• 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\( \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt \Omega } { { x^2}yz} dv \) =\( {7 \over 3} \)

  • 2022-05-29 问题

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

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

  • 2022-05-31 问题

    在其定义区间上连续的函数是( )。 A: \(f(x) = \left\{ {\matrix{ {x\quad ,{\rm{0}} \le x \le {\rm{1}}} \cr {1 - x\quad ,1 < x \le 2} \cr } } \right.\) B: \(f(x) = \left\{ {\matrix{ {x\quad ,0 < x \le 1 } \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) C: \(f(x) = \left\{ {\matrix{ {x\;\quad ,0 \le x < 1} \cr {0\;\quad \quad ,x = 1} \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) D: \(f(x) = \left\{ {\matrix{ { { 1 \over {x - 1}}\quad ,0 \le x \le 1} \cr {0\quad ,1 \le x \le 2} \cr } } \right.\)

    在其定义区间上连续的函数是( )。 A: \(f(x) = \left\{ {\matrix{ {x\quad ,{\rm{0}} \le x \le {\rm{1}}} \cr {1 - x\quad ,1 < x \le 2} \cr } } \right.\) B: \(f(x) = \left\{ {\matrix{ {x\quad ,0 < x \le 1 } \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) C: \(f(x) = \left\{ {\matrix{ {x\;\quad ,0 \le x < 1} \cr {0\;\quad \quad ,x = 1} \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) D: \(f(x) = \left\{ {\matrix{ { { 1 \over {x - 1}}\quad ,0 \le x \le 1} \cr {0\quad ,1 \le x \le 2} \cr } } \right.\)

  • 2022-05-29 问题

    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\( \int\!\!\!\int\limits_D {\left| { { x^2} + {y^2} - 1} \right|} d\sigma \) = \( {\pi \over 4} - {1 \over 2} \) 。

  • 2022-06-19 问题

    函数\(f(x) = \left\{ {\matrix{ { { x^2} - 1\;, - 1 \le x < 0} \cr {x\;\quad \;,0 \le x < 1} \cr {2 - x\;\quad ,1 \le x \le 2} \cr } } \right.\)在\(x =\)( )处间断。______

    函数\(f(x) = \left\{ {\matrix{ { { x^2} - 1\;, - 1 \le x < 0} \cr {x\;\quad \;,0 \le x < 1} \cr {2 - x\;\quad ,1 \le x \le 2} \cr } } \right.\)在\(x =\)( )处间断。______

  • 2022-10-26 问题

    设\(D = \left\{ {(x,y)\left| { { x^2} + {y^2} \le 9,x \ge 0,y \ge 0} \right.} \right\}\),则\(\int\!\!\!\int\limits_D {(x + 3y)} d\sigma = \)______

    设\(D = \left\{ {(x,y)\left| { { x^2} + {y^2} \le 9,x \ge 0,y \ge 0} \right.} \right\}\),则\(\int\!\!\!\int\limits_D {(x + 3y)} d\sigma = \)______

  • 2022-06-19 问题

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

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

  • 2022-06-09 问题

    计算\(\int\!\!\!\int\limits_\sum { { x^2}dydz + {y^2}dzdx + {z^2}} dxdy\),其中\(\sum\)为长方体\(\Omega \)的整个表面外侧,\(\Omega = \{ (x,y,z)|0 \le x \le a,0 \le y \le b,0 \le z \le c\} \)。 A: \((a + b + c)abc\) B: \((a -b + c)abc\) C: \((a + b -c)abc\) D: \((a - b - c)abc\)

    计算\(\int\!\!\!\int\limits_\sum { { x^2}dydz + {y^2}dzdx + {z^2}} dxdy\),其中\(\sum\)为长方体\(\Omega \)的整个表面外侧,\(\Omega = \{ (x,y,z)|0 \le x \le a,0 \le y \le b,0 \le z \le c\} \)。 A: \((a + b + c)abc\) B: \((a -b + c)abc\) C: \((a + b -c)abc\) D: \((a - b - c)abc\)

  • 2022-06-15 问题

    函数$y = \arcsin (2x + 1)<br/>$的定义域为 ( ). A: $\{ \left. x \right| - 1 \le x \le 0\} <br/>$ B: $\{ \left. x \right| - \frac{1}{2} \le x \le 0\} <br/>$ C: $\{ \left. x \right|x \ge - \frac{1}{2}\} <br/>$ D: ${\rm{\{ }}\left. x \right|x \le 0\}<br/>$

    函数$y = \arcsin (2x + 1)<br/>$的定义域为 ( ). A: $\{ \left. x \right| - 1 \le x \le 0\} <br/>$ B: $\{ \left. x \right| - \frac{1}{2} \le x \le 0\} <br/>$ C: $\{ \left. x \right|x \ge - \frac{1}{2}\} <br/>$ D: ${\rm{\{ }}\left. x \right|x \le 0\}<br/>$

  • 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: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

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