Line IntegralsDr. E. Jacobs

Introduction

Applications of integration to physics and engineering require an extensionof the integral called a line integral. Line integrals are necessary to expressthe work done along a path by a force. Line integrals are needed to describecirculation of fluids. They are also important in describing the relationshipbetween electric and magnetic fields. The name line integral is somewhatof a misnomer because it refers to adding up quantities along curved pathsas well as along straight lines. You have already encountered the notion ofa line integral when you learned about arclength in Calculus II and III.

Review of ArclengthIf a curve C is described by the parametric equations

x = x(t) and y = y(t) for a ≤ t ≤ b

then the length along curve C is given by:

Length(C) =∫ b

a

√(dx

dt

)2

+(

dy

dt

)2

dt

The equation is simple enough to use. For example, to find the lengthalong the segment of the circle x = 2 cos t, y = 2 sin t for 0 ≤ t ≤ π

8 we justcalculate:

Length =∫ π/8

0

√(x′(t))2 + (y′(t))2 dt =

∫ π/8

0

√4 sin2 t + 4 cos2 t dt =

π

4

The expression∫ b

a

√(x′(t))2 + (y′(t))2 dt can be understood with either a

geometric or physical interpretation. The usual geometric interpretation isto split up the curve C into small segments of length ds.

The length of a typical segment is approximated as a hypotenuse of a righttriangle of base dx and height dy

and therefore, by the Pythagorean Theorem:

ds =√

(dx)2 + (dy)2 =

√(dx

dt

)2

+(

dy

dt

)2

dt

If we use the notation∫

Cds to symbolize the limit of the sum of all the

segments of length ds as t ranges from a to b then it becomes clearer that

the length along C is represented by∫

Cds =

∫ b

a

√(dxdt

)2+

(dydt

)2

dt.

For a more physical interpretation, we consider a particle moving along Cwith it’s position after t seconds given by ~r = 〈x(t), y(t)〉. The velocity ofthe particle is ~v = d~r

dt = 〈x′(t), y′(t)〉 and so the speed of the particle is:

|~v| =√

(x′(t))2 + (y′(t))2

We can look at the formula :

ds = |~v| dt =√

(x′(t))2 + (y′(t))2 dt

as:change in distance = (speed)(change in time)

therefore, if we look at the length of curve C,∫

Cds as the sum of all the

(speed)(time) expressions, so

Total distance traveled =∫ b

a

|~v| dt =∫ b

a

√(x′(t))2 + (y′(t))2 dt

In three dimensions, the velocity vector has three coordinates

~v = 〈x′(t), y′(t), z′(t)〉

and the speed is |~v| =√

(x′(t))2 + (y′(t))2 + (z′(t))2 so the total distancealong C is:

∫

C

ds =∫ b

a

|~v| dt =∫ b

a

√(x′(t))2 + (y′(t))2 + (z′(t))2 dt

The notion of adding quantities up over a curve C applies not only todistance but to other physical quantities as well. Suppose, for example, wehad electric charge distributed along this path C with a charge density atpoint ~r = 〈x, y, z〉 of f(x, y, z) coulombs per meter. Then, a small segmentof length ds meters would hold a charge of f(x, y, z) ds coulombs. The totalcharge along this path would be the limit of the sum of such quantities,which would give us:

Total Charge along C =∫

C

f(x, y, z) ds

The computation of such an integral is similar to an arc length calculation.For example, suppose C is a helical segment, given by

~r = 〈x(t), y(t), z(t)〉 = 〈π cos t, π sin t, t〉 0 ≤ t ≤ π

Let us suppose that the linear charge density at any point (x, y, z) on thiscurve is given by f(x, y, z) = 1

x2+y2+z2 coulomb per meter. This wouldmean that as we move along the helix from (π, 0, 0) at t = 0 to (−π, 0, π)

at t = π, the charge density gets smaller and smaller. The density at~r = 〈π cos t, π sin t, t〉 is:

f(x, y, z) =1

x2 + y2 + z2=

1π2 cos2 t + π2 sin2 t + t2

=1

π2 + t2

The charge (in coulombs) along a segment of length ds is:

f(x, y, z) ds = f(x, y, z)|~v| dt

=1

π2 + t2

√(x′(t))2 + (y′(t))2 + (z′(t))2 dt

=1

π2 + t2

√(−π sin t)2 + (π cos t)2 + 1 dt

=√

π2 + 1π2 + t2

dt

Therefore, the total charge is:

∫

C

f(x, y, z) ds =∫ π

0

√π2 + 1

π2 + t2dt =

√π2 + 1

πtan−1 1 =

√π2 + 1

4

Work

Work is defined as the integral of force over distance.

W =∫ b

a

F (x) dx

This definition is only applicable if a force F is pushing something alonga straight line. You might expect that if a force is acting along a curve Cthen the definition of work needs to be extended to be:

W =∫

C

F (x, y, z) ds

since ds represents the change in distance along the curve.

However, there is a complication. Suppose the force is directed at an angleθ to the path C.

It is only the tangential component of the force that is doing work alongthe path. Therefore, we should be using |~F| cos θ in our integral. If ~T is theunit tangent vector, then |~F| cos θ = ~F • ~T and the expression for the workdone along path C is:

W =∫

C

~F • ~T ds

The unit tangent vector ~T can be found by dividing the tangent vector~v = d~r

dt by its length.

