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Line integrals

Figure 11:
\begin{figure}
\epsfysize =2in
\centerline{\epsffile{fig10.eps}}
\end{figure}
Consider a two-dimensional function $f(x,y)$ which is defined for all $x$ and $y$. What is meant by the integral of $f$ along a given curve from $P$ to $Q$ in the $x$-$y$ plane? We first draw out $f$ as a function of length $l$ along the path (see Fig. 11). The integral is then simply given by
\begin{displaymath}
\int_{P}^Q f(x,y) dl = {\rm Area under the curve}.
\end{displaymath} (60)

As an example of this, consider the integral of $f(x,y)= x y$ between $P$ and $Q$ along the two routes indicated in Fig. 12. Along route 1 we have $x=y$, so $dl= \sqrt{2}  dx$. Thus,

\begin{displaymath}
\int_P^Q x y dl = \int_0^1 x^2 \sqrt{2}  dx = \frac{\sqrt{2}}{3}.
\end{displaymath} (61)

The integration along route 2 gives
$\displaystyle \int_P^Q x y dl$ $\textstyle =$ $\displaystyle \left.\int_0^1 x y dx\right\vert _{y=0}+\left.\int_0^1 x y  dy
\right\vert _{x=1}$  
  $\textstyle =$ $\displaystyle 0 + \int_0^1 y dy = \frac{1}{2}.$ (62)

Note that the integral depends on the route taken between the initial and final points.

Figure 12:
\begin{figure}
\epsfysize =2in
\centerline{\epsffile{fig11.eps}}
\end{figure}

The most common type of line integral is that where the contributions from $dx$ and $dy$ are evaluated separately, rather that through the path length $dl$:

\begin{displaymath}
\int_P^Q \left[ f(x,y) dx + g(x,y) dy\right].
\end{displaymath} (63)

As an example of this, consider the integral

\begin{displaymath}
\int_P^Q \left[ y^3 dx + x dy\right]
\end{displaymath} (64)

along the two routes indicated in Fig. 13. Along route 1 we have $x=y+1$ and $dx=dy$, so
\begin{displaymath}
\int_P^Q = \int_{0}^1\left[y^3 dy + (y+1) dy\right] = \frac{7}{4}.
\end{displaymath} (65)

Along route 2,
\begin{displaymath}
\int_P^Q = \left.\int_1^2 y^3 dx\right\vert _{y=0} + \left.\int_0^1 x dy\right\vert _{x=2} = 2.
\end{displaymath} (66)

Again, the integral depends on the path of integration.
Figure 13:
\begin{figure}
\epsfysize =2in
\centerline{\epsffile{fig12.eps}}
\end{figure}

Suppose that we have a line integral which does not depend on the path of integration. It follows that

\begin{displaymath}
\int_P^Q \left(f dx + g dy\right) = F(Q) - F(P)
\end{displaymath} (67)

for some function $F$. Given $F(P)$ for one point $P$ in the $x$-$y$ plane, then
\begin{displaymath}
F(Q) = F(P) + \int_P^Q \left(f dx + g dy\right)
\end{displaymath} (68)

defines $F(Q)$ for all other points in the plane. We can then draw a contour map of $F(x,y)$. The line integral between points $P$ and $Q$ is simply the change in height in the contour map between these two points:
\begin{displaymath}
\int_P^Q \left(f dx + g dy\right) = \int_P^Q dF(x,y) = F(Q) - F(P).
\end{displaymath} (69)

Thus,
\begin{displaymath}
dF(x,y) = f(x,y) dx + g(x,y) dy.
\end{displaymath} (70)

For instance, if $F=x y^3$ then $dF= y^3 dx + 3 x  y^2 dy$ and
\begin{displaymath}
\int_P^Q \left(y^3 dx + 3 x y^2 dy\right) = \left[x y^3\right]_P^Q
\end{displaymath} (71)

is independent of the path of integration.

It is clear that there are two distinct types of line integral. Those which depend only on their endpoints and not on the path of integration, and those which depend both on their endpoints and the integration path. Later on, we shall learn how to distinguish between these two types.


next up previous
Next: Vector line integrals Up: Vectors Previous: Vector calculus
Richard Fitzpatrick 2006-02-02