Consider the function y=f ⁣(x)y=f\!\left(\mathbf x\right), where the domain DRnD\subseteq\mathbb R^n, and the function is differentiable everywhere.

According to some well-known theories, we can derive that the normal vector of the graph of the function at (x0,f ⁣(x0))\left(\mathbf x_0,f\!\left(\mathbf x_0\right)\right) is (f ⁣(x0),1)\left(\nabla f\!\left(\mathbf x_0\right),-1\right).

This gives us an idea that, in fact a conservative field consists of normal vectors of its potential function (a scalar function).

We also know that a scalar function can be derived from its gradient by integrating it along an arbitrary path (what exactly the path is is not important because it is a conservative field, so you can choose one as long as it can make the calculation easy). Here it can come into our minds that we can derive a multi-variable function from its normal vectors.

The method is to make the last component of the normal vectors be 1-1 and then calculate the integral of the rest components.

I am sorry that the passage is too brief, but I need to have some rest after experiencing several continuous tests today and yesterday. Bless me!