# Ex 8.2,1 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Last updated at Aug. 20, 2021 by Teachoo

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Ex 8.2 , 1 Find the area of circle 4๐ฅ^2+4๐ฆ^2=9 which is interior to the parabola x2 = 4๐ฆ Given equations 4๐ฅ^2+4๐ฆ^2=9 ๐ฅ^2=4๐ฆ We can write (1) as 4๐ฅ^2+4๐ฆ^2=9 ("4" ๐ฅ^2+4๐ฆ^2)/4=9/4 ๐ฅ^2+๐ฆ^2=9/4 ๐ฅ^2+๐ฆ^2=(3/2)^2 Comparing with ใ(๐ฅโโ)ใ^2+ใ (๐ฆโ๐)ใ^2=๐^2 , It is a circle with radius (r) = a & center = (h, k) = (0, 0) Also, ๐ฅ^2=4๐ฆ 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) 4๐ฅ^2+4๐ฆ^2=9 โฆ(1) ๐ฅ^2=4๐ฆ โฆ(2) Putting value of ๐ฅ^2 from (2) in (1) 4๐ฅ^2+4๐ฆ^2=9 4(4๐ฆ)+4๐ฆ^2=9 16๐ฆ+4๐ฆ^2โ9=0 ใ4๐ฆใ^2+16๐ฆโ9=0 ใ4๐ฆใ^2+18๐ฆโ2๐ฆโ9=0 2y(2y+9)โ1(2๐ฆ+9)=0 (2y โ1)(2๐ฆ+9)=0 Hence, ๐ฆ=1/2 & ๐ฆ=(โ9)/2 Putting values of y in (2) For ๐=๐/๐ ๐ฅ^2=4๐ฆ ๐ฅ^2=4 ร 1/2 ๐ฅ^2=2 ๐ฅ=ยฑโ2 Hence, ๐ฅ "= " โ2 & ๐ฅ" ="โโ2 For ๐=(โ๐)/๐ ๐ฅ^2=4๐ฆ ๐ฅ^2=4 ร((โ9)/( 2)) ๐ฅ^2=โ2 ร9 ๐ฅ^2=โ18 As square cannot be negative, x has no real value Hence the points are A=(โโ2 , 1/2) & C=(โ2 , 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 โ Area OCD Area BCDO Area BCDO = โซ_0^(โ2)โใ๐ฆ ๐๐ฅใ y โ equation of circle 4x2 + 4y2 = 9 4y2 = 9 โ 4x2 y2 = 9/4 โ x2 y = ยฑโ(9/4 " โ x2" ) Since BCDO is above x โ axis, we take y positive โด y = โ(9/4 " โ x2" ) So Area BCDO = โซ_0^(โ2)โโ(9/4 " โ x2" ) dx = โซ_0^(โ2)โโ((3/2)^2 " โ " "x" ^2 ) dx We know that โซ1โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅใ =๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ๐ฅ/๐+๐ใ Here a = 3/2 = [๐ฅ/2 โ((3/2)^2 " โ " "x" ^2 )+(3/2)^2/2 sin^(โ1)โกใ๐ฅ/((3/2) )ใ ]_0^โ2 = [๐ฅ/2 โ((3/2)^2 " โ " "x" ^2 )+9/8 sin^(โ1)โกใ2๐ฅ/3ใ ]_0^โ2 = [โ2/2 โ(9/4 " โ " ใ"(" โ2 ")" ใ^2 ) + 9/8 sin^(โ1)โก((2โ2)/3) ] โ [0/2 โ(9/4 " โ " "0" ^2 )+9/8 sin^(โ1)โก((2ร0)/3) ] = โ2/2 โ(9/4 " โ " 2)+9/8 sin^(โ1)โกใ((2โ2)/3)โ9/8 sin^(โ1)โกใ(0)ใ ใ = [๐ฅ/2 โ((3/2)^2 " โ " "x" ^2 )+(3/2)^2/2 sin^(โ1)โกใ๐ฅ/((3/2) )ใ ]_0^โ2 = [๐ฅ/2 โ((3/2)^2 " โ " "x" ^2 )+9/8 sin^(โ1)โกใ2๐ฅ/3ใ ]_0^โ2 = [โ2/2 โ(9/4 " โ " ใ"(" โ2 ")" ใ^2 ) + 9/8 sin^(โ1)โก((2โ2)/3) ] โ [0/2 โ(9/4 " โ " "0" ^2 )+9/8 sin^(โ1)โก((2ร0)/3) ] = โ2/2 โ(9/4 " โ " 2)+9/8 sin^(โ1)โกใ((2โ2)/3)โ9/8 sin^(โ1)โกใ(0)ใ ใ So, Area OCD = โซ_0^(โ2)โ๐ฆ dx = โซ_0^(โ2)โ๐ฅ^2/4 dx = 1/4 โซ_0^(โ2)โ๐ฅ^2 dx = 1/4 [๐ฅ^3/3]_0^โ2 = 1/4 [(โ2)^3/3โ0^3/3] = 1/4 [(โ2รโ2รโ2)/3โ0] = 1/4 [(2โ2)/3] = โ2/6 So Area BOC = Area BCDO โ Area OCD = โ2/4 + 9/8 sin^(โ1)โก((2โ2)/3) โ โ2/6 = โ2/4 โ โ2/6+9/8 sin^(โ1)โก((2โ2)/3) = โ2/12 +9/8 sin^(โ1)โก((2โ2)/3) Area required = Area ABCO = 2 ร Area BOC = 2 ร [โ2/12+9/8 sin^(โ1)โก((2โ2)/3) ] = โ2/6+9/4 sin^(โ1)โก((2โ2)/3)

Ex 8.2

Ex 8.2,1
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Chapter 8 Class 12 Application of Integrals (Term 2)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.