$\frac{3}{2} \pi$

#### 解析：

$L$ 为正向圆周 $x^{2}+y^{2}=2$ 在第一象限中的部分, 用参数式可表示为
$$\left\{\begin{array}{l} x=\sqrt{2} \cos \theta, \\ y=\sqrt{2} \sin \theta, \end{array} \quad \theta: 0 \rightarrow \frac{\pi}{2} .\right.$$

\begin{aligned} &\int_{L} x d y-2 y d x=\int_{0}^{\frac{\pi}{2}}[\sqrt{2} \cos \theta d \sqrt{2} \sin \theta-2 \sqrt{2} \sin \theta d \sqrt{2} \cos \theta] \\ &=\int_{0}^{\frac{\pi}{2}}[\sqrt{2} \cos \theta \cdot \sqrt{2} \cos \theta+2 \sqrt{2} \sin \theta \cdot \sqrt{2} \sin \theta] d \theta \\ &=\int_{0}^{\frac{\pi}{2}}\left[2 \cos ^{2} \theta+4 \sin ^{2} \theta\right] d \theta=\int_{0}^{\frac{\pi}{2}}\left[2\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+2 \sin ^{2} \theta\right] d \theta \\ &=\int_{0}^{\frac{\pi}{2}}\left[2+2 \sin ^{2} \theta\right] d \theta=\int_{0}^{\frac{\pi}{2}} 2 d \theta+\int_{0}^{\frac{\pi}{2}} 2 \sin ^{2} \theta d \theta=\left.2 \theta\right|_{0} ^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}(1-\cos 2 \theta) d \theta \\ &=\pi+\left.\theta\right|_{0} ^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 \theta d 2 \theta=\frac{3 \pi}{2}-\left.\frac{1}{2} \sin 2 \theta\right|_{0} ^{\frac{\pi}{2}} \\ &=\frac{3 \pi}{2}-\frac{1}{2}(\sin \pi-\sin 0)=\frac{3 \pi}{2}-0=\frac{3 \pi}{2} \end{aligned}