- 6.2.1 Calculate a scalar line integral along a curve.
- 6.2.2 Calculate a vector line integral along an oriented curve in space.
- 6.2.3 Use a line integral to compute the work done in moving an object along a curve in a vector field.
- 6.2.4 Describe the flux and circulation of a vector field.
We are familiar with single-variable integrals of the form where the domain of integration is an interval Such an interval can be thought of as a curve in the xy-plane, since the interval defines a line segment with endpoints and —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.
Scalar Line Integrals
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 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 at P. (We can do this because all the points in the curve are in the domain of ) We multiply by the arc length of the piece add the product 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 be a smooth curve in space given by the parameterization Let be a function with a domain that includes curve To define the line integral of the function over we begin as most definitions of an integral begin: we chop the curve into small pieces. Partition the parameter interval into n subintervals of equal width for where and (Figure 6.12). Let be a value in the ith interval Denote the endpoints of by Points Pi divide curve into pieces with lengths respectively. Let denote the endpoint of for Now, we evaluate the function at point for Note that is in piece and therefore is in the domain of Multiply by the length of which gives the area of the “sheet” with base and height This is analogous to using rectangles to approximate area in a single-variable integral. Now, we form the sum 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 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 along s
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 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 be a function with a domain that includes the smooth curve that is parameterized by The scalar line integral of along is
if this limit exists and are defined as in the previous paragraphs). If C is a planar curve, then C can be represented by the parametric equations and If C is smooth and is a function of two variables, then the scalar line integral of along C is defined similarly as
if this limit exists.
If is a continuous function on a smooth curve C, then always exists. Since is defined as a limit of Riemann sums, the continuity of is enough to guarantee the existence of the limit, just as the integral exists if g is continuous over
Before looking at how to compute a line integral, we need to examine the geometry captured by these integrals. Suppose that for all points on a smooth planar curve Imagine taking curve and projecting it “up” to the surface defined by thereby creating a new curve that lies in the graph of (Figure 6.13). Now we drop a “sheet” from down to the xy-plane. The area of this sheet is If for some points in then the value of is the area above the xy-plane less the area below the xy-plane. (Note the similarity with integrals of the form
From this geometry, we can see that line integral does not depend on the parameterization 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 where is the upper half of the unit circle.
The integrand is Figure 6.14 shows the graph of curve C, and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width and length 2. Therefore,
To see that using the definition of line integral, we let be a parameterization of C. Then, for any number in the domain of r. Therefore,
Find the value of where is the curve parameterized by
Note that in a scalar line integral, the integration is done with respect to arc length s, which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate to an integral with a variable of integration that is t.
Let for be a parameterization of Since we are assuming that is smooth, is continuous for all in In particular, and exist for all in According to the arc length formula, we have
If width is small, then function is almost constant over the interval Therefore,
and we have
See Figure 6.15.
In other words, as the widths of intervals shrink to zero, the sum converges to the integral Therefore, we have the following theorem.
Evaluating a Scalar Line Integral
Let be a continuous function with a domain that includes the smooth curve with parameterization Then
Although we have labeled Equation 6.6 as an equation, it is more accurately considered an approximation because we can show that the left-hand side of Equation 6.6 approaches the right-hand side as In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum. Since
we obtain the following theorem, which we use to compute scalar line integrals.
Scalar Line Integral Calculation
Let be a continuous function with a domain that includes the smooth curve C with parameterization Then
if C is a planar curve and is a function of two variables.
Note that a consequence of this theorem is the equation In other words, the change in arc length can be viewed as a change in the t domain, scaled by the magnitude of vector
Evaluating a Line Integral
Find the value of integral where is part of the helix parameterized by
To compute a scalar line integral, we start by converting the variable of integration from arc length s to t. Then, we can use Equation 6.8 to compute the integral with respect to t. Note that and
Notice that Equation 6.8 translated the original difficult line integral into a manageable single-variable integral. Since
Evaluate where C is the curve with parameterization
Independence of Parameterization
Find the value of integral where is part of the helix parameterized by Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.
As with the previous example, we use Equation 6.8 to compute the integral with respect to t. Note that and
so we have
Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.
Evaluate line integral where is the line with parameterization Reparameterize C with parameterization recalculate line integral and notice that the change of parameterization had no effect on the value of the integral.
