Area under curve

Chapter 8 Class 12 Application of Integrals (Term 2)
Concept wise

### Transcript

Example 1 Find the area enclosed by the circle π₯2 + π¦2 = π2Given 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^πβγπ ππγ 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^2βπ₯^2 ) Since AOBA lies in 1st Quadrant y = β(π^πβπ^π ) Now, Area of circle = 4 Γ β«1_0^πβγπ¦ ππ₯γ = 4 Γ β«1_0^πβγβ(π^2βπ₯^2 ) ππ₯γ = 4[π/π β(π^πβπ^π )+π^π/π γ"sin" γ^(βπ) π/π]_π^π = 4[π/2 β(π^2βπ^2 )+π^2/2 γ"sin" γ^(β1) π/πβ0/2 β(π^2β0)β0^2/2 γ"sin" γ^(β1) (0)] = 4[0+π^2/2 γ"sin" γ^(β1) (1)β0β0] = 4.π^2/2. π/2 = ππ^π Using: β(π^2βπ₯^2 )dx = 1/2 β(π^2βπ₯^2 ) + π^2/2 γ"sin" γ^(β1) π₯/4 + c

#### 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.