next up previous contents
Next: About this document ... Up: Integration and Vector Fields. Previous: Stoke's theorem.

Conservative Fields, Potential Function.

Definition 8.41   Let $\overrightarrow{F} $ be a vector field defined over an open region $\mathcal{D}$ in the 3-dimensional space. If, for any two points A and B in $\mathcal{D}$, the work integral $\int_A^B \overrightarrow{F}\cdot
d \overrightarrow{r} $ is independent of the path from A to B in $\mathcal{D}$, we say that $\overrightarrow{F} $ is a conservative field on $\mathcal{D}$.

Example 8.42   Take $\overrightarrow{F} =x \overrightarrow{i} + y \overrightarrow{j} + z \overrightarrow{k} $.

Definition 8.43   If $\overrightarrow{F} $ is the gradient field over $\mathcal{D}$ of some scalar function f(x,y,z), the function f is called a potential (function) for $\overrightarrow{F} $ over $\mathcal{D}$.

Note that a potential function is not uniquely defined.

Example 8.44   Let $\overrightarrow{F} = yz \overrightarrow{i} + xz \overrightarrow{j} + xy \overrightarrow{k} $. Then $\overrightarrow{F} = \overrightarrow{\nabla} f$, where $f(x,y,z)=xyz+K, \; K \in \mathbb{R} $.

Theorem 8.45   Let $\overrightarrow{F} =M \overrightarrow{i} + N \overrightarrow{j} + P \overrightarrow{k} $ be a vector field whose components M,N,P are continuous over a domain $\mathcal{D}$ in the space. Then there exists a function f such that $\overrightarrow{\nabla } f = \overrightarrow{F} $ (i.e. a potential for $\overrightarrow{F} $ ) if, and only if, $\overrightarrow{F} $ is a conservative vector field.

In this case, for every two points $A,B \in \mathcal{D}$, we have:

\begin{displaymath}\int_A^B \overrightarrow{F}\cdot d \overrightarrow{r} = f(B) - f(A).
\end{displaymath}

Theorem 8.46 (Component test)   Let $\overrightarrow{F} =M \overrightarrow{i} + N \overrightarrow{j} + P \overrightarrow{k} $ be a vector field whose components M,N,P have continuous first partial derivatives over a domain $\mathcal{D}$ in the space.

The field $\overrightarrow{F} $ is conservative if, and only if

\begin{displaymath}\frac {\partial P }{\partial y } = \frac {\partial N }{\parti...
...{\partial N }{\partial x } = \frac {\partial M }{\partial y }.
\end{displaymath}

Example 8.47   Take $\overrightarrow{F} =\underbrace{x}_{M} \overrightarrow{i} + \underbrace{y}_{N} \overrightarrow{j} + \underbrace{z}_{P} \overrightarrow{k} $.

1.
We have:

\begin{displaymath}\begin{cases}
\frac {\partial P }{\partial y } = 0 = \frac {\...
...partial x } = 0 = \frac {\partial M }{\partial y }.
\end{cases}\end{displaymath}

Hence, the field $\overrightarrow{F} $ is conservative.
2.
Let's find a potential for $\overrightarrow{F} $: we look for a function f(x,y,z) such that

\begin{displaymath}\overrightarrow{\nabla } f = x \overrightarrow{i} + y \overrightarrow{j} + z \overrightarrow{k}\end{displaymath}

i.e.

\begin{displaymath}\frac {\partial f }{\partial x } = x \; , \; \frac {\partial ...
...l y } = y \;
\text{and} \; \frac {\partial f }{\partial z }=z
\end{displaymath}

We have:

\begin{displaymath}f(x,y,z)= \frac 12 x^2 +C_1(y,z) = \frac 12 y^2 + C_2(x,z) = \frac 12 z^2 + C_3(x,y)
\end{displaymath}

where C1, C2, C3 are functions of two variables. It follows that:

\begin{displaymath}f(x,y,z)= \frac 12 ( x^2 +y^2 + z^2) +C \; , \; C \in \mathbb{R} .
\end{displaymath}

3.
Take A=(0,0,0) and B=(1,2,3). We will compute the integral $\int_A^B \overrightarrow{F}\cdot
d \overrightarrow{r} $ using two different paths:
(a)
On the segment AB:

A parametrization of the path is: $\overrightarrow{r} = t \overrightarrow{i} + 2t \overrightarrow{j} + 3t
\overrightarrow{k}\quad , \quad 0 \leq t \leq 1 .$

Thus:

  • $d \overrightarrow{r} = ( \overrightarrow{i} + 2 \overrightarrow{j} + 3
\overrightarrow{k} ) \; dt$;
  • $\overrightarrow{F} = t \overrightarrow{i} + 2t \overrightarrow{j} + 3t
\overrightarrow{k} $.

Now
\begin{align*}\overrightarrow{F}\cdot d \overrightarrow{r} & = (t +4t + 9t) \; dt \\
I &= \int_0^1 14t \; dt = [7t^2]_0^1=7.
\end{align*}

(b)
On a curve: Take $\overrightarrow{r} = t^2 \overrightarrow{i} + 2t^3 \overrightarrow{j} + 3t \overrightarrow{k}\; , \; 0 \leq t \leq 1.$

Thus

  • $d \overrightarrow{r} = ( 2t \overrightarrow{i} + 6t^2 \overrightarrow{j} + 3 \overrightarrow{k} ) \; dt$
  • $\overrightarrow{F} = t^2 \overrightarrow{i} + 2t^3 \overrightarrow{j} + 3t \overrightarrow{k} $

Now
\begin{align*}\overrightarrow{F}\cdot d \overrightarrow{r} & = (2t^3+12t^5+9t) \...
...\; dt = \left[ \frac 12 t^4 + 2 t^6 + \frac 92 t^2 \right]_0^1 = 7.
\end{align*}


next up previous contents
Next: About this document ... Up: Integration and Vector Fields. Previous: Stoke's theorem.
Noah Dana-Picard
2001-05-30
a