Check sibling questions

Example 7 - Find area lying above x-axis, included b/w circle

Example 7 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 7 - Chapter 8 Class 12 Application of Integrals - Part 3 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 4 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 5 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 6 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 7 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 8 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 9 Example 7 - Chapter 8 Class 12 Application of Integrals - Part 10

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Transcript

Question 5 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.