\(\left\{ {\left( {x,y} \right)\left| {2 \le {x^2} + {y^2} \le 4} \right.} \right\}\)是闭区域.
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\( \int\!\!\!\int\limits_D {\left| {y - {x^2}} \right|} d\sigma \) = \( { { 45} \over {16}} \) 。