设D是由\( 0 \le x \le 1 \) ，\( 0 \le y \le 1 \) 所围区域，则\( \int\!\!\!\int\limits_D {\left| { { x^2} + {y^2} - 1} \right|} d\sigma \) = \( {\pi \over 4} - {1 \over 2} \) 。

设D：\(0 \le x \le \pi ,0 \le y \le {\pi \over 2}\)，则\(\int\!\!\!\int\limits_D {sinxcosydxdy} \)的值为______

（接上题）（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

求向量$A = xi + yj + zk$通过闭区域$\Omega = \left\{ {\left( {x,y,z} \right)\left| {0 \le x \le 1,0 \le y \le 1,0 \le z \le 1} \right.} \right\}$的边界曲面流向外侧的通量。 A: 2 B: 3 C: 4 D: 5

设\( \Omega \) 是由\( 1 \le x \le 2 \) ，\( 0 \le y \le 1 \) ，\( 0 \le z \le 2 \) 所围区域，则\( \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt \Omega } { { x^2}yz} dv \) =\( {7 \over 3} \)

对于任意随机事件$A$，则$0\le P(A)\le 1$.

设\(D\)是由\( 0 \le x \le 1 \) ，\( 0 \le y \le 1 \) 所围区域，则\( \int\!\!\!\int\limits_D {x{y^2}} dxdy \) = \( {1 \over 6} \) 。

已知 $X$ 和 $Y$ 的联合密度函数为 $f(x,y)=$ $\begin{cases} cxy,& 0\le x\le 1, 0\le y\le 1,\\ 0,& \text{其他},\end{cases}$，则$c=$______ ， $P\{X

设 $D=\{(x,y)|x^2+y^2\le 4, y\ge 0\}$，则二重积分 $\iint_D xy^2dxdy=$______ .