Multiple Integrals Notes PDF

UNIT – IV MULTIPLE INTEGRALS 4.1Introduction The mathematical modeling of any engineering problem which leads to the formation of differential equation of more than one variable has its solution by the integration in terms of those variables...

UNIT – IV MULTIPLE INTEGRALS 4.1Introduction The mathematical modeling of any engineering problem which leads to the formation of differential equation of more than one variable has its solution by the integration in terms of those variables the need of the solution in an integral where many variables are involved motivated the study of integral calculus of several variables. In this chapter all the basic concepts related to the methods to approach such integrals are discussed. 4.2Double integration in Cartesian co – ordinates Let 𝑓( 𝑥, 𝑦 ) be a single valued function and continuous in a region R bounded by a closed curve C. Let the region R be subdivided in any manner into n sub regions 𝑅1 , 𝑅2 , 𝑅3 , ∙ ∙ ∙ , 𝑅𝑛 of areas 𝐴1 , 𝐴2 , 𝐴3 , ∙ ∙ ∙ , 𝐴𝑛.Let (𝑥𝑖 , 𝑦𝑗 ) be any point in the sub region𝑅𝑖. Then consider the sum formed by multiplying the area of each sub – region by the value of the function 𝑓( 𝑥, 𝑦 ) at any point of the sub – region and adding up the products which we denote ∑𝑛1 𝑓(𝑥𝑖 , 𝑦𝑗 )𝐴𝑖 The limit of this sum ( if it exists) as 𝑛 → ∞ in such a way that each 𝐴𝑖 → 0 is defined as the double integral of 𝑓( 𝑥, 𝑦 ) over the region R. Thus lim ∑𝑛1 𝑓(𝑥𝑖 , 𝑦𝑗 )𝐴𝑖 = ∬𝑅 𝑓(𝑥, 𝑦) 𝑑𝐴 𝑛 →∞ The above integral can be given as ∬𝑅 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥 𝑜𝑟 ∬𝑅 𝑓 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 Evaluation of Double Integrals 𝑦 𝑥 To evaluate ∫𝑦 1 ∫𝑥 1 𝑓(𝑥, 𝑦)𝑑𝑥 𝑑𝑦 we first integrate f(x, y) with respect to x partially, 0 0 that is treating y as a constant temporarily, between 𝑥0 and 𝑥1. The resulting function got after the inner integration and substitution of limits will be function of y. Then we integrate this function of with respect to y between the limits 𝑦0 and 𝑦1 as used. Region of Integration 𝑏 𝑓 (𝑥) Case (i) Consider the integral ∫𝑎 ∫𝑓 2(𝑥) 𝑓 (𝑥, 𝑦)𝑑𝑦 𝑑𝑥 Given that y varies from 𝑦 = 1 𝑓1 (𝑥) to 𝑦 = 𝑓2 (𝑥) x varies from 𝑥 = 𝑎 𝑡𝑜 𝑥 = 𝑏. We get the region R by 𝑦 = 𝑓1 (𝑥), 2 𝑦 = 𝑓2 (𝑥 ), 𝑥 = 𝑎 , 𝑥 = 𝑏. The points A, B, C, D are obtained by solving the intersecting curves. Here the region divided into vertical strips (𝑑𝑦 𝑑𝑥). 𝑑 𝑓 (𝑦) Case (ii) Consider the integral ∫𝑐 ∫𝑓 2(𝑦) 𝑓(𝑥, 𝑦)𝑑𝑥 𝑑𝑦 1 Here varies from 𝑥 = 𝑓1 (𝑦) to 𝑥 = 𝑓2 (𝑦) and y varies from 𝑦 = 𝑐 𝑡𝑜 𝑦 = 𝑑 ∴ the region is bounded by 𝑥 = 𝑓1 (𝑦), 𝑥 = 𝑓2 (𝑦), 𝑦 = 𝑐 , 𝑦 = 𝑑. The points P, Q, R, S are obtained by solving the intersecting curves. Here the region divided into horizontal strips (𝑑𝑥 𝑑𝑦). Problems based on Double Integration in Cartesian co-ordinates Example: 4.1 𝟏 𝟐 Evaluate ∫𝟎 ∫𝟏 𝒙(𝒙 + 𝒚)𝒅𝒚𝒅𝒙 Solution: 1 2 1 2 ∫0 ∫1 𝑥(𝑥 + 𝑦)𝑑𝑦𝑑𝑥 = ∫0 ∫1 (𝑥 2 + 𝑥𝑦)𝑑𝑦𝑑𝑥 2 1 𝑥𝑦 2 =∫0 [𝑥 2 𝑦 + ] 𝑑𝑥 2 1 1 𝑥 =∫0 [(2𝑥 2 + 2𝑥) − (𝑥 2 + 2)] 𝑑𝑥 1 𝑥 =∫0 [2𝑥 2 + 2𝑥 − 𝑥 2 − 2)] 𝑑𝑥 3  1 3 =∫0 [𝑥 2 + 2 𝑥] 𝑑𝑥 1 𝑥3 3 𝑥2 1 3 13 =[ + ] =( + ) − (0 + 0) = 3 2 2 0 3 4 12 Example: 4.2 𝒂 𝒃 Evaluate ∫𝟎 ∫𝟎 𝒙𝒚(𝒙 − 𝒚)𝒅𝒚𝒅𝒙 Solution: 𝒂 𝒃 𝑎 𝑏 ∫𝟎 ∫𝟎 𝑥𝑦(𝑥 − 𝑦)𝑑𝑦𝑑𝑥 = ∫0 ∫0 (𝑥 2 𝑦 − 𝑥𝑦 2 )𝑑𝑦𝑑𝑥 𝑎 𝑥 2𝑦 2 𝑏 𝑥𝑦 3 =∫0 [ − ] 𝑑𝑥 2 3 0 𝑎 𝑏 2𝑥 2 𝑏3 𝑥 =∫0 [( − ) − (0 − 0)] 𝑑𝑥 2 2 𝑎 𝑏 2𝑥 3 𝑏3 𝑥 2 =[( − )] 6 6 0 𝑎 3𝑏 2 𝑎 2𝑏3 =( − ) − (0 − 0 ) 6 6 𝑎 2𝑏 2 = (𝑎 − 𝑏 ) 6 Example: 4.3 𝒂 𝒃 𝒅𝒙𝒅𝒚 Evaluate ∫𝟐 ∫𝟐 𝒙𝒚 Solution: 𝑎 𝑏 𝑑𝑥𝑑𝑦 𝑎 1 𝑏 ∫2 ∫2 = ∫2 [𝑦 𝑙𝑜𝑔𝑥] 𝑑𝑦 𝑥𝑦 2 𝑎1 =∫2 𝑦 (𝑙𝑜𝑔𝑏 − 𝑙𝑜𝑔2)𝑑𝑦 𝑎1 𝑏 𝑎 = ∫2 𝑙𝑜𝑔 ( ) 𝑑𝑦 [∵ 𝑙𝑜𝑔 = 𝑙𝑜𝑔𝑎 − 𝑙𝑜𝑔𝑏] 𝑦 2 𝑏 𝑏 𝑎1 𝑏 =𝑙𝑜𝑔 2 ∫2 𝑑𝑦 = 𝑙𝑜𝑔 2 [log 𝑦]𝑎2 𝑦 𝑏 𝑏 𝑎 = 𝑙𝑜𝑔 2 [𝑙𝑜𝑔𝑎 − 𝑙𝑜𝑔2] =[𝑙𝑜𝑔 2] [𝑙𝑜𝑔 2 ] Example: 4.4 𝟏 𝟑 Evaluate ∫𝟎 ∫𝟐 (𝒙𝟐 + 𝒚𝟐 )𝒅𝒙𝒅𝒚 Solution: 1 3 1 𝑥3 3 ∫0 ∫2 (𝑥 2 + 𝑦 2 )𝑑𝑥𝑑𝑦 = ∫0 [ 3 + 𝑦 2 𝑥] 𝑑𝑦 2 1 33 23 =∫0 [( 3 + 3𝑦 2 ) − ( 3 + 2𝑦 2 )] 𝑑𝑦 1 8 =∫0 [9 + 3𝑦 2 − 3 − 2𝑦 2 ] 𝑑𝑦 4  1 1 19 19𝑦 𝑦3 =∫0 [ 3 + 𝑦 2 ] 𝑑𝑦 =[ + ] 3 3 0 19 1 20 =[ + ] = 3 3 3 Example: 4.5 𝟑 𝟐 Evaluate ∫𝟎 ∫𝟎 𝒆𝒙+𝒚 𝒅𝒚𝒅𝒙 Solution: 3 2 3 2 3 2 ∫0 ∫0 𝑒 𝑥+𝑦 𝑑𝑦𝑑𝑥 = ∫0 ∫0 𝑒 𝑥 𝑒 𝑦 𝑑𝑦𝑑𝑥 =[∫0 𝑒 𝑥 𝑑𝑥] [∫0 𝑒 𝑦 𝑑𝑦] =[𝑒 𝑥 ]30 [𝑒 𝑦 ]20 = [𝑒 3 − 𝑒 0 ][𝑒 2 − 𝑒 0 ] =[𝑒 3 − 1][𝑒 2 − 1] Note: If the limits are variable, then check the given problem is in the correct form Rule: (i) The limits for the inner integral are functions of , then the first integral is with respect to 𝑦 (ii) The limits for the inner integral are functions of , then the first integral is with respect to 𝑥 Example: 4.6 𝒂 √ 𝒂𝟐 −𝒙𝟐 Evaluate ∫𝟎 ∫𝟎 𝒅𝒙𝒅𝒚 Solution: The given integral is in incorrect form Thus the correct form is 𝑎 √𝑎 2 −𝑥 2 𝑎 2−𝑥 2 𝑎 ∫0 ∫0 𝑑𝑦𝑑𝑥 =∫0 [𝑦]√𝑎 0 𝑑𝑥 =∫0 [√𝑎2 − 𝑥 2 ]𝑑𝑥 𝑎 𝑥 𝑎2 𝑥 = [ √𝑎2 − 𝑥 2 + 𝑠𝑖𝑛−1 ] 2 2 𝑎 0 𝑎2 𝜋 =[(0 + 𝑠𝑖𝑛−1 1) − (0 + 0)] [∵ 𝑠𝑖𝑛−1 1 = , 𝑠𝑖𝑛−1 0 = 0] 2 2 𝑎2 𝜋 𝜋𝑎 2 = ( 2) = 2 4 Example: 4.7 𝒂 √ 𝒂𝟐 −𝒙𝟐 Evaluate ∫𝟎 ∫𝟎 𝒚(𝒙𝟐 + 𝒚𝟐 )𝒅𝒙𝒅𝒚 Solution: The given integral is in incorrect form Thus the correct form is 𝑎 √𝑎 2 −𝑥 2 𝑎 √𝑎 2 −𝑥 2 ∫0 ∫0 𝑦(𝑥 2 + 𝑦 2 )𝑑𝑦𝑑𝑥 = ∫0 ∫0 (𝑥 2 𝑦 + 𝑦 3 )𝑑𝑦𝑑𝑥 5  √𝑎 2 −𝑥 2 𝑎 𝑥 2𝑦 2 𝑦4 =∫0 [ 2 + ] 𝑑𝑥 4 0 2 𝑎 𝑥 2 (𝑎 2−𝑥 2) (𝑎 2−𝑥 2) =∫0 [ + ] 𝑑𝑥 2 4 𝑎 𝑎 2𝑥 2 𝑥4 𝑎4 𝑥4 2𝑎 2 𝑥 2 =∫0 [ − + + − ] 𝑑𝑥 2 2 4 4 4 𝑎 𝑎2 𝑥 3 𝑥5 𝑎 4𝑥 𝑥5 2𝑎 2 𝑥 3 =[ − 10 + + 20 − ] 6 4 12 0 𝑎 −𝑥 5 𝑎 4𝑥 𝑥5 =[ 10 + + 20] 4 0 −𝑎 5 𝑎5 𝑎5 =[ 10 + + 20 ] 4 𝑎5 = 5 Example: 4.8 𝟏 √𝒙 Evaluate ∫𝟎 ∫𝒙 𝒙𝒚(𝒙 + 𝒚)𝒅𝒙𝒅𝒚 Solution: The given integral is in incorrect form Thus the correct form is 1 √ x √ 1 x ∫0 ∫x xy(x + y)dydx = ∫0 ∫x (x 2 y + xy 2 )dydx 𝑥 1 𝑥 2𝑦 2 𝑥𝑦 3 √ =∫0 [ + ] 𝑑𝑥 2 3 𝑥 3 1 𝑥 𝑥 ⁄2 𝑥2 𝑥3 =∫0 [(𝑥 2 2 + 𝑥 ) − (𝑥 2 +𝑥 )] 𝑑𝑥 3 2 3 5 1 𝑥3 𝑥 ⁄2 5 =∫0 [ 2 + − 6 𝑥 4 ] 𝑑𝑥 3 7 1 𝑥4 𝑥 ⁄2 5 𝑥5 =[ 8 + 3(7⁄ ) − 6 ] 2 5 0 1 2 1 3 =(8 + 21 − 6) − (0 + 0 − 0) = 56 Example: 4.9 𝟏 √𝟏+𝒙𝟐 𝒅𝒙𝒅𝒚 Evaluate ∫𝟎 ∫𝟎 𝟏+𝒙𝟐 +𝒚𝟐 Solution: The given integral is in incorrect form Thus the correct form is 1 √1+𝑥 2 𝑑𝑦𝑑𝑥 1 √1+𝑥 2 𝑑𝑦𝑑𝑥 ∫0 ∫0 = ∫0 ∫0 2 1+𝑥 2 +𝑦 2 (√1+𝑥 2 ) +𝑦 2 6  1 1 𝑦 √1+𝑥 2 =∫0 [√1+𝑥 2 𝑡𝑎𝑛−1 (√1+𝑥 2 )] 𝑑𝑥 0 1 1 𝜋 =∫0 [√1+𝑥 2 𝑡𝑎𝑛−1 (1) − 0] 𝑑𝑥 [∵ 𝑡𝑎𝑛 −1 (1) = ] 4 1 1 𝜋 𝜋 1 1 =∫0 𝑑𝑥 = ∫ 𝑑𝑥 [𝑡𝑎𝑛 −1 (0) = 0] √1+𝑥 2 4 4 0 √1+𝑥 2 𝜋 1 = [𝑙𝑜𝑔[𝑥 + √1 + 𝑥 2 ]] 4 0 𝜋 = 4 log(1 + √2) Example: 4.10 𝟒 𝒙𝟐 𝒚⁄ Evaluate ∫𝟎 ∫𝟎 𝒆 𝒙 𝒅𝒚𝒅𝒙 Solution: The given integral is in correct form 𝑦 𝑥2 4 𝑥2 𝑦 4 𝑒 ⁄𝑥 ∫0 ∫0 𝑒 ⁄𝑥 𝑑𝑦𝑑𝑥 = ∫0 [ 1⁄ ] 𝑑𝑥 𝑥 0 4 𝑒𝑥 1 =∫0 [(1⁄ ) − (1⁄ )] 𝑑𝑥 𝑥 𝑥 4 4 =∫0 [𝑥𝑒 𝑥 − 𝑥 ]𝑑𝑥 =∫0 𝑥(𝑒 𝑥 − 1) 𝑑𝑥 4 𝑥2 =[𝑥 (𝑒 𝑥 − 𝑥 ) − (1) (𝑒 𝑥 − )] (by Bernoulli’s formula) 2 0 16 =[4(𝑒 4 − 4) − (𝑒 4 − ) − (0 − 1)] 2 =4𝑒 4 − 16 − 𝑒 4 + 8 + 1 =3𝑒 4 − 7 Example: 4.11 1 𝑥 Sketch roughly the region of integration for ∫0 ∫0 𝑓(𝑥, 𝑦) 𝑑𝑦𝑑𝑥 Solution: 1 𝑥 Given ∫0 ∫0 𝑓(𝑥, 𝑦) 𝑑𝑦𝑑𝑥 𝑥 varies from 𝑥 = 0 to 𝑥 = 1 𝑦 varies from 𝑦 = 0 to 𝑦 = 𝑥 7 Example: 4.12 𝒂 √𝒂𝟐 −𝒙𝟐 Shade the region of integration ∫𝟎 ∫√𝒂𝒙−𝒙𝟐 𝒅𝒙𝒅𝒚 Solution: 𝑎 √𝑎 2 −𝑥 2 ∫0 ∫√𝑎𝑥−𝑥 2 𝑑𝑦𝑑𝑥 is the correct form 𝑥 limit varies from 𝑥 = 0 to 𝑥 = 𝑎 𝑦 limit varies from 𝑦 = √𝑎𝑥 − 𝑥 2 to 𝑦 = √𝑎2 − 𝑥 2 i.e., 𝑦 2 = 𝑎𝑥 − 𝑥 2 to 𝑦 2 = 𝑎2 − 𝑥 2 i.e., 𝑦 2 + 𝑥 2 = 𝑎𝑥 to 𝑦 2 + 𝑥 2 = 𝑎2 𝑎 𝑎 𝑥 2 + 𝑦 2 = 𝑎𝑥 is a circle with centre ( , 0) and radius 2 2 𝑥 2 + 𝑦 2 = 𝑎2 is a circle with centre (0,0) and radius 𝑎 Exercise 4.1 Evaluate the following integrals 1 𝑥2 26 1.