Question

Find the area enclosed by the curve x = 3 cost, y = 2 sint.

Answer 1

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