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Chapter 4

Vectors and Tensors

In this chapter, we will switch our definitions of vectors to be objects that have a lengthand direction. Later, we will show that these vectors can be represented as column vectorsin the matrix algebra domain and that the vector operations to be described in this chapterdo have strong correspondence with matrix operations.

We will restrict our discussion to real vectors existing in the three dimensional space,and we will represent the vectors by an arrow in which the length of the arrow describes itsmagnitude and the arrowhead describes its direction. Symbolically, we include an underlineto distinguish the vectors treated in this chapter to those treated in the previous chapter,e.g. instead of v, we will denote the vector by v.

4.1 Free Vectors and Bound Vectors

There are two types of vectors. The first type is called the bound vector in which the vectoris strongly fixed to a specified point. The other type is called the free vector in which thevector could be moved in parallel during operations such as addition, dot products or crossproducts (see Figure 4.1). Ironically, free vectors are used more prevalently on rigid domains,while bound vectors are used more predominantly on flowing or flexible domains. Consideringthat we will be developing operations on flow fields, we need to insist on using bound vectorsas much as possible. Thus, we base our operations on bound vectors as a default. However,there will be occasions in which we might change the character of the vector to be free,specially when using free vectors significantly improves computational efficiency. (Whensuch change occurs, we will include a special mention.)

109

110 c© 2006 Tomas Co, all rights reserved

Figure 4.1: Two types of vectors: bound and free.

Figure 4.2: Addition of bound vectors.

4.2 Vector Operations

There are several operations involving vectors. We will begin with the general proceduresand then follow it up with alternative procedures based on unit vectors in the next section.

4.2.1 General Operations

1. Vector Addition.

• Notation: a + b = c

• Procedure: Assuming a and b are bound to the same point, i.e. their tailscoincide at point O, construct a parallelogram with a and b as two adjacent sides.Then c = a + b is the vector whose tail is at point O and its arrowhead is at thepoint directly opposite point O in the parallelogram. (see Figure 4.2).

In case a and b are free vectors, an alternative procedure can be used to obtain thesum. First, attach the tail of b to the arrowhead of a. Then c is the vector whosetail coincides with the tail of a and whose arrowhead coincides with arrowhead ofb (see Figure 4.3).

• Property: Vector addition is commutative.

2. Magnitude.

c© 2006 Tomas Co, all rights reserved 111

Figure 4.3: Addition of free vectors.

Figure 4.4: The norm of a vector.

• Notation: ‖v‖• Procedure: Measure the length of the arrow (see Figure 4.4).

3. Scalar Product.

• Notation: α v = w

• Procedure: Obtain the magnitude of v and multiply with the scalar factorα, then w becomes a vector whose length equal α‖v‖ and points in the directionof v (see Figure 4.5).

4. Dot Product. (also known as Inner Product)

• Notation: a · b = α

• Procedure: Obtain the length of projection of a on the direction of b, i.e.‖a‖ cos(θ), where θ is the included angle between a and b. Then multiply thisvalue with the length of b. Then α = ‖a‖‖b‖ cos(θ) . (see Figure 4.6).

• Property: Inner Products are commutative.

5. Cross Product.

Figure 4.5: Scalar product of a vector.


Figure 4.6: Dot product of vector a and b.

Figure 4.7: Cross product of vector a and b.

• Notation: a × b = c

• Procedure: Take the area of a parallelogram formed by a and b joined at thetails, i.e. area=‖a‖‖b‖ sin(θ), where θ is the angle from a to b in the counterclock-wise manner. Then c is simply a vector whose magnitude is equal to this areaand points in the direction perpendicular to the plane containing a and b using aright hand screw system (see Figure 4.7).

• Property: Cross products are anti-commutative, i.e. a × b = −b × a.

4.2.2 Vector Operations based on Unit Vectors

Consider a coordinate system having a basis of three mutually perpendicular axes. Thenunit vectors are simply the vectors of unit length each pointing in the direction of positivepart of the axes. For the Cartesian coordinate system, we will follow the convention shownin Figure 4.8. Based on the cartesian coordinate system, the unit vectors are then denotedby δx, δy, and δz.

Based on the general operations described in the previous section, the dot product andcross products among the unit vectors are given by:


Figure 4.8: The Cartesian coordinate system with three axis perpendicular to each other.

Figure 4.9: Vector V as a linear combination of unit vectors.

δx · δx = δy · δy = δz · δz = 1 (4.1)

δx · δy = δx · δz = δyδz = δy · δx = δz · δx = δzδy = 0 (4.2)

δx × δx = δy × δy = δz × δz = 0 (4.3)

δx × δy = δz ; δy × δz = δx ; δz × δx = δy

δy × δx = − δz ; δz × δy = − δx ; δx × δz = − δy (4.4)

Having identified the unit vectors, any vector can now be represented as a linearcombination of the unit vectors as shown in Figure 4.9, i.e.

V = vxδx + vyδy + vzδz (4.5)

The coefficient vx is the magnitude of the projection of V on δx, i.e. vx = V · δx.Likewise, vy = V · δy and vz = V · δz.

Based on (4.5) and the identities given in (4.1)-(4.4), the vector operations can beperformed in the following alternative manner:

1. Vector Addition.

v + w = (vxδx + vyδy + vzδz) + (wxδx + wyδy + wzδz)

= (vx + wx)δx + (vy + wy)δy + (vz + wz)δz


2. Norm.

‖v‖ = ‖vxδx + vyδy + vzδz‖

=√

v2x + v2

y + v2z

3. Scalar Product

α v = α(vxδx + vyδy + vzδz)

= αvxδx + αvyδy + αvzδz

4. Dot Product

v · w = (vxδx + vyδy + vzδz) · (wxδx + wyδy + wzδz)

= (vxδx) · (wxδx) + (vxδx) · (wyδy) + (vxδx) · (wzδz) +

(vyδy) · (wxδx) + (vyδy) · (wyδy) + (vyδy) · (wzδz) +

(vzδz) · (wxδx) + (vzδz) · (wyδy) + (vzδz) · (wzδz)

= (vxwx)(δx · δx) + (vxwy)(δx · δy) + (vxwz)(δx · δz) +

(vywx)(δy · δx) + (vywy)(δy · δy) + (vywz)(δy · δz) +

(vzwx)(δz · δx) + (vzwy)(δz · δy) + (vzwz)(δz · δz)

= vxwx + vywy + vzwz

5. Cross Product

v × w = (vxδx + vyδy + vzδz) × (wxδx + wyδy + wzδz)

= (vxδx) × (wxδx) + (vxδx) × (wyδy) + (vxδx) × (wzδz) +

(vyδy) × (wxδx) + (vyδy) × (wyδy) + (vyδy) × (wzδz) +

(vzδz) × (wxδx) + (vzδz) × (wyδy) + (vzδz) × (wzδz)

= (vxwx)(δx × δx) + (vxwy)(δx × δy) + (vxwz)(δx × δz) +

(vywx)(δy × δx) + (vywy)(δy × δy) + (vywz)(δy × δz) +

(vzwx)(δz × δx) + (vzwy)(δz × δy) + (vzwz)(δz × δz)

= (vxwy)δz − (vxwz)δy − (vywx)δz + (vywz)δx

+ (vzwx)δy − (vzwy)δx

= (vywz − vzwy)δx + (vzwx − vxwz)δy + (vxwy − vywx)δz

As a mnemonic, others have used the determinant formula for the cross product,

v × w = det

δx δy δz

vx vy vz

wx wy wz

(4.6)


For most people, this formula is easier to remember. Nonetheless, it is important toremember that (4.6) is just a memory tool and not the definition. (The distinctionbetween a memory tool and definition is most crucial later when we perform operationssuch as “curl” of vector fields.)

