By using intergration prove that the area of a circle of radius r units is `pir^2` square units.

Explanation

The equations of a circle of radius r units with its centre at the origin is
`x^2 + y^2 =r^2 …(i) `
This equation contaions only even powers of y.
So the curve is symmetrical about the x-axis .Also the above equation contains only even powers of x . So the curve is symmetrical about the y- axies
Now `x^2+ y^2=r^2rArr y = sqrt(r^2 -x^2)`
Area of the circle `=4 xx` (area OABO)
`=4xxunderset(0)overset(r)inty dx = 4 xx underset(0)overset(r)intsqrt(r^2-x^2)dx`
`= 4xxoverset(pi//2)underset(0)intr^2cos^2theta d theta` ["putting"x=r sin `theta`]
`=4r^2.underset(0)overset(pi//2)int((1+cos 2 theta))/(2)d theta= 2r^2.[ theta + (sin 2 theta)/(2)]_0 ^(pi//2)`
`(pir^2) ` sq units
Hence , the required area is `(pi r^2)` sq units