Find the area of the region bounded by the ellipse `x^2 / a^2 + y^2 / b^2 = 1`

Explanation

The given equation contains only even power of y
So, the cuve is symmetrical about the x-axis.
Also , the given equations contains only even powes of x.
So, the curve is symmetrival about the y-axies ltbgt A rough sketche of the ellipse can be drawn,as showm in the figure .
Now `x^2/a^2+y^2/b^2=1 rArr y= b/a . sqrt(a^2 -x^2)`
Area of the given elliopse
`= 4xx` (area OBCO)
`= 4xx underset(0)overset(a)int"y dx " = 4xx underset(0)overset(a)intb/a.sqrt(a^2 -x^2)dx`
`=(4b)/(a).underset(0)overset(pi//2)int a^2 cos^2 theta d theta` [ putting x = a sin `theta " so that dx =a cos " theta d theta]`
`=4r^2.underset(0)overset(pi//2)int((1+cos2 theta ))/2d theta =(2ab).[theta+(sin2 theta)/(2)]_0^(pi//2)`
`= (pi ab)` sq units
Hence , the area of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=" 1 is pi "ab ` sq units