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Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

Last updated at Dec. 12, 2019 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«_0^πβγπ¦ ππ₯γ 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^2βπ₯^2)/π^2 π¦^2=π^2/π^2 (π^2βπ₯^2 ) β΄ π¦=Β±β(π^2/π^2 (π^2βπ₯^2 ) ) π¦=Β±π/π β((π^2βπ₯^2 ) ) Since OAB is in 1st quadrant, value of y is positive β΄ π¦=π/π β(π^2βπ₯^2 ) Area of ellipse = 4 Γ β«_0^πβγπ¦.ππ₯γ = 4β«_0^πβπ/π β(π^2βπ₯^2 ) ππ₯ = 4π/π β«_0^πβγβ(π^2βπ₯^2 ) ππ₯" " γ = 4π/π [π₯/2 β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ ]_0^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 = 4π/π Γ π^2/2 sin^(β1)β‘(1) = 2ππ Γsin^(β1)β‘(1) = 2ππ Γ π/2 = πππ β΄ Required Area = π ππ square units