Example 1 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 29, 2021 by Teachoo

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Example 1 Find the area enclosed by the circle π₯2 + π¦2 = π2Given
π₯^2 + π¦^2= π^2
This is a circle with
Center = (0, 0)
Radius = π
Since radius is a,
OA = OB = π
A = (π, 0)
B = (0, π)
Now,
Area of circle = 4 Γ Area of Region OBAO
= 4 Γ β«1_π^πβγπ π πγ
Here,
y β Equation of Circle
We know that
π₯^2 + π¦^2 = π^2
π¦^2 = π^2β π₯^2
y = Β± β(π^2βπ₯^2 )
Since AOBA lies in 1st Quadrant
y = β(π^πβπ^π )
Now,
Area of circle = 4 Γ β«1_0^πβγπ¦ ππ₯γ
= 4 Γ β«1_0^πβγβ(π^2βπ₯^2 ) ππ₯γ
= 4[π/π β(π^πβπ^π )+π^π/π γ"sin" γ^(βπ) π/π]_π^π
= 4[π/2 β(π^2βπ^2 )+π^2/2 γ"sin" γ^(β1) π/πβ0/2 β(π^2β0)β0^2/2 γ"sin" γ^(β1) (0)]
= 4[0+π^2/2 γ"sin" γ^(β1) (1)β0β0]
= 4.π^2/2. π/2
= π π^π
Using: β(π^2βπ₯^2 )dx = 1/2 β(π^2βπ₯^2 ) + π^2/2 γ"sin" γ^(β1) π₯/4 + c

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.