What is the area between two circles, radius one, that go through each other's centres?
Not sure if this requires calculus or not. I couldn't work it out. *
referring to the attached pic:
both triangles ABC and DBC can be proved equilateral and congruent to each other.
the area of sector ABC
= (1/6)(pi)(1^2)
= pi/6
and the area of triangle ABC
= (1/2)(AB)(BC)(sin60)
= (1/2)(1)(1)(3^0.5 / 2)
= 3^0.5 / 4
the difference between these two areas is the area of each one red region (in attachment), let A be it.
hence, the total area of overlapped region of the two circles
= area of triangle ABC x 2 + 4A
= 3^0.5 /4 x 2 + 4(pi/6 - 3^0.5 / 4)
= 1.2284 sq. units
1. The area of the sector (That's the area with the blue borderline) is
$\displaystyle a_{sector} = \dfrac16 \cdot \pi \cdot 1^2 = \dfrac16 \cdot \pi$
2. The area of the right triangle (That's the lightgrey area) is
$\displaystyle a_{triangle} = \dfrac12 \cdot \dfrac12 \cdot \dfrac12 \cdot \sqrt{3} = \dfrac18 \cdot \sqrt{3}$
3. The white area is
$\displaystyle a_{quart} = a_{sector} - a_{triangle} = \dfrac16 \pi - \dfrac18 \sqrt{3}$
4. The area in question is
$\displaystyle a = 4 \cdot a_{quart} = \dfrac23 \pi - \dfrac12 \sqrt{3} \approx 1.22837$