Ex 8.2, 1 - Find area of circle 4x2 + 4y2 =9 interior to parabola - Ex 8.2

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Ex 8.2 , 1 Find the area of circle 4๐‘ฅ^2+4๐‘ฆ^2=9 which is interior to the parabola x2 = 4๐‘ฆ Given equations 4๐‘ฅ^2+4๐‘ฆ^2=9 ๐‘ฅ^2=4๐‘ฆ We can write (1) as 4๐‘ฅ^2+4๐‘ฆ^2=9 ("4" ๐‘ฅ^2+4๐‘ฆ^2)/4=9/4 ๐‘ฅ^2+๐‘ฆ^2=9/4 ๐‘ฅ^2+๐‘ฆ^2=(3/2)^2 Comparing with ใ€–(๐‘ฅโˆ’โ„Ž)ใ€—^2+ใ€– (๐‘ฆโˆ’๐‘˜)ใ€—^2=๐‘Ÿ^2 , It is a circle with radius (r) = a & center = (h, k) = (0, 0) Also, ๐‘ฅ^2=4๐‘ฆ This is equation of a parabola with a vertical axis Step 1: Make the figure Required Area = Area ABCO Finding point of intersection A and C Solving (1) and (2) 4๐‘ฅ^2+4๐‘ฆ^2=9 โ€ฆ(1) ๐‘ฅ^2=4๐‘ฆ โ€ฆ(2) Putting value of ๐‘ฅ^2 from (2) in (1) 4๐‘ฅ^2+4๐‘ฆ^2=9 4(4๐‘ฆ)+4๐‘ฆ^2=9 16๐‘ฆ+4๐‘ฆ^2โˆ’9=0 ใ€–4๐‘ฆใ€—^2+16๐‘ฆโˆ’9=0 ใ€–4๐‘ฆใ€—^2+18๐‘ฆโˆ’2๐‘ฆโˆ’9=0 2y(2y+9)โˆ’1(2๐‘ฆ+9)=0 (2y โˆ’1)(2๐‘ฆ+9)=0 Hence, ๐‘ฆ=1/2 & ๐‘ฆ=(โˆ’9)/2 Putting values of y in (2) For ๐’š=๐Ÿ/๐Ÿ ๐‘ฅ^2=4๐‘ฆ ๐‘ฅ^2=4 ร— 1/2 ๐‘ฅ^2=2 ๐‘ฅ=ยฑโˆš2 Hence, ๐‘ฅ "= " โˆš2 & ๐‘ฅ" ="โˆ’โˆš2 For ๐’š=(โˆ’๐Ÿ—)/๐Ÿ ๐‘ฅ^2=4๐‘ฆ ๐‘ฅ^2=4 ร—((โˆ’9)/( 2)) ๐‘ฅ^2=โˆ’2 ร—9 ๐‘ฅ^2=โˆ’18 As square cannot be negative, x has no real value Hence the points are A=(โˆ’โˆš2 , 1/2) & C=(โˆš2 , 1/2) Step 3: Finding Area Area required = Area ABCO Since ABCO is symmetric in y โ€“ axis, Area ABCO = 2 ร— Area BOC Area BOC = Area BCDO โˆ’ Area OCD Area BCDO Area BCDO = โˆซ_0^(โˆš2)โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— y โ†’ equation of circle 4x2 + 4y2 = 9 4y2 = 9 โˆ’ 4x2 y2 = 9/4 โˆ’ x2 y = ยฑโˆš(9/4 " โˆ’ x2" ) Since BCDO is above x โˆ’ axis, we take y positive โˆด y = โˆš(9/4 " โˆ’ x2" ) So Area BCDO = โˆซ_0^(โˆš2)โ–’โˆš(9/4 " โˆ’ x2" ) dx = โˆซ_0^(โˆš2)โ–’โˆš((3/2)^2 " โˆ’ " "x" ^2 ) dx We know that โˆซ1โ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =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 = 3/2 = [๐‘ฅ/2 โˆš((3/2)^2 " โˆ’ " "x" ^2 )+(3/2)^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/((3/2) )ใ€— ]_0^โˆš2 = [๐‘ฅ/2 โˆš((3/2)^2 " โˆ’ " "x" ^2 )+9/8 sin^(โˆ’1)โกใ€–2๐‘ฅ/3ใ€— ]_0^โˆš2 = [โˆš2/2 โˆš(9/4 " โˆ’ " ใ€–"(" โˆš2 ")" ใ€—^2 ) + 9/8 sin^(โˆ’1)โก((2โˆš2)/3) ] โˆ’ [0/2 โˆš(9/4 " โˆ’ " "0" ^2 )+9/8 sin^(โˆ’1)โก((2ร—0)/3) ] = โˆš2/2 โˆš(9/4 " โˆ’ " 2)+9/8 sin^(โˆ’1)โกใ€–((2โˆš2)/3)โˆ’9/8 sin^(โˆ’1)โกใ€–(0)ใ€— ใ€— = [๐‘ฅ/2 โˆš((3/2)^2 " โˆ’ " "x" ^2 )+(3/2)^2/2 sin^(โˆ’1)โกใ€–๐‘ฅ/((3/2) )ใ€— ]_0^โˆš2 = [๐‘ฅ/2 โˆš((3/2)^2 " โˆ’ " "x" ^2 )+9/8 sin^(โˆ’1)โกใ€–2๐‘ฅ/3ใ€— ]_0^โˆš2 = [โˆš2/2 โˆš(9/4 " โˆ’ " ใ€–"(" โˆš2 ")" ใ€—^2 ) + 9/8 sin^(โˆ’1)โก((2โˆš2)/3) ] โˆ’ [0/2 โˆš(9/4 " โˆ’ " "0" ^2 )+9/8 sin^(โˆ’1)โก((2ร—0)/3) ] = โˆš2/2 โˆš(9/4 " โˆ’ " 2)+9/8 sin^(โˆ’1)โกใ€–((2โˆš2)/3)โˆ’9/8 sin^(โˆ’1)โกใ€–(0)ใ€— ใ€— So, Area OCD = โˆซ_0^(โˆš2)โ–’๐‘ฆ dx = โˆซ_0^(โˆš2)โ–’๐‘ฅ^2/4 dx = 1/4 โˆซ_0^(โˆš2)โ–’๐‘ฅ^2 dx = 1/4 [๐‘ฅ^3/3]_0^โˆš2 = 1/4 [(โˆš2)^3/3โˆ’0^3/3] = 1/4 [(โˆš2ร—โˆš2ร—โˆš2)/3โˆ’0] = 1/4 [(2โˆš2)/3] = โˆš2/6 So Area BOC = Area BCDO โˆ’ Area OCD = โˆš2/4 + 9/8 sin^(โˆ’1)โก((2โˆš2)/3) โ€“ โˆš2/6 = โˆš2/4 โˆ’ โˆš2/6+9/8 sin^(โˆ’1)โก((2โˆš2)/3) = โˆš2/12 +9/8 sin^(โˆ’1)โก((2โˆš2)/3) Area required = Area ABCO = 2 ร— Area BOC = 2 ร— [โˆš2/12+9/8 sin^(โˆ’1)โก((2โˆš2)/3) ] = โˆš2/6+9/4 sin^(โˆ’1)โก((2โˆš2)/3)

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