4.3 Dyads

Dyads are one type of operators on vectors. The dyad requires two defining vectors asits arguments, say d1 and d2, and it is often denoted by writing both defining vectors sideby side, e.g. d1d2. The dyad operation consists of a sequence of procedures and is strictlydependent on the order in which the dyad is written. For instance, d1d2 is not the same asd2d1. The first vector in the dyad is called the antecedent and the second vector in thedyad is called the consequent.

Consider the dyad, d1d2 that is operating on vector v to yield another vector w.Symbolically, we have (d1d2)(v) = w. The procedure of the dyad operators consists of thefollowing steps: first, one takes the dot product between d2 and v to yield a scalar valueα = d2 · v. Then this value is used for the scalar product (α d1) to produce a vector w.Thus w is a vector whose length is (d2 · v)‖d1‖ and is pointing in the same direction as d1.( Based on this procedure, other use the dot product notation to remind them of the dyadoperation, i.e. (d1d2)(V ) = d1d2 · V . )

Among the different possible dyads, an important set of nine dyads are the unit dyadsformed by the unit vectors:

δxδx, δxδy, δxδz

δyδx, δyδy, δyδz

δzδx δzδy, δzδz

4.4 Tensors

We will define tensors as simply an operator which is a linear combination of all nine unitdyads.1

τ = τxxδxδx + τxyδxδy + τxzδxδz + . . . + τzzδzδz (4.7)

We denote tensors by including double underlines to distinguish it from vectors.

Example 4.1 Stress Tensors

1More accurately, the tensors defined above is called the second order tensor and vectors are also calledfirst order tensors. This suggests that there are higher order tensors. However, we will not delve into tensorsthat are higher than second order.


Figure 4.10: The stress vector with respect to δz as a sum of the normal stress and shearstress.

The stress tensor is actually an indicator for the state of stress at a point. Sincestress is defined as force per unit area, the stress at a particular point has meaningonly with respect to a plane. In three dimensional space, the plane S attached toa point p is automatically defined by a unit normal vector, say n, that is boundto point p. With respect to plane S, we can determine the amount of stressalong the direction n which is normal to the plane, called the normal stress.The other stress is directed parallel to the plane, and is called the shear stress.Taking the sum of the normal stress and the shear stress will simply give thestress vector at that point, which we can denote as τ (p,n).

Let T be the stress tensor given by

T = Txxδxδx + Txyδxδy + Txzδxδz

+Tyxδyδx + Tyyδyδy + Tyzδyδz

+Tzxδzδx + Tzyδzδy + Tzzδzδz

To understand the coefficients of T , consider the x–y plane at say point p asshown in Figure 4.10. The unit normal vector to this plane is δz. When thetensor T operates on δz (based on dyad operations discussed in the previoussection),

τ (p,δz) = T δz = Txzδx + Tyzδy + Tzzδz

Since Tzz is the term along δz, it gives the value of the normal stress to the x–yplane. On the other hand, the other two terms, Txz and Tyz are simply the x andy components of the shear stress, acting across the x–y plane.

For the general case,

n = nxδx + nyδy + nzδz


Then the stress vector at a point p with respect to the plane specified by a normalvector n, is obtained by having T operate on n,

τ (p,n) = T n (4.8)

= (Txxδxδx + Txyδxδy + . . . + Tyzδyδz + Tzzδzδz)(nxδx + nyδy + nzδz)

= (Txxnx + Txyny + Txznz)δx

+(Tyxnx + Tyyny + Tyznz)δy

+(Tzxnx + Tzyny + Tzznz)δz

The discussion leading to equation (4.8) is simply a statement of the relationshipbetween the stress tensor and the stress vector. Equation (4.8) is also known asCauchy’s fundamental theorem for stress. It was included here as an example toshow how the tensors as a linear combination of unit dyads are seen as operatorson vectors. A fuller development of the relationship (4.8) involves the applicationof Newton’s laws of motion.2

♦♦♦

There is a special tensor, called the unit tensor defined by

δ = δxδx + δyδy + δzδz (4.9)

When this tensor operates on any vector, the result is the same vector.

4.4.1 Tensor Operations

When applying tensors, the operations of the unit dyads are simply combined linearly. More-over, as shown in the previous example, when the vector being operated on by the tensorsare also represented as a linear combination of unit vectors, simpler formulas can result. Wesummarize these operations below.

1. Tensor Addition.

τ + σ =∑

i=x,y,z

∑

j=x,y,z

(τij + σij)δiδj

2. Scalar Multiplication of a Tensor.

ατ =∑

i=x,y,z

∑

j=x,y,z

(ατij)δiδj

2See D. C. Leigh, Nonlinear Continuum Mechanics, McGraw Hill, 1968, for a derivation of (4.8).


3. Inner Product of Two Tensors.

τ · σ =∑

i=x,y,z

∑

j=x,y,z

(

∑

k=x,y,z

τikσkj

)

δiδj

4. Inner Product of A Tensor with a Vector.

τ · v =∑

i=x,y,z

(

∑

j=x,y,z

τijvj

)

δi

5. Double Dot Product.

τ : σ =∑

i=x,y,z

∑

j=x,y,z

τijσji

4.4.2 Correspondence with Matrix Operations

If the coordinate system is the orthogonal Cartesian coordinate system that is fixed in space,then the tensor and vector operation can be represented by matrices by using the following:

δx =

100

δy =

010

δz =

001

Once this vector is represented by column vectors, a close correspondence is availablebetween tensors and matrices. This is shown in Table 4.1

4.5 Vector Differential Operations

In this section, we explore differential operations on vectors. First, we will focus on takingtime derivative of a vector bound at a fixed point p, i.e. Vp = Vp(t). Then, we investigatespatial differential operators of a vector field, V = V (x, y, z).

4.5.1 Vector Function of One Variable.

Consider a vector fixed at a point that is a function of a single variable, e.g. t,

V (t) = Vx(t)δx + Vy(t)δy + Vz(t)δz (4.10)

We can extend the same definition used for derivatives of real functions and apply itto vectors while using vector addition and scalar products,


Table 4.1: Correspondence between Tensors and Matrices.

