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Question
Find the area of the region bounded by the ellipse `x^2/4 + y^2/9 = 1.`
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Solution
The given ellipse is `x^2/4 + y^2/9 = 1`
Since the given curve is symmetrical about both axes.
∴ Area of ellipse = 4 areas (OABO)
∴ Requied area = 4 Area (OABO) = `4 int_0^3 x dy` ...[by taking horizontal strips]
`4 int_0^3 2/3 sqrt (9 - y^2) dx` `...[x^2/4 + y^2/9 = 1 ⇒ x^2/4 = 1 - y^2/9 ⇒ x = 2/3 sqrt (9 - y^2) (∵ x > 0)]`
`= 4 xx 2/3 [y/2 sqrt (9 - y^2) + 9/2 sin^-1 y/3]_0^3`
`= 4 xx 2/3 [(3/2 (0) + 9/2 sin^-1 (1)) - (0 - 0)]`
`= 4 xx 2/3 [9/2 (pi/2)]`
`= 4 xx (3pi)/2`
= 6π square units
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