Now that we can evaluate line integrals, we can use them to calculate arc length. If then
Therefore, is the arc length of
Calculating Arc Length
A wire has a shape that can be modeled with the parameterization Find the length of the wire.
The length of the wire is given by where C is the curve with parameterization r. Therefore,
Find the length of a wire with parameterization
Vector Line Integrals
The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let
be a continuous vector field in that represents a force on a particle, and let C be a smooth curve in contained in the domain of How would we compute the work done by in moving a particle along C?
To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve C; such a specified direction is called an orientation of a curve. The specified direction is the positive direction along C; the opposite direction is the negative direction along C. When C has been given an orientation, C is called an oriented curve (Figure 6.16). The work done on the particle depends on the direction along the curve in which the particle is moving.
A closed curve is one for which there exists a parameterization such that and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain
Let be a parameterization of C for such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along C. Divide the parameter interval into n subintervals of equal width. Denote the endpoints of by Points Pi divide C into n pieces. Denote the length of the piece from Pi−1 to Pi by For each i, choose a value in the subinterval Then, the endpoint of is a point in the piece of C between and Pi (Figure 6.17). If is small, then as the particle moves from to along C, it moves approximately in the direction of the unit tangent vector at the endpoint of Let denote the endpoint of Then, the work done by the force vector field in moving the particle from to Pi is so the total work done along C is
Letting the arc length of the pieces of C get arbitrarily small by taking a limit as gives us the work done by the field in moving the particle along C. Therefore, the work done by F in moving the particle in the positive direction along C is defined as
which gives us the concept of a vector line integral.
The vector line integral of vector field F along oriented smooth curve C is
if that limit exists.
With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once by the parameterization, the value of the line integral is unchanged. With vector line integrals, the orientation of the curve does matter. If we think of the line integral as computing work, then this makes sense: if you hike up a mountain, then the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, then Earth’s gravitational force does positive work on you. In other words, reversing the path changes the work value from negative to positive in this case. Note that if C is an oriented curve, then we let −C represent the same curve but with opposite orientation.
As with scalar line integrals, it is easier to compute a vector line integral if we express it in terms of the parameterization function r and the variable t. To translate the integral in terms of t, note that unit tangent vector T along C is given by (assuming Since as we saw when discussing scalar line integrals, we have
Thus, we have the following formula for computing vector line integrals:
Because of Equation 6.9, we often use the notation for the line integral
If then dr denotes vector differential
Find the value of integral where is the semicircle parameterized by and
Notice that this is the negative of the answer in Example 6.18. It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.
Let C be an oriented curve and let −C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:
That is, reversing the orientation of a curve changes the sign of a line integral.
Let be a vector field and let C be the curve with parameterization for Which is greater: or
Another standard notation for integral is In this notation, P, Q, and R are functions, and we think of dr as vector To justify this convention, recall that Therefore,
If then which implies that Therefore
Finding the Value of an Integral of the Form
Find the value of integral where C is the curve parameterized by
As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t. In this case, Equation 6.10 allows us to make this change:
Find the value of where is the curve parameterized by
We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve. To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves such that the endpoint of is the starting point of (Figure 6.19). When curves satisfy the condition that the endpoint of is the starting point of we write their union as
The next theorem summarizes several key properties of vector line integrals.
Properties of Vector Line Integrals
Let F and G be continuous vector fields with domains that include the oriented smooth curve C. Then
- where k is a constant
- Suppose instead that C is a piecewise smooth curve in the domains of F and G, where and are smooth curves such that the endpoint of is the starting point of Then
Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C, then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation Finally, if are intervals, then
which is analogous to property iv.
Using Properties to Compute a Vector Line Integral
Find the value of integral where C is the rectangle (oriented counterclockwise) in a plane with vertices and where (Figure 6.20).
Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let represent the side from to let represent the side from to let represent the side from to and let represent the side from to (Figure 6.20). Then,
We want to compute each of the four integrals on the right-hand side using Equation 6.8. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):
Notice that the value of this integral is positive, which should not be surprising. As we move along curve C1 from left to right, our movement flows in the general direction of the vector field itself. At any point along C1, the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C1, and the line integral will have positive value.
The calculations for the three other line integrals are done similarly:
Thus, we have
Calculate line integral where F is vector field and C is a triangle with vertices and oriented counterclockwise.
Applications of Line Integrals
Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.