Vector Notations Matrix Representation

a = axδx + ayδy + azδz a = (ax, ay, az)T

a + b a + b = (ax + bx, ay + by, az + bz)T

αv α v = (αvx, αvy, αvz)T

a · b aT b

ab abT

τ =∑

i=x,y,z

∑

j=x,y,z τijδiδj τ = (τij)

τ + σ τ + σ = (τij + σij)

ατ α τ = (ατij)

τ · σ τ σ = (∑

k=x,y,z τikσkj)

τ · v τ v = (∑

j=x,y,z τxjvj,∑

j=x,y,z τyjvj,∑

j=x,y,z τzjvj)T

τ : σ Trace(τ σ)


Figure 4.11: Definition of dV/dt.

dV

dt= lim

∆t→0

1

∆t(V (t + ∆t) − V (t))

= lim∆t→0

1

∆t

([

Vx(t + ∆t)δx + Vy(t + ∆t)δy + Vz(t + ∆t)δz

]

−[

Vx(t)δx + Vy(t)δy + Vz(t)δz

])

=dVx

dtδx +

dVy

dtδy +

dVz

dtδz (4.11)

To illustrate the definition geometrically, (4.11) is shown in Figure 4.11. Note that thevector dV/dt does not necessarily point in the same direction as V .

Applying (4.11) to different operations, the following identities follow immediately:

1. Derivative of vector sum.

d

dt(U(t) + V (t)) =

dU

dt+

dV

dt(4.12)

2. Derivative of scalar products.

d

dt(φ(t)V (t)) =

dφ

dtV + φ

dV

dt(4.13)

3. Derivative of dot products.

d

dt(U(t) · V (t)) =

dU

dt· V + U · dV

dt(4.14)

4. Derivative of cross products.

d

dt(U(t) × V (t)) =

dU

dt× V + U × dV

dt(4.15)


4.5.2 Differential operators on Vector Fields

In this section, we will consider both a scalar field, φ, and a vector field V in which both fieldsare defined as function of its position in a coordinate system, e.g. φ(x, y, z) and V (x, y, z),respectively.

4.5.2.1 Gradients

Consider a three dimensional scalar field φ(x, y, z). An example of a scalar field is thetemperature distribution in the three dimension space, T (x, y, z). Usually the gradation ofscalar fields are used to determine the driving forces during transport of mass, momentum orenergy. Thus, it is important to generate a flow field based on the gradations to determinethe direction and magnitude of the driving forces. The flow field, or vector field, that isgenerated by the gradation of a scalar field is called the gradient vector field. At eachpoint (x, y, z), there corresponds a vector attached to the point given by

gradient(φ(x, y, z)) =∂φ

∂xδx +

∂φ

∂yδy +

∂φ

∂zδz (4.16)

This definition is further simplified if we introduce a differential operator called the gradientoperator, or grad for short, denoted by ∇ and defined by

∇ = δx

∂

∂x+ δy

∂

∂y+ δz

∂

∂z(4.17)

Thus,

gradient(φ(x, y, z)) = ∇φ(x, y, z)

To graphically illustrate the definition of gradient, consider the special case in whichthe scalar field is defined only on the x–y plane. As shown in Figure 4.12, we need to takethe partial derivatives with respect to each dimension. Each of these partial derivativesdetermine the magnitude of the x or y components which are then added vectorially to yieldthe gradient vector at that point.

The gradient vector is also sometimes used to determine the rate of change in φ alonga path s in the domain. At any point, the differential of φ(x, y, z) is given by

dφ(x, y, z) =∂φ

∂xdx +

∂φ

∂ydy +

∂φ

∂zdz

Along the path s, the rate of change in φ is obtained by dividing by ds,

dφ(x, y, z)

ds=

∂φ

∂x

dx

ds+

∂φ

∂y

dy

ds+

∂φ

∂z

dz

ds(4.18)


Figure 4.12: Evaluation of ∇φ(x, y).

The right hand side of the equation can now be factored into a dot product,

dφ(x, y, z)

ds= (∇φ) ·

(

dx

dsδx +

dy

dsδy +

dz

dsδz

)

= (∇φ) · t (4.19)

where

t =dx

dsδx +

dy

dsδy +

dz

dsδz

is a unit vector tangent to the path s at the particular point (x, y, z) in the path as shownin Figure 4.13. The term dφ/ds is known as the directional derivative along the directiont.

By considering (4.19) and the definition of a dot product,

dφ

ds= ‖∇φ‖ cos θ (4.20)

where θ is the angle between ∇φ and t. Thus, if the vector ∇φ is pointed at the samedirection as t, i.e. θ = 0 and cos θ = 1, we obtain the maximum rate of increase in the scalarfield at that particular point. In other words, at a particular point (xo, yo, zo), the vector∇φ at this point indicates both the maximum rate of increase of φ within its differentialneighborhood and the direction under which this maximum is achieved.

Another consequence of (4.20) is that when θ = 90o, dφ/ds = 0, i.e. φ(x, y, z) isconstant in the direction perpendicular to the gradient vector, ∇φ. To illustrate, if φ =φ(x, y), a contour map such as shown in Figure 4.14 also indicates partial gradient field


Figure 4.13: The unit tangent vector along path s.

information. For instance, the gradient vectors at contour lines are directed normal to thecontour curves. Also, if the distance between two successive contour levels is farther at apoint, the gradient vector has a smaller magnitude at that point.

4.5.2.2 Divergence

Consider now a vector field V , where attached at a point (x, y, z) is a vector

V (x, y, z) = Vx(x, y, z)δx + Vy(x, y, z)δy + Vz(x, y, z)δz

Figure 4.15 shows a vector field for a two dimensional domain. The vectors at differentpoints will generally have different magnitudes and direction.

We will assume that the vector field is continuous, i.e. that the vector componentsVx(x, y, z), Vy(x, y, z) and Vz(x, y, z) have finite first order partial derivatives. The diver-gence of a vector field at a point is then given as the sum of partial derivatives of thecomponents along their respective axes,

divergence(V (x, y, z)) =∂Vx

∂x+

∂Vy

∂y+

∂Vz

∂z(4.21)

In terms of the ∇ operator, the divergence is written as ∇ · V ,

∇ · V =

(

δx

∂

∂x+ δy

∂

∂y+ δz

∂

∂z

)

·(

Vxδx + Vyδy + Vzδz

)


Figure 4.14: Contour maps for a two dimensional scalar field. Note that ‖∇φ‖ at Q is greaterthan ‖∇φ‖ at P .

Figure 4.15: A vector field in two dimensional domain.


Figure 4.16: Evaluation of the divergence of a vector field at a point.

= δx ·∂Vx

∂xδx + δx ·

∂Vy

∂xδy + δx ·

∂Vz

∂xδz

+δy ·∂Vx

∂yδx + δy ·

∂Vy

∂yδy + δy ·

∂Vz

∂yδz

+δz ·∂Vx

∂zδx + δz ·

∂Vy

∂zδy + δz ·

∂Vz

∂zδz

=∂Vx

∂x+

∂Vy

∂y+

∂Vz

∂z

Based on this definition, the evaluation of divergence is shown in Figure 4.16. Notethat the divergence of a vector field will yield a scalar field. This is in direct to contrast togradients, for which the gradient of a scalar field will result in a vector field.

In physical terms, if the vector field is specifically a velocity field, the divergence atthe point of a velocity field is a measure of how much a property such as concentration ormass diverges at that point due to the changes in velocities along each axis. In fact, thecontinuity equation (or differential mass balance) is given by

∇ · (ρ(x, y, z)V (x, y, z)) = 0

where ρ is the density and V is the velocity field, both of which are dependent on the position(x, y, z). If ∇ · (ρV ) > 0 at some point P , then we say that P is a source point. On theother hand, if ∇ · (ρV ) < 0 at point Q, then Q is said to be a sink point.


