Consider the unit sphere $x^2+y^2+z^2=1$ and a vector field $$(x(y^2-z^2), -y(x^2+2z^2), z(x^2+2y^2)).$$ on the sphere. Show that this vector field is conservative on the sphere and find a potential function of it.
I know how to show the vector field is conservative. One can simply show that the line integral of this vector field over any simple closed curve on the sphere is 0. However, I am not sure how to find a potential function of this vector field.
On
To make sure we're on the same page, in this answer I will use the spherical coordinates where $\phi$ is the polar angle (the latitude down from the North pole), and $\theta$ is the azimuthal angle (the longitude from the positive $x$ axis), and points will be labelled $(r,\theta,\phi)$.
Once you've shown that the field $\vec{F}$ is conservative on the sphere, it follows that the integral of the field over any curve on the sphere is path independent, and you can get your potential $\Phi$ from this; for any curve $P$ from a point $A$ to a point $B$, we have,
$$ \int_P \vec{F}\cdot d\ell =\int_A^B \nabla\Phi\cdot d\ell = \Phi(B) - \Phi(A) $$
Since the potential is determined up to a constant, we choose $\Phi$ such that $\Phi(1,0,0)=0$ (that is, the potential is $0$ at the North pole). Then for any point $P=(r,\theta,\phi)$ on the sphere, we can form the path $C$ from the pole to $P$ by first travelling down along $\phi$ (we will call this path $C_\phi$), then over along $\theta$ (we will call this path $C_\theta$). These curves can be parameterized by,
$$C_\phi = (1,0,t):0 \le t\le\phi$$ $$C_\theta = (1,t,\phi):0\le t\le\theta$$
Then,
$$ \begin{align*} \Phi(r,\theta,\phi) &= \int_C \vec{F}\cdot d\ell \\ &= \int_{C_\phi} \vec{F}\cdot d\ell + \int_{C_\theta} \vec{F}\cdot d\ell \end{align*} $$
From here, you would jsut convert your expression for the vector field into spherical coordinates, then take the dot product with each of the parameterized curves to make them an integral in $t$. This should not be any more conceptual work, but is a boatload of very tedious and annoying algebra, so I have graciously left it to the reader.
In general, the gradient vector field $\nabla f$ on a regular surface with parametrization $\textbf{x}(u, v)$ and first fundamental form $E=\textbf{x}_u\cdot\textbf{x}_u$, $G=\textbf{x}_v\cdot\textbf{x}_v$ and $F=\textbf{x}_u\cdot\textbf{x}_v=0$ is $$\frac{f_u}{E}\textbf{x}_u+\frac{f_v}{G}\textbf{x}_v.$$ As suggested by the other answer here, we parametrize the sphere with spherical coordinates $\textbf{x}(u, v)=(\sin u\cos v, \sin u\sin v, \cos u)$. Then $E=1$, $G=\sin^2u$ and $F=0$. Expressing the given vector field in terms of $u$ and $v$ and equating it with $\displaystyle f_u\textbf{x}_u+\frac{f_v}{\sin^2u}\textbf{x}_v$, we have $$\begin{cases} f_u&=-\cos u(\sin u+\sin u\sin^2v)\\ f_v&=-\sin^2u\sin v\cos v\end{cases}$$ We then have a desired potential function $\displaystyle f=-\frac{\sin^2u\sin^2v+\sin^2u}{2}$. In terms of $x$ and $y$, $$f=-\frac{x^2+2y^2}{2}.$$.