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Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

Last updated at Dec. 12, 2019 by Teachoo

Example 8 In Fig AOBA is the part of the ellipse 9π₯2+π¦2=36 in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB Here, Area Area Required = Area ABC Area ABC Area ABC = Area OACB β Area β OAB Area OACB Area OACB = β«_0^2βπ¦ππ₯ Here, π¦ β Equation of Ellpise 9π₯2+π¦2=36 π¦2=36β9π₯2 π¦="Β±" β(36β9π₯^2 ) As OACB is in 1st quadrant, Value of π¦ will be positive β΄ π¦=β(36β9π₯^2 ) Area OACB = β«_0^2βπ¦ππ₯ = β«_0^2ββ(36β9π₯^2 ) = β«_0^2ββ(9(36/9βπ₯^2)) = β«_0^2ββ(9(4βπ₯^2)) = β«_0^2ββ(3^2 (2^2βπ₯^2)) = 3β«_0^2βγβ(2^2βπ₯^2 ) ππ₯ γ = 3[π₯/2 β(2^2βπ₯^2 )+2^2/2 sin^(β1)β‘γπ₯/2γ ]_0^2 = 3 [2/2 β(2^2β2^2 )+2 sin^(β1)β‘γ2/2β0/2γ β(2^2β0^2 )β2 sin^(β1)β‘0 ] It is of form β«1βγβ(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )γ+π^2/2 γπ ππγ^(β1)β‘γπ₯/π+πγ Replacing a by 2 , we get = 3 [0+2 sin^(β1)β‘γ1β0β0γ ] = 3 [2 sin^(β1)β‘1 ] = 3 Γ 2 Γ π/2 = 3π Area OAB Area OAB =β«_0^2βπ¦ππ₯ Here, π¦ β equation of line AB Equation of line between A(2, 0) & B(0, 6) is (π¦ β 0)/(π₯ β 2)=(6 β 0)/(0 β 2) π¦/(π₯β2)=6/(β2) Eq. of line b/w (x1, y1) & (x2, y2) is (π¦ β π¦1)/(π₯ β π₯1)=(π¦2 β π¦1)/(π₯2 β π₯1) π¦/(π₯ β 2)=β3 π¦=β3(π₯β2) Area OBC =β«_0^2βγπ¦ ππ₯γ =β«_0^2βγβ3(π₯β2) ππ₯γ = β3β«_0^2βγ(π₯β2) ππ₯γ = β3 [π₯^2/2 β2π₯ ]_0^2 = β 3 [2^2/2β2 Γ2β0^2/2β2 Γ0] = β3 [β2] = 6 Area Required = Area OACB β Area β OAB =ππ βπ square units