Figure 4.17: Evaluation of the curl of a vector field at a point.

4.5.2.3 Curl

Due to the distribution in a vector field, neighboring vectors will affect the differential volumeat a point to twirl or rotate. Figure 4.17 shows the effects of vectors that lie in the x–y planeat a fixed z level. Next we define Czδz to be the vector to describe its tendency to rotatebased on the rate of increase of the y-component along the x-axis and the rate of increase ofthe x-components along the y-axis. Using the right-hand screw convention, the total effectis given by

Cz = lim∆→0

(

Vy(x + ∆x, y, z) − Vy(x, y, z)

∆x− Vx(x, y + ∆y, z) − Vy(x, y, z)

∆y

)

=∂Vy

∂x− ∂Vx

∂y

Following the same analysis in the x–z plane, we get Cyδy

Cy =∂Vx

∂z− ∂Vz

∂x

and in the y–z plane, we have Cxδx, where

Cx =∂Vz

∂y− ∂Vy

∂z


Adding the three vectors yields a vector known as the Curl of the vector field at the point(x, y, z),

Curl(V (x, y, z)) =

(

∂Vz

∂y− ∂Vy

∂z

)

δx +

(

∂Vx

∂z− ∂Vz

∂x

)

δy +

(

∂Vy

∂x− ∂Vx

∂y

)

δz (4.22)

Using the ∇ operator, the curl is written as the cross product of ∇ with V ,

∇× V =

(

δx

∂

∂x+ δy

∂

∂y+ δz

∂

∂z

)

×(

Vxδx + Vyδy + Vzδz

)

= δx ×∂Vx

∂xδx + δx ×

∂Vy

∂xδy + δx ×

∂Vz

∂xδz

+δy ×∂Vx

∂yδx + δy ×

∂Vy

∂yδy + δy ×

∂Vz

∂yδz

+δz ×∂Vx

∂zδx + δz ×

∂Vy

∂zδy + δz ×

∂Vz

∂zδz

=

(

∂Vz

∂y− ∂Vy

∂z

)

δx +

(

∂Vx

∂z− ∂Vz

∂x

)

δy +

(

∂Vy

∂x− ∂Vx

∂y

)

δz

= Curl(V )

Since under the cartesian coordinates the unit vectors have constant directions, thefollowing mnemonic will still work:

∇× V = det

δx δy δz

∂∂x

∂∂y

∂∂z

Vx Vy Vz

However, it is crucial to note that the determinant formula will not yield the correct curl for-mula when using cylindrical coordinate systems or spherical coordinate systems, because atboth these coordinate systems the unit vectors are pointing at different directions dependingon the position in the vector field. (These coordinate systems will be discussed further insection 4.6 and 4.7).

When the vector field V is specifically a velocity field, ∇ × V is also known as thevorticity. If a solid particle of differential volume is positioned at a point P in the velocityfield, then that particle will experience an angular velocity Ω whose direction (based on theright-hand screw rule) will point in the same direction as ∇ × V . However, as one can seefrom Figure 4.17, the z-component of the angular velocity Ωz is the average of the two effectsin Cz, i.e. Ωz = (1/2)Cz. This means that to obtain the magnitude of the angular velocity,we just need to divide the curl by 2, thus

Ω(x, y, z) =1

2∇× V


4.5.2.4 Laplacian

Taking the gradient of a scalar field, φ(x, y, z) will result in a vector field, ∇φ. After takingthe divergence of this resulting vector field, we obtain the operation known as the Laplacian,

Laplacian(φ(x, y, z)) = ∇ · (∇φ) (4.23)

=∂2φ

∂x2+

∂2φ

∂y2+

∂2φ

∂z2(4.24)

In some books, the Laplacian notation for φ is further simplified as ∇2φ.

As a simple physical interpretation, consider the heat conduction process in a threedimensional solid. The energy balance is

Rate of Change in Energy = Accumulation or Negative Divergence of Energy Flow

With constant heat conductivity, density and heat capacity, the divergence of energy flowcan be replaced by a proportionality to the divergence of the temperature gradient. Thus,

∂T

∂t= α∇2T

where α = k/(ρCp), k is thermal conductivity, ρ is density and Cp is the heat capacity.

4.5.2.5 Vector Differential Identities

Let φ(x, y, z) be a scalar field, V and U are vector fields. We list below some of the identitiesthat can be easily obtained by applying the definition directly to both sides of the equations.

∇ · φV = φ∇ · V + V · ∇φ (4.25)

∇× φV = φ∇× V + (∇φ) × V (4.26)

∇ · (U × V ) = V · ∇ × U − U · ∇ × V (4.27)

∇× (U × V ) = V · ∇U − U · ∇V + U∇ · V − V ∇ · U (4.28)

∇(U · V ) = U · ∇V + V · ∇U + U × (∇× V ) + V × (∇× U) (4.29)

∇×∇φ = 0 (4.30)

∇ · ∇ × V = 0 (4.31)

∇× (∇× V ) = ∇(∇ · V ) − ∇2V (4.32)


Example 4.2 . Proof of (4.29)

∇(U · V ) = δx

(

vx∂ux

∂x+ vy

∂uy

∂x+ vz

∂uz

∂x+ ux

∂vx

∂x+ uy

∂vy

∂x+ uz

∂vz

∂x

)

+δy

(

vx∂ux

∂y+ vy

∂uy

∂y+ vz

∂uz

∂y+ ux

∂vx

∂y+ uy

∂vy

∂y+ uz

∂vz

∂y

)

+δz

(

vx∂ux

∂z+ vy

∂uy

∂z+ vz

∂uz

∂z+ ux

∂vx

∂z+ uy

∂vy

∂z+ uz

∂vz

∂z

)

(4.33)

We can expand U × (∇× V ) to be

U × (∇× V ) = δx

(

uy∂vy

∂x− uy

∂vx

∂y− uz

∂vx

∂z+ uz

∂vz

∂x

)

+δy

(

uz∂vz

∂y− uz

∂vy

∂z− ux

∂vy

∂x+ ux

∂vx

∂y

)

+δz

(

ux∂vx

∂z− ux

∂vz

∂x− uy

∂vz

∂y+ uy

∂vy

∂z

)

(4.34)

and combined with (U · ∇) V ,

(U · ∇) V =

(

ux∂

∂x+ uy

∂

∂y+ uz

∂

∂z

)

(

vxδx + vyδy + vzδz

)

= δx

(

ux∂vx

∂x+ uy

∂vx

∂y+ uz

∂vx

∂z

)

+δy

(

ux∂vy

∂x+ uy

∂vy

∂y+ uz

∂vy

∂z

)

+δz

(

ux∂vz

∂x+ uy

∂vz

∂y+ uz

∂vz

∂z

)

(4.35)

to yield,

U × (∇× V ) + (U · ∇) V = δx

(

ux∂vx

∂x+ uy

∂vy

∂x+ uz

∂vz

∂x

)

+δy

(

ux∂vx

∂y+ uy

∂vy

∂y+ uz

∂vz

∂y

)

+δz

(

ux∂vx

∂z+ uy

∂vy

∂z+ uz

∂vz

∂z

)

(4.36)


By interchanging the role of U and V ,

V × (∇× U) + (V · ∇) U = δx

(

vx∂ux

∂x+ vy

∂uy

∂x+ vz

∂uz

∂x

)

+δy

(

vx∂ux

∂y+ vy

∂uy

∂y+ vz

∂uz

∂y

)

+δz

(

vx∂ux

∂z+ vy

∂uy

∂z+ vz

∂uz

∂z

)

(4.37)

By adding the left hand sides of (4.36) and (4.37) we get the left hand side of(4.33). Thus we prove the identity given in (4.29).

