Is it possible to reverse a gradient ($\vec{\nabla}$) operation? In calculus, the antiderivative (indefinite integral) can be considered as the reverse operation of a derivative.
A gradient yields a vector. Is there a similar way of reversing gradient, as you do with derivatives?
 A: I prefer a simpler one-step solution: Choose the curve $\vec\gamma(t) = t\vec x$ on $t\in[0,1]$, with endpoints $\vec\gamma(0)=\vec 0$ and $\vec\gamma(1) = \vec x$, and then the gradient theorem implies that
$$\begin{align}
f(\vec x) &= f(\vec 0) + \int_0^1 \nabla f(\vec\gamma(t))\cdot\vec\gamma'(t)\,\mathrm dt \\
&= f(\vec 0) + \int_0^1 \nabla f(t\vec x)\cdot\vec x\,\mathrm dt.
\end{align}$$
The unknown $f(\vec 0)$ plays the role of the constant of integration.
For example, if we have $\nabla f(x,y)=\begin{bmatrix}3x^2 - 3y^2 + 2x\\2y - 6xy\end{bmatrix}$, then
$$\nabla f(tx,ty)\cdot(x,y) = 3 t^2 (x^3-3xy^2)+2 t (x^2+y^2)$$
from which we easily get
$$f(x,y)=f(0,0)+x^3-3xy+x^2+y^2.$$
The $\log x$ term in M. Vinay's answer complicates things a little as the function is undefined at $\vec 0$. It is straightforward to get around this by taking a different starting point, say $\vec x_0=(1,0)$ and choosing the curve $\vec\gamma(t)=\vec x_0+t(\vec x-\vec x_0)$.
A: M. Vinay's answer is good. For more than 2 variables, the process can be chained. For $$F(x,y,z) = f_x x + f_y y + f_z z = \nabla f$$ we can integrate with respect to $x$
$$f(x,y,z)=\int f_x dx + g(y, z)$$
where $g$ is a function only in terms of $y$ and $z$. Then differentiating with respect to $y$
$$f_y(x,y,z)=\frac{\partial}{\partial y} \int f_x dx \ + g_y(y,z) $$
The first term is common and we are left with $g_y(y,z)$, which can be integrated to get $$g(y,z) = \int g_y(y,z) dy + h(z)$$
$$f(x,y,z)=\int f_x dx + \int g_y(y,z) dy + h(z)$$
Then repeating the process by differentiating $f$ with respect to $z$ and then integrating, we get all the terms of $f$:
$$f(x,y,z)=\int f_x (x,y,z) dx + \int g_y(y,z) dy + \int h_z(z) dz$$
A: Yes, it is possible. I will give an example to demonstrate the general procedure. Consider $f(x, y) = x^3 - 3xy^2 + x^2 + y^2 + \log x$. Then $\nabla f = \dfrac{\partial f}{\partial x}\hat{i} + \dfrac{\partial f}{\partial y}\hat{j} = \left(3x^2 - 3y^2 + 2x + \dfrac{1}{x}\right)\hat{i} + (2y -6xy)\hat{j}$.
To reverse this, we look at each component individually. We know that
$\dfrac{\partial f}{\partial x} = 3x^2 - 3y^2 + 2x + \dfrac{1}{x}\\
\dfrac{\partial f}{\partial y} = 2y - 6xy
$
Therefore:
$\begin{align}
\displaystyle f(x,y) & = \int \dfrac{\partial f}{\partial x}\, dx\\
& = \int 3x^2 - 3y^2 + 2x + \dfrac{1}{x}\, dx\\
& = x^3 - 3xy^2 + x^2 + \log x + u(y)
\end{align}$
What is that $u(y)$? It's the "constant" of integration, of course.When we differentiate $f$ with respect to $x$ partially, any term of $f$ not containing $x$ is a constant - this includes terms containing only $y$.
Now to determine $u(y)$, we look at $\dfrac{\partial f}{\partial y}$. We could integrate this with respect to $y$, similar to what we did with $\dfrac{\partial f}{\partial x}$. Then some of the terms after the integration will be common. The terms that are not common are those that constitute $u(y)$. So we need to look for terms of $\dfrac{\partial f}{\partial y}$ that do not contain $x$, and integrate them. Here, $\dfrac{\partial f}{\partial y}  = 2y - 6xy$, and the only term not containing $x$ is $2y$. Therefore:
$\displaystyle u(y) = \int 2y\, dy = y^2 + C$.
Thus, $\boxed{f(x, y) = x^3 - 3xy^2 + x^2 + \log x + y^2 + C}$.
In general:
$$f(x, y) = \int \dfrac{\partial f}{\partial x} dx + \int \left[\text{terms of $\dfrac{\partial f}{\partial y}$ that do not contain $x$}\right]\, dy$$
A: In general, if the $\nabla$ operator is expressed in some orthogonal coordinates $\mathbf{q}=(q^1,q^2,q^3)$, the gradient of a scalar function $\varphi(\mathbf{q})$ will be given by
$$\nabla \varphi(\mathbf{q})=\frac{\hat{\mathbf{e}}_{i}}{h_{i}} \frac{\partial \varphi}{\partial q^{i}} \tag{1}$$
And a line element will be
$$d \boldsymbol{\ell}=h_{i} d q^{i} \hat{\mathbf{e}}_{i}\tag{2}$$
So the dot product between these two vectors is
$$\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}=(\frac{\hat{\mathbf{e}}_{i}}{h_{i}} \frac{\partial \varphi}{\partial q^{i}} )·(h_{i} d q^{i} \hat{\mathbf{e}}_{i})=\frac{\partial \varphi}{\partial q^{i}}dq^i \tag{3}$$
Therefore, $\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}$ yelds the total derivative of $\varphi(\mathbf{q})$:
$$\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}=d\varphi(\mathbf{q})\tag{4}$$
So you can recover the $\varphi(\mathbf{q})$ function at the right of the nabla operator just by evaluating the line integral of $(4)$:
$$
\varphi(\mathbf{q}_2)-\varphi(\mathbf{q}_1) =\int_\gamma \nabla \varphi(\mathbf{q})·d \boldsymbol{\ell} \tag{5}
$$
