Check sibling questions

Example 10 - Find area enclosed between two circles: x2+y2=4

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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.