발산과 회전

(a) Since ${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(1+yz)+\frac{\partial }{\partial y}(2y+zx)+\frac{\partial }{\partial z}(3z^2+xy)=2+6z\ne 0$, $\mathbf{F}$ fails the screening test and cannot have a vector potential.

(b) The vector field $\mathbf{A}=A_1\hat{\boldsymbol{\imath}}+A_2\hat{\boldsymbol{\jmath}}$ is a vector potential for $\mathbf{G}$ if and only if $\mathbf{G}={ \boldsymbol{\nabla} }\times\mathbf{A}$, which is the case if and only if

$$\begin{aligned} -\frac{\partial A_2}{\partial z}&= yz\quad & \iff&\quad& &A_2=-\frac{1}{2} yz^2+B_2(x,y)\\ \frac{\partial A_1}{\partial z}&=zx & \iff&\quad & &A_1=\phantom{-}\frac{1}{2} xz^2+B_1(x,y)\\ \frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y} &=xy & \iff& & & \frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy \end{aligned}$$

There are infinitely many solutions to $\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy$. In fact $B_2$ is completely arbitrary. If one chooses $B_2=0$, then $B_1=-\frac{1}{2} xy^2$ does the job. If one chooses $B_1=0$, then $B_2=\frac{1}{2} x^2y$ does the job. Thus two solutions are $\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}}$ and $\mathbf{A}=\frac{1}{2} xz^2\hat{\boldsymbol{\imath}}+\frac{1}{2}(x^2-z^2)y\hat{\boldsymbol{\jmath}}$.