~T =~v|~v|

We already know from before that ds = |~v| dt so:

~T ds =~v|~v| |~v| dt = ~v dt

Therefore, if the initial point on the curve occurs at t = a and the finalpoint is at t = b, the integral for work is given by:

W =∫ b

a

~F • ~v dt

Note that ~v dt = d~rdt dt = d~r, so the integral for work can be written more

succinctly as:

W =∫

C

~F • d~r

This is called a line integral whether the path C is a straight line or a curve.

Example: Let’s consider the same helical segment C as before:

~r = 〈x(t), y(t), z(t)〉 = 〈π cos t, π sin t, t〉 0 ≤ t ≤ π

and let’s consider the work along this path by the force:

~F = (z − π)~i = 〈z − π, 0, 0〉

The force is parallel to the x-axis at all points and it is shrinking to 0 as weapproach the last point (−π, 0, π) on the helical segment. At any point(x, y, z) = (π cos t, π sin t, t), the force is ~F = 〈t− π, 0, 0〉 and d~r is givenby:

d~r =d~rdt

dt = 〈−π sin t, π cos t, 1〉 dt

Therefore,

~F • d~r = 〈t− π, 0, 0〉 • 〈−π sin t, π cos t, 1〉 dt = π(π − t) sin t dt

and the work can be easily calculated using integration by parts:

W =∫

C

~F • d~r = π

∫ π

0

(π − t) sin t dt = π2

There are two interesting things to be observed about this integral. As youlearned in Calculus I, reversing the limits of integration changes the answerby a negative sign:

π

∫ 0

π

(π − t) sin t dt = −π2

We interpret this as reversing direction along the path. If we start witht = π then we are beginning at the point (−π, 0, π) and if we stop witht = 0 then we are ending up at (π, 0, 0). We use the notation −C to denotethe path C traversed in the opposite direction. Therefore,

∫

−C

~F • d~r = −∫

C

~F • d~r

The second interesting thing to point out is the effect of changing the pathbetween (π, 0, 0) and (−π, 0, π). Consider the following path :

〈x, y, z〉 = 〈π − 2πt, 0, 2πt− πt2〉 for 0 ≤ t ≤ 1

Let’s call this path Ω. Ω is a parabolic segment connecting (π, 0, 0) to(−π, 0, π).

At any point along this path:

~F • d~r = 〈z − π, 0, 0〉 • d~rdt

dt

= 〈2πt− πt2 − π, 0, 0〉 • 〈−2π, 0, 2π − 2πt〉 dt = 2π(πt2 + π − 2πt) dt

Therefore, the work done by ~F along Ω is:

∫

Ω

~F • d~r = 2π

∫ 1

0

(πt2 + π − 2πt) dt =23π3

Note that∫

C~F • d~r is not the same as

∫Ω

~F • d~r even though C and Ω havethe same starting point and the same ending point. We say that the integralis path dependent. An important result we will encounter later is that, for

certain vector fields, called conservative vector fields, the path connecting apoint P to a point Q will not affect the answer. Since the path does makesa difference for ~F = (z−π)~i, this would be an example of a nonconservativevector field.

Example:Let ~F = 〈y − x, z − y, x− z〉. Integrate this vector field over the straightline segment L connecting (0, 0, 1) to (1, 2, 2).

Solution:The straight line L can be described by the equation:

~r = 〈0, 0, 1〉+ t〈1, 2, 1〉 = 〈t, 2t, 1 + t〉 for 0 ≤ t ≤ 1

At any point on this line, ~F is given by:

~F = 〈y−x, z− y, x− z〉 = 〈2t− t, (1+ t)−2t, t− (1+ t)〉 = 〈t, 1− t, −1〉

The vector d~r is given by:

d~r =d~rdt

dt = 〈1, 2, 1〉 dt

We can now calculate the dot product ~F • d~r:

~F • d~r = 〈t, 1− t, −1〉 • 〈1, 2, 1〉 dt = (t + 2(1− t)− 1) dt = (1− t) dt

The line integral is therefore equal to:∫

L

~F • d~r =∫ 1

0

(1− t) dt =12

It is not always necessary to use a parametric representation of the path weintegrate over. In general, d~r = d~r

dt dt =⟨

dxdt , dy

dt , dzdt

⟩dt = 〈dx, dy, dz〉.

If ~F = 〈F1, F2, F3〉 then ~F • d~r = F1 dx + F2 dy + F3 dz and so the lineintegral can be written as:

∫

L

~F • d~r =∫

L

F1 dx + F2 dy + F3 dz

So, for example, the line integral we just did above can be written as:∫

L

~F • d~r =∫

L

(y − x) dx + (z − y) dy + (x− z) dz

In the case of the line we just considered, where x = t, y = 2t and z = 1+ t,we have:

y = 2x z = 1 + x

Therefore,∫

L

~F • d~r =∫

L

(y − x) dx + (z − y) dy + (x− z) dz

=∫ 1

0

(2x− x) dx + ((1 + x)− 2x) 2 dx + (x− (1 + x)) dx

=∫ 1

0

(1− x) dx

=12

It all comes to the same thing, but there are times that a description of thepath in terms of x, y and z directly might be more convenient. For example,suppose L1 denotes the straight line segment from (0, 0, 1) to (1, 2, 1) andL2 denotes the staight line segment from (1, 2, 1) to (1, 2, 2) and let L3 bethe combined path along L1 followed by L2.

The work done along the combined path L3 is the sum of the work donealong L1 and the work done along L2.

