$解：∵△ABC是等腰三角形，BC=\sqrt{2}$
$∴AC=\sqrt{2}，AB=2，∠DAB=45°$
$∵△ABC旋转后与△ADE重合$
$∴S_{△ABC}=S_{△ADE}，AB=AD=2，$
$AC=AE=\sqrt{2}，∠CAE=∠DAB=45°$
$∴S_{阴影部分}=S_{扇形BAD}-S_{△ABC}+S_{△ADE}-S_{扇形CAE}$
$=S_{扇形BAD}-S_{扇形CAE}$
$=\frac {45\pi ×2^2}{360}-\frac {45\pi ×( \sqrt{2} ) ^2}{360}$
$=\frac {\pi}{4}$
$∴线段BC在上述旋转过程中所扫过的面积为\frac {\pi}{4}.$
