Question

The area of the region enclosed between the circles $x^2 + y^2 = 4$ and $x^2 + (y - 2)^2 = 4$ is:

• A. $\frac{4}{3}(2\pi - 3\sqrt{3})$
• B. $\frac{2}{3}(4\pi - 3\sqrt{3})$
• C. $\frac{4}{3}(2\pi - \sqrt{3})$
• D. $\frac{2}{3}(2\pi - 3\sqrt{3})$

Hint:

Symmetry plays a huge role in area between curves. Always check if the area is composed of identical parts.

Answer 1

The Correct Option is B

The circles intersect where $x^2+y^2 = x^2+(y-2)^2$.
$y^2 = y^2 - 4y + 4 \implies 4y = 4 \implies y = 1$.
Intersection points are $(\pm\sqrt{3}, 1)$.
Consider the sector of the circle $x^2+y^2=4$ from angles corresponding to intersection.
Center $(0,0)$, radius 2. Chord is at $y=1$. Distance $d=1$.
Angle $\alpha$: $\cos \alpha = 1/2 \implies \alpha = \pi/3$. Total angle $2\alpha = 2\pi/3$.
Area of circular segment = Area(Sector) - Area(Triangle).
$A_{seg} = \frac{1}{2}r^2(2\alpha) - \frac{1}{2}r^2\sin(2\alpha) = \frac{1}{2}(4)(\frac{2\pi}{3}) - \frac{1}{2}(4)(\frac{\sqrt{3}}{2})$.
$A_{seg} = \frac{4\pi}{3} - \sqrt{3}$.
The total area is composed of two such identical segments (one from each circle).
Total Area = $2(\frac{4\pi}{3} - \sqrt{3}) = \frac{8\pi}{3} - 2\sqrt{3}$.
Factor out 2/3: $\frac{2}{3}(4\pi - 3\sqrt{3})$.