Independence of Path

Solving $A\vec{x} = \vec{b}$

A region $D$ in $\mathbb{R}^n$ is connected iff any 2 points in $D$ can be joined by a piecewise smooth curve contained entirely in $D$.

Simply Connected Region

A connected open region $D$ in the plane is simply connected iff every simple closed curve contained in $D$ encloses only points of $D$.

In the plane, $D$ is simply connected iff it has no holes.

Fundamental Theorem of Line Integrals

Suppose $C$ be a piecewise smooth curve parametrized by $\boldsymbol{r}(t)$, $a \le t \le b$.

If $\phi$ is a scalar potential of $\boldsymbol{F}$ on a region containing $C$,

Equivalence of Conservative Fields

Let $\boldsymbol{F}(x,y,z) = f\hat{i} + g\hat{j} + h\hat{k}$ be continuous on an open connected region $D \subseteq \mathbb{R}^3$. The following are equivalent:

• $\boldsymbol{F}$ is conservative on $D$
• $\displaystyle\int_C \boldsymbol{F} \cdot \text{d}\boldsymbol{r} = 0$ for every piecewise smooth closed curve $C$ in $D$
• For any 2 points $P$, $Q$ in $D$, $\displaystyle\int_C \boldsymbol{F} \cdot \text{d}\boldsymbol{r}$ is independent of the piecewise smooth curve $C$ in $D$ from $P$ to $Q$

Conservative Field Test

Let $P(x,y)$ and $Q(x,y)$ have continuous first-order partial derivatives on an open, simply connected region $D$ in the plane. The vector field

is conservative on $D$ iff, for each point in $D$,

The region must be simply connected. On a region with a hole, $\partial P/\partial y = \partial Q/\partial x$ can hold everywhere without $\boldsymbol{F}$ being conservative.