📝 题目
＊3．求下列向量场 $\boldsymbol{A}$ 的散度： （1） $\boldsymbol{A}=\left(x^{2}+y z\right) \boldsymbol{i}+\left(y^{2}+x z\right) \boldsymbol{j}+\left(z^{2}+x y\right) \boldsymbol{k}$ ； （2）$A=\mathrm{e}^{x y} i+\cos (x y) j+\cos \left(x z^{2}\right) k$ ； （3）$A=y^{2} i+x y j+x z k$ ．
💡 答案与解析
[AI解答]
[AI解答]
**（1）** 向量场 $$ \boldsymbol{A} = (x^{2}+yz)\boldsymbol{i} + (y^{2}+xz)\boldsymbol{j} + (z^{2}+xy)\boldsymbol{k} $$ 散度公式为 $$ \operatorname{div}\boldsymbol{A} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$ 其中 $$ P = x^{2}+yz,\quad Q = y^{2}+xz,\quad R = z^{2}+xy $$ 计算偏导数： $$ \frac{\partial P}{\partial x} = 2x,\quad \frac{\partial Q}{\partial y} = 2y,\quad \frac{\partial R}{\partial z} = 2z $$ 因此 $$ \operatorname{div}\boldsymbol{A} = 2x + 2y + 2z = 2(x+y+z) $$
**（2）** 向量场 $$ \boldsymbol{A} = e^{xy}\boldsymbol{i} + \cos(xy)\boldsymbol{j} + \cos(xz^{2})\boldsymbol{k} $$ 令 $$ P = e^{xy},\quad Q = \cos(xy),\quad R = \cos(xz^{2}) $$ 计算偏导数： $$ \frac{\partial P}{\partial x} = y e^{xy} $$ $$ \frac{\partial Q}{\partial y} = -x \sin(xy) $$ $$ \frac{\partial R}{\partial z} = -2xz \sin(xz^{2}) $$ 因此 $$ \operatorname{div}\boldsymbol{A} = y e^{xy} - x \sin(xy) - 2xz \sin(xz^{2}) $$
**（3）** 向量场 $$ \boldsymbol{A} = y^{2}\boldsymbol{i} + xy\boldsymbol{j} + xz\boldsymbol{k} $$ 令 $$ P = y^{2},\quad Q = xy,\quad R = xz $$ 计算偏导数： $$ \frac{\partial P}{\partial x} = 0,\quad \frac{\partial Q}{\partial y} = x,\quad \frac{\partial R}{\partial z} = x $$ 因此 $$ \operatorname{div}\boldsymbol{A} = 0 + x + x = 2x $$
**最终答案：** （1）$\displaystyle \operatorname{div}\boldsymbol{A} = 2(x+y+z)$ （2）$\displaystyle \operatorname{div}\boldsymbol{A} = y e^{xy} - x \sin(xy) - 2xz \sin(xz^{2})$ （3）$\displaystyle \operatorname{div}\boldsymbol{A} = 2x$
难度：★☆☆☆☆