12.类似地，计算二重积分iintlimits_(x^2+y^2leqslant 1)|x^2+y^2-(x+y)/(sqrt(2))|dx dy.

将积分区域转换为极坐标系，其中 $x = r\cos\theta$，$y = r\sin\theta$，$dx\,dy = r\,dr\,d\theta$。原积分变为 \[ \iint\limits_{r \leq 1} \left| r^2 - r\cos\left(\theta - \frac{\pi}{4}\right) \right| r\,dr\,d\theta. \] 令 $\phi = \theta - \frac{\pi}{4}$，则 \[ \int_0^{2\pi} \int_0^1 \left| r^2 - r\cos\phi \right| r\,dr\,d\phi. \] 对 $r$ 积分得 \[ \begin{cases} \frac{1}{4} - \frac{\cos\phi}{3} & \text{若 } \cos\phi \leq 0, \\ \frac{1}{4} - \frac{\cos\phi}{3} + \frac{\cos^4\phi}{6} & \text{若 } \cos\phi > 0. \end{cases} \] 对 $\phi$ 积分并利用对称性，最终结果为 \[ \boxed{\frac{9\pi}{16}}. \]