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Area between curve and curve
Misc 13
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important
Misc 4 Important
Misc 11 Important
Ex 8.2,1 Important Deleted for CBSE Board 2023 Exams You are here
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Example 15 Important
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Area between curve and curve
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)