Example 1

Suppose γ ⊂ R² is the circular arc oriented counterclockwise from (5, 0) to (−4, 3). Using the definition of a line integral,

$${\displaystyle {\begin{aligned}\int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}((5\sin t)(-5\sin t)+(5\cos t)(5\cos t))\,\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\left(-\sin ^{2}t+\cos ^{2}t\right)\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\cos(2t)\mathrm {d} t\ =\ \left.{\tfrac {25}{2}}\sin(2t)\right|_{0}^{\pi -\tan ^{-1}\!\left({\tfrac {3}{4}}\right)}\\[.5em]&={\tfrac {25}{2}}\sin \left(2\pi -2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\\[.5em]&=-{\tfrac {25}{2}}\sin \left(2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\ =\ -{\frac {25(3/4)}{(3/4)^{2}+1}}=-12.\end{aligned}}}$$

This result can be obtained much more simply by noticing that the function ${\displaystyle f(x,y)=xy}$ has gradient ${\displaystyle \nabla f(x,y)=(y,x)}$, so by the Gradient Theorem:

$${\displaystyle \int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y=\int _{\gamma }\nabla (xy)\cdot (\mathrm {d} x,\mathrm {d} y)\ =\ xy\,|_{(5,0)}^{(-4,3)}=-4\cdot 3-5\cdot 0=-12.}$$

Example 2

For a more abstract example, suppose γ ⊂ Rⁿ has endpoints p, q, with orientation from p to q. For u in Rⁿ, let |u| denote the Euclidean norm of u. If α ≥ 1 is a real number, then

$${\displaystyle {\begin{aligned}\int _{\gamma }|\mathbf {x} |^{\alpha -1}\mathbf {x} \cdot \mathrm {d} \mathbf {x} &={\frac {1}{\alpha +1}}\int _{\gamma }(\alpha +1)|\mathbf {x} |^{(\alpha +1)-2}\mathbf {x} \cdot \mathrm {d} \mathbf {x} \\&={\frac {1}{\alpha +1}}\int _{\gamma }\nabla |\mathbf {x} |^{\alpha +1}\cdot \mathrm {d} \mathbf {x} ={\frac {|\mathbf {q} |^{\alpha +1}-|\mathbf {p} |^{\alpha +1}}{\alpha +1}}\end{aligned}}}$$

Here the final equality follows by the gradient theorem, since the function f(x) = |x|^α+1 is differentiable on Rⁿ if α ≥ 1.

If α < 1 then this equality will still hold in most cases, but caution must be taken if γ passes through or encloses the origin, because the integrand vector field |x|^{α − 1}x will fail to be defined there. However, the case α = −1 is somewhat different; in this case, the integrand becomes |x|⁻²x = ∇(log |x|), so that the final equality becomes log |q| − log |p|.

Note that if n = 1, then this example is simply a slight variant of the familiar power rule from single-variable calculus.

Example 3

Suppose there are n point charges arranged in three-dimensional space, and the i-th point charge has charge Q_i and is located at position p_i in R³. We would like to calculate the work done on a particle of charge q as it travels from a point a to a point b in R³. Using Coulomb's law, we can easily determine that the force on the particle at position r will be

$${\displaystyle \mathbf {F} (\mathbf {r} )=kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}}$$

Here |u| denotes the Euclidean norm of the vector u in R³, and k = 1/(4πε₀), where ε₀ is the vacuum permittivity.

Let γ ⊂ R³ − {p₁, ..., p_n} be an arbitrary differentiable curve from a to b. Then the work done on the particle is

$${\displaystyle W=\int _{\gamma }\mathbf {F} (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\int _{\gamma }\left(kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}\right)\cdot \mathrm {d} \mathbf {r} =kq\sum _{i=1}^{n}\left(Q_{i}\int _{\gamma }{\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}\cdot \mathrm {d} \mathbf {r} \right)}$$

Now for each i, direct computation shows that

$${\displaystyle {\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}=-\nabla {\frac {1}{\left|\mathbf {r} -\mathbf {p} _{i}\right|}}.}$$

Thus, continuing from above and using the gradient theorem,

$${\displaystyle W=-kq\sum _{i=1}^{n}\left(Q_{i}\int _{\gamma }\nabla {\frac {1}{\left|\mathbf {r} -\mathbf {p} _{i}\right|}}\cdot \mathrm {d} \mathbf {r} \right)=kq\sum _{i=1}^{n}Q_{i}\left({\frac {1}{\left|\mathbf {a} -\mathbf {p} _{i}\right|}}-{\frac {1}{\left|\mathbf {b} -\mathbf {p} _{i}\right|}}\right)}$$

We are finished. Of course, we could have easily completed this calculation using the powerful language of electrostatic potential or electrostatic potential energy (with the familiar formulas W = −ΔU = −qΔV). However, we have not yet defined potential or potential energy, because the converse of the gradient theorem is required to prove that these are well-defined, differentiable functions and that these formulas hold (see below). Thus, we have solved this problem using only Coulomb's Law, the definition of work, and the gradient theorem.

