# Example 2 - Chapter 8 Class 12 Application of Integrals

Last updated at April 16, 2024 by Teachoo

Examples

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Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 2 Find the area enclosed by the ellipse π₯^2/π^2 +π¦^2/π^2 =1 We have to find Area Enclosed by ellipse Since Ellipse is symmetrical about both x-axis and y-axis β΄ Area of ellipse = 4 Γ Area of OAB = 4 Γ β«_π^πβγπ π πγ We know that , π₯^2/π^2 +π¦^2/π^2 =1 π¦^2/π^2 =1βπ₯^2/π^2 π¦^2/π^2 =(π^2βγ π₯γ^2)/π^2 π¦^2=π^2/π^2 (π^2βπ₯^2 ) π¦=Β±β(π^2/π^2 (π^2βπ₯^2 ) ) π=Β±π/π β((π^πβπ^π ) ) Since OAB is in 1st quadrant, value of y is positive β΄ π=π/π β(π^πβπ^π ) Area of ellipse = 4 Γ β«_0^πβγπ¦.ππ₯γ = 4β«_π^πβπ/π β(π^πβπ^π ) π π = 4π/π β«_0^πβγβ(π^2βπ₯^2 ) ππ₯" " γ = ππ/π [π/π β(π^πβπ^π )+π^π/π γπππγ^(βπ)β‘γπ/πγ ]_π^π = 4π/π [(π/2 β(π^2βπ^2 )+π^2/2 sin^(β1)β‘γπ/πγ )β(0/2 β(π^2β0)βπ^2/2 sin^(β1)β‘(0) )] = 4π/π [0+π^2/2 sin^(β1)β‘γ(1)γβ0β0] It is of form β(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ = ππ/π Γ π^π/π γπ¬π’π§γ^(βπ)β‘(π) = 2ππ Γsin^(β1)β‘(1) = 2ππ Γ π/2 = π ππ β΄ Required Area = π ππ square units