举一反三
- 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\(D = \left\{ {(x,y)\left| { { x^2} + {y^2} \le 9,x \ge 0,y \ge 0} \right.} \right\}\)，则\(\int\!\!\!\int\limits_D {(x + 3y)} d\sigma = \)______
- 下列方程中，不是全微分方程的为（ ）。 A: \(\left( {3{x^2} + 6x{y^2}} \right)dx + \left( {6{x^2}y + 4{y^2}} \right)dy = 0\) B: \({e^y}dx + \left( {x \cdot {e^y} - 2y} \right)dy = 0\) C: \(y\left( {x - 2y} \right)dx - {x^2}dy = 0\) D: \(\left( { { x^2} - y} \right)dx - xdy = 0\)
- 函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( ) A: $\left\{(x,y)\left|~x+y\geq<br/>0\right.\right\}$; B: $\left\{(x,y)\left|~x+y\geq<br/>1~\text{或}~x+y\leq<br/>-1 \right.\right\}$; C: $\left\{(x,y)\left|~x+y\geq<br/>1\right.\right\}$; D: $\left\{(x,y)\left|~x+y\geq<br/>\dfrac{4}{~\pi^2~}\right.\right\}$．
- 设\(D\)是由\( - 1 \le x \le 1 \) ，\( 0 \le y \le 2 \) 所围区域，则\( \int\!\!\!\int\limits_D {\left| {y - {x^2}} \right|} d\sigma \) = \( { { 45} \over {16}} \) 。
内容
- 0
曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为（ ） A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
- 1
设D是由\( 0 \le x \le 1 \) ，\( 0 \le y \le 1 \) 所围区域，则\( \int\!\!\!\int\limits_D {\left| { { x^2} + {y^2} - 1} \right|} d\sigma \) = \( {\pi \over 4} - {1 \over 2} \) 。
- 2
设\(D = \left\{ {(x,y)\left| { { x^2} + {y^2} \le 4,x \ge 0,y \ge 0} \right.} \right\}\)，则\(\int\!\!\!\int\limits_D {(x + y)} d\sigma = \) A: \(0\) B: \( { { 8} \over 3}\) C: \( { { 16} \over 3}\) D: \( { { 32} \over 3}\)
- 3
设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\)，则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)
- 4
若\({y_1}\left( x \right), {y_2}\left( x \right)\)都是\(y' + P\left( x \right)y = Q\left( x \right)\)的特解，且 \({y_1}\left( x \right), {y_2}\left( x \right)\) 线性无关，则通解可表为\(y\left( x \right) = {y_1}\left( x \right) + C\left[ { { y_1}\left( x \right) - {y_2}\left( x \right)} \right]\)。