6.3 Conservative vector fields  (Page 7/16)

Suppose that $f$ is a potential function. Then, $\text{∇}f=\text{F}$ and therefore

To integrate this function with respect to x, we can use u -substitution. If $u=x^2+y^2,$ then $\frac{du}{2}=xdx,$ so

for some function $h(y).$ Therefore,

Since $f$ is a potential function for F ,

Since $f(x,y)=\frac{G}{\sqrt{x^2+y^2}}+h(y),$ $f_y$ also equals $\frac{\text{−}Gy}{{\left(x^2+y^2\right)}^{3\text{/}2}}+h^{\prime }\left(y\right).$

Therefore,

which implies that $h^{\prime }(y)=0.$ Thus, we can take $h(y)$ to be any constant; in particular, we can let $h(y)=0.$ The function

is a potential function for the gravitational field F . To confirm that $f$ is a potential function, note that

Note that the domain of F is all of ${ℝ}^2$ and ${ℝ}^3$ is simply connected. Therefore, we can use [link] to determine whether F is conservative. Let

Since $Q_z=x^{2}y$ and $R_y=0,$ the vector field is not conservative.

Note that the domain of F is the part of ${ℝ}^2$ in which $y>0.$ Thus, the domain of F is part of a plane above the x -axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use [link] to determine whether F is conservative. Let

Then $P_y=\frac{x}{y}=Q_x$ and thus F is conservative.