# Example 7 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 12, 2019 by Teachoo

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Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Important Deleted for CBSE Board 2022 Exams You are here

Example 8 Important Deleted for CBSE Board 2022 Exams

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Chapter 8 Class 12 Application of Integrals (Term 2)

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Last updated at Dec. 12, 2019 by Teachoo

Example 7 Find the area lying above x-axis and included between the circle π₯2 +π¦2=8π₯ and inside of the parabola π¦2=4π₯ Since equation of circle is of form (π₯βπ)^2+(π¦βπ)^2=π^2 , We convert our equation π₯^2+π¦^2=8π₯ π₯^2β8π₯+π¦^2=0 π₯^2β2 Γ4 Γπ₯+π¦^2=0 π₯^2β2 Γ4 Γπ₯+4^2β4^2+π¦^2=0 (π₯β4)^2+π¦^2=4^2 So, Circle has center (4 , 0) & Radius =4 We need to find Area OPQC Point P is point of intersection of circle and parabola Finding Point P Equation of circle is π₯^2+π¦^2=8π₯ Putting π¦^2=4π₯ π₯^2+4π₯=8π₯ π₯^2=8π₯β4π₯ π₯^2=4π₯ π₯^2β4π₯=0 π₯(π₯β4)=0 So, π₯=0 & π₯=4 For π = 0 π¦^2=4π₯=4 Γ 0=0 π¦=0 So, point is (0, 0) For π = 4 π¦^2=4π₯=4 Γ4=4^2 π¦=4 So, point is (4, 4) So, π₯=0 & π₯=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that π₯-coordinate same as that of center (4, 0) β΄ P lies above point C So, we need to change the figure New figure Area Required Area Required = Area OPC + Area PCQ Area OPC Area OPC = β«_0^4βγπ¦ ππ₯γ Here, y β Equation of parabola y2 = 4x y = Β± β4π₯ y = Β± 2βπ₯ Since OPC is in 1st quadrant, value of y is positive y = 2βπ₯ β΄ Area OPC = β«_0^4βγ2βπ₯γ ππ₯ = 2 β«_0^4βπ₯^(1/2) ππ₯ = 2 [π₯^(1/2 + 1)/(1/2 + 1)]_0^4 = 2 [π₯^(3/2)/(3/2)]_0^4 = 2 Γ 2/3 [(4)^(3/2)β(0)^(3/2) ] = 4/3 [8β0] = 32/3 Area PCQ Area PCQ = β«_4^8βγπ¦ ππ₯γ Here, y β Equation of circle x2 + y2 = 8x y2 = 8x β x2 y = Β± β(8π₯βπ₯^2 ) Since PCQ is in 1st quadrant, value of y is positive y = β(8π₯βπ₯^2 ) β΄ Area PCQ = β«_4^8ββ(8π₯βπ₯^2 ) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯)) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯+16β16)) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯+16)β(β16)) ππ₯ = β«_4^8ββ(16β(π₯^2β8π₯+16)) ππ₯ = β«_4^8ββ(16β(π₯β4)^2 ) ππ₯ = β«_4^8ββ(4^2β(π₯β4)^2 ) ππ₯ = [((π₯ β 4))/2 β(4^2βγ(π₯β4)γ^2 )+4^2/2 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 It is of form β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Here, a = 4, x = x β 4 = [((π₯ β 4))/2 β(16β(π₯^2β8π₯+4^2))+16/2 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 = [((π₯ β 4))/2 β(β(π₯^2β8π₯))+8 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 = [((8 β 4))/2 β(β(8^2β8(8)))+8 γπ ππγ^(β1)β‘γ ((8 β 4))/4γ ] β [((4 β 4))/2 β(β(4^2β8(4)))+8 γπ ππγ^(β1)β‘γ ((4 β 4))/4γ ] = [4/2 β0+8 γπ ππγ^(β1)β‘γ 1γ ] β [0+8 γπ ππγ^(β1)β‘γ 0γ ] = 8 γπ ππγ^(β1)β‘γ 1γ β 8 γπ ππγ^(β1)β‘γ 0γ = 8(π/2) β 8 Γ 0 = 4π As γπ ππγ^(β1)β‘γ 1γ = π/2 & γπ ππγ^(β1)β‘γ 0γ = 0 Thus, Area Required = Area OPC + Area PCQ = 32/3 + 4π = π/π (8 + 3π ) square units