设$L$是平面$x\cos\alpha+y\cos\beta+z\cos\gamma=1$上的闭曲线, 它所包围的区域面积为$S$, 则曲线积分$$\oint_L (z\cos\beta-y\cos\gamma)dx+(x\cos\gamma-z\cos\alpha)dy+(y\cos\alpha-x\cos\beta)dz=S,$$其中$L$依正向进行。
举一反三
- (4)$A$矢量的方向余弦(与三个坐标轴的夹角余弦)的大小是： A: $cos\alpha=3/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ B: $cos\alpha=4/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ C: $cos\alpha=2/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ D: $cos\alpha=3/\sqrt{14},cos\beta=9/\sqrt{14},cos\gamma=3/\sqrt{14}$
- 以\( (2,2,1) \)为起点，以\( (1,3,0) \)为终点的向量的方向余弦为（ ）. A: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = {1 \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) B: \( \cos \alpha = {1 \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) C: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) D: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = {1 \over {\sqrt 3 }} \)
- \(设f(x,y,z)=\frac{x\cos y+y\cos z+z\cos x}{1+\cos x+\cos y+\cos z},则df|_{(0,0,0)}=(\,)\)
- 曲线积分$$\int_{(0,0}^{(x,y)}(2x\cos y-y^2\sin x)dx+(2y\cos x-x^2\sin y)dy=$$ A: $y^2\cos x+x^2\cos y$ B: $x^2\cos x+y^2\cos y$ C: $x^2\sin y+y^2\sin x$ D: $x^2\sin x+y^2\sin y$
- \((\cos\alpha \cos\beta, \cos\alpha \sin\beta, \sin\alpha),(1,1,0),(1,2,1)\)张成六面体体积最大为___. A: \(\sqrt{3}\) B: \(2\sqrt{3}\) C: \(\sqrt{6}\)