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प्रश्न
Find the area enclosed by the curve x = 3 cost, y = 2 sint.
उत्तर
Eliminating t as follows:
x = 3 cost
y = 2 sint
⇒ `x/3` = cos t
`y/2` = sin t
We obtain `x^2/9 + y^2/4` = 1.
Which is the equation of an ellipse.
From the figure in the question, we get
The required area = `4 int_0^3 2/3 sqrt(9 - x^2) "d"x`
= `8/3 [x/2 sqrt(9 - x^2) + 9/2 sin^-1 x/3]_0^3`
= 6π sq.units
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