∫0 ∫0 (𝑥 2 + 𝑦 2 ) 𝑑𝑦𝑑𝑥 Ans: 105 8  1 1 1 2. ∫0 ∫𝑥 (𝑥 2 + 𝑦 2 ) 𝑑𝑥𝑑𝑦 Ans: 3 𝑎 𝑎 𝑥 𝜋𝑎 3. ∫0 ∫𝑦 𝑑𝑥𝑑𝑦 Ans: 𝑥 2 +𝑦 2 4 2 3 4. ∫1 ∫1 (𝑥𝑦 2 ) 𝑑𝑥𝑑𝑦 Ans: 13 3 √4−𝑦 241 5. ∫0 ∫1 (𝑥 + 𝑦) 𝑑𝑥𝑑𝑦 Ans: 60 1 √2−𝑥 2 𝑥 1 6. ∫0 ∫𝑥 𝑑𝑦𝑑𝑥 Ans: 1 − √𝑥 2+𝑦 2 √2 1 𝑥 1 7. ∫0 ∫0 𝑒 𝑥+𝑦 𝑑𝑦𝑑𝑥 Ans: (𝑒 − 1)2 2 3 3𝑥+3 32 8.∫−1 ∫𝑥 2 𝑑𝑦𝑑𝑥 Ans: 3 2 𝑥+2 9 9. ∫−1 ∫𝑥 2 𝑑𝑦𝑑𝑥 Ans: 2 𝑎 ⁄√2 𝑦 𝑎4 10. ∫0 ∫0 (𝑦 2 ) 𝑑𝑦𝑑𝑥 Ans: 32 (𝜋 + 2) 4.3 Double integration in Polar co-ordinates Consider the integral 𝜃 2 𝑟 2 ∫𝜃1 ∫𝑟1 𝑓 (𝑟, 𝜃)𝑑𝑟 𝑑𝜃 which is in polar form. This integral is bounded over the region by the straight line 𝜃 = 𝜃1 , 𝜃 = 𝜃2 and the curves 𝑟 = 𝑟1 , 𝑟 = 𝑟2.To evaluate the integral, we first integrate with respect to r between the limits 𝑟 = 𝑟1 and 𝑟 = 𝑟2 (treating 𝜃 as a constant).The resulting expression is then integrated with respect to 𝜃 between the limits 𝜃 = 𝜃1 and 𝜃 = 𝜃2. Geometrically, AB and CD are the curves 𝑟 = 𝑓1 (𝜃) and 𝑟 = 𝑓2 (𝜃) bounded by the lines 𝜃 = 𝜃1 and 𝜃 = 𝜃2 so that ABCD is the region of integration. PQ is a wedge of angular thickness 𝛿𝜃. 𝑟 Then ∫𝑟 2 𝑓(𝑟, 𝜃)𝑑𝑟 indicates that the integration is performed along PQ(i.e., 𝑟 varies , 𝜃 1 constant) and the integration with respect to 𝜃 9  𝜃 2 ∫𝜃1 𝑓(𝑟, 𝜃)𝑑𝜃 means rotation of the strip PQ from AC to BD Problems based on double integration in Polar Co-ordinates Example: 4.13 𝝅⁄ 𝒔𝒊𝒏𝜽 Evaluate ∫𝟎 𝟐 𝒓𝒅𝜽𝒅𝒓 ∫𝟎 Solution: 𝜋⁄ 𝑠𝑖𝑛𝜃 2 Given ∫0 ∫0 𝑟𝑑𝜃𝑑𝑟 𝜋⁄ 𝑠𝑖𝑛𝜃 2 = ∫0 ∫0 𝑟𝑑𝑟𝑑𝜃 (Correct form) 𝜋⁄ 𝒓𝟐 𝑠𝑖𝑛𝜃 𝜋⁄ (𝑠𝑖𝑛𝜃)2 2[ ] 2[ = ∫𝟎 𝑑𝜃 = ∫0 − 0] 𝑑𝜃 𝟐 𝟎 2 1 𝜋⁄ 2 = 2 ∫0 𝑠𝑖𝑛2 𝜃𝑑𝜃 1 1 𝜋 𝜋 = 2.2. 2 = 8 Example: 4.14 𝝅 𝒔𝒊𝒏𝜽 Evaluate ∫𝟎 ∫𝟎 𝒓𝒅𝒓𝒅𝜽 Solution: 𝝅 𝒔𝒊𝒏𝜽 Given ∫𝟎 ∫𝟎 𝒓𝒅𝒓𝒅𝜽 𝜋 𝑟 2 𝑠𝑖𝑛𝜃 = ∫0 [ 2 ] 𝑑𝜃 0 𝜋 𝑠𝑖𝑛 2 𝜃 = ∫0 𝑑𝜃 2 1 𝜋 1−𝑐𝑜𝑠2𝜃 = 2 ∫0 [ 2 ] 𝑑𝜃 1 𝑠𝑖𝑛2𝜃 𝜋 = 4 [𝜃 − ] 2 0 1 = 4 [(𝜋 − 0) − (0 − 0)] 𝜋 = 4 Example: 4.15 𝝅 𝒂 Evaluate ∫𝟎 ∫𝟎 𝒓𝒅𝒓𝒅𝜽 Solution: 𝜋 𝑎 Given ∫0 ∫0 𝑟𝑑𝑟𝑑𝜃 𝜋 𝑟2 𝑎 = ∫0 [ 2 ] 𝑑𝜃 0 10  𝜋 𝑎2 = ∫0 𝑑𝜃 2 𝑎2 = [𝜃]𝜋0 2 𝜋𝑎 2 = 2 Example: 4.16 𝝅⁄𝟐 𝟐𝒄𝒐𝒔𝜽 Evaluate ∫−𝝅⁄𝟐 ∫𝟎 𝒓𝟐 𝒅𝜽𝒅𝒓 Solution: 𝜋⁄2 2𝑐𝑜𝑠𝜃 Given ∫−𝜋⁄2 ∫0 𝑟 2 𝑑𝜃𝑑𝑟 𝜋⁄2 2𝑐𝑜𝑠𝜃 =∫−𝜋⁄2 ∫0 𝑟 2 𝑑𝑟𝑑𝜃 (correct form) 2𝑐𝑜𝑠𝜃 𝜋⁄2 𝑟3 =∫−𝜋⁄2 [ 3 ] 𝑑𝜃 0 𝜋⁄2 (2𝑐𝑜𝑠𝜃) 3 =∫−𝜋⁄2 [ − 0] 𝑑𝜃 3 8 𝜋⁄2 = ∫ 𝑐𝑜𝑠 3 𝜃 𝑑𝜃 3 −𝜋⁄2 8 𝜋⁄2 = 3 (2) ∫0 𝑐𝑜𝑠 3 𝜃 𝑑𝜃 16 2 32 = [. 1] = 3 3 9 Example: 4.17 𝝅⁄𝟐 𝒂 Evaluate ∫𝟎 ∫𝒂(𝟏−𝒄𝒐𝒔𝜽) 𝒓𝟐 𝒅𝜽𝒅𝒓 Solution: 𝜋⁄2 𝑎 Given ∫0 ∫𝑎(1−𝑐𝑜𝑠𝜃) 𝑟 2 𝑑𝜃𝑑𝑟 𝜋⁄2 𝑟 3 𝑎 = ∫0 [ ] 𝑑𝜃 3 𝑎(1−𝑐𝑜𝑠𝜃) 𝜋⁄2 𝑎 3 𝑎 3 (1−𝑐𝑜𝑠𝜃)3 = ∫0 [ − ] 𝑑𝜃 3 3 𝑎3 𝜋⁄2 = ∫0 [1 − (1 − 𝑐𝑜𝑠𝜃)3 ]d𝜃 3 𝑎3 𝜋⁄2 = ∫0 [1 − (1 − 3𝑐𝑜𝑠𝜃 + 3𝑐𝑜𝑠 2 𝜃 − 𝑐𝑜𝑠 3 𝜃)]d𝜃 3 𝑎3 𝜋⁄2 = ∫0 [3𝑐𝑜𝑠𝜃 + 3𝑐𝑜𝑠 2 𝜃 − 𝑐𝑜𝑠 3 𝜃)]𝑑𝜃 3 𝑎3 1𝜋 2 = [(3𝑠𝑖𝑛𝜃)𝜋0 ⁄2 − 3 + ] 3 22 3 𝑎3 𝜋 2 = [3 − 3 + ] 3 2 3 11  𝑎 3 36−9𝜋+8 = [ ] 3 12 𝑎3 = [44 − 9π] 36 Exercise 4.3 Evaluate the following integrals 𝜋⁄2 𝑎 16 𝑎 5 1. ∫0 ∫𝑎𝑐𝑜𝑠𝜃 𝑟 4 𝑑𝑟𝑑𝜃 Ans: (𝜋 − 15) 10 2𝜋 𝑎 𝜋𝑎 2 2. ∫0 ∫𝑎𝑠𝑖𝑛𝜃 𝑟𝑑𝑟𝑑𝜃 Ans: 4 𝜋⁄4 𝑐𝑜𝑠2𝜃 𝜋 3. ∫−𝜋 ⁄4 ∫0 𝑟𝑑𝑟𝑑𝜃 Ans: 8 𝜋⁄2 𝑎𝑐𝑜𝑠𝜃 𝑎3 4. ∫0 ∫0 𝑟√𝑎2 − 𝑟 2 𝑑𝑟𝑑𝜃 Ans: 18 (3𝜋 − 4) 𝜋⁄4 𝑎𝑠𝑖𝑛𝜃 𝑟 𝑎(𝜋−3) 5. ∫0 ∫0 𝑑𝑟𝑑𝜃 Ans: √𝑎 2 −𝑟 2 6 𝑏 𝜋⁄ 3𝜋𝑏2 6. ∫𝑏⁄ ∫0 2 𝑟𝑑𝜃𝑑𝑟 Ans: 16 2 4.4 Change of order of integration Change of order of integration is done to make the evaluation of integral easier The following are very important when the change of order of