题目

$$
\text{求二重积分：} \int_0^1dy\int_y^1 x\sqrt{2xy-y^2} dx
$$

解答

对于 $x\sqrt{x-1}$ 这个函数 直接积分 是不太好积的，考虑 极直互化 或 交换积分次序

本题的 积分区域 是一个 角型区域，考虑化 极坐标

$$
\begin{aligned}
&\int_0^\frac{\pi}{4} d\theta \int_0^\frac{1}{\cos\theta} r\cos\theta\sqrt{2r^2\cos\theta\sin\theta - r^2\sin^2\theta} rdr \\
=&
\int_0^{\frac{\pi}{4}} d\theta \int_0^\frac{1}{\cos\theta} r^3\cos\theta \sqrt{2\sin\theta\cos\theta - \sin^2\theta} dr \\
=&
\frac{1}{4}\int_0^\frac{\pi}{4} \cos\theta \sqrt{2\sin\theta\cos\theta - \sin^2\theta} \cdot \frac{1}{\cos^4\theta} d\theta \\
=&
\frac{1}{4}\int_0^\frac{\pi}{4} \sqrt{2\tan\theta - \tan^2\theta} \cdot \frac{1}{\cos^2\theta} d\theta \\
=&
\frac{1}{4}\int_0^\frac{\pi}{4} \sqrt{2\tan\theta - \tan^2\theta}d\tan\theta \\
=&
\frac{1}{4}\int_0^1 \sqrt{2u - u^2} du = \frac{1}{4}\int_{-\frac{\pi}{2}}^0 \cos^2t dt = \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos^2t dt \\
=& \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \quad (\text{点火公式})\\
=& \frac{\pi}{16}
\end{aligned}
$$