I have 3 questions regarding the solution to the following problem:
Let $$\vec{F} = M \hat i + N \, \hat j=\big(\sqrt{x^2+y^2}\big)^n (x \, \hat i + y \, \hat j)$$Whenever possible, find a potential function $f$ such that $\vec{F}=\nabla f$.
Let's start with the "whenever possible": $\vec{F}=\nabla f \implies \text{curl} \, \vec F = 0 \iff N_y-M_x = 0$. So $$n x y (x^2 + y^2)^{n/2 - 1} - n x y (x^2 + y^2)^{n/2 - 1}=0$$
But the LHS is undefined for $(x,y,n)=(0,0,\{0,2\})$, so a potential function is guaranteed to exist for $n\neq 0,2.$ Question 1: Is this reasoning correct? Or is the LHS $k-k=0$ with no holes?
Now to find $f$, the potential function. $f=\int M \, \mathrm dx$ (plus a $g(y)$ term):
$$f(x,y)=\int x(x^2+y^2)^{n/2} \, \mathrm dx = \int \frac 12 u^{n/2} \, \mathrm du = \frac{(x^2 + y^2)^{n/2 + 1}}{n + 2} + g(y)$$ And $$ f_y = N$$ $$\frac{\partial}{\partial y} \bigg [ \frac{(x^2 + y^2)^{n/2 + 1}}{n + 2} + g(y) \bigg] = y(x^2+y^2)^{n/2}$$ $$y(x^2+y^2)^{n/2} + g'(y) = y(x^2+y^2)^{n/2}$$ $$\implies g'(x)=0$$ Hence $$f(x,y)=\frac{(x^2 + y^2)^{n/2 + 1}}{n + 2}$$
Question 2: Was there an obvious way to see that $g'(x) = 0$ without integrating $M$ and then comparing?
Question 3: Looking at $f$ again, we see that we have the new restriction $n \neq -2$. How do we reconcile this new restriction with the ones we found at the beginning? Where did this new restriction come from?