integration takes place 1. If the limits of the inner integral is a function of x (or function of y) then the first integration should be with respect to y (or with respect to x ) 2. Draw the region of integration by using the given limits 3. If the integration is first with respect to x keeping y as a constant then consider the horizontal strip and find the new limits accordingly 4. If the integration is first with respect to y keeping x a constant then consider the vertical strip and find the new limits accordingly 5. After find the new limits evaluate the inner integral first and then the outer integral Problems Example:4.18 𝒂 𝒂 Change the order of integration in ∫𝟎 ∫𝒙 𝒇(𝒙, 𝒚)𝒅𝒚 𝒅𝒙 Solution: Given 𝑦: 𝑥 → 𝑎 𝑥: 0 → 𝑎 The region is bounded by 𝑦 = 𝑥, 𝑦 = 𝑎, 𝑥 = 0 and 𝑥 = 𝑎 12 𝑥 axis limit represents the horizontal strip and 𝑦 axis limit represents vertical 𝑥: 0 → 𝑦 𝑦: 0 → 𝑎 By changing the order we get 𝑎 𝑦 ∫ ∫ 𝑓 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦 0 0 Example: 4.19 𝟏 𝒙 Change the order of integration ∫𝟎 ∫𝟎 𝒇(𝒙, 𝒚)𝒅𝒚 𝒅𝒙 Solution: Given 𝑦: 0 → 𝑥 𝑥: 0 → 1 The region is bounded by 𝑦 = 0, 𝑦 = 𝑥, 𝑥 = 0, 𝑥 = 1 𝑥: 𝑦 → 1 𝑦: 0 → 1 13  By changing the order we get 1 1 ∫ ∫ 𝑓 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦 0 𝑦 Example: 4.20 𝒂 𝒂 Change the order of integration and hence evaluate it ∫𝟎 ∫𝒙 (𝒙𝟐 +𝒚𝟐 )𝒅𝒚 𝒅𝒙 Solution: It is correct form, given order is 𝑑𝑦𝑑𝑥 given 𝑦: 𝑥 → 𝑎 𝑥: 0 → 𝑎 the region is bounded by 𝑦 = 𝑥, 𝑦 = 𝑎 𝑥 = 0 and 𝑥 = 𝑎 𝑥 axis limit represent the horizontal strip 𝑦 axis limit represents vertical path changed order is 𝑑𝑥 𝑑𝑦 𝑥: 0 → 𝑦 𝑦: 0 → 𝑎 𝑎 𝑦 𝑎 𝑥3 𝑦 ∫0 ∫0 (𝑥 2 +𝑦 2 )𝑑𝑥 𝑑𝑦 = ∫0 [ 3 + 𝑦 2 𝑥] dy 0 𝑎 𝑦3 = ∫0 [ 3 + 𝑦 3 ] dy 𝑎 𝑦4 𝑦4 𝑎4 𝑎4 𝑎4 = [12 + ] = + = 4 0 12 4 3 Example: 4.21 𝟒𝒂 𝟐√𝒂𝒙 Change the order of integration for∫𝟎 ∫𝒙𝟐 ⁄𝟒𝒂 𝒙𝒚 𝒅𝒚 𝒅𝒙 Solution: It is correct form Given order is 𝑑𝑦𝑑𝑥 𝑥2 Given 𝑦: 4𝑎 → 2√𝑎𝑥 14  𝑥: 0→ 4𝑎 The region is bounded by 𝑥 2 = 4𝑎𝑦 , 𝑦 2 = 4a𝑥 𝑥 = 0 and 𝑥 = 4𝑎 Changed order is 𝑑𝑥𝑑𝑦 draw a horizontal strip 𝑦2 𝑥: 4𝑎 → 2√𝑎𝑦 𝑦 ∶ 0 → 4𝑎 2 𝑎𝑦 4𝑎 2√𝑎𝑦 4𝑎 𝑥 2𝑦 √ ∫0 ∫𝑦 2⁄ 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫0 [ ]𝑦2 𝑑𝑦 4𝑎 2 4𝑎 4𝑎 (2√𝑎𝑦)2𝑦 𝑦2 𝑦 = ∫0 { − [4𝑎]2 2 } 𝑑𝑦 2 4𝑎 4𝑎𝑦 𝑦5 = ∫0 [( ) 𝑦 − 32𝑎2 ] 𝑑𝑦 2 4𝑎 4𝑎𝑦 3 𝑦6 =[ − 192𝑎2 ] 6 0 4𝑎(4𝑎)3 (4𝑎)6 = − 192𝑎2 6 128𝑎 4 4096 = − 𝑎4 3 192 64𝑎 4 = 3 Example: 4.22 𝒂 𝒃 √𝒃𝟐 −𝒚𝟐 Change the order of integration of ∫𝟎 ∫𝟎𝒃 𝒙𝒚𝒅𝒙𝒅𝒚 and hence evaluate it Solution: It is correct form Given order is 𝑑𝑥𝑑𝑦 𝑎 Given 𝑥 ∶ 0→ 𝑏 √𝑏2 − 𝑦 2 𝑦∶ 0 → 𝑏 15  𝑎 𝑥2 𝑦2 The region is bounded by 𝑥 = 0, 𝑥 = 𝑏 √𝑏2 − 𝑦 2 ⇒ 𝑎2 + 𝑏2 = 1 𝑦 = 0, 𝑦 = 𝑏 Changed order is 𝑑𝑦𝑑𝑥 Draw the vertical strip 𝑏 𝑦 ∶ 0 → 𝑎 √𝑎2 − 𝑥 2 𝑥∶ 0→ 𝑎 𝑏 𝑏 √𝑎 2 −𝑥 2 √𝑎 2−𝑥 2 𝑎 𝑎 𝑥𝑦 2 𝑎 ∫0 ∫0𝑎 𝑥𝑦 𝑑𝑦 𝑑𝑥 = ∫0 [ 2 ] 𝑑𝑥 0 𝑏 2 2 2 𝑎 [𝑎√𝑎 −𝑥 ] 𝑥 = ∫0 𝑑𝑥 2 𝑏2 𝑎 = 2𝑎2 ∫0 𝑥(𝑎2 − 𝑥 2 ) 𝑑𝑥 𝑏2 𝑎 = 2𝑎2 ∫0 (𝑥𝑎2 − 𝑥 3 )dx 𝑎 𝑏2 𝑎 2𝑥 2 𝑥4 = 2𝑎2 [ − ] 2 4 0 𝑏2 𝑎4 𝑎4 = 2𝑎2 [ 2 − ] 4 𝑎4 𝑎4 𝑎2 𝑎4 = 𝑏2 [2𝑎2 − ] = 𝑏2 [ − ] 4 2 4 𝑎 2𝑏 2 = 8 Example: 4.23 𝟏 𝟐−𝒙 Change the order of integration and hence evaluate ∫𝟎 ∫𝒙𝟐 𝒙𝒚 𝒅𝒚 𝒅𝒙 Solution: It is correct form Given order is 𝑑𝑦𝑑𝑥 16  Given 𝑦 ∶ 𝑥 2 → 2 − 𝑥 𝑥∶ 0 →1 The region is bounded by = 𝑥 2 , 𝑦 + 𝑥 = 2 𝑥 = 0, 𝑥 = 1 Now divide the region in to two parts i.e. R1 and R2 Changed order is 𝑑𝑥𝑑𝑦 Draw horizontal strip For Region R1 Limits are 𝑥: 0 → √𝑦 𝑦: 0 → 1 1 √𝑦 1 𝑥 2 𝑦 √𝑦 ∫0 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫0 [ 2 ] 𝑑𝑦 0 1 (√𝑦)2𝑦 = ∫0 𝑑𝑦 2 1 𝑦2 = ∫0 dy 2 1 𝑦3 = 0 = 1/6 For region R2 Limits are 𝑥 ∶ 0 → 2 − 𝑦 𝑦∶ 1→2 2 2−𝑦 2 2−𝑦 𝑥 2𝑦 ∫ ∫ 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫ [ ] 𝑑𝑦 1 0 1 2 0 2 (2−𝑦)2𝑦 =∫1 𝑑𝑦 2 17  2 (4−4𝑦+𝑦 2 )𝑦 = ∫1 𝑑𝑦 2 2 1 4𝑦 2 4𝑦 3 𝑦4 = [ − + ] 2 2 3 4 1 1 32 4 1 = 2 [8 − + 4 − 2 + 3 − 4] 3 5 = 24 R = R1 + R2 1 5 = 6 + 24 3 =8 Example: 4.24 𝟏 𝟐−𝒚 Change the order of integration in ∫𝟎 ∫𝒚 𝒙𝒚 𝒅𝒙 𝒅𝒚 and hence evaluates Solution: It is correct form 𝑥 ∶ 𝑦 → 2−𝑦 𝑦: 0 → 1 The region is bounded by 𝑥 = 𝑦 , 𝑥 + 𝑦 = 2 𝑦 = 0 , 𝑦 = 1 Now divide the region into two parts ie. R1 and R2 Changed order is 𝑑𝑦𝑑𝑥 Draw horizontal strip For region R1 Limits are 𝑥: 0→ 1 18  𝑦: 0 → 𝑥 1 2−𝑦 1 𝑥 ∫0 ∫𝑦 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫0 ∫0 𝑥𝑦 𝑑𝑦 𝑑𝑥 1 𝑥𝑦 2 𝑥 = ∫0 [ ] 𝑑𝑥 2 0 1 𝑥3 = ∫0 [3−0] 𝑑𝑥 1 1 1 1 𝑥4 = ∫0 𝑥 3 𝑑𝑥 = [ ] 2 2 4 0 1 1 = 8 [𝑥 4 ]10 = [ 1 − 0] 8 1 =8 For region R2 𝑥: 1 → 2 𝑦: 0 → 2 − 𝑥 2 2−𝑥 2 𝑥𝑦 2 2−𝑥 ∫1 ∫0 𝑥𝑦 𝑑𝑦 𝑑𝑥 = ∫1 [ ] 𝑑𝑥 2 0 2 𝑥 (2−𝑥)2 = ∫1 [ − 0] 𝑑𝑥 2 1 2 𝑥(4+𝑥 2− 4𝑥) = 2 ∫1 𝑑𝑥 2 1 2 = 2 ∫1 (4𝑥 + 𝑥 3 − 4𝑥 2 ) 𝑑𝑥 2 1 𝑥2 𝑥4 𝑥3 = 2 [4 + −4 ] 2 4 3 1 2 1 𝑥4 𝑥3 = [2𝑥 2 + −4 ] 2 4 3 1 1 16 4 1 4 = [(8 + − (8)) − (2 + − )] 2 4 3 4 3 1 32 1 4 = 2 [8 + 4 − − 2 − 4 + 3] 3 1 5 5 = 2 = 24 ⇒ R = R1 + R2 1 5 = 8 + 24 1 =3 Example: 4.25 ∞ ∞ 𝒆−𝒚 Change the order of integration ∫𝟎 ∫𝒙 𝒅𝒚 𝒅𝒙 and hence evaluate it 𝒚 Solution: 19  It is correct form Given order is 𝑑𝑦 𝑑𝑥 Given 𝑦 ∶ 𝑥 → ∞ 𝑥∶ 0→ ∞ Changed order is 𝑑𝑥𝑑𝑦 Draw a horizontal strip 𝑥∶ 0→𝑦 𝑦∶ 0→ ∞ ∞ 𝑦 𝑒 −𝑦 ∞ 𝑥 𝑦 ∫0 ∫0 𝑑𝑥 𝑑𝑦 = ∫𝑜 [𝑒 −𝑦 𝑦 ] 𝑑𝑦 𝑦 0 ∞ 𝑒 −𝑦 ∞ = ∫𝑜 𝑒 −𝑦 𝑑𝑦 = [ ] 1 0 = −[𝑒 −∞ − 𝑒 0 ] = 1 Example: 4.26 𝒂 𝒂 𝒙 Change the order of integration I = ∫𝟎 ∫𝒚 𝒅𝒙 𝒅𝒚 and the evaluate it 𝒙𝟐 +𝒚𝟐 Solution: It is correct form Given order 𝑑𝑥𝑑𝑦 𝑥∶ 𝑦→𝑎 𝑦∶ 0→ 𝑎 The region is bounded by 𝑥 = 𝑦 , 𝑥 = 𝑎 𝑦 = 0,𝑦 = 𝑎 20  Changed order is 𝑑𝑦 𝑑𝑥 Draw a vertical strip 𝑦∶ 0→𝑥 𝑥∶ 0→ 𝑎 𝑥 𝑎 𝑥 𝑥 𝑎 𝑡𝑎𝑛 −1 𝑦 ∫0 ∫0 dy dx = ∫0 𝑥 [ (𝑥 )] 𝑑𝑥 𝑥 2 +𝑦 2 𝑥 0 𝑎 𝑥 = ∫0 [𝑡𝑎𝑛−1 (𝑥) − 𝑡𝑎𝑛−1 0] 𝑑𝑥 𝑎𝜋 = ∫0 𝑑𝑥 4 𝜋 𝑎 = [ 4 𝑥] 0 𝜋 = 𝑎 4 Example: 4.27 𝒂 √𝒂𝟐 −𝒙𝟐 Evaluate ∫𝟎 ∫𝟎 𝒙𝒚 𝒅𝒚 𝒅𝒙 by changing the order of integration Solution: It is correct form Given order 𝑑𝑦 𝑑𝑥 Given 𝑦 ∶ 0 → √𝑎2 − 𝑥 2 𝑥∶ 0→ 𝑎 the region is bounded by 𝑦 = 0, 𝑦 = √𝑎2 − 𝑥 2 𝑥= 0 , 𝑥=𝑎 21  changed order 𝑑𝑥𝑑𝑦 Draw horizontal strip 𝑥 ∶ 0 → √𝑎2 − 𝑦 2 𝑦∶ 0→ 𝑎 𝑎 √𝑎 2 −𝑥 2 𝑎 √𝑎 2 −𝑦 2 ∫0 ∫0 𝑥𝑦 𝑑𝑦 𝑑𝑥 =∫0 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦 √𝑎 2 −𝑦 2 𝑎 𝑥2 = [ ∫0 𝑦 2 ] 𝑑𝑦 0 1 𝑎 = 2 ∫0 𝑦(𝑎2 -𝑦 2 ) 𝑑𝑦 𝑎 1 𝑎2 𝑦 2 𝑦4 =2 [ − ] 2 4 0 1 𝑎2 𝑎4 = 2[ 2 − ] 4 𝑎4 = 8 Exercise: 4.4 Change the order of integration and hence evaluate the following 𝑎 √𝑎 2 −𝑥 2 𝜋𝑎 4 1.∫−𝑎 ∫0 (𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑑𝑥 Ans: 4 𝑎 2√𝑎𝑥 4 2. ∫0 ∫0 𝑥 2 𝑑𝑦 𝑑𝑥 Ans: 7 𝑎4 4 2√𝑥 16 3. ∫0 ∫𝑥2 𝑑𝑦 𝑑𝑥 Ans: 3 4 𝑎 𝑎+√𝑎 2 −𝑦 2 1 4. ∫0 ∫𝑎−√𝑎2 −𝑦 2 𝑑𝑥 𝑑𝑦 Ans: 2 ∞ 𝑦 𝑦2 1 5. ∫0 ∫0 𝑦𝑒 − 𝑥 Ans: 2 1 2−𝑦 1 6. ∫0 ∫𝑦 𝑥𝑦 𝑑𝑥𝑑𝑦 Ans: 3 22  1 2−𝑥 𝑥 7. ∫0 ∫𝑦 𝑑𝑦𝑑𝑥 Ans: log 4 − 1 𝑦 3 6⁄ 𝑥 8. ∫1 ∫0 𝑥 2 𝑑𝑦 𝑑𝑥 Ans: 24 𝑎 𝑎 𝑥 𝑎2 9. ∫0 ∫𝑦 𝑑𝑥 𝑑𝑦 Ans: log(1 + √2) √𝑥 2+𝑦 2 2 4 2√𝑦 28 10. ∫1 ∫2⁄ 𝑑𝑥 𝑑𝑦 Ans: −2 log 4 𝑦 3 4.5 Area enclosed by plane curves (Cartesian coordinates) Area = ∫ ∫ 𝒅𝒚 𝒅𝒙 (or) Area = ∫ ∫ 𝒅𝒙𝒅𝒚 Example: 4.28 Find the area enclosed by the curves y=2𝒙𝟐 and 𝒚𝟐 = 𝟒𝒙 Solution: Area = ∫ ∫ 𝑑𝑦 𝑑𝑥 𝑦 ∶ 2𝑥 2 → 2√𝑥 𝑥∶ 0→ 1 1 2 √𝑥 Area = ∫0 ∫2𝑥 2 𝑑𝑦 𝑑𝑥 1 = ∫0 [𝑦]22𝑥√𝑥2 𝑑𝑥 1 = ∫0 (2√𝑥 − 2𝑥 2 )𝑑𝑥 3 1 2𝑥 ⁄2 2𝑥 