∫

L3

~F • d~r =∫

L1

~F • d~r +∫

L2

~F • d~r

Along L1, y = 2x and z = 1 so dy = 2 dx and dz = 0. Therefore:∫

L1

~F • d~r =∫

L1

(y − x) dx + (z − y) dy =∫ 1

0

(2− 3x) dx =12

Along L2, x = 1 and y = 2 at every point so dx = dy = 0.∫

L2

~F • d~r =∫

L2

(x− z) dz =∫ 2

1

(1− z) dz = −12

Therefore,∫

L3

~F • d~r =∫

L1

~F • d~r +∫

L2

~F • d~r =12− 1

2= 0

Please note that even though L and L3 have the same initial point andthe same final point,

∫L

~F • d~r is not the same as∫

L3~F • d~r, so ~F is not a

conservative vector field.

Example:Consider the path from (2, 0) to (0, 1) along the upper portion of the ellipsex2

4 + y2 = 1. Let’s call this path C1 because a little later we are going toconsider a different path, C2, connecting these two points.

Consider the following vector field:

~F = 〈−x + y, x− 2y〉and calculate the line integral:

∫

C1

~F • d~r

Let’s begin by describing the path with the equation y = 12

√4− x2. It

follows that dy = −x2√

4−x2 dx

~F • d~r = (−x + y) dx + (x− 2y) dy

= (−x +12

√4− x2) dx + (x−

√4− x2)

( −x

2√

4− x2

)dx

= −x

2dx +

2− x2

√4− x2

dx

To integrate over the path, we must be careful to start at x = 2 and endup at x = 0 if we want to integrate in the direction specified.

∫

C1

~F • d~r = −∫ 0

2

x

2dx +

∫ 0

2

2− x2

√4− x2

dx = 1−∫ 2

0

2− x2

√4− x2

dx

The remaining integral is somewhat tedious, but if you use trigonometricsubstitution with x = 2 sin θ, then you will find that

∫ 2

02−x2√4−x2 dx = 0 and

therefore, ∫

C1

~F • d~r = 1

The problem is actually much simpler if we use the parametric representa-tion of this ellipse:

x = 2 cos t y = sin t∫

C1

~F • d~r =∫

C1

(−x + y) dx + (x− 2y) dy

=∫ π/2

0

(−2 cos t + sin t)(−2 sin t dt) + (2 cos t− 2 sin t)(cos t dt)

=∫ π/2

0

(2 sin t cos t + 2(cos2 t− sin2 t)) dt

=∫ π/2

0

(sin 2t + 2 cos 2t) dt

= 1

Now, just for comparison, lets integrate from (2, 0) to (0, 1) along a differentpath. Let C2 be the elliptical path along the other three quarters of theellipse.

Since the parametric representation gives us the easiest integral, let’s usethat. To traverse path C2, we simply vary t from 2π to π

2 .

∫

C2

~F • d~r =∫ π/2

2π

(sin 2t + 2 cos 2t) dt = 1

Thus the choice of path has made no difference. Does this mean that theintegral of ~F from (2, 0) to (0, 1) is path independent? The problem is that,for all we know, there may be some other path, call it C3 from (2, 0) to(0, 1) where the answer for

∫C3

~F • d~r comes out differently. How can weresolve this? The answer lies in a line integral called a closed loop integral.

Closed Loop Integrals

Let’s use the same vector field ~F and the same ellipse as the last prob-lem, but this time, let’s integrate around the entire ellipse. Thus, if westart at (2, 0), we will end there as well. We can use the same parametricrepresentation of the ellipse and simply adjust the limits of integration.

x = 2 cos t y = sin t 0 ≤ t ≤ 2π

Let’s call this path C. When the starting point is the same as the endpoint,the path is called a closed loop and we use the symbol

∮for the integral.

∮

C

~F • d~r =∫ 2π

0

(sin 2t + 2 cos 2t) dt = 0

The integral of a vector field around a closed loop is called the circulation.It is not a coincidence that the circulation is 0. This a reliable feature ofpath independent integrals. In general, suppose we have two different pathsconnecting point P to point Q. Call these paths C1 and C2.

If we reverse direction along path C2, we get a closed loop C.

∮

C

~F • d~r =∫

C1

~F • d~r +∫

−C2

~F • d~r =∫

C1

~F~•d~r−∫

C2

~F • d~r

Thus, if the integral of a vector field is path independent and∫

C1~F • d~r

equals∫

C2~F • d~r then the circulation

∮C

~F • d~r is 0. Conversely, if thecirculation is always zero, no matter what closed loop we are considering,then the integrals along two different paths C1 and C2 connecting points Pto Q will always come out to be the same as each other.

The circulation of a vector field is of interest for other reasons. For example,in physics the concept of circulation comes up in Ampere’s Law which relatesthe current I along a wire to the magnetic field B around the wire.

If C is a closed loop around the wire, the circulation of the magnetic fieldB is proportional to the current I.∮

C

~B • d~r = µ0I

Another place we need this concept is in the study of fluids. If ~v is thevelocity vector field of a fluid, one of the starting points of fluid theory isthat the circulation is zero.

∮

C

~v • d~r = 0

Circulation Per Unit Area

Closely associated with the circulation of a vector field is the circulationper unit area. More precisely, if ~F is a vector field and C is a closed loopthen the circulation per unit area is defined as:

1Area

∮

C

~F • d~r

where the area referred to is the area inside the closed loop. Now, supposewe shrink the closed loop down and make the area smaller.

