Solve all your doubts with Teachoo Black (new monthly pack available now!)

Ex 8.2

Ex 8.2,1
Important
Deleted for CBSE Board 2023 Exams
You are here

Ex 8.2, 2 Deleted for CBSE Board 2023 Exams

Ex 8.2, 3 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 6 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at Dec. 29, 2021 by Teachoo

Ex 8.2, 1 Find the area of circle 4π₯^2+4π¦^2=9 which is interior to the parabola x2 = 4π¦ Given Circle and a Parabola Circle 4π₯^2+4π¦^2=9 π₯^2+π¦^2=9/4 π^π+π^π=(π/π)^π Hence, Center = (0, 0) Radius = 3/2 Parabola π₯^2=4π¦ This is a parabola with vertical axis Our figure looks like Required Area = Area ABCO Finding point of intersection A and C Solving (1) and (2) 4π₯^2+4π¦^2=9 β¦(1) π₯^2=4π¦ β¦(2) Putting value of π₯^2 from (2) in (1) 4π₯^2+4π¦^2=9 4(4π¦)+4π¦^2=9 16π¦+4π¦^2β9=0 γππγ^π+πππβπ=π γ4π¦γ^2+18π¦β2π¦β9=0 2y(2y+9)β1(2π¦+9)=0 (ππ βπ)(ππ+π)=π Hence, π=π/π & π=(βπ)/π Putting values of y in (2) For π=π/π π₯^2=4π¦ π₯^2=4 Γ 1/2 π₯^2=2 π=Β±βπ Hence, π₯ "= " β2 & π₯" ="ββ2 For π=(βπ)/π π₯^2=4π¦ π₯^2=4 Γ((β9)/( 2)) π₯^2=β2 Γ9 π₯^2=β18 As square cannot be negative, x has no real value Hence the points are A=(ββπ , π/π) & C=(βπ , π/π) Finding Area Area required = Area ABCO Since ABCO is symmetric in y β axis, Area ABCO = 2 Γ Area BOC Area BOC = Area BCDO β Area OCD Area BCDO Area BCDO = β«_0^(β2)βγπ¦ ππ₯γ y β Equation of circle 4x2 + 4y2 = 9 4y2 = 9 β 4x2 y2 = 9/4 β x2 y = Β±β(9/4 " β x2" ) Since BCDO is above x β axis, we take positive value of y β΄ y = β(π/π " β x2" ) Area BCDO = β«_π^(βπ)ββ(π/π " β x2" ) dx = β«_0^(β2)ββ((3/2)^2 " β " "x" ^2 ) dx = [π/π β((π/π)^π " β " "x" ^π )+(π/π)^π/π γπππγ^(βπ)β‘γπ/((π/π) )γ ]_π^βπ = [π₯/2 β((3/2)^2 " β " "x" ^2 )+9/8 sin^(β1)β‘γ2π₯/3γ ]_0^β2 = [β2/2 β(9/4 " β " γ"(" β2 ")" γ^2 ) + 9/8 sin^(β1)β‘((2β2)/3) ] β [0/2 β(9/4 " β " "0" ^2 )+9/8 sin^(β1)β‘((2Γ0)/3) ] We know that β«1βγβ(π^2βπ₯^2 ) ππ₯γ =π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γπ₯/π+πγ Putting a = 3/2 = β2/2 β(9/4 " β " 2)+9/8 sin^(β1)β‘γ((2β2)/3)β9/8 sin^(β1)β‘γ(0)γ γ = β2/2Γβ(1/4)+9/8 sin^(β1)β‘γ((2β2)/3)β9/8Γ0γ = βπ/π+ π/π sinβ1((πβπ)/π) Area OCD Area OCD = β«_0^(β2)βπ¦ dx y β Equation of parabola x2 = 4y y = π^π/π So, Area OCD = β«_0^(β2)βπ¦ dx = β«_π^(βπ)βπ^π/π dx = 1/4 β«_0^(β2)βπ₯^2 dx = π/π [π^π/π]_π^βπ = 1/4 [(β2)^3/3β0^3/3] = 1/4 [(β2Γβ2Γβ2)/3β0] = π/π [(πβπ)/π] = βπ/π Now, Area BOC = Area BCDO β Area OCD = βπ/π + π/π γπππγ^(βπ)β‘((πβπ)/π) β βπ/π = β2/4 β β2/6+9/8 sin^(β1)β‘((2β2)/3) = βπ/ππ +π/π γπππγ^(βπ)β‘((πβπ)/π) Required Area = Area ABCO = 2 Γ Area BOC = 2 Γ [βπ/ππ+π/π γπππγ^(βπ)β‘((πβπ)/π) ] = βπ/π+π/π γπππγ^(βπ)β‘((πβπ)/π)