Find the area of region enclosed between two curves (x-6)^2+y^2=36 and x^2+y^2=36
Answers
Required area is (24π - 18√3) sq. units
Step-by-step Explanation:
The given circles are
(x - 6)² + y² = 6² ..... (1)
x² + y² = 6² ..... (2)
The centre of the circle (1) is at (6, 0) and its radius is 6 units, and the centre of the circle (2) is at (0, 0) and its radius is 6 units.
From (1) and (2), we write
x² - 12x + 36 + y² = 36
x² + y² = 36
On subtraction, we get
12x = 36
or, x = 3
Then y = ± 3√3
So the point of intersections are (3, 3√3) and (3, - 3√3)
We consider the point (3, 3√3) to be B. Then the coordinates of the point A be (3, 0).
Now we proceed to find the area of the closed region ABCOA.
The OCB curve is given by the circle (1).
Therefore the area of the region ABCOA is
= ∫₀³ √{6² - (x - 6)²} dx
= [ (x - 6)√{6² - (x - 6)²}/2 + 6²/2 * sin⁻¹(x - 6)/6]₀³
= [ (x - 6)/2 * √(12x - x²) + 18 * sin⁻¹(x - 6)/6]₀³
= {- 3/2 * √27 + 18 * sin⁻¹(- 1/2)} - {18 * sin⁻¹(- 1)}
= - (9√3)/2 + 18 (- π/6) - 18 (- π/2)
= - (9√3)/2 - 3π + 9π
= 6π - (9√3)/2 sq. units
Due to symmetry, the area of the enclosed region by two circles is
= 4 * {6π - (9√3)/2} sq. units
= (24π - 18√3) sq. units
