Total Area of Three Tangent Circles and the Region They Bounded

Consider the scenario where three circles with a radius of 2 are mutually tangent, forming a unique shape. The question at hand is to calculate the total area of these circles and the region bounded by them. This is a fascinating geometry problem that splits into two distinct areas: the area of the circles and the area of the region they enclose.

Step 1: Area of the Circles

The area of a single circle can be calculated using the formula:

[ A pi r^2 ]

Given that the radius of each circle is 2, we substitute:

[ A pi 2^2 pi cdot 4 4pi ]

Since there are three such circles, the total area of the circles is:

[ text{Total area of circles} 3 cdot 4pi 12pi ]

Step 2: Area of the Region Bounded by the Circles

When three circles of equal radii are mutually tangent, they form a shape known as a Reuleaux triangle. The centers of these circles create an equilateral triangle with each side equal to the circle's diameter (4 in this case).

Area of the Equilateral Triangle

The area of an equilateral triangle can be calculated with the formula:

[ A_{text{triangle}} frac{sqrt{3}}{4} s^2 ]

Substituting the side length (4 units) gives us:

[ A_{text{triangle}} frac{sqrt{3}}{4} cdot 4^2 frac{sqrt{3}}{4} cdot 16 4sqrt{3} ]

Area of the Circular Segments

Each corner of the triangle creates a circular segment, with the angle at the center of each circle subtended by the segment being 60 degrees (or (frac{pi}{3}) radians).

The area of one segment is calculated as follows:

Area of the sector of the circle:

[ A_{text{sector}} frac{1}{2} r^2 theta frac{1}{2} cdot 4 cdot frac{pi}{3} frac{2pi}{3} ]

Area of the triangle formed by the radii and the chord:

[ A_{text{triangle}} frac{1}{2} cdot r cdot r cdot sinleft(frac{pi}{3}right) frac{1}{2} cdot 4 cdot frac{sqrt{3}}{2} 2sqrt{3} ]

Thus, the area of one segment is:

[ A_{text{segment}} A_{text{sector}} - A_{text{triangle}} frac{2pi}{3} - 2sqrt{3} ]

Since there are three such segments, the total area of the segments is:

[ A_{text{segments}} 3 left( frac{2pi}{3} - 2sqrt{3} right) 2pi - 6sqrt{3} ]

Step 3: Total Area of the Bounded Region

The total area of the region bounded by the circles is the sum of the area of the triangle and the areas of the segments:

[ A_{text{bounded}} 4sqrt{3} 2pi - 6sqrt{3} 2pi - 2sqrt{3} ]

The final step is to combine the area of the circles and the bounded region:

[ text{Total area} 12pi (2pi - 2sqrt{3}) 14pi - 2sqrt{3} ]

Therefore, the final answer is:

[ boxed{14pi - 2sqrt{3}} ]