤ interval c x b, the curve y = f(x) is below the x-axis then we write the A area bcb ∫∫∫ydx=+ ydx ydx yfx = ( ) B aac or in otherwords, if x-changes sign from a to b, y = f(x) changes sign at Fig. 3.3 some interval point x = c (say), then the area for x from a to c and Y c to b, are calculated seperately and then their numerical value are added (see Fig. 3.4) Similarly, the result can be extended if y changes sign at more A than one intermediate point in the interval [a, b]. (iv) Area of the region bounded between two continuous curves f(x) xa = +ve and g(x) on [a, b] and the vertical lines x = a, x = b is given by area b xc = M Afxgxdx=−∫[() ()] , X a O L N ≥ –ve xb = where f(x) g(x) in [a, b]. (see Fig. 3.5) area In the region under consideration, representative rectangle is ∆ shown with height: f(xi) – g(xi), width: x and P(xi g(xi)); Q(xi f(xi)) If f(x) ≥ g(x) in [a, c] and f(x) ≤ g(x) in [c, b], then we write the area as B Fig. 3.4 cb Afxgxdxgxfxdx=−∫∫[() ()] +− [() ()] ac as shown in Fig. 3.6. Y fx() ∆x gx() fx() Y Q f()x i gx() PP gx()i X X a xa = xc = xb = O xi b Fig. 3.5 Fig. 3.6 202 Engineering Mathematics through Applications Example 1: (i) Find the area bounded by the parabola y2 = 4ax and its latus-ractum. (ii) Show that the area cut off from a parabola by any double ordinate is two-third of the corresponding rectangle obtained by the double ordinates and its distance from the vertex. Solution (i): For the parabola y2 = 4ax, let the double ordinate PP' be x = c Since the curve is symmetrical about X-axis, therefore, for part of the curve above X-axis, y be taken as positive, i.e. yax= 4 Thus, the area bounded by the parabola with its double ordinate PMP’ (i.e. Latus ractum) yax2 = 4 Y (4aa , 4 ) = Area P’APP’ = 2 Area PAMP B c P 3 cc 113x28x = 0 ==2244∫∫ydx ax dx = a222 = a c 3 00 3 A M 20 xc = 11 (ii) Again for P(x, y), yMPax==442 = acac =22 so that 11 C PP'==2 MP 4 a22 c P´ Now the area of the rectangle formed by the double ordinate Fig. 3.7 (PMP’) and its distance from the vertex A (i.e. AM) = PP' × AM ( 11) 13 =⋅=44ac22 c ac 22 Now the area cut of from the parabola by double ordinates, i.e. area P’APP’ 8213() 11 2 ==ac224 ac 22 = area of the rectangle formed by PP' and AM. 33 3 Hence, the area formed by the parabola and its latus ractum is two third of the area of the rectangle formed by the double ordinates with its distance from the vertex, A. Note: Vice versa, the area of the rectangle is 3 times the area bounded by the parabola with latus ractum. 2 Y Example 2: Find the area between the curve x2y2 = a2(y2 – x2) and its asymptotes. Solution: The given curve x2y2 = a2(y2 – x2) is symmetrical about both the axis and at the origin, x = y = ±x as the tangents. Further, x = ± a are the two y Ba(– , 0) O asymptotes parallel to Y-axis. The curve no where X Aa(, 0) y intersects with the axis except at (0, 0). Whence the = – curve does not enclose area with its abscissa or x ordinates (see Fig. 3.8). Due to its symmetry about both the axis, the whole area between the curve and its asymptotes is xa = – x = a a aax ==44∫ydx dx θ ∫22− (Putting x = a sin ) 0 0ax Fig. 3.8 Geometrical Applications of Integration 203 π π θ 2 =⋅2 asin ⋅θθ=∫22 θθ= 4sin4∫aadaθadcos 4 0acos 0 Example 3: Find the area of the curve a2y2 = x3(2a – x). Solution: Without going into geometrical details of the curve, the area enclosed by it in the first quadrant is Y 3 22aax2 A==∫∫ydx2 a− x dx 00a (i) The curve intersects X-axis at x = 0, x = 2a. O X´ X (ii) Axis of X is the tangent at the origin. (0, 0) (,a 0) Aa(2 , 0) dy== dy =∞=3 3a (iii) 0at xa2, at x a ,0 dx dx 2 2 Put x = 2a sin2θ π 3 Fig. 3.9 (2a sin2 θ )2 A =⋅∫2 2aa−θ⋅θθθ 2 sin2 4 a sin cos d 0 a π =θθθ∫2 16ad24 sin cos 2 0 ⋅π π2 ==2 31 a 16a 6422⋅⋅ 2 π (pp−−…−−… 1)( 3) ( qq 1)( 3) π using ∫2sinpqdθθθ= cos 0 ()(2)2pqpq++− Hence the total area is πa2. Example 4: For the curve y2(a – x) = x2(a + x) [NIT Kurukshetra, 2004] (i) Find the area of the loop Y (ii) Area of the portion bounded by the curve and its asymptote. x = a Solution: The curve passing through the origin is x O = symmetrical about X-axis and has x = a as its A y X´ X y asymptote. It intersects the axis of X at (–a, 0) as shown (–a , 0) = – in Fig. 3.10. x For the area of the loop, x varies from –a to 0. Further, the loop is symmetrical about X-axis Y´ 00ax++ 0 ax Fig. 3.10 ∴ Area of the loop== 2∫∫ydx=2 x dx 2 ∫ x dx (on rationalization) −−aaax− − aax22− 002 =ax+ x 2∫∫dx dx −−aaax22−− ax 22 00−1 222−− =−222+aax() 2()∫∫ax a x dx dx −−aaax22− 204 Engineering Mathematics through Applications 000−1 =−−−+222 21 −− 22 2()()adxadx∫∫∫xa x dx a x −−−aaaax22− 0 02221 −−− =−−+222 2 1xxaxa −+ 1 x 2()sinaa∫da x dx sin −a aa22 −a 00 12220 axxax− − =−2(−+a22 )2 sin 1 − ax − −a22aa −a 2 ()21a − () − −π =200sin(1)00, −−a+() − −−− where sin1 (−= 1) 2 2 πaa22 a 2 =−+2a2 =⋅−+π=π− 2 [ 4 ] ( 4) numerically 44 2 Alternately: 0 ax+ Axdx=2,∫ Put x = a sinθ, dx = acos θdθ −aax− −π Limits for x = 0, θ = 0; x = –a, θ= 2 0 +θ =θa(1 sin ) θθ Aa2sin∫−π ad cos a(1−θ sin ) 2 0 +θ+θ =θ1sin1sin θθ 2sin∫−π aad cos 1sin1sin−θ+θ 2 00θθ+θ =θ+θθ222sin cos (1 sin ) θ= 2ad∫∫−πda2 −π [sin sin ] 1sin−θ2 22 0 θθ0 =θ+−θ−θ+−221sin2θ= 2sin(1cos2)a∫−π da2 cos 224−π 2 2 aa22 =π−=(4)(4)numerically −π 22 (b) Area between the curve and its asymptote aa+ ==ax 22∫∫ydx x dx 00ax− This integrand is same as in the case (a) simply limits are 0 to a.
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