♦♦♦

4.6 Cylindrical Coordinate System

The cylindrical coordinate system is a coordinate system which can be derived from therectangular coordinate system using the following nonlinear transformations:

r =√

x2 + y2 (4.38)

θ = tan−1(y/x) (4.39)

z = z (4.40)

This transformation is invertible, i.e. the rectangular coordinate can be recovered whengiven the cylindrical coordinates,

x = r cos θ (4.41)

y = r sin θ (4.42)

z = z (4.43)

The cylindrical coordinate system is shown in Figure 4.18

A summary of the relationships between cylindrical and rectangular coordinates isgiven in Table 4.2. Although these relationships can be obtained by mathematical manipu-lations, it is still instructive to see the geometric implications.

At a point (r, θ, z), the pair of unit vectors δr and δθ is just the pair of unit vectors δx

and δy rotated counter clockwise by an angle θ. By defining a rotation operator,3

Rr→c =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

(4.44)

3 Note that the rotation operator in (4.44) would rotate a vector clockwise by an angle θ. However, sincewe are rotating the reference axis, this operator would do the reverse, i.e. rotate the axes counterclockwise.


Table 4.2: Relationship between Rectangular and Cylindrical Coordinates

Rectangular Cylindrical

∂∂x

= cos θ ∂∂r

−(

sin θr

)

∂∂θ

∂∂y

= sin θ ∂∂r

+(

cos θr

)

∂∂θ

∂∂z

= ∂∂z

∂∂r

= cos θ ∂∂x

+ sin θ ∂∂y

∂∂θ

= −r sin θ ∂∂x

+ r cos θ ∂∂y

∂∂z

= ∂∂z

δx = cos θδr − sin θδθ

δy = sin θδr + cos θδθ

δz = δz

δr = cos θδx + sin θδy

δθ = − sin θδx + cos θδy

δz = δz

V = Vxδx + Vyδy + Vzδz V = Vrδr + Vθδθ + Vzδz

Vx = Vr cos θ − Vθ sin θVy = Vr sin θ + Vθ cos θVz = Vz

Vr = Vx cos θ + Vy sin θVθ = −Vx sin θ + Vy cos θVz = Vz

∂δi/∂j = 0for i, j = x, y, z

∂δr/∂θ = δθ

∂δθ/∂θ = −δr

zero for all other cases

∇ = δx∂∂x

+ δy∂∂y

+ δz∂∂z

∇ = δr∂∂r

+ δθ1r

∂∂θ

+ δz∂∂z


Figure 4.18: Cylindrical coordinate system.

we have(δr, δθ, δz)

t = Rr→c

(

δx, δy, δz

)t(4.45)

And since Rr→c is orthogonal,(

δx, δy, δz

)t= Rt

r→c (δr, δθ, δz)t (4.46)

Applying (4.45) and (4.46) to a vector in rectangular coordinates,

V = Vxδx + Vyδy + Vzδz

= (Vx, Vy, Vz)(δx, δy, δz)t

= (Vx, Vy, Vz)Rtr→c(δr, δθ, δz)t (4.47)

Thus,(Vr, Vθ, Vz) = (Vx, Vy, Vz)R

tr→c (4.48)

or(Vr, Vθ, Vz)

t = Rr→c(Vx, Vy, Vz)t (4.49)

To obtain the relationship between the partial differential operators in each represen-tation, the chain rule yields

∂∂r

∂∂θ

∂∂z

=

∂x∂r

∂y∂r

∂z∂r

∂x∂θ

∂y∂θ

∂z∂θ

∂x∂z

∂y∂z

∂z∂z

∂∂x

∂∂y

∂∂z


=

cos θ sin θ 0−r sin θ r cos θ 0

0 0 1

∂∂x

∂∂y

∂∂z

= Dr→cRr→c

∂∂x

∂∂y

∂∂z

(4.50)

where, Dr→c = diag(1, r, 1).

To derive the ∇ operator in cylindrical coordinates.

∇ = (δx, δy, δz)

(

∂

∂x,

∂

∂y,

∂

∂z

)t

= [(δr, δθ, δz)Rr→c]

[

Rtr→cD

−1r→c

(

∂

∂r,

∂

∂θ,

∂

∂z

)t]

= (δr,1

rδθ, δz)

(

∂

∂r,

∂

∂θ,

∂

∂z

)t

(4.51)

Since the unit vectors δr and δθ are dependent on θ, the gradient formulas such ascurls, divergences and Laplacian are more complicated.

The partial derivative of δr with respect to θ can be interpreted geometrically usingthe definition

∂δr

∂θ= lim

∆θ→0

δr(r, θ + ∆θ) − δr(r, θ)

∆θ(4.52)

Since (4.52) requires the subtraction of two vectors positioned at different points, weneed to treat the unit vectors as free vectors. As shown in the Figure 4.19, the limit yieldsa vector that points in the same direction as δθ. The norm of this limit vector is

∣

∣

∣

∣

lim∆θ→0

δr(r, θ + ∆θ) − δr(r, θ)

∆θ

∣

∣

∣

∣

= lim∆θ→0

2|δr| sin ∆θ/2

∆θ= 1

thus,∂δr

∂θ= δθ (4.53)

Similarly, for ∂δθ/∂θ,

∂δθ

∂θ= lim

∆θ→0

δθ(r, θ + ∆θ) − δθ(r, θ)

∆θ(4.54)


Figure 4.19: Unit vectors along r at different θ positions.

The length of this limit vector is∣

∣

∣

∣

lim∆θ→0

δθ(r, θ + ∆θ) − δθ(r, θ)

∆θ

∣

∣

∣

∣

= lim∆θ→0

2|δθ| sin ∆θ/2

∆θ= 1

As shown in Figure 4.20, the limit yields a vector that points in opposite direction of δr.

Figure 4.20: Unit vectors along θ at different θ positions.

Thus,∂δθ

∂θ= − δr (4.55)

Due to identities (4.53) and (4.55), the gradients, curls and Laplacian in cylindricalcoordinates need to be performed using the basic definitions.

