【答案】 \(12\pi - \frac{16}{3}\)

由题可知，积分区域\(D =\{(x,y)|(x - 2)^2 + y^2 \leq 2^2, x^2 + (y - 2)^2 \leq 2^2\}\)

对应图形为右图所示，

显然积分区域关于\(y = x\)对称

记\(D_1 =\{(x,y)| x^2 + (y - 2)^2 \leq 2^2, y \leq x\}\)

\(D_2 =\{(x,y)|(x - 2)^2 + y^2 \leq 2^2, y \geq x\}\)

故\(I = \iint_D (x - y)^2 dxdy = \iint_{D_1} (x - y)^2 dxdy + \iint_{D_2} (x - y)^2 dxdy\)

\(= 2\iint_{D_1} (x - y)^2 dxdy\)

令\(x = r\cos\theta, y = r\sin\theta\) 则在积分区域\(D_1\)上有 \(\begin{cases}0 \leq r \leq 4\cos\theta \\ 0 \leq \theta \leq \frac{\pi}{4} \end{cases}\)

则\(\iint_{D_1} (x - y)^2 dxdy = \iint_{D_1} (x^2 + y^2 - 2xy) dxdy = \int_{0}^{\frac{\pi}{4}} d\theta \int_{0}^{4\cos\theta} (r^2 - 2r^2 \cos\theta\sin\theta) dr\)

\(= \int_{0}^{\frac{\pi}{4}} d\theta \int_{0}^{4\cos\theta} (r^3 - 2r^3 \cos\theta\sin\theta) dr = 6\pi - \frac{8}{3}\)

故\(I = 2(6\pi - \frac{8}{3}) = 12\pi - \frac{16}{3}\) 。