Examples

Chapter 8 Class 12 Application of Integrals
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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₯βπ)^2+(π¦βπ)^2=π^2 , We convert our equation π₯^2+π¦^2=8π₯ π₯^2β8π₯+π¦^2=0 π₯^2β2 Γ4 Γπ₯+π¦^2=0 π₯^2β2 Γ4 Γπ₯+4^2β4^2+π¦^2=0 (π₯β4)^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₯β4π₯ π₯^2=4π₯ π₯^2β4π₯=0 π₯(π₯β4)=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 Γ4=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«_0^4βγπ¦ ππ₯γ Here, y β Equation of parabola y2 = 4x y = Β± β4π₯ y = Β± 2βπ₯ Since OPC is in 1st quadrant, value of y is positive y = 2βπ₯ β΄ Area OPC = β«_0^4βγ2βπ₯γ ππ₯ = 2 β«_0^4βπ₯^(1/2) ππ₯ = 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 [8β0] = 32/3 Area PCQ Area PCQ = β«_4^8βγπ¦ ππ₯γ Here, y β Equation of circle x2 + y2 = 8x y2 = 8x β x2 y = Β± β(8π₯βπ₯^2 ) Since PCQ is in 1st quadrant, value of y is positive y = β(8π₯βπ₯^2 ) β΄ Area PCQ = β«_4^8ββ(8π₯βπ₯^2 ) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯)) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯+16β16)) ππ₯ = β«_4^8ββ(β(π₯^2β8π₯+16)β(β16)) ππ₯ = β«_4^8ββ(16β(π₯^2β8π₯+16)) ππ₯ = β«_4^8ββ(16β(π₯β4)^2 ) ππ₯ = β«_4^8ββ(4^2β(π₯β4)^2 ) ππ₯ = [((π₯ β 4))/2 β(4^2βγ(π₯β4)γ^2 )+4^2/2 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 It is of form β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Here, a = 4, x = x β 4 = [((π₯ β 4))/2 β(16β(π₯^2β8π₯+4^2))+16/2 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 = [((π₯ β 4))/2 β(β(π₯^2β8π₯))+8 γπ ππγ^(β1)β‘γ ((π₯ β 4))/4γ " " ]_4^8 = [((8 β 4))/2 β(β(8^2β8(8)))+8 γπ ππγ^(β1)β‘γ ((8 β 4))/4γ ] β [((4 β 4))/2 β(β(4^2β8(4)))+8 γπ ππγ^(β1)β‘γ ((4 β 4))/4γ ] = [4/2 β0+8 γπ ππγ^(β1)β‘γ 1γ ] β [0+8 γπ ππγ^(β1)β‘γ 0γ ] = 8 γπ ππγ^(β1)β‘γ 1γ β 8 γπ ππγ^(β1)β‘γ 0γ = 8(π/2) β 8 Γ 0 = 4π As γπ ππγ^(β1)β‘γ 1γ = π/2 & γπ ππγ^(β1)β‘γ 0γ = 0 Thus, Area Required = Area OPC + Area PCQ = 32/3 + 4π = π/π (8 + 3π) square units

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.