Question

Find the area of the region enclosed between the two curves x² + y² = 9 and (x − 3)² + y² = 9.

Answer 1

Let the two curves be named as y₁ and y₂ where
\[y_1 : \left( x - 3 \right)^2 + y^2 = 9 . . . . . \left( 1 \right)\]
\[ y_2 : x^2 + y^2 = 9 . . . . . \left( 2 \right)\]
The curve x² + y² = 9 represents a circle with centre (0, 0) and the radius is 3.
The curve (x − 3)² + y² = 9 represents a circle with centre (3, 0)  and has a radius 3.
To find the intersection points of two curves equate them.
On solving (1) and (2) we get

\[x = \frac{3}{2}\text{ and }y = \pm \frac{3\sqrt{3}}{2}\]
Therefore, intersection points are

\[\left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right)\text{ and }\left( \frac{3}{2}, - \frac{3\sqrt{3}}{2} \right)\]
Now, the required area (OABO) =2 [area(OACO) +area (CABC)]
Here,

\[\text{ Area }\left( OACO \right) = \int_0^\frac{3}{2} Y_1 d x\]

\[= \int_0^\frac{3}{2} \sqrt{9 - \left( x - 3 \right)^2} d x\] 

And

\[\text{ Area }\left( CABC \right) = \int_\frac{3}{2}^3 \left| Y_2 \right| d x\]

\[= \int_\frac{3}{2}^3 \sqrt{9 - x^2}dx\]
Thus the required area is given by,
A = 2 [area(OACO) +area(CABC)]

\[2\left( \int_0^\frac{3}{2} \sqrt{9 - \left( x - 3 \right)^2} d x + \int_\frac{3}{2}^3 \sqrt{9 - x^2} d x \right)\]

\[= 2 \left[ \frac{\left( x - 3 \right)}{2}\sqrt{9 - \left( x - 3 \right)^2} + \frac{9}{2} \sin^{- 1} \left( \frac{x - 3}{3} \right) \right]_0^\frac{3}{2} + 2 \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{- 1} \left( \frac{x}{3} \right) \right]^3_\frac{3}{2}\]

\[= 2\left[ \frac{\frac{3}{2} - 3}{2}\sqrt{9 - \left( \frac{3}{3} - 3 \right)^2} + \frac{9}{2} \sin^{- 1} \left( \frac{\frac{3}{2} - 3}{3} \right) - \frac{0 - 3}{2}\sqrt{9 - \left( 0 - 3 \right)^2} - \frac{9}{2} \sin^{- 1} \left( \frac{0 - 3}{3} \right) \right] + 2\left[ \frac{3}{2}\sqrt{9 - 3^2} + \frac{9}{2} \sin^{- 1} \left( \frac{3}{3} \right) - \left( \frac{3}{4}\sqrt{9 - \frac{9}{4}} \right) - \frac{9}{2} \sin^{- 1} \left( \frac{\frac{3}{2}}{3} \right) \right]\]

\[= 2\left[ - \frac{9\sqrt{3}}{8} - \frac{9\pi}{12} + \frac{9\pi}{4} \right] + 2\left[ \frac{9\pi}{4} - \frac{9\sqrt{3}}{8} - \frac{9\pi}{12} \right]\]

\[= - \frac{18\sqrt{3}}{8} - \frac{18\pi}{12} + \frac{18\pi}{4} + \frac{18\pi}{4} - \frac{18\sqrt{3}}{8} - \frac{18\pi}{12}\]

\[= \frac{- 36\sqrt{3}}{8} - \frac{36\pi}{12} + \frac{36\pi}{4}\]

\[= - \frac{9\sqrt{3}}{2} - 3\pi + 9\pi\]

\[= 6\pi - \frac{9\sqrt{3}}{2}\]
Hence the required area is \[6\pi - \frac{9\sqrt{3}}{2}\] square units.