设有界单连通闭区域$D$的边界正向曲线为$C$,$P(x,y),Q(x,y)$在该闭区域$D$上存在一阶连续偏导数,则用格林公式将曲线积分$\displaystyle\oint_CP(x,y)dx+Q(x,y)dy$化为二重积分的形式为____
A. $\displaystyle \iint_D(\frac{\partial{Q}}{\partial{x}}+\frac{\partial{P}}{\partial{y}})d\sigma$
B. $\displaystyle \iint_D(\frac{\partial{P}}{\partial{x}}-\frac{\partial{Q}}{\partial{y}})d\sigma$
C. $\displaystyle \iint_D(Q-P)d\sigma$
D. $\displaystyle \iint_D(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}})d\sigma$
查看答案
函数 $f(x,y)$ 在点 $(x_0,y_0)$ 处连续且可偏导, 是 $f(x,y)$ 在点 $(x_0,y_0)$ 可微的 .
A. 充分必要条件;
B. 必要但非充分条件;
C. 充分但非必要条件;
D. 既非充分也非必要条件.
三重积分$\displaystyle I=\iiint_\Omega(x^2+y^2+z^2)\mathrm{d}v$,其中$\Omega$是由$z^2=x^2+y^2$与$z=-1$围成的区域,则$I$可化为____
A. $\displaystyle\int_0^{2\pi}\mathrm{d}\theta\int_0^1\rho \mathrm{d}\rho\int_0^\rho(\rho^2+z^2)\mathrm{d}z$
B. $\displaystyle\int_0^{2\pi}\mathrm{d}\theta\int_0^1\rho \mathrm{d}\rho\int_\rho^{-1}(\rho^2+z^2)\mathrm{d}z$
C. $\displaystyle4\int_0^{\frac{\pi}{2}}d\theta\int_0^1\rho d\rho\int_{-1}^{-\rho}(\rho^2+z^2)dz$
D. $\displaystyle\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^1\rho \mathrm{d}\rho\int_\rho^{-1}(\rho^2+z^2)\mathrm{d}z$
设有界单连通闭区域$D$的边界正向曲线为$C$,$P(x,y),Q(x,y)$在该闭区域$D$上存在一阶连续偏导数,则用格林公式将曲线积分$\displaystyle\oint_CP(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y$化为二重积分的形式为____
A. $\displaystyle \iint_D(\frac{\partial{Q}}{\partial{x}}+\frac{\partial{P}}{\partial{y}})\mathrm{d}\sigma$
B. $\displaystyle \iint_D(\frac{\partial{P}}{\partial{x}}-\frac{\partial{Q}}{\partial{y}})\mathrm{d}\sigma$
C. $\displaystyle \iint_D(Q-P)\mathrm{d}\sigma$
D. $\displaystyle \iint_D(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}})\mathrm{d}\sigma$
设$P(x,y),Q(x,y)$在单连通区域$D$内存在一阶连续偏导数,则曲线积分$\displaystyle\int_CQ(x,y)\mathrm{d}x-P(x,y)\mathrm{d}y$在$D$内积分与路经$C$无关的充分必要条件是____
A. $\displaystyle \frac{\partial{Q}}{\partial{x}}=\frac{\partial{P}}{\partial{y}}$
B. $\displaystyle -\frac{\partial{P}}{\partial{x}}=\frac{\partial{Q}}{\partial{y}}$
C. $\displaystyle \frac{\partial{Q}}{\partial{x}}=-\frac{\partial{P}}{\partial{y}}$
D. $\displaystyle \frac{\partial{Q}}{\partial{y}}=\frac{\partial{P}}{\partial{x}}$
微信支付 
支付宝支付