8.设双纽线 (({x)^2+(y)^2)}^2=(2)^2((x)^2-(y)^2) 围成的平面区域记为D,则二重积分 iint ((x)^2+(y)^2)dsigma = ()-|||-(A)π )2π (C)3π (D)4π

步骤 1：转换为极坐标
将双纽线方程转换为极坐标形式。设 $x = r\cos\theta$ 和 $y = r\sin\theta$，则有：
\[
(r^2)^2 = 2^2(r^2\cos^2\theta - r^2\sin^2\theta)
\]
化简得：
\[
r^4 = 4r^2(\cos^2\theta - \sin^2\theta)
\]
\[
r^2 = 4\cos(2\theta)
\]
步骤 2：确定积分区域
双纽线的极坐标方程为 $r = 2\sqrt{\cos(2\theta)}$，其中 $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$。由于双纽线关于x轴对称，我们只需计算上半部分，然后乘以2。
步骤 3：计算二重积分
将二重积分转换为极坐标形式：
\[
\iint_D (x^2 + y^2) d\sigma = \iint_D r^2 d\sigma = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{2\sqrt{\cos(2\theta)}} r^2 \cdot r dr d\theta
\]
\[
= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{2\sqrt{\cos(2\theta)}} r^3 dr d\theta
\]
\[
= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left[ \frac{r^4}{4} \right]_0^{2\sqrt{\cos(2\theta)}} d\theta
\]
\[
= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(2\sqrt{\cos(2\theta)})^4}{4} d\theta
\]
\[
= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{16\cos^2(2\theta)}{4} d\theta
\]
\[
= 4 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(2\theta) d\theta
\]
\[
= 4 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1 + \cos(4\theta)}{2} d\theta
\]
\[
= 2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (1 + \cos(4\theta)) d\theta
\]
\[
= 2 \left[ \theta + \frac{\sin(4\theta)}{4} \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}
\]
\[
= 2 \left( \frac{\pi}{4} + 0 - \left( -\frac{\pi}{4} + 0 \right) \right)
\]
\[
= 2 \left( \frac{\pi}{4} + \frac{\pi}{4} \right)
\]
\[
= 2 \cdot \frac{\pi}{2}
\]
\[
= \pi
\]