Suppose that a piece of wire is modeled by curve C in space. The mass per unit length (the linear density) of the wire is a continuous function We can calculate the total mass of the wire using the scalar line integral The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by for some point in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral
Calculating the Mass of a Wire
Calculate the mass of a spring in the shape of a curve parameterized by with a density function given by kg/m (Figure 6.21).
To calculate the mass of the spring, we must find the value of the scalar line integral where C is the given helix. To calculate this integral, we write it in terms of t using Equation 6.8:
Therefore, the mass is kg.
Calculate the mass of a spring in the shape of a helix parameterized by with a density function given by kg/m.
When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the work done by a vector field representing a force is a standard use of vector line integrals. Recall that if an object moves along curve C in force field F, then the work required to move the object is given by
Flux and Circulation
We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.
Let C be a plane curve and let F be a vector field in the plane. Imagine C is a membrane across which fluid flows, but C does not impede the flow of the fluid. In other words, C is an idealized membrane invisible to the fluid. Suppose F represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing C?
Recall that the line integral of F along C is —in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector and instead compute integral we determine the flux across C. To be precise, the definition of integral is the same as integral except the T in the Riemann sum is replaced with N. Therefore, the flux across C is defined as
where and are defined as they were for integral Therefore, a flux integral is an integral that is perpendicular to a vector line integral, because N and T are perpendicular vectors.
If F is a velocity field of a fluid and C is a curve that represents a membrane, then the flux of F across C is the quantity of fluid flowing across C per unit time, or the rate of flow.
More formally, let C be a plane curve parameterized by Let be the vector that is normal to C at the endpoint of and points to the right as we traverse C in the positive direction (Figure 6.23). Then, is the unit normal vector to C at the endpoint of that points to the right as we traverse C.
The flux of F across C is line integral
We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see Equation 6.9).
Calculating Flux across a Curve
Let F be a vector field and let C be a smooth curve with parameterization Let The flux of F across C is
Flux across a Curve
Calculate the flux of across a unit circle oriented counterclockwise (Figure 6.24).
To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization The normal vector to a unit circle is Therefore, the flux is
Calculate the flux of across the line segment from to where the curve is oriented from left to right.
Let be a two-dimensional vector field. Recall that integral is sometimes written as Analogously, flux is sometimes written in the notation because the unit normal vector N is perpendicular to the unit tangent T. Rotating the vector by 90° results in vector Therefore, the line integral in Example 6.21 can be written as
Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field F along an oriented closed curve is called the circulation of F along C. Circulation line integrals have their own notation: The circle on the integral symbol denotes that C is “circular” in that it has no endpoints. Example 6.18 shows a calculation of circulation.
To see where the term circulation comes from and what it measures, let v represent the velocity field of a fluid and let C be an oriented closed curve. At a particular point P, the closer the direction of v(P) is to the direction of T(P), the larger the value of the dot product The maximum value of occurs when the two vectors are pointing in the exact same direction; the minimum value of occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation measures the tendency of the fluid to move in the direction of C.
Let be the vector field from Example 6.16 and let C represent the unit circle oriented counterclockwise. Calculate the circulation of F along C.
We use the standard parameterization of the unit circle: Then, and Therefore, the circulation of F along C is
Notice that the circulation is positive. The reason for this is that the orientation of C “flows” with the direction of F. At any point along the circle, the tangent vector and the vector from F form an angle of less than 90°, and therefore the corresponding dot product is positive.
In Example 6.25, what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by Then
Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of F.
Calculate the circulation of along a unit circle oriented counterclockwise.
Calculate the work done on a particle that traverses circle C of radius 2 centered at the origin, oriented counterclockwise, by field Assume the particle starts its movement at
The work done by F on the particle is the circulation of F along C: We use the parameterization for C. Then, and Therefore, the circulation of F along C is
The force field does zero work on the particle.
Notice that the circulation of F along C is zero. Furthermore, notice that since F is the gradient of F is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.
Calculate the work done by field on a particle that traverses the unit circle. Assume the particle begins its movement at
Section 6.2 Exercises
True or False? Line integral is equal to a definite integral if C is a smooth curve defined on and if function is continuous on some region that contains curve C.
True or False? Vector functions and define the same oriented curve.
True or False?
True or False? A piecewise smooth curve C consists of a finite number of smooth curves that are joined together end to end.