3 =[ − ] 3/2 3 0 3 1 4𝑥 ⁄2 2𝑥 3 =[ − ] 3 3 0 4 2 2 =3−3 =3 23 Example: 4.29 Find the area between the parabola y2 =4ax and x2 = 4ay Solution: Area =∫ ∫ 𝑑𝑦 𝑑𝑥 𝑥2 𝑦 ∶ 4𝑎 → 2√𝑎𝑥 𝑥 ∶ 0 → 4𝑎 4𝑎 2√𝑎𝑥 = ∫0 ∫𝑥2 𝑑𝑦𝑑𝑥 4𝑎 4𝑎 = ∫0 [𝑦]2𝑥2√𝑎𝑥 𝑑𝑥 4𝑎 4𝑎 𝑥2 = ∫0 (2 √𝑎𝑥 − 4𝑎) 𝑑𝑥 4𝑎 3 2√𝑎 𝑥 ⁄2 𝑥3 = [ 3 − ] ⁄2 12𝑎 0 4 3⁄ (4𝑎)3 = 3 √𝑎 (4a) 2 − 12𝑎 32𝑎 2 16𝑎 2 = − 3 3 16𝑎 2 = 3 Example: 4.30 𝐱𝟐 𝐲𝟐 Find the area of ellipse 𝐚𝟐 + =1 𝐛𝟐 Solution: 24 Area =4 ∬ dx dy a 𝑥 ∶ 0 → b √b 2 − y 2 𝑦 ∶ 0 → 𝑎b a 𝑏 √b2 −y2 Area = 4∫0 ∫0b 𝑑𝑦 𝑑𝑥 a b √b2−y2 = 4∫0 [x]b0 dy 𝑏 a = 4 ∫0 [b √b 2 − y 2 − 0] 𝑑𝑦 𝑏 4𝑎 𝑏2 𝑦 𝑦 = [ 𝑠𝑖𝑛−1 ( ) + √𝑏2 − 𝑦 2 ] 𝑏 2 𝑏 2 0 4𝑎 𝑏2 𝜋 = [( 2 + 0) − 0] 𝑏 2 4𝑎𝑏 𝑏 2 𝜋 = 𝑏 2 2 = 𝜋𝑎𝑏 Example: 4.31 Evaluate ∬ 𝐱𝐲 𝐝𝐱 𝐝𝐲 over the positive quadrant of the circle 𝐱 𝟐 + 𝐲 𝟐 = 𝟏 Solution: 𝑥 ∶ 0 → √1 − y 2 𝑦∶ 0→ 1 25  1 √1−y2 ∬ xy dx dy = ∫0 ∫0 xy dx dy 2 1 𝑥 2𝑦 √1−𝑦 = ∫0 [ 2 ] 𝑑𝑦 0 1 1 = 2 ∫0 (√1 − 𝑦 2 )2 𝑦 𝑑𝑦 1 1 = 2 ∫0 (1 − 𝑦 2 ) 𝑦 𝑑𝑦 1 1 = 2 ∫0 (𝑦 − 𝑦 3 ) 𝑑𝑦 1 1 𝑦2 𝑦4 = 2[2 − ] 4 0 1 1 1 1 1 1 = 2 [2 − 4] = [] = 2 4 8 Example: 4.32 Find the smaller of the area bounded by 𝒚 = 𝟐 − 𝒙 and 𝐱 𝟐 + 𝐲 𝟐 = 𝟒 Solution: Area= ∬ dy dx 𝑦 ∶ 2 − 𝑥 → √4 − x 2 𝑥∶ 0→ 2 2 √4−x2 = ∫0 ∫2−x dy dx 2 2 = ∫0 [y]√4−x 2−x dx 2 = ∫0 [√4 − x 2 − (2 − x)]dx 2 x 22 x x2 = [ √4 − x 2 + sin−1 ( ) − 2x + ] 2 2 2 2 0 4 π 42 = 0 + 2 (2) − 4 + 2 = π − 2 square unit 26 Example: 4.33 Evaluate ∬ 𝐱𝐲 𝐝𝐱𝐝𝐲 over the positive quadrant for which 𝒙 + 𝒚 ≤ 𝟏 Solution: 𝑥 ∶ 0 → 1−𝑦 𝑦∶ 0→ 1 1 1−y ∬ xy dxdy = ∫ ∫ xy dx dy 0 0 1 𝑥 2𝑦 1−𝑦 = ∫0 ( ) 𝑑𝑦 2 0 1 1−𝑦 2 𝑦 = ∫0 𝑑𝑦 2 1 1 = 2 ∫0 (𝑦 2 − 2𝑦 2 + 𝑦 3 )𝑑𝑦 1 1 𝑦2 2𝑦 3 𝑦4 =2 [ 2 − + ] 3 4 0 1 1 2 1 1 16−8+3 1 = 2 [2 − 3 + 4] = 2 [ ] = 12 24 Example: 4.34 Using double integral find the area bounded by 𝐲 = 𝐱 and 𝐲 = 𝐱2 Solution: 27 Area = ∬ dy dx 𝑦 ∶ 𝑥2 → 𝑥 𝑥∶ 0→ 1 1 x = ∫0 ∫x2 dy dx 1 = ∫0 [y]xx2 dx 1 = ∫0 (x − x 2 )dx 1 x2 x3 = [2 + 3] 0 1 1 1 =2−3 =6 Example: 5.35 Evaluate ∬(𝐱 𝟐 + 𝐲 𝟐 )𝐝𝐱𝐝𝐲 where A is area bounded by the curves 𝐱 𝟐 =y, x=1 and x=2 about x axis Solution: 𝑦 ∶ 0 → 𝑥2 𝑥∶1→ 2 2 x2 ∬(x 2 + y 2 )dxdy = ∫1 ∫0 (x 2 + y 2 )dydx x2 2 y3 = ∫1 [x 2 y + ] dx 3 0 2 x6 = ∫1 ( x 4 + 3 )dx 2 x5 x7 =[ 5 + 21] 1 25 27 1 1 = [ 5 + 21 − 5 − 21] 1286 = 105 28 Example: 4.36 Find the area enclosed by the curves 𝒚 = 𝒙𝟐 and 𝒙 + 𝒚 − 𝟐 = 𝟎 Solution: Given y=𝑥 2 and 𝑥 + 𝑦 − 2 = 0 x 0 1 2 −1 −2 Y=2−𝑥 2 1 0 3 4 x 0 1 −1 2 -2 𝑦 = 𝑥2 0 1 1 4 4 Area = ∬ dy dx 𝑦 ∶ 𝑥2 → 2 − 𝑥 𝑥 ∶ −2 → 1 1 2−x 1 ∫−2 ∫x2 dy dx = ∫−2[y]2−x x2 dx 1 = ∫−2(2 − 𝑥 − 𝑥 2 )𝑑𝑥 1 𝑥2 𝑥3 = [2𝑥 − − ] 2 3 −2 1 1 4 8 27 = [2 − 2 − 3] –[−4 − 2 + 3] = 6 Example: 4.37 Find by double integration the area lying between the parabola y=4x-𝐱 𝟐 and the line y=x Solution: Given 𝑦 = 4𝑥 − 𝑥 2 and 𝑦 = 𝑥 29 x 0 1 2 −1 −2 3 𝑦 = 4𝑥 − 𝑥 2 0 3 4 −5 −12 3 Area= ∬ 𝑑𝑦 𝑑𝑥 y : 𝑥 → 4𝑥 − 𝑥 2 x : 0→ 3 3 4𝑥−𝑥 2 3 2 ∫0 ∫𝑥 𝑑𝑦 𝑑𝑥 = ∫0 [𝑦]4𝑥−𝑥 𝑥 𝑑𝑥 3 = ∫0 (4𝑥 − 𝑥 2 − 𝑥 )𝑑𝑥 3 = ∫0 (3𝑥 − 𝑥 2 )𝑑𝑥 3 3𝑥 2 𝑥3 27 27 9 =[ − ] = − =2 2 3 0 2 3 Exercise: 4.5 1. Evaluate ∬ xdy dx over the region between the parabola y 2 = x and the lines 𝑥 + 𝑦 = 2, 4 𝑥 = 0 and 𝑥 = 1 Ans: 15 x2 y2 𝜋𝑎𝑏3 2. Evaluate ∬ y 2 dxdy over the area of ellipse a2 + b2 =1 Ans: 4 16√2 3. Find the area between the curve 𝑦 2 = 4 − 𝑥 and the line 𝑦 2 = 𝑥 Ans: 3 4.6Area Enclosed by Plane Curves [Polar co-ordinates] Area= ∬ 𝐫𝐝𝐫𝐝𝛉 Problems Based on Area Enclosed by Plane Curves Polar Coordinates Example: 4.38 Find using double integral, the area