• 2022-06-19 问题

    设D:\(0 \le x \le 1,0 \le y \le 1\),由二重积分的几何意义及性质可知\(\int\!\!\!\int\limits_D 3 d\sigma \) =______ 。

    设D:\(0 \le x \le 1,0 \le y \le 1\),由二重积分的几何意义及性质可知\(\int\!\!\!\int\limits_D 3 d\sigma \) =______ 。

  • 2022-05-29 问题

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

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

  • 2022-06-14 问题

    曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

    曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

  • 2022-05-29 问题

    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

    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

  • 2022-06-17 问题

    求=函数[tex=3.857x1.214]3rosTj+le/zLUOkB0oXFpw==[/tex]的 [tex=0.643x0.786]/he/ol8BkDuTTL9yMPtH4Q==[/tex] 阶导数.

    求=函数[tex=3.857x1.214]3rosTj+le/zLUOkB0oXFpw==[/tex]的 [tex=0.643x0.786]/he/ol8BkDuTTL9yMPtH4Q==[/tex] 阶导数.

  • 2022-06-06 问题

    Le livre sur le bureau est à moi, _____ sur la chaise est ______. A: celui / le mien B: celle / le tien C: celui / le tien D: celle / le mien

    Le livre sur le bureau est à moi, _____ sur la chaise est ______. A: celui / le mien B: celle / le tien C: celui / le tien D: celle / le mien

  • 2022-07-02 问题

    (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: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

    (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: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

  • 2022-07-02 问题

    (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: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

  • 2022-06-06 问题

    ___ père de Marie travaille ____ bureau. A: / ; au B: Le, au C: Le; à le D: Le ,à un

    ___ père de Marie travaille ____ bureau. A: / ; au B: Le, au C: Le; à le D: Le ,à un

  • 2022-05-29 问题

    Le petit Nicolas le chocolat.

    Le petit Nicolas le chocolat.

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