True or False? If C is given by then
For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path.
from (0, 1, 0) to (1, 0, 0)
[T] Evaluate where C is the right half of circle and is traversed in the clockwise direction.
[T] Evaluate where C is the line segment from to (1, 2).
For the following exercises, find the work done.
Find the work done by vector field on a particle moving along a line segment that goes from to
Find the work done by a person weighing 150 lb walking exactly one revolution up a circular, spiral staircase of radius 3 ft if the person rises 10 ft.
Find the work done by force field on a particle as it moves along the helix from point to point
Find the work done by vector field in moving an object along path C, the straight line which joins points (1, 0) and (0, 1).
Find the work done by force in moving an object along curve where
Find the mass of a wire in the shape of a circle of radius 2 centered at (3, 4) with linear mass density
For the following exercises, evaluate the line integrals.
Evaluate where and C is the part of the graph of from to
Evaluate where is the helix
Evaluate over the line segment from to
Let C be the line segment from point (0, 1, 1) to point (2, 2, 3). Evaluate line integral
[T] Use a computer algebra system to evaluate the line integral where C is the arc of the parabola from (−5, −3) to (0, 2).
[T] Use a computer algebra system to evaluate the line integral over the path C given by where
[T] Use a CAS to evaluate line integral over path C given by where
Evaluate line integral where C lies along the x-axis from
[T] Use a CAS to evaluate where C is
[T] Use a CAS to evaluate where C is
In the following exercises, find the work done by force field F on an object moving along the indicated path.
C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 1)
Let F be vector field Compute the work of integral where C is the path
Compute the work done by force along path where
Evaluate where and C is the segment of the unit circle going counterclockwise from to (0, 1).
Force acts on a particle that travels from the origin to point (1, 2, 3). Calculate the work done if the particle travels:
- along the path along straight-line segments joining each pair of endpoints;
- along the straight line joining the initial and final points.
- Is the work the same along the two paths?
Find the work done by vector field on a particle moving along a line segment that goes from (1, 4, 2) to (0, 5, 1).
How much work is required to move an object in vector field along the upper part of ellipse from (2, 0) to
A vector field is given by Evaluate the line integral of the field around a circle of unit radius traversed in a clockwise fashion.
Evaluate the line integral of scalar function along parabolic path connecting the origin to point (1, 1).
Find along C: from (0, 0) to (1, 3).
Find along C: from (0, 0) to (1, 3).
For the following exercises, use a CAS to evaluate the given line integrals.
[T] Evaluate where C is represented by
[T] Evaluate line integral where, is the arc of curve from to
[T] Evaluate the integral where is a triangle with vertices (0, 1, 2), (1, 0, 3), and
[T] Evaluate line integral where is curve from (1, 0) toward
[T] Evaluate line integral where is the right half of circle
[T] Evaluate where and
Evaluate where and
C is any path from to (5, 1).
Find the line integral of over path C defined by from point (0, 0, 0) to point (2, 4, 8).
Find the line integral of where C is ellipse from
For the following exercises, find the flux.
Compute the flux of across a line segment from (0, 0) to (1, 2).
Let and let C be curve Find the flux across C.
Let and let C be curve Find the flux across C.
Let and let C: Calculate the flux across C.
Let Calculate flux F orientated counterclockwise across curve C:
Find the line integral of where C consists of two parts: and is the intersection of cylinder and plane from (0, 4, 3) to is a line segment from to (0, 1, 5).
A spring is made of a thin wire twisted into the shape of a circular helix Find the mass of two turns of the spring if the wire has constant mass density.
A thin wire is bent into the shape of a semicircle of radius a. If the linear mass density at point P is directly proportional to its distance from the line through the endpoints, find the mass of the wire.
An object moves in force field counterclockwise from point (2, 0) along elliptical path to and back to point (2, 0) along the x-axis. How much work is done by the force field on the object?
Find the work done when an object moves in force field along the path given by
If an inverse force field F is given by where k is a constant, find the work done by F as its point of application moves along the x-axis from
David and Sandra plan to evaluate line integral along a path in the xy-plane from (0, 0) to (1, 1). The force field is David chooses the path that runs along the x-axis from (0, 0) to (1, 0) and then runs along the vertical line from (1, 0) to the final point (1, 1). Sandra chooses the direct path along the diagonal line from (0, 0) to (1, 1). Whose line integral is larger and by how much?