Keep making the area smaller.

If we take the limit as the area goes to zero, we are shrinking the closed loopdown to a point. The limit, if it exists, will be referred to as the circulationper unit area at a point.

limArea→0

1Area

∮

C

~F • d~r

If a vector field is conservative, then the circulation per unit area is goingto be zero, no matter how small the area, and therefore the circulation perunit area at a point can be expected to be zero at all points. The conversealso turns out to be true, but we will discuss this later when we get to aresult called Stokes’ Theorem.

Derivative Formula for Circulation Per Unit Area

There is a simple formula for finding the circulation per unit area that is interms of the partial derivatives of the coordinates of the vector field. Thereare two preliminary facts from calculus that we need to get this formula.The first is the Mean Value Theorem of Integrals. If f(x) is continuouson the interval [a, b], then there is some point c in this interval with theproperty that: ∫ b

a

f(x) dx = f(c)(b− a)

We also need the limit definition of the derivative:

f ′(x) = limh→0

f(x + h)− f(x)h

Please note that there are other similar limits that also come out the sameway:

f ′(x) = limh→0

f(x)− f(x− h)h

as well as:

f ′(x) = limh→0

f(x + h

2

)− f(x− h

2

)

h

You can use l’Hopital’s Rule to see that each of these limits come out equalto f ′(x). Note that the very last limit can be written as:

f ′(x) =f

(x + ∆x

2

)− f(x− ∆x

2

)

∆x+ E where lim

∆x→0E = 0

Let us begin our analysis of the circulation per unit area by considering aclosed loop parallel to the yz plane. For simplicity, we will take this loopto be a rectangular path C. There is a point (x, y, z) in the center of thisloop.

We will call the four segments of this rectangular path C1, C2, C3 and C4.The integral around the rectangular loop C is the sum of the line integralsalong each of the four segments:∮

C

~F • d~r =∫

C1

~F • d~r +∫

C2

~F • d~r +∫

C3

~F • d~r +∫

C4

~F • d~r

We will assume that the four corners of this rectangle have the coordinates:(x, y − ∆y

2, z − ∆z

2

) (x, y +

∆y

2, z − ∆z

2

)

(x, y +

∆y

2, z +

∆z

2

) (x, y − ∆y

2, z +

∆z

2

)

Let’s start with path C1 and C3

∫

C1

~F • d~r =∫

C1

F1 dx + F2 dy + F3 dz =∫

C1

F3 dz

This is true because x and y are constant along C1 so dx = dy = 0 onthis path. By the Mean Value Theorem for integrals, there is a point Pwith P =

(x, y + ∆y

2 , z∗)

for z∗ in the interval[z − ∆z

2 , z + ∆z2

]so that

∫C1

F3 dz = F3(P)∆z. By continuity, F3(P) = F3

(x, y + ∆y

2 , z)

+E1 wherelim

∆z→0E1 = 0 so

∫

C1

~F • d~r = F3

(x, y +

∆y

2, z

)∆z + E1∆z

Let’s write this more concisely as:

∫

C1

~F • d~r ≈ F3

(x, y +

∆y

2, z

)∆z

In the same way, we write the integral along C3 as:

∫

C3

~F • d~r ≈ −F3

(x, y − ∆y

2, z

)∆z

Please note the negative sign in front of F3 this time. This is because onC3, we start at z + ∆z

2 and end at z − ∆z2 so the direction is the opposite

of C1. Now, add these integrals together:∫

C1

~F • d~r +∫

C3

~F • d~r ≈ F3

(x, y +

∆y

2, z

)∆z − F3

(x, y − ∆y

2, z

)∆z

=

F3

(x, y + ∆y

2 , z)

∆z − F3

(x, y − ∆y

2 , z)

∆y

∆y ∆z

≈ ∂F3

∂y∆y ∆z

Next, we repeat this argument for paths C2 and C4

and the result is:∫

C2

~F • d~r +∫

C4

~F • d~r ≈ −∂F2

∂z∆y ∆z

Add the results together to get an approximation for∮

~F • d~r.∮

~F • d~r ≈ ∂F3

∂y∆y ∆z − ∂F2

∂z∆y ∆z

The error in this approximation goes to 0 as ∆y and ∆z go to 0. Since thearea inside C is Area= ∆y ∆z, the circulation per unit area is:

1Area

∮~F • d~r ≈ ∂F3

∂y− ∂F2

∂z

Therefore,

limArea→0

1Area

∮~F • d~r =

∂F3

∂y− ∂F2

∂z

So, the significance of the expression:

∂F3

∂y− ∂F2

∂z

is that it represents the circulation per unit area at (x, y, z) in a planeparallel to the yz plane. What about the other planes? An analysis similarto what we have just done shows that:

∂F1

∂z− ∂F3

∂x

represents the circulation per unit area at (x, y, z) in a plane parallel to thexz plane. Finally, the expression:

∂F2

∂x− ∂F1

∂y

represents the circulation per unit area at (x, y, z) in a plane parallel to thexy plane. We can package all these quantities very conveniently in a vectorcalled the curl defined as:

curl ~F =(

∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k

Thus, ~i • curl ~F is the circulation per unit area in the plane perpendicularto the ~i vector, ~j • curl ~F is the circulation per unit area in the planeperpendicular to the~j vector and ~k • curl ~F is the circulation per unit areain the plane perpendicular to the ~k vector. More generally, it can be shownthat if ~n is any unit vector then ~n • curl ~F is the circulation per unit areain the plane perpendicular to ~n. There is a nice formula using determinantsfor the curl. We write the expression ∂F3

