若D为 x^2+y^2≤1 的上半部分,计算二重积分 I=∫∫y^2√(1-x^2dxdy)

首先，根据对称性，可以将积分区域限制在第一象限中，然后进行变量替换。

令 $x = r\cos\theta, y = r\sin\theta$，则有 $J = \begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r$，于是有

$$
\begin{aligned}
I &= \int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\left(r\sin^2\theta\sqrt{1-r^2\cos^2\theta}\right)r\mathrm{d}r\mathrm{d}\theta\\
&= \int_{0}^{\frac{\pi}{2}}\int_{0}^{1}r^3\sin^2\theta\sqrt{1-r^2\cos^2\theta}\mathrm{d}r\mathrm{d}\theta
\end{aligned}
$$

接下来，对于固定的 $\theta$，考虑对 $r$ 进行积分，令 $t = r^2\cos^2\theta$，则有 $\mathrm{d}t = 2r\cos^2\theta\mathrm{d}r$，于是有

$$
\begin{aligned}
I &= \int_{0}^{\frac{\pi}{2}}\int_{0}^{1}r^3\sin^2\theta\sqrt{1-r^2\cos^2\theta}\mathrm{d}r\mathrm{d}\theta\\
&= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos^2\theta}\frac{1}{2}\sin^2\theta\sqrt{1-t}\mathrm{d}t\mathrm{d}\theta\\
&= \int_{0}^{\frac{\pi}{2}}\left[-\frac{1}{3}(1-t)^{\frac{3}{2}}\sin^2\theta\right]_{0}^{\cos^2\theta}\mathrm{d}\theta\\
&= \int_{0}^{\frac{\pi}{2}}\frac{1}{3}(1-\cos^6\theta)\sin^2\theta\mathrm{d}\theta\\
&= \frac{1}{3}\int_{0}^{\frac{\pi}{2}}\sin^2\theta\mathrm{d}\theta - \frac{1}{3}\int_{0}^{\frac{\pi}{2}}\sin^2\theta\cos^6\theta\mathrm{d}\theta\\
&= \frac{1}{6}\int_{0}^{\frac{\pi}{2}}(1-\cos 2\theta)\mathrm{d}\theta - \frac{1}{24}\int_{0}^{\frac{\pi}{2}}\sin^2\theta(1-\sin^2\theta)^3\mathrm{d}\sin\theta\\
&= \frac{1}{6}\left[\frac{\pi}{2}-\frac{1}{2}\sin 2\theta\right]_{0}^{\frac{\pi}{2}} - \frac{1}{24}\int_{0}^{1}(u-u^4)^3\mathrm{d}u\\
&= \frac{\pi}{12} - \frac{1}{24}\int_{0}^{1}(u^3-3u^2+3u-1)^3\mathrm{d}u\\
&= \frac{\pi}{12} - \frac{1}{24}\int_{0}^{1}(u^9-9u^8+36u^7-84u^6+126u^5-126u^4+84u^3-36u^2+9u-1)\mathrm{d}u\\
&= \frac{\pi}{12} - \frac{1}{24}\left[\frac{1}{10}u^{10}-\frac{9}{8}u^8+\frac{9}{2}u^7-\frac{21}{2}u^6+\frac{63}{10}u^5-\frac{42}{5}u^4+\frac{21}{2}u^3-\frac{9}{4}u^2+\frac{9}{10}u\right]_{0}^{1}\\
&= \frac{\pi}{12} - \frac{11}{320}
\end{aligned}
$$

因此，原式的值为 $\frac{\pi}{12} - \frac{11}{320}$。