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


Transcript

Example 7 Find the area lying above x-axis and included between the circle π‘₯2 +𝑦2=8π‘₯ and inside of the parabola 𝑦2=4π‘₯ Since equation of circle is of form (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘)^2=π‘Ÿ^2 , We convert our equation π‘₯^2+𝑦^2=8π‘₯ π‘₯^2βˆ’8π‘₯+𝑦^2=0 π‘₯^2βˆ’2 Γ—4 Γ—π‘₯+𝑦^2=0 π‘₯^2βˆ’2 Γ—4 Γ—π‘₯+4^2βˆ’4^2+𝑦^2=0 (π‘₯βˆ’4)^2+𝑦^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 π‘₯^2+𝑦^2=8π‘₯ Putting 𝑦^2=4π‘₯ π‘₯^2+4π‘₯=8π‘₯ π‘₯^2=8π‘₯βˆ’4π‘₯ π‘₯^2=4π‘₯ π‘₯^2βˆ’4π‘₯=0 π‘₯(π‘₯βˆ’4)=0 So, π‘₯=0 & π‘₯=4 For 𝒙 = 0 𝑦^2=4π‘₯=4 Γ— 0=0 𝑦=0 So, point is (0, 0) For 𝒙 = 4 𝑦^2=4π‘₯=4 Γ—4=4^2 𝑦=4 So, point is (4, 4) So, π‘₯=0 & π‘₯=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that π‘₯-coordinate same as that of center (4, 0) ∴ 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 = ∫_0^4▒〖𝑦 𝑑π‘₯γ€— Here, y β†’ Equation of parabola y2 = 4x y = Β± √4π‘₯ y = Β± 2√π‘₯ Since OPC is in 1st quadrant, value of y is positive y = 2√π‘₯ ∴ Area OPC = ∫_0^4β–’γ€–2√π‘₯γ€— 𝑑π‘₯ = 2 ∫_0^4β–’π‘₯^(1/2) 𝑑π‘₯ = 2 [π‘₯^(1/2 + 1)/(1/2 + 1)]_0^4 = 2 [π‘₯^(3/2)/(3/2)]_0^4 = 2 Γ— 2/3 [(4)^(3/2)βˆ’(0)^(3/2) ] = 4/3 [8βˆ’0] = 32/3 Area PCQ Area PCQ = ∫_4^8▒〖𝑦 𝑑π‘₯γ€— Here, y β†’ Equation of circle x2 + y2 = 8x y2 = 8x – x2 y = Β± √(8π‘₯βˆ’π‘₯^2 ) Since PCQ is in 1st quadrant, value of y is positive y = √(8π‘₯βˆ’π‘₯^2 ) ∴ Area PCQ = ∫_4^8β–’βˆš(8π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ = ∫_4^8β–’βˆš(βˆ’(π‘₯^2βˆ’8π‘₯)) 𝑑π‘₯ = ∫_4^8β–’βˆš(βˆ’(π‘₯^2βˆ’8π‘₯+16βˆ’16)) 𝑑π‘₯ = ∫_4^8β–’βˆš(βˆ’(π‘₯^2βˆ’8π‘₯+16)βˆ’(βˆ’16)) 𝑑π‘₯ = ∫_4^8β–’βˆš(16βˆ’(π‘₯^2βˆ’8π‘₯+16)) 𝑑π‘₯ = ∫_4^8β–’βˆš(16βˆ’(π‘₯βˆ’4)^2 ) 𝑑π‘₯ = ∫_4^8β–’βˆš(4^2βˆ’(π‘₯βˆ’4)^2 ) 𝑑π‘₯ = [((π‘₯ βˆ’ 4))/2 √(4^2βˆ’γ€–(π‘₯βˆ’4)γ€—^2 )+4^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 ((π‘₯ βˆ’ 4))/4γ€— " " ]_4^8 It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 π‘₯/π‘Ž+𝑐〗 Here, a = 4, x = x – 4 = [((π‘₯ βˆ’ 4))/2 √(16βˆ’(π‘₯^2βˆ’8π‘₯+4^2))+16/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 ((π‘₯ βˆ’ 4))/4γ€— " " ]_4^8 = [((π‘₯ βˆ’ 4))/2 √(βˆ’(π‘₯^2βˆ’8π‘₯))+8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 ((π‘₯ βˆ’ 4))/4γ€— " " ]_4^8 = [((8 βˆ’ 4))/2 √(βˆ’(8^2βˆ’8(8)))+8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 ((8 βˆ’ 4))/4γ€— ] – [((4 βˆ’ 4))/2 √(βˆ’(4^2βˆ’8(4)))+8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 ((4 βˆ’ 4))/4γ€— ] = [4/2 √0+8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1γ€— ] – [0+8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 0γ€— ] = 8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1γ€— – 8 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 0γ€— = 8(πœ‹/2) – 8 Γ— 0 = 4πœ‹ As 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 1γ€— = πœ‹/2 & 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖 0γ€— = 0 Thus, Area Required = Area OPC + Area PCQ = 32/3 + 4πœ‹ = πŸ’/πŸ‘ (8 + 3𝝅) 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 12 years. He provides courses for Maths and Science at Teachoo.