# Ex 8.1, 5 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.1, 5 Find the area of the region bounded by the ellipse 𝑥24+ 𝑦29=1 Equation of given Ellipse is :- 𝑥24+ 𝑦29=1 𝑥2 22+ 𝑦2 32=1 Area of Ellipse = Area of ABCD = 2 × [Area of BCD] = 2 × −22𝑦 .𝑑𝑥 We know that 𝑥24+ 𝑦29=1 𝑦29=1− 𝑥24 𝑦29= 4 − 𝑥24 𝑦2= 94 4− 𝑥2 Taking Square Root on Both Sides 𝑦=± 94 4−4 𝑥2 𝑦 =± 32 4− 𝑥2 Since , BCD is above x-axis ∴ 𝑦= 32 4− 𝑥2 Area of Ellipse = 2 × −22𝑦 .𝑑𝑥 = 2 × −22 32 4− 𝑥2 𝑑𝑥 = 3 −22 22− 𝑥2 𝑑𝑥 = 3 𝑥2 22− 𝑥2+ 222 sin−1 𝑥2−22 = 3 22 22− 22− −42 22− −22+2 sin−1 22− sin−1 −22 = 3 1.0+1.0+2 sin−1 1−2 𝒔𝒊𝒏−𝟏 −𝟏 = 3 0+2 sin−1 1−2(− 𝒔𝒊𝒏−𝟏 𝟏) = 3 0+2 sin−1 1+2 sin−1 1 = 3 4 𝒔𝒊𝒏−𝟏 𝟏 = 3 × 4 × 𝝅𝟐 = 6π ∴ Area of Ellipse = 6π square units

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