<< Chapter < Page | Chapter >> Page > |
We are familiar with single-variable integrals of the form ${\int}_{a}^{b}f\left(x\right)dx,$ where the domain of integration is an interval $\left[a,b\right].$ Such an interval can be thought of as a curve in the xy -plane, since the interval defines a line segment with endpoints $\left(a,0\right)$ and $\left(b,0\right)$ —in other words, a line segment located on the x -axis. Suppose we want to integrate over any curve in the plane, not just over a line segment on the x -axis. Such a task requires a new kind of integral, called a line integral.
Line integrals have many applications to engineering and physics. They also allow us to make several useful generalizations of the Fundamental Theorem of Calculus. And, they are closely connected to the properties of vector fields, as we shall see.
A line integral gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let’s look at scalar line integrals first.
A scalar line integral is defined just as a single-variable integral is defined, except that for a scalar line integral, the integrand is a function of more than one variable and the domain of integration is a curve in a plane or in space, as opposed to a curve on the x -axis.
For a scalar line integral, we let C be a smooth curve in a plane or in space and let $f$ be a function with a domain that includes C . We chop the curve into small pieces. For each piece, we choose point P in that piece and evaluate $f$ at P. (We can do this because all the points in the curve are in the domain of $f.$ ) We multiply $f\left(P\right)$ by the arc length of the piece $\text{\Delta}s,$ add the product $f\left(P\right)\text{\Delta}s$ over all the pieces, and then let the arc length of the pieces shrink to zero by taking a limit. The result is the scalar line integral of the function over the curve.
For a formal description of a scalar line integral, let $C$ be a smooth curve in space given by the parameterization $\text{r}\left(t\right)=\u27e8x\left(t\right),y\left(t\right),z\left(t\right)\u27e9,$ $a\le t\le b.$ Let $f\left(x,y,z\right)$ be a function with a domain that includes curve $C.$ To define the line integral of the function $f$ over $C,$ we begin as most definitions of an integral begin: we chop the curve into small pieces. Partition the parameter interval $\left[a,b\right]$ into n subintervals $\left[{t}_{i-l},{t}_{i}\right]$ of equal width for $\text{l}\le i\le n,$ where ${t}_{0}=a$ and ${t}_{n}=b$ ( [link] ). Let ${t}_{i}^{*}$ be a value in the i th interval $\left[{t}_{i-\text{l}},{t}_{i}\right].$ Denote the endpoints of $\text{r}\left({t}_{0}\right),\text{r}\left({t}_{1}\right)\text{,\u2026},\text{r}\left({t}_{n}\right)$ by ${P}_{0}\text{,\u2026},{P}_{n}.$ Points P _{i} divide curve $C$ into $n$ pieces ${C}_{1},{C}_{2}\text{,\u2026},{C}_{n,}$ with lengths $\text{\Delta}{s}_{1},\text{\Delta}{s}_{2}\text{,\u2026},\text{\Delta}{s}_{n},$ respectively. Let ${P}_{i}^{*}$ denote the endpoint of $\text{r}({t}_{i}^{*})$ for $1\le i\le n.$ Now, we evaluate the function $f$ at point ${P}_{i}^{*}$ for $1\le i\le n.$ Note that ${P}_{i}^{*}$ is in piece ${C}_{1},$ and therefore ${P}_{i}^{*}$ is in the domain of $f.$ Multiply $f\left({P}_{i}^{*}\right)$ by the length $\text{\Delta}{s}_{1}$ of ${C}_{1},$ which gives the area of the “sheet” with base ${C}_{1},$ and height $f\left({P}_{i}^{*}\right).$ This is analogous to using rectangles to approximate area in a single-variable integral. Now, we form the sum $\sum _{i=1}^{n}f\left({P}_{i}^{*}\right)\text{\Delta}{s}_{i}}.$ Note the similarity of this sum versus a Riemann sum; in fact, this definition is a generalization of a Riemann sum to arbitrary curves in space. Just as with Riemann sums and integrals of form ${\int}_{a}^{b}g\left(x\right)dx},$ we define an integral by letting the width of the pieces of the curve shrink to zero by taking a limit. The result is the scalar line integral of $f$ along $C.$
Notification Switch
Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?