设平面有界区域 D 位于第一象限, 由曲线 $x^2+y^2-x y=1, x^2+y^2-x y=2$
与直线 $y=\sqrt{3} x, y=0$ 围成, 计算 $\iint_D \frac{1}{3 x^2+y^2} d x d y$.
【答案】 本题目采用极坐标进行计算
$$
\begin{aligned}
& \iint_D \frac{1}{3 x^2+y^2} d \sigma \\
& =\int_0^{\frac{\pi}{3}} \frac{1}{\left(3 \cos ^2 \theta+\sin ^2 \theta\right)} \cdot \ln r \mid \frac{\sqrt{\frac{2}{1-\sin \theta \cos \theta}}}{\sqrt{1-\cos \theta \cos \theta}} d \theta=\int_0^{\frac{\pi}{3}} \frac{1}{\left(3 \cos ^2 \theta+\sin ^2 \theta\right)} \cdot \ln \sqrt{2} d \theta \\
& =\frac{1}{2} \ln 2 \cdot \int_0^{\frac{\pi}{3}} \frac{1}{\left(3+\tan ^2 \theta\right) \cdot \cos ^2 \theta} d \theta=\ln 2 \cdot \int_0^{\frac{\pi}{3}} \frac{1}{\left(3+\tan ^2 \theta\right)} d \tan \theta \\
& =\left.\frac{1}{2} \ln 2 \cdot \frac{1}{\sqrt{3}} \arctan \frac{\tan \theta}{\sqrt{3}}\right|_0 ^{\frac{\pi}{3}}=\frac{\pi}{8 \sqrt{3}} \ln 2 . \\
&
\end{aligned}
$$


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