So above is a hypothetical problem I gave myself to figure out if I could do potential differences properly using Newton's Gravitation Formula and I wasn't 100% that my Math was correct (I'm basically self-taught on line integrals), so I wanted some feedback on that. The problem is that there is some mass $M$ evenly distributed in a circle with radius 1 m at some point in 2-space. What is the gravitational potential difference between the surface of this 2D "planet" and the point $(x_1, y_1)$? (Pretend there are subscripts in my diagram). For convenience purposes, I centered the "planet" at the origin.
$$V_g = \int_C \vec{F_g}(x,y) \cdot d\vec{s}$$ $$\vec{F_g}(x,y) = \frac{GM}{x_1^2 + y_1^2}\boldsymbol{\hat{\textbf{r}}}$$ $$V_g = \int_C \frac{GM}{x_1^2 + y_1^2}\boldsymbol{\hat{\textbf{r}}} \cdot d\vec{s}$$ $$ = \int_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1}\frac{GM}{x^2 + \sin^2(\tan^{-1}\frac{y_1}{x_1})}dx + \int_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\frac{GM}{x_1^2 + y^2}dy$$ $$=GM\bigg(\int_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1}\frac{1}{x^2 + \sin^2(\tan^{-1}\frac{y_1}{x_1})}dx + \int_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\frac{1}{x_1^2 + y^2}dy\bigg)$$ $$=GM\bigg(\bigg[\frac{1}{\sin(\tan^{-1}\frac{y_1}{x_1})}\tan^{-1}\bigg(\frac{x}{\sin(\tan^{-1}\frac{y_1}{x_1})}\bigg)\bigg|_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1} \bigg] + \bigg[ \frac{1}{x_1}\tan^{-1}\bigg(\frac{y}{x_1}\bigg)\bigg|_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\bigg]\bigg)$$
Also, yes, I realize I unnecessarily punished myself using Cartesian coordinates, so here's the whole thing in polar coordinates.
$$V_g = \int_1^{r_1} \frac{GM}{r^2}dr = \int_1^{r_1} GMr^{-2}dr = \frac{-GM}{r}\bigg|_1^{r_1} = GM - \frac{GM}{r_1}$$
