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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
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams You are here
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. 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