∂y − ∂F2∂z as:

∂F3

∂y− ∂F2

∂z=

∣∣∣∣∂∂y

∂∂z

F2 F3

∣∣∣∣

Use this notation for each coordinate of the curl

curl ~F =~i∣∣∣∣

∂∂y

∂∂z

F2 F3

∣∣∣∣−~j∣∣∣∣

∂∂x

∂∂z

F1 F3

∣∣∣∣ + ~k∣∣∣∣

∂∂x

∂∂y

F1 F2

∣∣∣∣

This can be written more compactly as a three by three determinant:

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

This is commonly written as ∇ × ~F since the three by three determinanthas the same algebraic form as the cross product. When a vector field ~Fis conservative, the circulation per unit area is zero at every point and inevery plane, regardless of the orientation of the plane. Thus, we can tellif a vector field ~F is conservative simply by checking if ∇ × ~F is the zerovector.

Example:Calculate the curl of ~F = 〈−x + y, x− 2y, 0〉

∇ × ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−x + y x− 2y 0

∣∣∣∣∣∣=~i(0)−~j(0) + ~k(1− 1)= 〈0, 0, 0〉

This verifies our observation that the integral of ~F seemed to be path inde-pendent.

Example:Calculate the curl of ~F = 〈y − x, z − y, x− z〉

∇ × ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y − x z − y x− z

∣∣∣∣∣∣=~i(0− (−1))−~j(1− 0) + ~k(0− 1)= 〈1, −1, −1〉

Since curl ~F is not the zero vector, this verifies our observation that the valueof the line integral between two points might change if the path connectingthe two points changes.

The Potential Function

Let us suppose that ~F represents force so that∫

C~F • d~r is work. If ~F is

a conservative vector field then the integral is path independent. That is,if C connects point P0 to Q then the value of

∫C

~F • d~r would remain thesame even if we changed the path C connecting P0 to Q.

Since the path doesn’t matter, we will use the notation:

∫ Q

P0

~F • d~r

Suppose we took a side trip through point P to get from P0 to Q. If Γ isa path from P0 to P and Ω is a path from P to Q then:

∫

C

~F • d~r =∫

Γ

~F • d~r +∫

Ω

~F • d~r

Equivalently: ∫ Q

P0

~F • d~r =∫ P

P0

~F • d~r +∫ Q

P

~F • d~r

Now, let us consider the work done from point P0 = (x0, y0, z0) to P =(x, y, z). If we regard the initial point (x0, y0, z0) as a constant and thefinal point (x, y, z) as the variable then the work done is a function of(x, y, z) and will be denoted by φ(x, y, z)

φ(x, y, z) =∫ P

P0

~F • d~r =∫ (x,y,z)

(x0,y0,z0)

~F • d~r

φ(x, y, z) will be referred to as the potential function for ~F. The integralsum

∫ Q

P0~F • d~r =

∫ P

P0~F • d~r +

∫ Q

P~F • d~r that we considered before can be

written as:

φ(Q) = φ(P) +∫ Q

P

~F • d~r

and therefore, ∫ Q

P

~F • d~r = φ(Q)− φ(P)

Thus, the integral of a conservative vector field between any two pointscan be expressed as the change in its potential function. Therefore, givena vector field ~F it would be convenient to know how to get the formulafor the potential function φ. Of course, we do have an integral definitionφ(x, y, z) =

∫ P

P0~F • d~r and this could be used. However, if φ is obtained

by performing an antiderivative operation on the coordinates of ~F it wouldmake sense that we should be able to obtain the coordinates of ~F by per-forming some derivative operations on φ. Let’s begin by looking at thepartial derivative of φ(x, y, z) with respect to x.

∂φ

∂x= lim

h→0

φ(x + h, y, z)− φ(x, y, z)h

As we have just seen, φ(x + h, y, z) − φ(x, y, z) would result from a lineintegral.

φ(x + h, y, z)− φ(x, y, z) =∫ (x+h,y,z)

(x,y,z)

~F • d~r

=∫ (x+h,y,z)

(x,y,z)

F1 dx + F2 dy + F3 dz

Since the line integral of this vector field is path independent, we might aswell take the path to be a straight line from (x, y, z) to (x + h, y, z). Thismeans that y and z are constant along the path, so dy = dz = 0 and we areleft with:

φ(x + h, y, z)− φ(x, y, z) =∫ (x+h,y,z)

(x,y,z)

F1 dx

By the Mean Value Theorem of Integrals, there is a value x∗ somewherein the interval [x, x + h] such that

∫ (x+h,y,z)

(x,y,z)F1 dx = F1(x∗, y, z)h, and

therefore:

φ(x + h, y, z)− φ(x, y, z) =∫ (x+h,y,z)

(x,y,z)

F1 dx = F1(x∗, y, z)h

In exactly the same manner, we can show that:

φ(x, y + h, z)− φ(x, y, z) =∫ (x,y+h,z)

(x,y,z)

F2 dy = F2(x, y∗, z)h

φ(x, y, z + h)− φ(x, y, z) =∫ (x,y,z+h)

(x,y,z)

F3 dz = F3(x, y, z∗)h

where y∗ is some value in [y, y + h] and z∗ is some value in [z, z + h]. If wenow divide by h, we get:

φ(x + h, y, z)− φ(x, y, z)h

= F1(x∗, y, z)

φ(x, y + h, z)− φ(x, y, z)h

= F2(x, y∗, z)

φ(x, y, z + h)− φ(x, y, z)h

= F3(x, y, z∗)

If we take the limit as h goes to 0, we are left with:

∂φ

∂x= F1

∂φ

∂y= F2

∂φ

∂z= F3

This implies that ~F is the gradient of the potential function.

