Question

Using integration find the area of the region bounded between the two circles x² + y² = 9 and (x – 3)² + y² = 9.

Answer 1

Let us consider the diagram

Here, x² + y² = 9   ...(i)

(x – 3)² + y² = 9   ...(ii)

are two circles with centres (0, 0) and (3, 0) respectively.

After solving (i) and (ii),

9 – 6x = 0

⇒ `x = 3/2`

Then required area

= `2int_0^(3//2) sqrt(9 - (x - 3)^2) dx + 2int_(3//2)^3 sqrt(9 - x^2) dx`

= `2[((x - 3))/2 sqrt(9 - (x - 3)^2) + 9/2 sin^-1  ((x - 3))/3]_0^(3//2) + 2[x/2 sqrt(9 - x^2) + 9/2 sin^-1  x/3]_(3//2)^3`

Area = `2{[(-3)/4 sqrt(9 - 9/4) + 9/2 sin^-1 ((-1)/2)] - [0 + 9/2 sin^-1 (-1)] + [0 + 9/2 sin^-1 1] - [3/4 sqrt(9 - 9/4) + 9/2 sin^-1  1/2]}`

⇒ Area = `2{- (9sqrt(3))/8 - 9/2 xx π/6 + 9/2 xx π/2 + 9/2 xx π/2 - (9sqrt(3))/8 - 9/2 xx π/6}`

⇒ Area = `2{-2 xx (9sqrt(3))/8 - 9 xx π/6 + 9/2 xx π}`

⇒ Area = `2{3π - (9sqrt(3))/4}` sq. units