# Finding a potential function for $\vec F=\frac{-y}{x^2+y^2}\vec i+\frac{x}{x^2+y^2}\vec j$ on $\Omega=\mathbb{R}^2- \{(x,y)\,|\,y=0,\,x\geq0 \}$

In one of my Calculus III classes the professor presented the following vector field, defined on the set $$S$$ of all points $$(x,y) \neq(0,0)$$ : $$\bbox[6px,border:1px solid black] { \vec{F}=\frac{-y}{x^2+y^2}\vec{i} + \frac{x}{x^2+y^2}\vec{j} }$$

Although $$\vec{F}$$ is not a gradient on $$S$$, it is a gradient on the set $$\Omega=\mathbb{R}^2-\left\{(x,y)\,|\,y=0, \, x\geq0\right\}$$, i.e, all points in the xy-plane except those on the positive x-axis. In the class, my professor wrote a potential function for $$\vec{F}$$ on $$\Omega$$ using the following functions:

\bbox[6px,border:1px solid black] { \begin{alignat}{0} \Psi_1(x,y)=-\arctan\left(\frac x y \right), \,y\neq0 &\text{and} &\Psi_2(x,y)=\arctan\left(\frac y x \right), \,x\neq0 \end{alignat} }

The first is a potential of $$\vec F$$ on $$\Omega^+=\{(x,y)\in\mathbb{R}^2\,|\,y>0\}$$ and on $$\Omega^-=\{(x,y)\in\mathbb{R}^2\,|\,y<0\}$$.

The last one is a potential of $$\vec F$$ too, but on $$\Omega_+=\{(x,y)\in\mathbb{R}^2\,|\,x>0\}$$ and on $$\Omega_-=\{(x,y)\in\mathbb{R}^2\,|\,x<0\}$$

The potential function wrote by my professor: $$\bbox[8px,border:1px solid black] { \phi (x,y) = \begin{cases} \Psi_1(x,y) & \text{if (x,y) \in R_1} \\[2ex] \Psi_2(x,y)+\frac\pi 2 & \text{if (x,y) \in R_2} \\[2ex] \Psi_1(x,y)+2\pi & \text{if (x,y) \in R_3} \\ \end{cases} }$$ (Note: I remember that my professor started with the argument that since $$\nabla\Psi_1=\vec F = \nabla\Psi_2$$, then $$\Psi_1-\Psi_2=k, \,k \in \mathbb{R}$$)

My doubts:

• I've tried a lot, but still have not figured out how to build $$\phi (x,y)$$ using $$\Psi_1$$ and $$\Psi_2$$. So, how to find this specific potential function of $$\vec F$$ ?
• Why can't I simply write potential functions of $$\vec{F}$$ in the following ways \phi_A (x,y) = \begin{alignat}{0} \begin{cases} \Psi_1(x,y) & \text{if (x,y) \in R_1} \\[2ex] \Psi_2(x,y) & \text{if (x,y) \in R_2} \\[2ex] \Psi_1(x,y) & \text{if (x,y) \in R_3} \\ \end{cases} &\text{or} &\phi_B(x,y)= \begin{cases} \Psi_1(x,y) & \text{if y>0} \\[2ex] 0 & \text{if y=0 and x<0} \\[2ex] \Psi_1(x,y) & \text{if y<0} \\ \end{cases} \end{alignat} since $$\nabla\phi_A=\nabla\phi_B=\vec F$$ on $$\Omega$$ ?
• Exactly how are $R_1$, $R_2$ and $R_3$ defined? May 12, 2019 at 14:37
• Hint: What values do $\Psi_1$ and $\Psi_2$ have on the four halfaxes? Are $\phi_A$ and $\phi_B$ continuous? May 12, 2019 at 14:39
• Using polar coordinates $(r,\alpha)$, where $\alpha$ is the counterclockwise angle from the positive x-axis, we have $\phi=\alpha - \pi/2$ on all of $\Omega.$ The need to split $\Omega$ into regions occurs when we want to express $\phi$ in terms of $x$ and $y$. May 12, 2019 at 15:32
• Since they aren't continuous, they are not potentials for all of $\Omega$. May 12, 2019 at 15:38
• In that case you have one patch covering all of the domain, which excludes origin. May 12, 2019 at 16:17