Line & Surface Integrals

Table of Contents
1. Line integrals - definition and parametrisation
2. Path dependence, conservative fields and potentials
3. Surface integrals and flux
4. Divergence of a vector field
5. Tutorial
6. Solutions to Tutorial Problems
7. Self Assessment Quiz
8. Solutions to Self Assessment Quiz
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We begin with special functions of vector fields and the differential operators that act on them. Recall that ordinary one-dimensional integrals represent sums. For example, the work done by a force on a particle moving along the x-axis from x = a to x = b is given by the line integral below.

To generalise to two or three dimensions, consider a force F acting on a particle which moves along a curve C in space. The work done when the particle traverses the curve C is the line integral of the force along that curve:

The path C may be open (different start and end points) or closed (returns to its starting point). Unlike ordinary single-variable integrals, a line integral in space may depend on the entire path taken, not only on the end points. There are important exceptions: for example, gravitational and electrostatic forces are path independent. Such forces are called conservative.

Along any smooth curve we can choose a single parameter (often time). If the particle position is r(t) and its velocity is ṙ(t), then the work along the curve can be written as

where ṙ(t) is the velocity at time t.

Line integrals - definition and parametrisation

If the vector field is given in component form, the line integral along C may be evaluated by parametrising the curve. Let r(t) = x(t) î + y(t) ĵ + z(t) k̂ for t1 ≤ t ≤ t2. Then

and the integral reduces to an integral over the parameter t. Parametrisation converts a geometric integral along C into a single-variable integral.

Example 1

Problem statement: Compute the work done by a two-dimensional force field

along three different paths joining the origin O(0,0) to the point P(2,1).

(i) Path 1: straight line connecting O to P with equation y = x/2.

Substitute y = x/2 in the first integral and x = 2y in the second to evaluate the integral.

(ii) Path 2: a parabola given by

The calculation proceeds as in case (i): substitute the parabola parametrisation into the integral and evaluate.

(iii) Path 3: two-segment path OQ then QP, where O(0,0) → Q(0,1) and Q(0,1) → P(2,1). On the segment OQ we have x = 0 so

and hence the integral over OQ reduces to

Along OQ, dx = 0 so the contribution from that segment is zero. From Q to P, dy = 0 and y = 1, so

These three computations show that the value of the line integral can be path dependent.

Example 2

Problem statement: Consider the force field

and evaluate its line integral along the first quadrant of the unit circle

taken anticlockwise from (1,0) to (0,1). In this quadrant both x and y are positive. Sketching the vector field shows that the force vectors point generally upward along the path, so the angle between the tangent to the path and the force is acute; therefore the integral is expected to be positive:

Parameterise the unit circle by

and substitute into the line integral to evaluate it.

Path dependence, conservative fields and potentials

In general the line integral depends on the chosen path. When the value depends on the path, the field is called nonconservative. If the line integral depends only on the end points and not on the specific path, the field is conservative. For a conservative field F, the line integral around any closed curve is zero:

If F depends only on the end points, it means there exists a scalar function φ such that

In that case

so the line integral between two points depends only on the values of φ at those points. Physicists usually define force as the negative gradient of potential:

Here φ is called the potential. A mathematician may call φ a potential even without the negative sign; in physics the negative sign makes F a force that points toward lower potential.

Surface integrals and flux

A surface integral is an integral of a vector field over a surface S. The flux of a vector field through S is

where n̂ is the unit normal to the surface (the outward normal when the surface is closed). The quantity

is called the flux of the vector field through S.

An open surface is bounded by a closed curve (for example, the rim of a cup or a butterfly net). The direction of the outward normal is fixed by the right-hand rule: if the boundary curve is traversed in the direction of rotation of a right-handed screw, the direction the screw moves gives the outward normal.

To define a surface integral we require a two-sided surface (an inside and an outside separated by an edge). Common surfaces are two-sided; one-sided surfaces such as a Möbius strip do not admit a unique normal at every point, so the usual surface integral is not defined for them.

A Möbius strip is formed by taking a strip of paper and joining the short edges after a half-turn so the strip has a single continuous side. On such a surface the normal cannot be chosen consistently; hence the standard notion of flux is not applicable.

Divergence of a vector field

The divergence of a vector field at a point measures the net flux per unit volume leaving an infinitesimal closed surface around that point. It is defined as the limit of the surface flux divided by the enclosed volume as the volume shrinks to zero:

This defines divergence as a pointwise scalar function in the region where the vector field is defined. Intuitively, divergence tells how much the field locally diverges (positive divergence) or converges (negative divergence) at a point.

For a finite volume bounded by a closed surface, imagine partitioning the volume into many small volumes. The internal faces between adjacent small volumes have normals that are oppositely directed on the two halves and their flux contributions cancel. Only the flux through the outer boundary remains. Therefore the flux through a closed surface equals the sum of fluxes of all constituent elemental volumes, which leads to the Divergence Theorem.

