Question

Represent the area of the region determined by the intersection of the circles x

2 + y
2 = 4

and (x − 2)2 + (y − 2)2 = 4 by an integral. Do not evaluate the integral.

Answer 1

I think you mean the circles

x² + y² = 4

(x - 2)² + (y - 2)² = 4

Notice that both have radius 2, but the first one is centered at the origin (0, 0) while the second one is centered at (2, 2). The two circles intersect at the points (2, 0) and (0, 2), which means the bounded region between them is contained entirely in the first quadrant.

If we solve both equations above for y, we would find two solutions in each case:

x² + y² = 4 ⇒ y = ± √(4 - x²)

(x - 2)² + (y - 2)² = 4 ⇒ y = 2 ± √(4 - (x - 2)²)

The positive/negative root corresponds to the upper/lower semicircle the make up the full circle. The semicircles we're interested in are

y = + √(4 - x²)

y = 2 - √(4 - (x - 2)²)

that is, the upper half of the circle centered at the origin, and the lower half of the circle centered at (2, 2).

So, in standard Cartesian coordinates, the bounded region (call it R) is the set

`R = \left\{(x, y) : 0 \le x \le 2 \text{ and } 2-√(4-(x-2)^2) \le y \le √(4-x^2) \right\}`

and the area is given by the double integral

`\displaystyle \iint_R dA = \boxed{\int_0^2 \int_(2-√(4-(x-2)^2))^(√(4-x^2)) dy \, dx}`