of the cardioid 𝐫 = 𝐚(𝟏 + 𝐜𝐨𝐬𝛉) [A.U 2011][A.U 2014][A.U 2015] Solution: 30  Area= ∬ 𝐫𝐝𝐫𝐝𝛉 The curve is symmetrical about the initial line θ varies from ∶ 0 → π r varies from: 0 → a(1 + cosθ) θ=π r=a(1+cosθ) Hence, required area = 2 ∫θ=0 ∫r=0 rdrdθ π r2 r=a(1+cosθ) π r=a(1+cosθ) = 2 ∫0 [ 2 ] dθ = ∫0 [r 2 ]r=0 dθ r=0 π π = ∫0 [a2 (1 + cosθ)2 − 0]dθ = a2 ∫0 [1 + cos 2 θ + 2cos θ]dθ π 1+cos2θ = a2 ∫0 [1 + + 2cosθ] dθ 2 a2 π a2 π = 2 ∫0 [2 + 1 + cos2θ + 4cosθ] dθ = 2 ∫0 [3 + cos2θ + 4cosθ] dθ a2 sin2θ π a2 = [3θ + + 4sinθ] = [(3π + 0 + 0) − (0 + 0 + 0)] 2 2 0 2 3 = 2 a2 π square units. Example: 4.39 Find the area of the cardioid 𝐫 = 𝐚(𝟏 − 𝐜𝐨𝐬𝛉) Solution: Area= ∬ 𝐫𝐝𝐫𝐝𝛉 The curve is symmetrical about the initial line. 31 θ varies from ∶ 0 → π r varies from: 0 → a(1 − cosθ) θ=π r=a(1−cosθ) Hence, required area = 2 ∫θ=0 ∫r=0 rdrdθ π r2 r=a(1−cosθ) π r=a(1−cosθ) = 2 ∫0 [ 2 ] dθ = ∫0 [r 2 ]r=0 dθ r=0 π π = ∫0 [a2 (1 − cosθ)2 − 0]dθ = a2 ∫0 [1 + cos 2 θ − 2cos θ]dθ π 1+cos2θ = a2 ∫0 [1 + − 2cosθ] dθ 2 a2 π a2 π = 2 ∫0 [2 + 1 + cos2θ − 4cosθ] dθ = 2 ∫0 [3 + cos2θ − 4cosθ] dθ a2 sin2θ π a2 = [3θ + − 4sinθ] = [(3π + 0 − 0) − (0 + 0 − 0)] 2 2 0 2 3 = 2 a2 π square units. Example: 4.40 Find the area of a circle of radius ‘a’ by double integration in polar co-ordinates. Solution: Area= ∬ 𝐫𝐝𝐫𝐝𝛉 The equation of circle with pole on the circle and diameter through the point as initial line is r = 2acosθ Area = 2 × upper area π θ= r=2acosθ = 2 ∫θ=02 ∫r=0 rdrdθ π = ∫02 [r 2 ]2acosθ 0 dθ π 1 π = 4a ∫0 cos 2 θdθ 2 2 = 4a2. 2. 2 = πa2 square units. 32 Example: 4.41 Find the area of the lemniscates 𝐫 𝟐 = 𝐚𝟐 𝐜𝐨𝐬𝟐𝛉 by double integration. [A.U R-08] Solution: Area= ∬ 𝒓𝒅𝒓𝒅𝜽 Area = 4 × area of upper half of one loop. π a√cos2θ = 4 ∫04 ∫0 rdrdθ π = 2 ∫0 (r 2 )a√cos2θ 4 0 dθ π = 2a2 ∫04 cos2θ dθ π 2 sin2θ 4 = 2a ( ) 2 0 = a2 square units. Example: 4.42 Find the area that lies inside the cardioid 𝐫 = 𝐚(𝟏 + 𝐜𝐨𝐬𝛉) and outside the circle 𝒓 = 𝒂, by double integration. [A.U 2014] Solution: Area= ∬ 𝐫𝐝𝐫𝐝𝛉 Both the curves are symmetric about the initial line. 33  𝜋 𝜃= 𝑟=𝑎(1+𝑐𝑜𝑠𝜃) Hence, the required area = 2 ∫𝜃=02 ∫𝑟=𝑎 𝑟𝑑𝑟𝑑𝜃 𝜋 𝑟=𝑎(1+𝑐𝑜𝑠𝜃) 𝑟2 = 2 ∫02 [ ] 𝑑𝜃 2 𝑟=𝑎 𝜋 𝑎 2 (1+𝑐𝑜𝑠𝜃)2 𝑎2 = 2 ∫02 [ − ] 𝑑𝜃 2 2 𝜋 = 𝑎2 ∫02 [1 + 𝑐𝑜𝑠 2 𝜃 + 2𝑐𝑜𝑠 𝜃 − 1]𝑑𝜃 𝜋 1+𝑐𝑜𝑠2𝜃 = 𝑎2 ∫02 [ + 2𝑐𝑜𝑠𝜃] 𝑑𝜃 2 𝜋 𝑎2 = 2 ∫02 [1 + 𝑐𝑜𝑠2𝜃 + 4𝑐𝑜𝑠𝜃] 𝑑𝜃 𝜋 𝑎2 𝑠𝑖𝑛2𝜃 2 = [𝜃 + + 4𝑠𝑖𝑛𝜃] 2 2 0 𝑎2 𝜋 = [( + 0 + 4) − (0 + 0 + 0)] 2 2 𝑎2 = (𝜋 + 8)square units. 4. Example: 4.43 Find the area inside the circle 𝐫 = 𝐚𝐬𝐢𝐧𝛉 and outside the cardioid 𝒓 = 𝒂(𝟏 − 𝐜𝐨𝐬𝛉) [A.U.Jan.2009] Solution: From the figure, we get π θ varies from ∶ 0 → 2 r varies from: a(1 − cosθ) → asinθ π θ= r=asinθ) The required area = ∫θ=02 ∫r=a(1−cosθ) rdrdθ 𝜋 𝑟=𝑎𝑠𝑖𝑛𝜃) 𝑟2 = ∫02 [ 2 ] 𝑑𝜃 𝑟=𝑎(1−𝑐𝑜𝑠𝜃) 𝜋 𝑎 2 𝑠𝑖𝑛 2 𝜃 𝑎 2 (1− 𝑐𝑜𝑠𝜃)2 = ∫02 [ − ] 𝑑𝜃 2 2 34  𝜋 𝑎2 = 2 ∫02[𝑠𝑖𝑛2 𝜃 − 1 − 𝑐𝑜𝑠 2 𝜃 + 2𝑐𝑜𝑠𝜃] 𝑑𝜃 𝜋 𝜋 𝜋 𝜋 𝑎2 2 2 = [∫0 𝑠𝑖𝑛 𝜃𝑑𝜃 − ∫0 𝑑𝜃 − ∫0 𝑐𝑜𝑠 𝜃𝑑𝜃 + 2 ∫0 𝑐𝑜𝑠𝜃𝑑𝜃 ] 2 2 2 2 2 𝜋 𝜋 𝜋 π 𝑎2 = [−[𝜃]02 + 2[𝑠𝑖𝑛𝜃]02 ] ∴ ∫02 𝑠𝑖𝑛2 𝜃𝑑𝜃 = ∫02 cos 2 θdθ 2 𝑎2 𝜋 1𝜋 = [− ( − 0) + 2(1 − 0)] =22 2 2 𝑎2 𝜋 = [2 − ] 2 2 𝑎2 = (4 − 𝜋) square units. 