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

Transcript

Example 10 Find the area of the region enclosed between the two circles: π‘₯2+𝑦2=4 and (π‘₯ –2)2+𝑦2=4 First we find center and radius of both circles π‘₯^2+ 𝑦^2 = 4 γ€–(π‘₯βˆ’0)γ€—^2 + γ€–(π‘¦βˆ’0)γ€—^2 = 2^2 Thus, Center = (0, 0) Radius = 2 (π‘₯βˆ’2)^2 + 𝑦^2 = 4 γ€–(π‘₯βˆ’2)γ€—^2 + γ€–(π‘¦βˆ’0)γ€—^2 = 2^2 Thus, Center = (2, 0) Radius = 2 Drawing figure Finding point of intersection, A & A’ Solving π‘₯^2 + 𝑦^2 = 4 …(1) (γ€–π‘₯βˆ’2)γ€—^2 + 𝑦^2 = 4 …(2) Comparing (1) & (2) π‘₯^2 + 𝑦^2 = γ€–"(x βˆ’ 2)" γ€—^2 + 𝑦^2 π‘₯^2 = γ€–"(x βˆ’ 2)" γ€—^2 π‘₯^2 βˆ’ γ€–"(x βˆ’ 2)" γ€—^2 = 0 (x βˆ’ (x βˆ’ 2) (x + (x βˆ’ 2)) = 0 (x βˆ’ x + 2) (x + x βˆ’ 2) = 0 2(2x βˆ’ 2) = 0 (2x βˆ’ 2) = 0/2 2x βˆ’ 2 = 0 2x = 2 x = 2/2 x = 1 Putting x = 1 in (1) π‘₯^2 + 𝑦^2 = 4 1 + 𝑦^2 = 4 𝑦^2 = 4 βˆ’ 1 𝑦^2 = 3 y = Β± √3 Hence A = (1, √3) & A’ = (1, β€“βˆš3) Also, point D = (1, 0) Area required Area required = Area ACA’D + Area OADA’ Area ACA’D Area ACA’D = 2 Area ADC = 2 ∫_1^2▒〖𝑦 𝑑π‘₯γ€— Here, π‘₯^2+𝑦^2=4 𝑦^2=4βˆ’π‘₯^2 𝑦=±√(4βˆ’π‘₯^2 ) Since Area ADC is in 1st quadrant, value of y will be positive y = √(4βˆ’π‘₯^2 ) Area ACA’D = 2 ∫_1^2β–’γ€–βˆš(4βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =2∫_1^2β–’γ€–βˆš(2^2βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =2[π‘₯/2 √(2^2βˆ’π‘₯^2 )+2^2/2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]_1^2 It is of form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )γ€—+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 Replacing a by 2 , we get =γ€–2[π‘₯/2 √(4βˆ’π‘₯^2 )+2 sin^(βˆ’1)⁑〖π‘₯/2γ€— ]γ€—_1^2 =2[2/2 √(4βˆ’2^2 )+2 sin^(βˆ’1)⁑〖2/2βˆ’[1/2 √(4βˆ’1^2 )+2 sin^(βˆ’1)⁑〖1/2γ€— ]γ€— ] =2[1.√(4βˆ’4)+2 sin^(βˆ’1)⁑〖1βˆ’[1/2 √(4βˆ’1)+2 sin^(βˆ’1)⁑〖1/2γ€— ]γ€— ] =2[1.0+2πœ‹/2⁑〖1βˆ’[1/2 √3+2πœ‹/6]γ€— ] =2[πœ‹βˆ’βˆš3/2βˆ’πœ‹/3] =2[2πœ‹/3βˆ’βˆš3/2] =πŸ’π…/πŸ‘βˆ’βˆšπŸ‘ Area OADA’ Area OADA’ =2 Γ— Area OAD = 2∫_0^1▒〖𝑦 𝑑π‘₯γ€— Here, (π‘₯βˆ’2)^2+𝑦^2=4 𝑦^2=4βˆ’(π‘₯βˆ’2)^2 𝑦=±√(4βˆ’(π‘₯βˆ’2)^2 ) Since OAD is in 1st quadrant, value of y will be positive 𝑦=√(4βˆ’(π‘₯βˆ’2)^2 ) Hence, Area OADA’ = 2∫_0^1β–’γ€–βˆš(4βˆ’(π‘₯βˆ’2)^2 ) 𝑑π‘₯γ€— Putting t = (π‘₯βˆ’2) Diff. w.r.t. π‘₯ 𝑑𝑑/𝑑π‘₯=1 𝑑𝑑 =𝑑π‘₯ So, Area OADA’ = 2∫_0^1β–’γ€–βˆš(4βˆ’(π‘₯βˆ’2)^2 ) 𝑑π‘₯γ€— =2∫_(βˆ’2)^(βˆ’1)β–’γ€–βˆš(4βˆ’π‘‘^2 ) 𝑑𝑑〗 =2∫_(βˆ’2)^(βˆ’1)β–’γ€–βˆš(2^2βˆ’π‘‘^2 ) 𝑑𝑑〗 It is of form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )γ€—+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 Replacing a with 2 & x with t, we get =2[𝑑/2 √(2^2βˆ’π‘‘^2 )+2^2/2 sin^(βˆ’1)⁑〖𝑑/2γ€— ]_(βˆ’2)^(βˆ’1) = 2[(βˆ’1)/( 2) √(2^2βˆ’(βˆ’1)^2 )+2 sin^(βˆ’1) (βˆ’1)/2]βˆ’2[(βˆ’2)/( 2) √(2^2βˆ’(βˆ’2)^2 )+2 sin^(βˆ’1) (βˆ’2)/2] = 2[(βˆ’1)/( 2) √(4βˆ’1)+2sin^(βˆ’1) ((βˆ’1)/2)]βˆ’2[βˆ’1√(4βˆ’4)+2sin^(βˆ’1) (βˆ’1)] = 2[(βˆ’1)/( 2)Γ—βˆš3+2 sin^(βˆ’1) ((βˆ’1)/2)]βˆ’2[0+2sin^(βˆ’1) (βˆ’1)] = 2[(βˆ’βˆš3)/( 2)+2 sin^(βˆ’1) ((βˆ’1)/2)]βˆ’2[2 sin^(βˆ’1) (βˆ’1)] = βˆ’βˆš3+4 sin^(βˆ’1) ((βˆ’1)/2)βˆ’4 sin^(βˆ’1) (βˆ’1) Using sin–1 (–x) = – sin–1 x = βˆ’βˆš3βˆ’4 sin^(βˆ’1) (1/2)+4 sin^(βˆ’1) (1) = βˆ’βˆš3βˆ’4 Γ—πœ‹/6+4Γ—πœ‹/2 = βˆ’βˆš3βˆ’2πœ‹/3+2πœ‹ = βˆ’βˆš3+4πœ‹/3 Therefore, Area required = Area ACA’D + Area OADA’ = (4πœ‹/3βˆ’βˆš3) + ("–" √3 " + " 4πœ‹/3) = πŸ–π…/πŸ‘βˆ’πŸβˆšπŸ‘ 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.