Line Integrals of Conservative Vector Fields

Gradient Theorem (Fundamental Theorem of Analysis for Line Integrals)

If $\mathbf{v}:D\to {\mathbb{R}}^n$ is the gradient field of a scalar field $f:D\to \mathbb{R}$ on an open subset $D\subseteq {\mathbb{R}}^n$, then its line integral over a piecewise continuously differentiable curve parameterisation $\gamma :[a;b]\to D$ can be calculated as

$${\int }_{\gamma }\mathbf{v}\cdot \mathrm{ds}=f(\gamma (b))-f(\gamma (a))$$

PROOF

By definition

$${\int }_{\gamma }\mathbf{v}\cdot \mathrm{ds}={\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}$$

The chain rule for scalar fields tells us that the integrand is the derivative of $f(\gamma (t))$:

$$\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))=\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)$$ $${\int }_a^b\nabla f(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))\mathrm{dt}$$

The fundamental theorem of real analysis then gives us

$${\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))\mathrm{dt}=f(\gamma (t)){|}_a^b=f(\gamma (b))-f(\gamma (a))$$

Theorem: Path Independence of Line Integrals of Conservative Vector Fields

A real vector field $\mathbf{v}:D\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ is conservative if and only if its line integrals over all simple curves with piecewise continuously differentiable reparameterisations ${\gamma }_1:[a;b]\to {\mathbb{R}}^n$ and ${\gamma }_2:[a;b]\to {\mathbb{R}}^n$ (i.e. ${\gamma }_1(a)={\gamma }_2(a)$ and ${\gamma }_1(b)={\gamma }_2(b)$) are equal.

$${\int }_{{\gamma }_1}\mathbf{v}\cdot \mathrm{ds}={\int }_{{\gamma }_2}\mathbf{v}\cdot \mathrm{ds}$$

PROOF

TODO

Theorem: Line Integrals of Conservative Vector Fields over Closed Curves

A continuous vector field $\mathbf{v}:D\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ is conservative if and only if its line integral over every closed curve $\mathcal{C}\subseteq D$ is always zero.

$${\oint }_{\mathcal{C}}\mathbf{v}\cdot \mathrm{ds}=0$$

PROOF

TODO