~F = 〈F1, F2, F3〉 =⟨

∂φ

∂x,

∂φ

∂y,

∂φ

∂z

⟩= ∇φ

We have just established that if the integral of ~F is path independent then~F must be the gradient of its potential function. The converse is also true.That is, if ~F = ∇φ for some scalar-valued function φ then the integral of ~Fbetween two points P and Q must be independent of the path connectingthose two points. To understand why this true, take any path C betweenP and Q. C will be described by parametric equations of the form:

x = x(t) y = y(t) z = z(t) for a ≤ t ≤ b

At any point t, φ will have the value φ(x(t), y(t), z(t)) on the curve. Bythe Chain Rule for Partial Derivatives,

d

dt(φ(x(t), y(t), z(t))) =

∂φ

∂x

dx

dt+

∂φ

∂y

dy

dt+

∂φ

∂z

dy

dt= ∇φ • d~r

dt= ~F • d~r

dt

Therefore,

∫

C

~F • d~r =∫ b

a

~F • d~rdt

dt =∫ b

a

d

dt(φ(x(t), y(t), z(t))) dt

= φ(x(b), y(b), z(b))− φ(x(a), y(a), z(a)) = φ(Q)− φ(P)

The final answer apparently comes out to φ(Q)− φ(P) regardless of whichpath C we have chosen. Notice that the line integral can now be writtenas: ∫ Q

P

∇φ • d~r = φ(Q)− φ(P)

This resembles the Fundamental Theorem of Calculus:∫ b

a

f ′(x) dx = f(b)− f(a)

so the formula∫ Q

P∇φ • d~r = φ(Q)− φ(P) is essentially a generalization of

the fundamental theorem to line integrals.

Example:

Calculate the potential function for ~F = 〈−x + y, x − 2y〉 and use yourresult to recalculate the line integral of ~F over C1, the elliptical path from(2, 0) to (0, 1) that we considered earlier.

Solution: If φ is the potential function, then ∂φ∂x = −x+y and ∂φ

∂y = x−2y.Take either one of these equations, it doesn’t matter much which one, andintegrate with respect to the appropriate variable. That is, either integrate−x + y with respect to x (holding y constant) or integrate x − 2y withrespect to y (holding x constant). Since it doesn’t matter which we startwith, let’s choose ∂φ

∂y = x− 2y. This implies that:

φ(x, y) =∫

(x− 2y) dy (holding x constant)

= xy − y2

You may notice that I haven’t written the usual constant of integration+C at the end of this antiderivative, so xy − y2 is not the most generalexpression for which ∂φ

∂y = x − 2y. Not only would xy − y2 + C also workfor any constant C, but the following expressions would also work:

xy − y2 + sin x xy − y2 + ex xy − y2 + ln x + C

More generally,xy − y2 + G(x)

Since x was being held constant in our integration, x may appear in our“constant” of integration.

If ∂φ∂y = x − 2y where the only condition on φ then xy − y2 + G(x) is the

answer for any function G(x). However, there is one more condition, namely∂φ∂x = −x + y and this will determine a specific expression for G(x)

∂

∂x(xy − y2 + G(x) = −x + y

y + G′(x) = −x + y so G′(x) = −x

If G′(x) = −x then G(x) = − 12x2 + C and therefore:

φ(x, y) = xy − y2 − 12x2 + C

Now, let’s use this to integrate ~F from (2, 0) to (0, 1). Since the integral ofthis ~F is path independent, we can get our answer directly with φ

∫ (0,1)

(2,0)

~F • d~=φ(0, 1)− φ(2, 0) = (−1 + C)− (−1222 + C) = 1

Notice how the +C cancels out. The +C will always cancel out when weevaluate φ(Q)− φ(P) so, from now on, we might as well choose C = 0 forcalculation of other potential functions.

Example: Find the potential function for ~F = 〈3y2 + y, x〉 and calculatethe line integral of this vector field over the straight line path from (0, 0) to(1, 4).

Solution: If there is a potential function φ then φ must satisfy the con-ditions:

∂φ

∂x= 3y2 + y

∂φ

∂y= x

If ∂φ∂y = x the φ(x, y) must have the form xy + G(x). If we now impose

the condition ∂φ∂x = 3y2 + y we get the equation y + G′(x) = 3y2 + y which

leads to the contradictory result that G′(x) = 3y2. G′(x) cannot equalan expression depending on the y variable. We conclude that there is nopotential function φ for this vector field.

We could have predicted this result had we calculated the curl of ~F.

∇× ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

3y2 + y x 0

∣∣∣∣∣∣= 〈0, 0, −6y − y〉

Had there been a function φ with ∇φ = ~F then:

∇× ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

∂φ∂x

∂φ∂y

∂φ∂z

∣∣∣∣∣∣

=⟨

∂2φ

∂y∂z− ∂2φ

∂z∂y,

∂2φ

∂z∂x− ∂2φ

∂x∂z,

∂2φ

∂x∂y− ∂2φ

∂y∂x

⟩

= 〈0, 0, 0〉

So, a good general rule is that if ∇ × ~F is not the zero vector then don’tbother looking for a function φ with ~F = ∇φ.