For example, the divergence formula in cylindrical coordinates is given by

∇ · v =

(

δr

∂

∂r+

1

rδθ

∂

∂θ+ δz

∂

∂z

)

· (Vrδr + Vθδθ + Vzδz)

= δr ·(

∂Vr

∂rδr +

∂Vθ

∂rδθ +

∂Vz

∂rδz

)


+1

rδθ ·

(

∂Vr

∂θδr +

∂Vθ

∂θδθ +

∂Vz

∂θδz

)

+1

rδθ ·

(

Vr∂δr

∂θ+ Vθ

∂δθ

∂θ

)

+δz ·(

∂Vr

∂zδr +

∂Vθ

∂zδθ +

∂Vz

∂zδz

)

=∂Vr

∂r+

1

r

∂Vθ

∂θ+

Vr

r+

∂Vz

∂z(4.56)

For the curl,

∇× v =

(

δr

∂

∂r+

1

rδθ

∂

∂θ+ δz

∂

∂z

)

× (Vrδr + Vθδθ + Vzδz)

= δr ×(

∂Vr

∂rδr +

∂Vθ

∂rδθ +

∂Vz

∂rδz

)

+1

rδθ ×

(

∂Vr

∂θδr +

∂Vθ

∂θδθ +

∂Vz

∂θδz

)

+1

rδθ ×

(

Vr∂δr

∂θ+ Vθ

∂δθ

∂θ

)

+δz ×(

∂Vr

∂zδr +

∂Vθ

∂zδθ +

∂Vz

∂zδz

)

=∂Vθ

∂rδz −

∂Vz

∂rδθ

+1

r

(

−∂Vr

∂θδz +

∂Vz

∂θδr

)

+1

rVθδz

+∂Vr

∂zδθ −

∂Vθ

∂zδr

=

(

1

r

∂Vz

∂θ− ∂Vθ

∂z

)

δr +

(

−∂Vz

∂rδθ +

∂Vr

∂zδθ

)

δθ +

(

∂Vθ

∂r

1

r

(

−∂Vr

∂θ+ Vθ

))

δz

=

(

1

r

∂Vz

∂θ− ∂Vθ

∂z

)

δr +

(

−∂Vz

∂rδθ +

∂Vr

∂zδθ

)

δθ +

(

1

r

∂

∂r(rVθ) −

1

r

∂Vr

∂θ

)

δz

(4.57)

A summary of important differential operations is conveniently listed in Bird, Stewartand Lightfoot, “Transport Phenomena”, J.Wiley, 1960, pp. 739B.


4.7 Spherical Coordinate System

The spherical coordinate system is a coordinate system which can be derived from therectangular coordinate system using the following nonlinear transformation:

r =√

x2 + y2 + z2 (4.58)

θ = tan−1(√

x2 + y2/z) (4.59)

φ = tan−1(y/x) (4.60)

whose inverse transformation is given by

x = r sin θ cos φ (4.61)

y = r sin θ sin φ (4.62)

z = r cos θ (4.63)

A diagram showing the relationship between the rectangular and spherical coordinate systemis shown in the following figure: A summary of important transformation between the two

Figure 4.21: The spherical coordinate system.


coordinate systems is given below:

Rectangular Spherical

∂∂x

= sin θ cos φ ∂∂r

+(

cos θ cos φr

)

∂∂θ

−(

sin φr sin θ

)

∂∂φ

∂∂y

= sin θ sin φ ∂∂r

+(

cos θ sin φr

)

∂∂θ

+(

cos φr sin θ

)

∂∂φ

∂∂z

= cos θ ∂∂r

−(

sin θr

)

∂∂θ

∂∂r

= sin θ cos φ ∂∂x

+ sin θ sin φ ∂∂y

+ cos θ ∂∂z

∂∂θ

= r cos θ cos φ ∂∂x

+ r cos θ sin φ ∂∂y

− r sin θ ∂∂z

∂∂φ

= −r sin θ sin φ ∂∂x

+ r sin θ cos φ ∂∂y

δx = sin θ cos φδr + cos θ cos φδθ

− sin φδφ

δy = sin θ sin φδr + cos θ sin φδθ

+ cos φδφ

δz = cos θδr − sin θδθ

δr = sin θ cos φδx + sin θ sin φδy

+ cos θδz

δθ = cos θ cos φδx + cos θ sin φδy

− sin θδz

δφ = − sin φδx + cos φδy

V = Vxδx + Vyδy + Vzδz V = Vrδr + Vθδθ + Vφδφ

Vx = sin θ cos φVr + cos θ cos φVθ

− sin φVφ

Vy = sin θ sin φVr + cos θ sin φVθ

+ cos φVφ

Vz = cos θVr − sin θVθ

Vr = sin θ cos φVx + sin θ sin φVy

+ cos θVz

Vθ = cos θ cos φVx + cos θ sin φVy

− sin θVz

Vφ = − sin φVx + cos φVy

∂δi/∂j = 0for i, j = x, y, z

∂δr/∂θ = δθ

∂δθ/∂θ = −δr

∂δr/∂φ = δφ sin θ∂δθ/∂φ = δφ cos θ∂δφ/∂φ = −δr sin θ − δθ cos θ

zero for all other cases

∇ = δx∂∂x

+ δy∂∂y

+ δz∂∂z

∇ = δr∂∂r

+ δθ1r

∂∂θ

+ δφ1

r sin θ∂∂φ

The geometric relationship among the unit vectors in rectangular and spherical coordinatesystems involve two consecutive rotations. At a point (r, θ, φ), the unit vectors are firstrotated φ radians counterclockwise along δx, δy plane which can be achieved by

Rrs1 =

cos φ sin φ 0− sin φ cos φ 0

0 0 1


Then another rotation of θ radians clockwise is applied on the three unit vectors along thetransformed δx, δz plane using Rrs2.

Rrs2 =

cos θ 0 − sin θ0 1 0

sin θ 0 cos φ

Finally, the vectors that were originally δx, δy and δz are now renamed δθ, δφ and δr,respectively. The relationship is thus given by

(δθ, δφ, δr)t = Rrs2Rrs1(δx, δy, δz)

t

Since the usual convention uses the specific order of (r, θ, φ), we could add another reorderingoperator

Ereord =

0 0 11 0 00 1 0

That is,(δr, δθ, δφ)

t = EreordRrs2Rrs1(δx, δy, δz)t = Rr→s(δx, δy, δz)

t

where

Rr→s =

sin θ cos φ sin θ sin φ cos θcos θ cos φ cos θ sin φ − sin θ− sin φ cos φ 0

(4.64)

Since Rr→s is orthogonal, the inverse relationship is given by

(δx, δy, δz)t = Rt

rs(δr, δθ, δφ)t

A vector in rectangular coordinates is given by

V = (Vx, Vy, Vz)(δx, δy, δz)t

= (Vx, Vy, Vz)Rtr→s(δr, δθ, δφ)

t

hence,(Vr, Vθ, Vφ)

t = Rr→s(Vx, Vy, Vz)t (4.65)

The partial differential operators between the rectangular and spherical coordinate systemis again obtained by using the chain rule,

∂∂r

∂∂θ

∂∂φ

=

∂x∂r

∂y∂r

∂z∂r

∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

∂∂x

∂∂y

∂∂z


=

sin θ cos φ sin θ sin φ cos θ

r cos θ cos φ r cos θ sin φ −r sin θ

−r sin θ sin φ r sin θ cos φ 0

∂∂x

∂∂y

∂∂z

= Dr→sRrs

∂∂x

∂∂y

∂∂z

(4.66)

where, Dr→s = diag(1, r, r sin θ).

