보존 벡터장

Let $\varphi$ be a potential for $\mathbf{F}$. Define $\phi=\varphi+ax+by+cz$. Then ${ \boldsymbol{\nabla} } \phi = { \boldsymbol{\nabla} }\varphi+(a,b,c)=\mathbf{F}+(a,b,c)$. So, $\mathbf{F}+(a,b,c)$ is also conservative.

(a) We shall show that $\mathbf{F}(x,y,z)$ is conservative by finding a potential for it. $\varphi(x,y,z)$ is a potential for this $\mathbf{F}$ if and only if

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y,z) &= x \\ \frac{\partial \varphi}{\partial y}(x,y,z) &= -2y \\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 3z \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y,z) = \frac{x^2}{2} + f(y,z)$$

Substituting this into the second equation gives

$$\frac{\partial f}{\partial y}(y,z) = -2y$$

which integrates to

$$f(y,z) = -y^2+ g(z)$$

Finally, substituting $\varphi(x,y,z) = \frac{x^2}{2} -y^2 + g(z)$ into the last equation gives

$$g'(z) = 3z$$

which integrates to

$$g(z) = \frac{3}{2} z^2 +C$$

with $C$ being an arbitrary constant. So, $\mathbf{F}(x,y,z)$ is conservative and $\varphi(x,y,z)=\tfrac{1}{2} x^2-y^2+\frac{3}{2}z^2$ is one allowed potential.

(b) The field $\mathbf{F}= F_1\,\hat{\boldsymbol{\imath}} + F_2\,\hat{\boldsymbol{\jmath}}$ can be conservative only if it passes the screening test

$$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$$

In this case

$$\frac{\partial F_1}{\partial y} =\frac{\partial}{\partial y}\Big(\frac{x}{x^2+y^2}\Big) =-\frac{2xy}{{(x^2+y^2)}^2}$$

is different from

$$\frac{\partial F_2}{\partial x} =\frac{\partial}{\partial x}\Big(\frac{-y}{x^2+y^2}\Big) =\frac{2xy}{{(x^2+y^2)}^2}$$

for all $(x,y)$ with $x$ and $y$ both nonzero. So $\mathbf{F}$ is not conservative.