Question 34 (Choice 2) Find the area of the ellipse π₯2 + 9π¦2 = 36 using integration
Equation of Ellipse is :-
π₯2 + 9π¦2 = 36
π₯^2/36+(9π¦^2)/36=1
π₯^2/36+π¦^2/4=1
π₯^2/6^2 +π¦^2/2^2 =1
Since Ellipse is symmetric along x and y-axis
Area of ellipse = Area of ABCD
= 4 Γ [Area Of OBC]
= 2 Γ β«_0^6βγπ¦.γ ππ₯
Finding y
π₯2 + 9π¦2 = 36
9π¦2 = 36 β π₯2
π¦^2=1/9 (36βπ₯^2 )
Taking square root on both sides
y = Β± β(1/9 (36βπ₯^2 ) )
y = Β± 1/3 β(36βπ₯^2 )
Since OBC is above x-axis
y will be positive
β΄ π=π/π β(ππβπ^π )
Area of ellipse = 4 Γ β«_π^πβγπ.γ π π
= 4 Γ β«_0^6βγ 1/3 β(36βπ₯^2 )γ ππ₯
= 4/3 β«_0^6ββ((6)^2βπ₯^2 ) ππ₯
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
Replacing a by 6 we get
= 4/3 [π₯/2 β((6)^2βπ₯^2 )+(6)^2/2 sin^(β1)β‘γ π₯/6γ ]_0^6
= 4/3 [6/2 β((6)^2β(6)^2 )+18 γ sinγ^(β1)β‘(6/6)β0/2 β((6)^2β(0)^2 )β18sin^(β1) (0/6)]
= 4/3 [18 sin^(β1) (1)]
= 4/3 Γ 18 Γπ/2 = 12π square units

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.