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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

Transcript

Example 7 Find the area lying above x-axis and included between the circle ๐‘ฅ2 +๐‘ฆ2=8๐‘ฅ and inside of the parabola ๐‘ฆ2=4๐‘ฅ Since equation of circle is of form (๐‘ฅโˆ’๐‘Ž)^2+(๐‘ฆโˆ’๐‘)^2=๐‘Ÿ^2 , We convert our equation ๐‘ฅ^2+๐‘ฆ^2=8๐‘ฅ ๐‘ฅ^2โˆ’8๐‘ฅ+๐‘ฆ^2=0 ๐‘ฅ^2โˆ’2 ร—4 ร—๐‘ฅ+๐‘ฆ^2=0 ๐‘ฅ^2โˆ’2 ร—4 ร—๐‘ฅ+4^2โˆ’4^2+๐‘ฆ^2=0 (๐‘ฅโˆ’4)^2+๐‘ฆ^2=4^2 So, Circle has center (4 , 0) & Radius =4 We need to find Area OPQC Point P is point of intersection of circle and parabola Finding Point P Equation of circle is ๐‘ฅ^2+๐‘ฆ^2=8๐‘ฅ Putting ๐‘ฆ^2=4๐‘ฅ ๐‘ฅ^2+4๐‘ฅ=8๐‘ฅ ๐‘ฅ^2=8๐‘ฅโˆ’4๐‘ฅ ๐‘ฅ^2=4๐‘ฅ ๐‘ฅ^2โˆ’4๐‘ฅ=0 ๐‘ฅ(๐‘ฅโˆ’4)=0 So, ๐‘ฅ=0 & ๐‘ฅ=4 For ๐’™ = 0 ๐‘ฆ^2=4๐‘ฅ=4 ร— 0=0 ๐‘ฆ=0 So, point is (0, 0) For ๐’™ = 4 ๐‘ฆ^2=4๐‘ฅ=4 ร—4=4^2 ๐‘ฆ=4 So, point is (4, 4) So, ๐‘ฅ=0 & ๐‘ฅ=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that ๐‘ฅ-coordinate same as that of center (4, 0) โˆด P lies above point C So, we need to change the figure New figure Area Required Area Required = Area OPC + Area PCQ Area OPC Area OPC = โˆซ_0^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, y โ†’ Equation of parabola y2 = 4x y = ยฑ โˆš4๐‘ฅ y = ยฑ 2โˆš๐‘ฅ Since OPC is in 1st quadrant, value of y is positive y = 2โˆš๐‘ฅ โˆด Area OPC = โˆซ_0^4โ–’ใ€–2โˆš๐‘ฅใ€— ๐‘‘๐‘ฅ = 2 โˆซ_0^4โ–’๐‘ฅ^(1/2) ๐‘‘๐‘ฅ = 2 [๐‘ฅ^(1/2 + 1)/(1/2 + 1)]_0^4 = 2 [๐‘ฅ^(3/2)/(3/2)]_0^4 = 2 ร— 2/3 [(4)^(3/2)โˆ’(0)^(3/2) ] = 4/3 [8โˆ’0] = 32/3 Area PCQ Area PCQ = โˆซ_4^8โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, y โ†’ Equation of circle x2 + y2 = 8x y2 = 8x โ€“ x2 y = ยฑ โˆš(8๐‘ฅโˆ’๐‘ฅ^2 ) Since PCQ is in 1st quadrant, value of y is positive y = โˆš(8๐‘ฅโˆ’๐‘ฅ^2 ) โˆด Area PCQ = โˆซ_4^8โ–’โˆš(8๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ)) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ+16โˆ’16)) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ+16)โˆ’(โˆ’16)) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(16โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ+16)) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(16โˆ’(๐‘ฅโˆ’4)^2 ) ๐‘‘๐‘ฅ = โˆซ_4^8โ–’โˆš(4^2โˆ’(๐‘ฅโˆ’4)^2 ) ๐‘‘๐‘ฅ = [((๐‘ฅ โˆ’ 4))/2 โˆš(4^2โˆ’ใ€–(๐‘ฅโˆ’4)ใ€—^2 )+4^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ((๐‘ฅ โˆ’ 4))/4ใ€— " " ]_4^8 It is of form โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=๐‘ฅ/2 โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ๐‘ฅ/๐‘Ž+๐‘ใ€— Here, a = 4, x = x โ€“ 4 = [((๐‘ฅ โˆ’ 4))/2 โˆš(16โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ+4^2))+16/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ((๐‘ฅ โˆ’ 4))/4ใ€— " " ]_4^8 = [((๐‘ฅ โˆ’ 4))/2 โˆš(โˆ’(๐‘ฅ^2โˆ’8๐‘ฅ))+8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ((๐‘ฅ โˆ’ 4))/4ใ€— " " ]_4^8 = [((8 โˆ’ 4))/2 โˆš(โˆ’(8^2โˆ’8(8)))+8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ((8 โˆ’ 4))/4ใ€— ] โ€“ [((4 โˆ’ 4))/2 โˆš(โˆ’(4^2โˆ’8(4)))+8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ((4 โˆ’ 4))/4ใ€— ] = [4/2 โˆš0+8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1ใ€— ] โ€“ [0+8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 0ใ€— ] = 8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1ใ€— โ€“ 8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 0ใ€— = 8(๐œ‹/2) โ€“ 8 ร— 0 = 4๐œ‹ As ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1ใ€— = ๐œ‹/2 & ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 0ใ€— = 0 Thus, Area Required = Area OPC + Area PCQ = 32/3 + 4๐œ‹ = ๐Ÿ’/๐Ÿ‘ (8 + 3๐…) square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.