There are a couple of things to note about this last vector field. The first isthat even though ~F = 〈3y2 + y, x〉 it doesn’t mean that we can’t calculatethe line integral along the straight line path from the origin to (1, 4). Let’scall this path Γ. Along Γ, y = 4x so,

∫

Γ

(3y2 + y) dx + x dy =∫ 1

0

(3(4x)2 + 4x) dx + x (4 dx)

=∫ 1

0

(48x3 + 8x

)dx

= 20

The second thing to note is even though ~F = 〈3y2+y, x〉 is not the gradientof any function, there is still a way to use potential functions to do the lineintegral. The vector field ~F can be split up in the following way:

~F = 〈3y2 + y, x〉 = 〈3y2, 0〉+ 〈y, x〉

The vector field 〈y, x〉 does indeed have a potential function. A simplecalculation shows that 〈y, x〉 = ∇φ where φ = xy. So, ~F has split upinto a nonconservative part and a conservative part. We can now do the

calculation of∫Γ

~F • d~r using this fact:∫

Γ

~F • d~r =∫

Γ

(〈3y2, 0〉+∇φ) • d~r

=∫

Γ

〈3y2, 0〉 • 〈dx, dy〉+∫

Γ

∇φ • d~r

=∫ 1

0

3(4x)2 dx + φ(1, 4)− φ(0, 0)

=∫ 1

0

48x3 dx + (1)(4)− (0)(0)

= 16 + 4 = 20

Next, let’s try a more three dimensional example.

Example:

Find the potential function φ(x, y, z) for ~F = k(x2+y2+z2)3/2 〈x, y, z〉 where

k is a constant. The significance of this vector field is that it is the sameformat as the gravitational force or electrostatic force.

Solution:

A straightforward, but slightly tedious calculation, shows that ∇× ~F = ~0except at the origin. So, we should be able to find a function φ(x, y.z) suchthat:

∂φ

∂x=

kx

(x2 + y2 + z2)3/2

∂φ

∂y=

ky

(x2 + y2 + z2)3/2

∂φ

∂z=

kz

(x2 + y2 + z2)3/2

Let’s start with the equation for ∂φ∂x . Integrate with respect to x to get an

initial general expression for φ

φ(x, y, z) =∫

kx

(x2 + y2 + z2)3/2dx

To calculate this integral, let u = x2 + y2 + z2 and remember that y and zare being held constant in this operation.

φ(x, y, z) =k

2

∫u−3/2 du = −ku−1/2 + G(y, z) =

−k√x2 + y2 + z2

+ G(y, z)

Next, we impose the condition

∂φ

∂y=

ky

(x2 + y2 + z2)3/2on φ =

−k√x2 + y2 + z2

+ G(y, z)

to obtain:

ky

(x2 + y2 + z2)3/2+

∂G

∂y(y, z) =

ky

(x2 + y2 + z2)3/2

and therefore ∂G∂y (y, z) = 0. From this we conclude that G is either constant

or depends on z alone. Thus,

φ(x, y, z) =−k√

x2 + y2 + z2+ H(z)

Finally, we impose the condition

∂φ

∂z=

kz

(x2 + y2 + z2)3/2on φ(x, y, z) =

−k√x2 + y2 + z2

+ H(z)

and we obtain:

kz

(x2 + y2 + z2)3/2+ H ′(x) =

kz

(x2 + y2 + z2)3/2

Therefore, H must be a constant. We might as well choose this constant tobe 0 and we are left with the potential function:

φ(x, y, z) =−k√

x2 + y2 + z2

The Differential

Recall from Calculus III that if φ = φ(x, y, z) then the differential of φ isgiven by:

dφ =∂φ

∂xdx +

∂φ

∂ydy +

∂φ

∂zdz

We can write this as a dot product:

dφ =∂φ

∂xdx +

∂φ

∂ydy +

∂φ

∂zdz =

⟨∂φ

∂x,

∂φ

∂y,

∂φ

∂z

⟩• 〈dx, dy, dz〉 = ∇φ • d~r

Therefore, the Fundamental Theorem for Line Integrals:

∫ Q

P

∇φ • d~r = φ(Q)− φ(P)

can be written more simply as:

∫ Q

P

dφ = φ(Q)− φ(P)

Let’s examine this relationship more closely. Suppose we have divided ourpath C into n subintervals.

The difference in potential between a typical point (xi, yi, zi) and the pointbefore it is:

φ(xi, yi, zi)− φ(xi−1, yi−1, zi−1)

Let’s use the notation φi to stand for φ(xi, yi, zi) The sum of all thesedifferences on C is:

n∑

i=1

(φi − φi−1)

If we write out the sum, we can see that most of the terms cancel:

n∑

i=1

(φi−φi−1) = (φ1−φ0)+(φ2−φ1)+(φ3−φ2)+· · ·+(φn−φn−1) = φn−φ0

If the endpoints of C are P and Q, then φn = φ(Q) and φ0 = φ(P)

n∑

i=1

(φi − φi−1) = φ(Q)− φ(P)

All this is saying is that the sum of all the interior changes of the potentialfunction is the change in the potential function from one endpoint to theother. That is, the sum of all the small interior changes is the net change.The formula

∫ Q

Pdφ = φ(Q) − φ(P) is simply the limit of the sum as the

number of points goes to infinity.