The ∇ operator in spherical coordinates can then be obtained as follows,

∇ = (δx, δy, δz)

(

∂

∂x,

∂

∂y,

∂

∂z

)t

=[

(δr, δθ, δφ)Rr→s

]

[

Rtr→sD

−1r→s

(

∂

∂r,

∂

∂θ,

∂

∂φ

)t]

= (δr,1

rδθ,

1

r sin θδz)

(

∂

∂r,

∂

∂θ,

∂

∂φ

)t

(4.67)

Just like the cylindrical coordinate systems, the unit vectors in spherical coordinatesystems are not fixed in space. Based on Figure 4.22,

∂δr

∂θ= δθ (4.68)

Figure 4.22: Unit vectors along r at different θ positions.

While, based on Figure 4.23,

∂δθ

∂θ= − δr (4.69)


Figure 4.23: Unit vectors along θ at different θ positions.

For the differential changes in φ, the rotation is obtained on a projection on a horizontalplane. Based on Figure 4.24, ∂δφ/∂φ will yield a unit vector that is directed inward alongthe horizontal plane. From Figure 4.25, we see that this vector is a vector combination of δr

and δθ, resulting in

∂δφ

∂φ= − cos δθ − sin θδr (4.70)

Figure 4.24: Unit vectors along φ at different φ positions projected on the horizontal plane.

Figure 4.25: The horizontal inward unit vector as sum of vectors in the r and θ directions.

By projecting δr onto the horizontal plane, we have the vectors shown in Figure 4.26


Figure 4.26: Changes in δr projected onto the horizontal plane.

Then the result becomes,∂δr

∂φ= sin θδφ (4.71)

Similarly, by projecting δr onto the horizontal plane,

∂δθ

∂φ= cos θδφ (4.72)

To obtain the divergence, we need to distribute the operations.

∇ · V =

(

δr

∂

∂r+

1

rδθ

∂

∂θ+

1

r sin θδφ

∂

∂φ

)

·(

Vrδr + Vθδθ + Vφδφ

)

= δr ·(

∂Vr

∂rδr +

∂Vθ

∂rδθ +

∂Vφ

∂rδφ

)

+1

rδθ ·

(

∂Vr

∂θδr +

∂Vθ

∂θδθ +

∂Vφ

∂θδφ

)

+1

rδθ ·

(

Vr∂δr

∂θ+ Vθ

∂δθ

∂θ

)

+1

r sin θδφ ·

(

∂Vr

∂φδr +

∂Vθ

∂φδθ +

∂Vφ

∂φδφ

)

+1

r sin θδφ ·

(

Vr∂δr

∂φ+ Vθ

∂δθ

∂φ+ Vφ

∂δφ

∂φ

)

=∂Vr

∂r+

1

r

∂Vθ

∂θ+ 2

V r

r+

1

r sin θ

∂Vφ

∂φ+

Vθ cos θ

r sin θ

=1

r2

∂(r2Vr)

∂r+

1

r sin θ

∂(Vθ sin θ)

∂θ+

1

r sin θ

∂Vφ

∂φ(4.73)

For the curl,


∇× V =

(

δr

∂

∂r+

1

rδθ

∂

∂θ+

1

r sin θδφ

∂

∂φ

)

×(

Vrδr + Vθδθ + Vφδφ

)

= δr ×(

∂Vr

∂rδr +

∂Vθ

∂rδθ +

∂Vφ

∂rδφ

)

+1

rδθ ×

(

∂Vr

∂θδr +

∂Vθ

∂θδθ +

∂Vφ

∂θδφ

)

+1

rδθ ×

(

Vr∂δr

∂θ+ Vθ

∂δθ

∂θ

)

+1

r sin θδφ ×

(

∂Vr

∂φδr +

∂Vθ

∂φδθ +

∂Vφ

∂φδφ

)

+1

r sin θδφ ×

(

Vr∂δr

∂φ+ Vθ

∂δθ

∂φ+ Vφ

∂δφ

∂φ

)

=∂Vθ

∂rδφ − ∂Vφ

∂rδθ −

1

r

∂Vr

∂θδφ +

1

r

∂Vφ

∂θδr +

Vθ

rδφ

+1

r sin θ

∂Vr

∂φδθ −

1

r sin θ

∂Vθ

∂φδr −

Vφ

rδθ +

Vφ cos θ

r sin θδr

=

(

+1

r sin θ

∂(Vφ sin θ)

∂θ− 1

r sin θ

∂Vθ

∂φ

)

δr +

(

−1

r

∂(rVφ)

∂r+

1

r sin θ

∂Vr

∂φ

)

δθ

+

(

1

r

∂(rVθ)

∂r− 1

r

∂Vr

∂θ

)

δφ (4.74)

A summary of other important differential operations is listed in Bird, Stewart andLightfoot, “Transport Phenomena”, J.Wiley, 1960, pp. 739C-D.

Chapter 5

Integral Theorems

In this chapter we discuss different integral operations such as line integral, surface integraland volume integral. Based on these integrals we have important theorems such as Green’sLemma, the Divergence Theorem and Stoke’s Theorem that are useful in several applicationsin vector analysis.

5.1 Line Integrals

Line integrals are specified by three components:

1. The path of integration, C(x, y, z), which is a continuous curve in (x, y, z).

2. The integrand, F (x, y, z), which is a scalar function.

3. The differential, which could either be dx, dy, dz or ds where

ds =√

dx2 + dy2 + dz2 (5.1)

Based on these three components, the line integrals are defined as the limit of a sum-mation of the products of the integrand F (x, y, z) and either of the subintervals, ∆x, ∆y,∆z or ∆s, along the path of integration.

∫

C

F (x, y, z)dx = lim∆xi→0,n→∞

n∑

i=1

F (xi, yi, zi)∆xi (5.2)

( line integral with respect to x

along the path C )

143


∫

C

F (x, y, z)ds = lim∆si→0,n→∞

n∑

i=1

F (xi, yi, zi)∆si (5.3)

( line integral with respect to the arc traversed, s,

along the path C )

Similar definitions apply to∫

cF (x, y, z)dy, and

∫

cF (x, y, z)dz.

Suppose the integrand F and integration path C are functions only of x and y, Figure5.1 illustrates the area interpretation of the line integrals. The line integral with respect tothe arc of the path,

∫

CF (x, y)ds is the area under the curve C. On the other hand, the line

integral with respect to x,∫

CF (x, y)dx is the area projected onto the plane formed by the

x-axis and the F -axis.

Figure 5.1: Area interpretation of line integrals.

Of course, this area interpretation applies only to segments of the integration pathwhere C(x, y) is single-valued with respect to x. This means that when evaluating∫

CF (x, y)dx, the integration path C will have to be partitioned such that within each

segment the subpath is single-valued with respect to x. For instance, in Figure 5.2, theintegration path from A to B will have to be partitioned into segment ADE, segment EFand segment FGB.

Thus, for the integration path shown in Figure 5.2, the line integral with respect to xis given by the sum of three line integrals, i.e.