4 Example: 4.44 Evaluate ∫𝑹 ∫ 𝒓𝟐 𝐬𝐢𝐧𝛉 𝐝𝐫𝐝𝛉 where R is the semi circle 𝐫 = 𝟐𝐚𝐜𝐨𝐬𝛉 above the initial line. Solution: π θ varies from ∶ 0 → 2 r varies from: 0 → 2acosθ Let I = ∫ ∫ 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝑟𝑑𝜃 𝜋 2𝑎𝑐𝑜𝑠𝜃 = ∫02 ∫0 (𝑟 2 𝑠𝑖𝑛𝜃)𝑑𝑟𝑑𝜃 𝜋 𝑟=2𝑎𝑐𝑜𝑠𝜃) 𝑟3 = ∫02 [ 3 𝑠𝑖𝑛𝜃] 𝑑𝜃 𝑟=0 𝜋 8𝑎 3 = ∫02 [ 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛𝜃 − 0] 𝑑𝜃 3 𝜋 8𝑎 3 = ∫02 3 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛𝜃 𝑑𝜃 35  𝜋 8𝑎 3 = 3 ∫02 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛𝜃 𝑑𝜃 𝜋 𝜋 8𝑎 3 3 8𝑎 3 𝑐𝑜𝑠 4𝜃 2 = ∫0 𝑐𝑜𝑠 𝜃 𝑑(𝑐𝑜𝑠𝜃) = 2 [ ] 3 3 4 0 8𝑎 3 1 −2𝑎 3 = [0 − ] = 3 4 3 Example: 4.45 Evaluate ∫ ∫ 𝐫√𝐚𝟐 − 𝐫 𝟐 𝐝𝐫𝐝𝛉 over the upper half of the circle 𝐫 = 𝐚𝐜𝐨𝐬𝛉. Solution: 𝜋 𝜃 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 ∶ 0 → 2 𝑟 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚: 0 → 𝑎𝑐𝑜𝑠𝜃 Let I = ∫ ∫ 𝑟√𝑎2 − 𝑟 2 𝑑𝑟𝑑𝜃 𝜋 𝑎𝑐𝑜𝑠𝜃 = ∫02 ∫0 𝑟√𝑎2 − 𝑟 2 𝑑𝑟𝑑𝜃 Put 𝑎2 − 𝑟 2 = 𝑡 2 -2rdr = 2tdt -r dr = t dt r→ 0 => t → 𝑎 r→ 𝑎 𝑐𝑜𝑠𝜃 => t → 𝑎𝑠𝑖𝑛𝜃 𝜋 𝑎𝑠𝑖𝑛𝜃 = ∫02 ∫0 √𝑡 2 (−𝑡𝑑𝑡)𝑑𝜃 𝜋 𝜋 𝑟=𝑎𝑠𝑖𝑛𝜃) 𝑎𝑠𝑖𝑛𝜃 2 𝑡3 = − ∫02 ∫0 𝑡 𝑑𝑡𝑑𝜃 = = − ∫02 [ 3 ] 𝑑𝜃 𝑟=𝑎 𝜋 1 = − 3 ∫02 [𝑡 3 ]𝑎𝑠𝑖𝑛𝜃 𝑎 𝑑𝜃 36  𝜋 1 = − 3 ∫02[𝑎3 𝑠𝑖𝑛3 𝜃 − 𝑎3 ] 𝑑𝜃 𝜋 𝑎3 = − 3 ∫0 [𝑠𝑖𝑛3 𝜃 − 1] 𝑑𝜃 2 𝜋 𝑎3 = 3 ∫02 [1 − 𝑠𝑖𝑛3 𝜃] 𝑑𝜃 𝜋 𝜋 𝑎3 𝑎3 = 3 [𝜃]02 − 3 ∫02 𝑠𝑖𝑛3 𝜃 𝑑𝜃 𝜋 𝑎3 𝜋 𝑎3 2 2 = [ − 0] − [. 1] [∴ ∫02 𝑠𝑖𝑛3 𝜃𝑑𝜃 =.1] 3 2 3 3 3 𝑎3 𝜋 𝑎3 2 = − 3 2 3 3 𝑎3 𝜋 2 = [ − ] 3 2 3 Exercise: 4.6 1. Evaluate ∬ rsinθ drdθ over the cardioid r = a(1 − cosθ) above the initial line. 4a2 Ans: 3 rdrdθ 2. Evaluate ∬ 2 2 over one loop of the lemniscates r 2 = a2 cos2θ √a +r π Ans: a (2 − 2 ) 3. Find by double integration the area bounded by the circles r = 2sinθ and r = 4𝑠𝑖𝑛𝜃 Ans:3 π πa2 4. Find the area outside r = 2acosθ and inside r = a(1 + cosθ) Ans: 2 4.7 Triple Integrals Triple integration in cartesian co-ordinates is defined over a region R is defined by ∭R f(x, y, z)dxdydz or ∭R f(x, y, z)dV or ∭R f(x, y, z)d(x, y, z). Type I – Problems on Triple Integrals Example: 4.46 𝐚 𝐛 𝐜 Evaluate ∫𝟎 ∫𝟎 ∫𝟎 (𝐱 + 𝐲 + 𝐳)𝐝𝐳𝐝𝐲𝐝𝐱 Solution: c a b c a b z2 ∫0 ∫0 ∫0 (x + y + z)dzdydx = ∫0 ∫0 [xz + yz + 2 ] dydx 0 a b c2 = ∫0 ∫0 (xc + yc + 2 ) dydx b a y2 c c2 y = ∫0 [xcy + 2 + 2 ] dx 0 37  a b2 c bc2 = ∫0 (xbc + + ) dx 2 2 a x2 bc b2 cx bc2x =[ + + ] 2 2 2 0 a2 bc ab2 c abc2 =[ + + ] 2 2 2 abc = (a + b + c) 2 Example: 4.47 𝟏 𝟏 𝟏−𝒙 Evaluate ∫𝟎 ∫𝒚𝟐 ∫𝟎 𝒙𝒅𝒛𝒅𝒙𝒅𝒚 Solution: 1 1 1−x 1 1 ∫0 ∫y2 ∫0 xdzdxdy = ∫0 ∫y2 [xz]1−x 0 dxdy 1 1 = ∫0 ∫y2 x(1 − x) dxdy 1 1 = ∫0 ∫y2 (x − x 2 ) dxdy 1 x2 1 x3 = ∫0 [ 2 − 3 ] 2 dy y 1 1 1 y4 y6 = ∫0 (2 − 3 − + ) dy 2 3 1 y y y5 y7 = [2 − 3 − 10 + 21] 0 1 1 1 1 = 2 − 3 − 10 + 21 105−70−21+10 24 4 = = 210 = 35 210 Example: 4.48 𝟏 𝟏−𝒙 𝒙+𝒚 Evaluate ∫𝟎 ∫𝟎 ∫𝟎 𝒆𝒛 𝒅𝒙𝒅𝒚𝒅𝒛 Solution: 1 1−𝑥 𝑥+𝑦 1 1−𝑥 𝑥+𝑦 ∫0 ∫0 ∫0 𝑒 𝑧 𝑑𝑥𝑑𝑦𝑑𝑧 = ∫0 ∫0 ∫0 𝑒 𝑧 𝑑𝑧𝑑𝑦𝑑𝑥 1 1−𝑥 = ∫0 ∫0 [𝑒 𝑧 ]0𝑥+𝑦 𝑑𝑦𝑑𝑥 1 1−𝑥 = ∫0 ∫0 (𝑒 𝑥+𝑦 − 1) 𝑑𝑦𝑑𝑥 1 = ∫0 [𝑒 𝑥+𝑦 − 𝑦]1−𝑥 0 𝑑𝑥 1 = ∫0 (𝑒 𝑥+1−𝑥 − 1 + 𝑥 − 𝑒 𝑥 ) 𝑑𝑥 1 = ∫0 (𝑒 − 1 + 𝑥 − 𝑒 𝑥 ) 𝑑𝑥 1 𝑥2 = [𝑒𝑥 − 𝑥 + − 𝑒𝑥] 2 0 38  1 = 𝑒−1+2−𝑒−0+0−0+1 1 =2 Example: 4.49 𝒂 √𝒂𝟐 −𝒚𝟐 √𝒂𝟐 −𝒙𝟐 −𝒚𝟐 Evaluate ∫𝟎 ∫𝟎 ∫𝟎 𝒅𝒛𝒅𝒙𝒅𝒚 Solution: a √a2 −y2 √a2 −x2 −y2 a √a2 −y2 2 −x2 −y2 ∫0 ∫0 ∫0 dzdxdy = ∫0 ∫0 [z]√a 0 dxdy a √a2 −y2 = ∫0 ∫0 √a2 − x 2 − y 2 dxdy a √a2 −y2 2 = ∫0 ∫0 √(√a2 − y 2 ) − x 2 dxdy a x = ∫0 [2 √(a2 − y 2 ) − x 2 + √a2 −y2 a2 −y2 −1 x sin ] dy 2 √a2 −y2 0 a a2 −y2 = ∫0 (0 + sin−1 1 − 0 − 0) dy 2 a a2 −y2 π = ∫0 ( ) 2 dy 2 π a = 4 ∫0 (a2 − y 2 ) dy a π y3 = 4 [a2 y − ] 3 0 π a3 = 4 [a3 − − 0] 3 π 2a3 = 4( ) 3 πa3 = 6 Example: 4.50 𝒍𝒐𝒈 𝒂 𝒙 𝒙+𝒚 Evaluate ∫𝟎 ∫𝟎 ∫𝟎 𝒆𝒙+𝒚+𝒛 𝒅𝒛𝒅𝒚𝒅𝒙 Solution: 𝑙𝑜𝑔 𝑎 𝑥 𝑥+𝑦 𝑙𝑜𝑔 𝑎 𝑥 𝑥+𝑦 ∫0 ∫0 ∫0 𝑒 𝑥+𝑦+𝑧 𝑑𝑧𝑑𝑦𝑑𝑥 = ∫0 ∫0 [𝑒 𝑥+𝑦+𝑧 ]0 𝑑𝑦𝑑𝑥 𝑙𝑜𝑔 𝑎 𝑥 = ∫0 ∫0 (𝑒 2(𝑥+𝑦) − 𝑒 𝑥+𝑦 ) 𝑑𝑦𝑑𝑥 𝑥 𝑙𝑜𝑔 𝑎 𝑒 2(𝑥+𝑦) = ∫0 [ − 𝑒 𝑥+𝑦 ] 𝑑𝑥