Proof of the converse

Suppose U is an open, path-connected subset of Rⁿ, and F : U → Rⁿ is a continuous and path-independent vector field. Fix some element a of U, and define f : U → R by

$${\displaystyle f(\mathbf {x} ):=\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }$$

Here γ[a, x] is any (differentiable) curve in U originating at a and terminating at x. We know that f is well-defined because F is path-independent. Let v be any nonzero vector in Rⁿ. By the definition of the directional derivative,

$${\displaystyle {\begin{aligned}{\frac {\partial f(\mathbf {x} )}{\partial \mathbf {v} }}&=\lim _{t\to 0}{\frac {f(\mathbf {x} +t\mathbf {v} )-f(\mathbf {x} )}{t}}\\&=\lim _{t\to 0}{\frac {\int _{\gamma [\mathbf {a} ,\mathbf {x} +t\mathbf {v} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} -\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot d\mathbf {u} }{t}}\\&=\lim _{t\to 0}{\frac {1}{t}}\int _{\gamma [\mathbf {x} ,\mathbf {x} +t\mathbf {v} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} \end{aligned}}}$$

To calculate the integral within the final limit, we must parametrize γ[x, x + tv]. Since F is path-independent, U is open, and t is approaching zero, we may assume that this path is a straight line, and parametrize it as u(s) = x + sv for 0 < s < t. Now, since u'(s) = v, the limit becomes

$${\displaystyle \lim _{t\to 0}{\frac {1}{t}}\int _{0}^{t}\mathbf {F} (\mathbf {u} (s))\cdot \mathbf {u} '(s)\,\mathrm {d} s={\frac {\mathrm {d} }{\mathrm {d} t}}\int _{0}^{t}\mathbf {F} (\mathbf {x} +s\mathbf {v} )\cdot \mathbf {v} \,\mathrm {d} s{\bigg |}_{t=0}=\mathbf {F} (\mathbf {x} )\cdot \mathbf {v} }$$

where the first equality is from the definition of the derivative with a fact that the integral is equal to 0 at t = 0, and the second equality is from the first fundamental theorem of calculus. Thus we have a formula for ∂_vf, (one of ways to represent the directional derivative) where v is arbitrary; for ${\displaystyle f(\mathbf {x} ):=\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }$ (see its full definition above), its directional derivative with respect to v is

$${\displaystyle {\frac {\partial f(\mathbf {x} )}{\partial \mathbf {v} }}=\partial _{\mathbf {v} }f(\mathbf {x} )=D_{\mathbf {v} }f(\mathbf {x} )=\mathbf {F} (\mathbf {x} )\cdot \mathbf {v} }$$

where the first two equalities just show different representations of the directional derivative. According to the definition of the gradient of a scalar function f, ${\displaystyle \nabla f(\mathbf {x} )=\mathbf {F} (\mathbf {x} )}$, thus we have found a scalar-valued function f whose gradient is the path-independent vector field F (i.e., F is a conservative vector field.), as desired.^[3]

Example of the converse principle

To illustrate the power of this converse principle, we cite an example that has significant physical consequences. In classical electromagnetism, the electric force is a path-independent force; i.e. the work done on a particle that has returned to its original position within an electric field is zero (assuming that no changing magnetic fields are present).

Therefore, the above theorem implies that the electric force field F_e : S → R³ is conservative (here S is some open, path-connected subset of R³ that contains a charge distribution). Following the ideas of the above proof, we can set some reference point a in S, and define a function U_e: S → R by

$${\displaystyle U_{e}(\mathbf {r} ):=-\int _{\gamma [\mathbf {a} ,\mathbf {r} ]}\mathbf {F} _{e}(\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }$$

Using the above proof, we know U_e is well-defined and differentiable, and F_e = −∇U_e (from this formula we can use the gradient theorem to easily derive the well-known formula for calculating work done by conservative forces: W = −ΔU). This function U_e is often referred to as the electrostatic potential energy of the system of charges in S (with reference to the zero-of-potential a). In many cases, the domain S is assumed to be unbounded and the reference point a is taken to be "infinity", which can be made rigorous using limiting techniques. This function U_e is an indispensable tool used in the analysis of many physical systems.