How to find a function whose curl is $(7e^y,8x^7 e^{x^8},0)$? How to find a function whose curl is $(7e^y,8x^7 e^{x^8},0)$?
I've tried several integration but can't find a trivial form. 
 A: Here's a way to do it without guessing, a la: https://www.physicsforums.com/threads/operation-that-will-undo-a-curl-operation.14580/
Let $F = (f,g,h)$. It turns out that one function is arbitrary, so we set $f=0$ arbitrarily. Then we choose $g$ and $h$ to meet the curl conditions. We actually have a lot of freedom in determining $g$ and $h$.
The first equation is
$f_z-h_x = 8x^7 e^{x^8}$,
so $$h = \int f_z\,dx - \int 8x^7 e^{x^8}\,dx = -e^{x^8} + C(y,z).$$
Second,
$g_x - f_y = 0$,
so $$g = \int f_y\,dx = K(y,z).$$ 
Finally, we must have 
$h_y - g_z = 7e^y$,
so 
$${\partial C(y,z)\over \partial y} - {\partial K(y,z) \over \partial z} = 7e^y.$$
Once again, we have considerable freedom, so we set $K(y,z)=0$ to get
$${\partial C(y,z) \over \partial y}=7e^y$$
or 
$$C(y,z) = \int 7e^y \,dy  = 7e^y + C^*(z),$$
where $C^*(z)$ is a function of just $z$, which we can set to zero. Thus
$$F = (f,g,h) = (0,0,-e^{x^8} + 7e^y).$$
A: $8x^7e^{x^8}$ looks a whole lot like $\frac{\partial}{\partial x}e^{x^8}$, so let's say that this contribution comes from the $z$-component of the original field.
Also, $7e^y$ is $\frac{\partial}{\partial y}7e^y$, so this could also come from the $z$-component of the original field. This takes care of every term in the curl, so using 
$$
(0,0,7e^y - e^{x^8})
$$
will work.
A: Since $\vec\nabla \times \vec F(x,y,z) $  has an x component that depends only on y  and a y component that depends only on x and no z component 
A good guess is that  $\vec F(x,y,z) $ takes the form 
$\vec F(x,y,z) = (0,0,g(x)+h(y)) $
$g(x) $ and $h(y) $ can be found easily by integration.
