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A diagram of a curve in quadrant one. Several points and segments are labeled. Starting at the left, the first points are P_0 and P_1. The segment between them is labeled delta S_1. The next points are P_i-1, P_i, and P_i+1. The segments connecting them are delta S_i and delta S_j+1. Point P_i starred and point P_i+1 starred are located on each segment, respectively. The last two points are P_n-1 and P_n, connected by segment S_n.
Curve C has been divided into n pieces, and a point inside each piece has been chosen.

You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths Δ s 1 , Δ s 2 ,… , Δ s n aren’t necessarily the same; in the definition of a single-variable integral, the curve in the x -axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.


Let f be a function with a domain that includes the smooth curve C that is parameterized by r ( t ) = x ( t ) , y ( t ) , z ( t ) , a t b . The scalar line integral    of f along C is

C f ( x , y , z ) d s = lim n i = 1 n f ( P i * ) Δ s i

if this limit exists ( t i * and Δ s i are defined as in the previous paragraphs). If C is a planar curve, then C can be represented by the parametric equations x = x ( t ) , y = y ( t ) , and a t b . If C is smooth and f ( x , y ) is a function of two variables, then the scalar line integral of f along C is defined similarly as

C f ( x , y ) d s = lim n i = 1 n f ( P i * ) Δ s i ,

if this limit exists.

If f is a continuous function on a smooth curve C , then C f d s always exists. Since C f d s is defined as a limit of Riemann sums, the continuity of f is enough to guarantee the existence of the limit, just as the integral a b g ( x ) d x exists if g is continuous over [ a , b ] .

Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that f ( x , y ) 0 for all points ( x , y ) on a smooth planar curve C . Imagine taking curve C and projecting it “up” to the surface defined by f ( x , y ) , thereby creating a new curve C that lies in the graph of f ( x , y ) ( [link] ). Now we drop a “sheet” from C down to the xy-plane. The area of this sheet is C f ( x , y ) d s . If f ( x , y ) 0 for some points in C , then the value of C f ( x , y ) d s is the area above the xy-plane less the area below the xy-plane. (Note the similarity with integrals of the form a b g ( x ) d x . )

A diagram in three dimensions. The original curve C in the (x,y) plane looks like a parabola opening to the left with vertex in quadrant 1. The surface defined by f(x,y) is shown always above the (x,y) plane. A curve on the surface directly above the original curve C is labeled as C’. A blue sheet stretches down from C’ to C.
The area of the blue sheet is C f ( x , y ) d s .

From this geometry, we can see that line integral C f ( x , y ) d s does not depend on the parameterization r ( t ) of C . As long as the curve is traversed exactly once by the parameterization, the area of the sheet formed by the function and the curve is the same. This same kind of geometric argument can be extended to show that the line integral of a three-variable function over a curve in space does not depend on the parameterization of the curve.

Finding the value of a line integral

Find the value of integral C 2 d s , where C is the upper half of the unit circle.

The integrand is f ( x , y ) = 2 . [link] shows the graph of f ( x , y ) = 2 , curve C , and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width π and length 2. Therefore, C 2 d s = 2 π .

A graph in three dimensions. There is a flat plane just above the (x,y) plane. The upper half of the unit circle in quadrants 1 and 2 of the (x,y) plane is raised up to form a semicircle sheet into the z-plane.
The sheet that is formed by the upper half of the unit circle in a plane and the graph of f ( x , y ) = 2 .

To see that C 2 d s = 2 π using the definition of line integral, we let r ( t ) be a parameterization of C . Then, f ( r ( t i ) ) = 2 for any number t i in the domain of r . Therefore,

C f d s = lim n i = 1 n f ( r ( t i * ) ) Δ s i = lim n i = 1 n 2 Δ s i = 2 lim n i = 1 n 2 Δ s i = 2 ( length of C ) = 2 π .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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