Denote the divergence of F evaluated at the location of the element by the expression in brackets in the figure. Summing over all elements and passing to the integral limit gives the Divergence Theorem (also called Gauss's theorem):

Cartesian expression for divergence

Because divergence is defined in the infinitesimal limit, compute it for an infinitesimal rectangular parallelepiped (dimensions Δx, Δy, Δz) aligned with the coordinate axes.

Consider the pair of faces perpendicular to the y-axis. The left face is at y and has outward normal -ĵ. The flux through that face is

The opposite face is at y + Δy with outward normal +ĵ; the flux there is

Subtract the flux from the left face from that of the right face and expand the field component in a Taylor series about y, retaining only first-order terms. This gives the net outward flux from these two faces as

By symmetry, the other two pairs of faces produce analogous terms. The total outward flux from the six faces is therefore

Dividing by the volume ΔV = Δx Δy Δz and taking the limit ΔV → 0 gives the Cartesian divergence:

Thus for a vector field F = Fx î + Fy ĵ + Fz k̂ the divergence in Cartesian coordinates is

Recall the gradient operator

Using this operator one can write

Physically, a vector field such as

plotted in the region -1 ≤ x ≤ 1, -1 ≤ y ≤ 1 shows vectors pointing away from the origin; this field has positive divergence at the origin (spread outward). A field with negative divergence would instead show vectors converging toward a point.

Tutorial

1. Calculate the line integral of the vector field
2. 
3. Calculate the line integral of the vector field along the following two paths joining the origin to the point P(1,1,1):
   (a) Along a straight line joining the origin to P.
   (b) Along a path parameterised by
4. 
5. Calculate the line integral of
6. 

   over a quarter circle in the upper half plane along the path connecting (3,0) to (0,3). What would be the result if the path was taken along the same circular path but in the reverse direction?

7. Calculate the line integral of the scalar function
8. 

   over the right half of the semicircle

   

   along the counterclockwise direction from (0,-2) to (0,+2).

9. A two dimensional force field is given by
10. 

    Find a potential function for this force field.

11. Calculate the flux of a constant vector field
12. 

    through the curved surface of a hemisphere of radius R whose base lies in the x-y plane.

13. Calculate the divergence of the position vector
14. 
15. Calculate the divergence of
16. 
17. Calculate the divergence of
18. 

Solutions to Tutorial Problems

1. (a) Along the straight line x = y = z substitute into the vector field and express the differential dr in terms of the single parameter; then integrate over the parameter interval.

Write the integrand as

(b) For the path parameterisation substitute x(t), y(t), z(t) into the integrand and evaluate the single-variable integral.

The line integral becomes

2. Parameterise the quarter circle by

so that

The integral reduces to

If the curve is traversed in the opposite direction, the sign of the integral reverses; numerically the integral becomes +18 in the reversed traversal given the values in the problem statement.

3. The path is along a circle of radius 2. Parameterise by

The line integral evaluates to

4.

The potential function is therefore

where C is an arbitrary constant.

5. Use spherical polar coordinates. The surface element on a sphere of radius R is

and the unit normal is radial, making an angle θ with the z-axis. Thus

6.

This important relation is used frequently.

7.

8.

Self Assessment Quiz

1. Find the line integral of the vector field
2. 

   from (0,0) to (1,1).

3. Calculate the line integral of the vector field
4. 

   along the following two paths joining the origin to P(1,1,1): 
   (a) Along a straight line joining the origin to P. 
   (b) Along a path parameterised by

   
5. From the result of Problem 2, can you conclude that the force is conservative? If so, determine a potential function for this vector field.
6. A potential function is given by
7. 

   where a, b and c are constants. Find the force field.

8. A conservative force field is given by
9. 

   Calculate the work done by the force in taking a particle from the origin to the point (1,1,2).

10. A force field is given by
11. 

    where r is the distance of the point from the origin. Calculate the divergence at a point other than the origin.

Solutions to Self Assessment Quiz

1. Substitute the parametrisation

and express the line integral as

2. (i) For the straight line path use

(ii) For the second path substitute the given parametrisation

3. It is not sufficient in general to test only two paths to conclude that a field is conservative. However, for this particular field both paths give the same integral and the field turns out to be conservative. Let the potential be

Equate components of the force to the negative gradients of φ:

From these relations the potential is

where C is an arbitrary constant (which may be set to zero). The line integral between two points is therefore

as expected for a conservative field.

4. Obtain the force components by differentiating the potential:

5. Since the force is conservative it can be written as the negative gradient of a scalar potential. Writing

one can show the potential is

The work done in taking the particle from the origin to (1,1,2) is

which evaluates to 9 - 0 = 9 units for the given potential.

6. Given

Because the function depends only on the radial distance r use the chain rule to obtain

and evaluate the divergence at points r ≠ 0.

Summary

This chapter introduced line integrals and surface integrals of vector fields, the notion of work as a line integral, path dependence and conservative fields, surface flux, and the divergence operator including the Divergence Theorem. Examples and exercises illustrate parametrisation methods, evaluation on different paths, finding potentials, and calculating flux and divergence in common coordinate systems.