Find the area of the region bounded by the ellipse `(x^2)/4+(y^2)/9=1`

Explanation

The given equation of the ellipse can be represented as `(x^2)/4+(y^2)/9=1`
the standard form of ellipse is given by `(x^2)/a^2 +(y^2)/b^2=1`
by comparing we get,
`a=2` and `b=3`
`=> y= 3sqrt( 1-x^4/(4) )......(i)`
It can be observed that the ellipse is symmetrical about x-axis and y-axis.
Area bounded by ellipse = `4xx` Area `OAB`
now
Area of` OAB` `= int_0^a ydx = int_0^2 ydx`
`= int_0^2 3sqrt( 1-x^4/(4) ) dx`
`= 3/2int_0^2 sqrt(4-x^2)` dx
`= 3/2[x/(2). sqrt(4-x^2) +4/(2) sin^-1 (x/2)]_0^2`
`= 3/2 [ (2pi)/2] = (3pi)/2`
Therefore, area bounded by the ellipse `= 4xx(3pi)/2= 6pi` sq. units