Question

Find the area of the circle x² + y² = 6² 

Answer 1

By the symmetry of the circle, required area of the circle is 4 times the area of the region OPQO.

For the region OPQO,

The limits of integration are x = 0 and x = 6.

Given equation of the circle is x² + y² = 6²  

∴ y² = 6² – x² 

∴ y = `+- sqrt(6^2 - x^2)`

∴ y = `sqrt(6^2 - x^2)`  ......[∵ In first quadrant, y > 0]

∴ Required area = 4 (area of the region OPQO)

= `4 xx int_0^6 y*"d"x`

= `4 xx int_0^6 sqrt(6^2 - x^2)  "d"x`

= `4[x/2 sqrt((6)^2 - x^2) + (6)^2/2 sin^-1 (x/6)]_0^6`

= `4{[6/2 sqrt((6)^2 - (6)^2) + (6)^2/2 sin^-1 (6/6)] - [0/2 sqrt((6)^2 - (0)^2) + (6)^2/2 sin^-1 (0/6)]}`

= `4{[0 + 36/2 sin^-1 (1)] - [0 + 0]}`

= `4(36/2 xx pi/2)`

= 36π sq.units