Find a vector field $G$ with curl ($G$) = $F$

Let $F(x, y, z) = (y, z, x^2)$ on $\mathbb{R}^3$. We know that $$y = \frac{\partial G_3}{ \partial y} - \frac{\partial G_2 }{\partial z}, \\ z = \frac{\partial G_1}{ \partial z} - \frac{\partial G_3 }{\partial x}, \\ x^2 = \frac{\partial G_2}{ \partial x} - \frac{\partial G_1 }{\partial y}.$$
How do I go further?

Solution 1:

Knowing that $\vec G(x,y,z)=(G_1(x,y,z),G_2(x,y,z),G_3(x,y,z))$ can only be determined up to a gradient of a scalar function, we can without loss of generality take $ G_1(x,y,z)=0$

$$ x^2 = \frac{\partial G_2}{ \partial x} - \frac{\partial G_1 }{\partial y}. \implies G_2(x,y,z) = \frac 13 x^3+f(y,z)$$

$$ z = \frac{\partial G_1}{ \partial z} - \frac{\partial G_3 }{\partial x}\implies G_3(x,y,z) =-xz+h(y)$$

for functions $f$ and $h$ to be determined by the remaining equation ...

$$ y = \frac{\partial G_3}{ \partial y} - \frac{\partial G_2 }{\partial z}\implies \frac{\partial h(y)}{\partial y}-\frac{\partial f(y,z)}{\partial z}=y $$

one solution that clearly works is $h(y)=\frac 12 y^2$ and $f(x,y)=0$

So $\vec G(x,y,z)=(0, \frac 13 x^3 , \frac 12 y^2 -xz)$ is a vector field satisfying $\vec\nabla \times \vec G(x,y,z)=(y, z, x^2)$