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Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at Dec. 12, 2019 by Teachoo

Example 10 Find the area of the region enclosed between the two circles: π₯2+π¦2=4 and (π₯ β2)2+π¦2=4 First we find center and radius of both circles π₯^2+ π¦^2 = 4 γ(π₯β0)γ^2 + γ(π¦β0)γ^2 = 2^2 Thus, Center = (0, 0) Radius = 2 (π₯β2)^2 + π¦^2 = 4 γ(π₯β2)γ^2 + γ(π¦β0)γ^2 = 2^2 Thus, Center = (2, 0) Radius = 2 Drawing figure Finding point of intersection, A & Aβ Solving π₯^2 + π¦^2 = 4 β¦(1) (γπ₯β2)γ^2 + π¦^2 = 4 β¦(2) Comparing (1) & (2) π₯^2 + π¦^2 = γ"(x β 2)" γ^2 + π¦^2 π₯^2 = γ"(x β 2)" γ^2 π₯^2 β γ"(x β 2)" γ^2 = 0 (x β (x β 2) (x + (x β 2)) = 0 (x β x + 2) (x + x β 2) = 0 2(2x β 2) = 0 (2x β 2) = 0/2 2x β 2 = 0 2x = 2 x = 2/2 x = 1 Putting x = 1 in (1) π₯^2 + π¦^2 = 4 1 + π¦^2 = 4 π¦^2 = 4 β 1 π¦^2 = 3 y = Β± β3 Hence A = (1, β3) & Aβ = (1, ββ3) Also, point D = (1, 0) Area required Area required = Area ACAβD + Area OADAβ Area ACAβD Area ACAβD = 2 Area ADC = 2 β«_1^2βγπ¦ ππ₯γ Here, π₯^2+π¦^2=4 π¦^2=4βπ₯^2 π¦=Β±β(4βπ₯^2 ) Since Area ADC is in 1st quadrant, value of y will be positive y = β(4βπ₯^2 ) Area ACAβD = 2 β«_1^2βγβ(4βπ₯^2 ) ππ₯γ =2β«_1^2βγβ(2^2βπ₯^2 ) ππ₯γ =2[π₯/2 β(2^2βπ₯^2 )+2^2/2 sin^(β1)β‘γπ₯/2γ ]_1^2 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 =γ2[π₯/2 β(4βπ₯^2 )+2 sin^(β1)β‘γπ₯/2γ ]γ_1^2 =2[2/2 β(4β2^2 )+2 sin^(β1)β‘γ2/2β[1/2 β(4β1^2 )+2 sin^(β1)β‘γ1/2γ ]γ ] =2[1.β(4β4)+2 sin^(β1)β‘γ1β[1/2 β(4β1)+2 sin^(β1)β‘γ1/2γ ]γ ] =2[1.0+2π/2β‘γ1β[1/2 β3+2π/6]γ ] =2[πββ3/2βπ/3] =2[2π/3ββ3/2] =ππ /πββπ Area OADAβ Area OADAβ =2 Γ Area OAD = 2β«_0^1βγπ¦ ππ₯γ Here, (π₯β2)^2+π¦^2=4 π¦^2=4β(π₯β2)^2 π¦=Β±β(4β(π₯β2)^2 ) Since OAD is in 1st quadrant, value of y will be positive π¦=β(4β(π₯β2)^2 ) Hence, Area OADAβ = 2β«_0^1βγβ(4β(π₯β2)^2 ) ππ₯γ Putting t = (π₯β2) Diff. w.r.t. π₯ ππ‘/ππ₯=1 ππ‘ =ππ₯ So, Area OADAβ = 2β«_0^1βγβ(4β(π₯β2)^2 ) ππ₯γ =2β«_(β2)^(β1)βγβ(4βπ‘^2 ) ππ‘γ =2β«_(β2)^(β1)βγβ(2^2βπ‘^2 ) ππ‘γ 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 with 2 & x with t, we get =2[π‘/2 β(2^2βπ‘^2 )+2^2/2 sin^(β1)β‘γπ‘/2γ ]_(β2)^(β1) = 2[(β1)/( 2) β(2^2β(β1)^2 )+2 sin^(β1) (β1)/2]β2[(β2)/( 2) β(2^2β(β2)^2 )+2 sin^(β1) (β2)/2] = 2[(β1)/( 2) β(4β1)+2sin^(β1) ((β1)/2)]β2[β1β(4β4)+2sin^(β1) (β1)] = 2[(β1)/( 2)Γβ3+2 sin^(β1) ((β1)/2)]β2[0+2sin^(β1) (β1)] = 2[(ββ3)/( 2)+2 sin^(β1) ((β1)/2)]β2[2 sin^(β1) (β1)] = ββ3+4 sin^(β1) ((β1)/2)β4 sin^(β1) (β1) Using sinβ1 (βx) = β sinβ1 x = ββ3β4 sin^(β1) (1/2)+4 sin^(β1) (1) = ββ3β4 Γπ/6+4Γπ/2 = ββ3β2π/3+2π = ββ3+4π/3 Therefore, Area required = Area ACAβD + Area OADAβ = (4π/3ββ3) + ("β" β3 " + " 4π/3) = ππ /πβπβπ square units