Green’s Theorem

We have seen that if ~F is a conservative vector field then∮

C~F • d~r = 0.

There is a very convenient formula, called Stokes’ Theorem, for∮

C~F•d~r that

is true whether ~F is conservative or not. We shall consider the special casewhere C is a closed loop in the xy plane and ~F = F1(x, y)~i+F2(x, y)~j. Theformula for

∮C

F1 dx+F2 dy for this special case is called Green’s Theorem.

Let’s begin with the fact that ∂F2∂x − ∂F1

∂y represents the circulation per unitarea at a point. Let D denote the region enclosed by C. We can divide Dinto n interior pieces.

Let’s focus on the ith interior piece Di

Let’s suppose that it’s boundary is Ci. The boundary will consist of atmost 4 segments. Let’s call the segments Ci1, Ci2, Ci3 and Ci4

If ∆Ai denotes the area of Di then if Pi is some point in the interior, then:

(∂F2

∂x(Pi)− ∂F1

∂y(Pi)

)∆Ai

will approximate the circulation around Di because we are multiplying thecirculation per unit area by the area. Therefore,

∮

Ci

~F • d~r ≈(

∂F2

∂x(Pi)− ∂F1

∂y(Pi)

)∆Ai

If we write∮

Ci

~F • d~r as the sum of the line integrals along its boundariesthen:

4∑

j=1

∫

Cij

~F • d~r ≈(

∂F2

∂x(Pi)− ∂F1

∂y(Pi)

)∆Ai

Add these up over the interior of Dn∑

i=1

4∑

j=1

∫

Cij

~F • d~r ≈n∑

i=1

(∂F2

∂x(Pi)− ∂F1

∂y(Pi)

)∆Ai

This expression simplifies dramatically if we notice certain cancellations.

Let’s look at two adjacent sections, say D1 and D2.

The circulation around D1 is:∫

C11

~F • d~r +∫

C12

~F • d~r +∫

C13

~F • d~r +∫

C14

~F • d~r

and the circulation around D2 is:∫

C21

~F • d~r +∫

C22

~F • d~r +∫

C23

~F • d~r +∫

C24

~F • d~r

If we were to add these integrals together, we get eight line integrals. How-ever, notice that

∫C12

~F • d~r = − ∫C24

~F • d~r because the two integralstraverse the same line segment but in opposite directions. Therefore theycancel and instead of getting eight line integrals, we will really only get sixline integrals once we have eliminated the terms that cancel. Now, let’s addanother section D3 and to illustrate the next cancellation more clearly, let’stake this next section to be directly above D1.

The circulation around D3 is:∫

C31

~F • d~r +∫

C32

~F • d~r +∫

C33

~F • d~r +∫

C34

~F • d~r

If we add this to the six line integrals we had before, there is some additionalcancellation since

∫C31

~F • d~r = − ∫C13

~F • d~r. Consequently, we cancel outthe integrals over all shared boundaries.

We continue in this manner of the entire region D. The only integrals thatwill not cancel will be the integrals along the outer boundary of D.

∮

C

~F • d~r =n∑

i=1

4∑

j=1

∫

Cij

~F • d~r

Therefore, ∮

C

~F • d~r ≈n∑

i=1

(∂F2

∂x(Pi)− ∂F1

∂y(Pi)

)∆Ai

The error in approximation goes to 0 as n goes to infinity and the sum onthe right approaches a double integral.

∮

C

~F • d~r =∫ ∫

D

(∂F2

∂x− ∂F1

∂y

)dA

This formula is known as Green’s Theorem. It says, essentially, that thesum of all the interior circulations will add up to the circulation around theouter boundary.

Example:Let C be the triangular loop from (0, 0) to (1, 0) to (1, 1) and finally backto (0, 0). Let ~F = 〈4y, 6x2〉. Use Green’s Theorem to evaluate

∮C

~F • d~r.Solution:

∮

C

~F • d~r =∫ 1

0

∫ x

0

(∂F2

∂x− ∂F1

∂y

)dy dx =

∫ 1

0

∫ x

0

(12x− 4) dy dx = 2

Calculation of Area with Green’s Theorem

If we choose special expressions F1 and F2, we can relate area to the lineintegrals. For example, if F1 = 0 and F2 = x then

∮C

F1 dx + F2 dy =∫∫D

(∂F2∂x − ∂F1

∂y

)dA becomes:

∮

C

x dy =∫ ∫

D1 dA = Area(D)

This is not the only choice for F1 and F2 that will result in area. Forexample, if F1 = y and F2 = 0, then Green’s Theorem becomes:∮

C

y dx =∫ ∫

D−1 dA = −Area(D)

Subtract these two integrals and the result is:

2Area(D) =∮

C

−y dx + x dy

Divide by 2 and we get:

Area(D) =12

∮

C

−y dx + x dy

Example:Use a line integral to find the area inside an ellipse.

Solution:The parametric equations x = a cos t, y = b sin t for 0 ≤ t ≤ 2π describethe ellipse. By Green’s Theorem:

Area(Ellipse) =12

∮

C

−y dx + x dy

=12

∫ 2π

0

(−b sin t)(−a sin t dt) + (a cos t)(b cos t dt)

=12

∫ 2π

0

ab(sin2 t + cos2 t

)dt

=ab

2

∫ 2π

0

dt

= πab