Question 5 - Chapter 8 Class 12 Application of Integrals (Important Question)

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Question 5 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

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.

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