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1. Chapter 8 Class 12 Application of Integrals
2. Serial order wise
3. Ex 8.2

Transcript

Ex 8.2 , 1 Find the area of circle 4𝑥^2+4𝑦^2=9 which is interior to the parabola x2 = 4𝑦 Given equations 4𝑥^2+4𝑦^2=9 𝑥^2=4𝑦 We can write (1) as 4𝑥^2+4𝑦^2=9 ("4" 𝑥^2+4𝑦^2)/4=9/4 𝑥^2+𝑦^2=9/4 𝑥^2+𝑦^2=(3/2)^2 Comparing with 〖(𝑥−ℎ)〗^2+〖 (𝑦−𝑘)〗^2=𝑟^2 , It is a circle with radius (r) = a & center = (h, k) = (0, 0) Also, 𝑥^2=4𝑦 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) 4𝑥^2+4𝑦^2=9 …(1) 𝑥^2=4𝑦 …(2) Putting value of 𝑥^2 from (2) in (1) 4𝑥^2+4𝑦^2=9 4(4𝑦)+4𝑦^2=9 16𝑦+4𝑦^2−9=0 〖4𝑦〗^2+16𝑦−9=0 〖4𝑦〗^2+18𝑦−2𝑦−9=0 2y(2y+9)−1(2𝑦+9)=0 (2y −1)(2𝑦+9)=0 Hence, 𝑦=1/2 & 𝑦=(−9)/2 Putting values of y in (2) For 𝒚=𝟏/𝟐 𝑥^2=4𝑦 𝑥^2=4 × 1/2 𝑥^2=2 𝑥=±√2 Hence, 𝑥 "= " √2 & 𝑥" ="−√2 For 𝒚=(−𝟗)/𝟐 𝑥^2=4𝑦 𝑥^2=4 ×((−9)/( 2)) 𝑥^2=−2 ×9 𝑥^2=−18 As square cannot be negative, x has no real value Hence the points are A=(−√2 , 1/2) & C=(√2 , 1/2) Step 3: Finding Area Area required = Area ABCO Since ABCO is symmetric in y – axis, Area ABCO = 2 × Area BOC Area BOC = Area BCDO − Area OCD Area BCDO Area BCDO = ∫_0^(√2)▒〖𝑦 𝑑𝑥〗 y → equation of circle 4x2 + 4y2 = 9 4y2 = 9 − 4x2 y2 = 9/4 − x2 y = ±√(9/4 " − x2" ) Since BCDO is above x − axis, we take y positive ∴ y = √(9/4 " − x2" ) So Area BCDO = ∫_0^(√2)▒√(9/4 " − x2" ) dx = ∫_0^(√2)▒√((3/2)^2 " − " "x" ^2 ) dx We know that ∫1▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥〗 =𝑥/2 √(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1)⁡〖𝑥/𝑎+𝑐〗 Here a = 3/2 = [𝑥/2 √((3/2)^2 " − " "x" ^2 )+(3/2)^2/2 sin^(−1)⁡〖𝑥/((3/2) )〗 ]_0^√2 = [𝑥/2 √((3/2)^2 " − " "x" ^2 )+9/8 sin^(−1)⁡〖2𝑥/3〗 ]_0^√2 = [√2/2 √(9/4 " − " 〖"(" √2 ")" 〗^2 ) + 9/8 sin^(−1)⁡((2√2)/3) ] − [0/2 √(9/4 " − " "0" ^2 )+9/8 sin^(−1)⁡((2×0)/3) ] = √2/2 √(9/4 " − " 2)+9/8 sin^(−1)⁡〖((2√2)/3)−9/8 sin^(−1)⁡〖(0)〗 〗 = [𝑥/2 √((3/2)^2 " − " "x" ^2 )+(3/2)^2/2 sin^(−1)⁡〖𝑥/((3/2) )〗 ]_0^√2 = [𝑥/2 √((3/2)^2 " − " "x" ^2 )+9/8 sin^(−1)⁡〖2𝑥/3〗 ]_0^√2 = [√2/2 √(9/4 " − " 〖"(" √2 ")" 〗^2 ) + 9/8 sin^(−1)⁡((2√2)/3) ] − [0/2 √(9/4 " − " "0" ^2 )+9/8 sin^(−1)⁡((2×0)/3) ] = √2/2 √(9/4 " − " 2)+9/8 sin^(−1)⁡〖((2√2)/3)−9/8 sin^(−1)⁡〖(0)〗 〗 So, Area OCD = ∫_0^(√2)▒𝑦 dx = ∫_0^(√2)▒𝑥^2/4 dx = 1/4 ∫_0^(√2)▒𝑥^2 dx = 1/4 [𝑥^3/3]_0^√2 = 1/4 [(√2)^3/3−0^3/3] = 1/4 [(√2×√2×√2)/3−0] = 1/4 [(2√2)/3] = √2/6 So Area BOC = Area BCDO − Area OCD = √2/4 + 9/8 sin^(−1)⁡((2√2)/3) – √2/6 = √2/4 − √2/6+9/8 sin^(−1)⁡((2√2)/3) = √2/12 +9/8 sin^(−1)⁡((2√2)/3) Area required = Area ABCO = 2 × Area BOC = 2 × [√2/12+9/8 sin^(−1)⁡((2√2)/3) ] = √2/6+9/4 sin^(−1)⁡((2√2)/3)

Ex 8.2 