∫

C

F (x, y)dx =

∫

[ADE]

F (x, y)dx +

∫

[EF ]

F (x, y)dx +

∫

[FGB]

F (x, y)dx (5.4)


Figure 5.2: A curve in which the projection of C onto x or y is not single valued.

This area interpretation is more difficult to see when the curve is three dimensional,i.e. when C = C(x, y, z). One possible visualization is to imagine a creature mining throughthe ground along a path C for substance A. The ground has different concentration distri-bution of A. Let F (x, y, z) be the amount of A gathered per unit length traveled. Then,as the creature travels along a differential path, ds, it accumulates an amount F (x, y, z)ds.The total amount of A gathered along the whole path C is then given by the line integral∫

CF (x, y, z)ds. On the other hand, the integral

∫

CF (x, y, z)dx is the total amount gathered

only along the x-direction traveled.

For the scenario of the mining creature given above, the line integral∫

CF (x, y, z)dx

does not appear to be as relevant as the line integral with respect to s. However, these lineintegrals will be quite important when we use line integrals to define and compute surfaceintegrals or volume integrals. For surface integrals, the differential surface will be often bedescribed by the products such as dxdy, dxdz or dydz. While for volume integrals, thedifferential volume will often be described by the product dxdydz.

5.1.1 The Path of Integration

The path of integration has a sectionally smooth and continuous curve that can either beopen or closed. For a closed path, the starting point of the path coincides with the end pointof the path. In either case, the direction of the path is crucial during integration.

There are several ways of describing this path. One method is to parameterize thechanging position of the curve by a real parameter, say t, starting with t = 0 and endingwith t = 1.1 Another method is to use explicit relationships. For instance, if the curve residein the (x, y) plane, then y = y(x) or x = x(y) can be used to describe the curve. Anothermethod is to use coordinate transformation, e.g. using cylindrical or spherical coordinates.

In some cases, the path may need to be partitioned into several non-overlapping curvesin order to facilitate computation. One requirement is that each partition should be smooth.

1A more general formulation would be to let the parameter start at t = a and end with t = b, whereb > a. However, using translation and scaling, it could be reduced back to a = 0 and b = 1.


Example 5.1

For the closed curve shown in Figure 5.3,

−8 −7 −6 −5 −4 −3 −2 −1 0 1 2−5

−4

−3

−2

−1

0

1

x

y a

b

c

d

Figure 5.3: A close path of integration in counterclockwise direction.

we have a closed elliptical curve described by

(

x + 3

2

)2

+ (y + 2)2 = 4 (5.5)

which starts at point a : (−7,−2) then moves through point b : (−3,−4) then point c : (1,−2)then point d : (−3, 0) and back to point a. We have the following possible descriptions:

1. Parameterized Form. First, simplify the equation to allow for easier parameteriza-tions. For instance, we can transform (5.5) into an equation of a unit circle by usingthe following variables:

X =x + 3

4and Y =

y + 2

2

Then (5.5) becomesX2 + Y 2 = 1

which we can parameterized as

X = − cos (2πt) and Y = − sin (2πt)

Note that we have included the negative signs so that the curve will start at point aand then move counterclockwise through points b, c and d and the back to a as the


value of t changes from 0 to 1. Thus, the parameterized curve, in terms of the originalvariables, is given by:

Path Cabcda : x = −3 − 4 cos(2πt)

y = −2 − 2 sin(2πt)

from t = 0 to t = 1

2. Explicit function of x. We need to partition the curve into two segments. The firstsegment is the lower half and the second segment is the upper half. However, since thedirection is crucial we need to indicate the start and end values of x. Thus we have forour second description of the path:

Path Cabcda = Cabc + Ccda

where

Cabc : y = −2 −

√

4 −(

x + 3

2

)2

from x = −7 to x = 1

Ccda : y = −2 +

√

4 −(

x + 3

2

)2

from x = 1 to x = −7

3. Explicit function of y. We need to partition the curve into three segments. Thefirst segment is the lower left half, the second segment is the right half and the thirdsegment is the upper left half. We need at least three segments to preserve the startand end points. Thus, we have for our third description of the path:

Path Cabcda = Cab + Cbcd + Cda

where

Cab : x = −3 − 2

√

4 − (y + 2)2 from y = −2 to y = −4

Cbcd : x = −3 + 2

√

4 − (y + 2)2 from y = −4 to y = 0

Cda : x = −3 − 2

√

4 − (y + 2)2 from y = 0 to y = −2

♦♦♦


5.1.2 Computation of Line Integrals

Depending on computational convenience, one could choose to use a path described in eitherthe parameterized form or the explicit function in x or y. However, in case the path doesnot lie in a flat plane, the parameterized form is usually the preferred method.

Option 1: Using Parameterized Form. In this case, the integrand is first transformedto be a function of the parameter t, i.e. with path C : x = x(t), y = y(t), z = z(t),

f(x(t), y(t), z(t)) = g(t) (5.6)

Second, the differentials are also transformed to be in terms of dt, i.e.

dx =dx

dtdt (5.7)

dy =dy

dtdt (5.8)

dz =dz

dtdt (5.9)

ds = dt

√

(

dx

dt

)2

+

(

dy

dt

)2

+

(

dz

dt

)2

(5.10)

Finally, since t varies from 0 to 1 in a straight path, the line integral now reduces to theordinary integral, i.e.

∫

C

f(x, y, z)dx =

∫ 1

0

(

g(t)dx

dt

)

dt (5.11)

∫

C

f(x, y, z)dy =

∫ 1

0

(

g(t)dy

dt

)

dt (5.12)

∫

C

f(x, y, z)dz =

∫ 1

0

(

g(t)dz

dt

)

dt (5.13)

∫

C

f(x, y, z)ds =

∫ 1

0

g(t)

√

(

dx

dt

)2

+

(

dy

dt

)2

+

(

dz

dt

)2

dt (5.14)

Option 2: Using Explicit Function Form. As mentioned above, this approach is usedonly if the path or subpath lie in flat planes. So let us just consider only the situation whereC is in the (x, y) plane. This option is just a special case of option 1. When one can describethe path of integration with y as an explicit function of x, i.e. y = y(x), then choose aparameterization such as

x = t and y = y(x) = y(t)

and we get back option 1. The only difference is that instead of starting at t = 0, we startat t = xstart. Likewise, instead of ending at t = 1, we end at t = xend.


Thus, we have

∫

C

f(x, y)dx =

∫ xend

xstart

f(x, y(x))dx (5.15)

∫

C

f(x, y)dy =

∫ xend

xstart

f(x, y(x))dy

dxdx (5.16)

∫

C

f(x, y)ds = ±∫ xend

xstart

f(x, y(x))

√

1 +

(

dy

dx

)2

dx (5.17)

Likewise, if x can be described as an explicit function of y, i.e. x = x(y), we have

∫

C

f(x, y)dx =

∫ yend

ystart

f(x(y), y)dx

dydy (5.18)

∫

C

f(x, y)dy =

∫ yend

ystart

f(x(y), y)dy (5.19)

∫

C

f(x, y)ds = ±∫ yend

ystart

f(x(y), y)

√

1 +

(

dx

dy

)2

dy (5.20)

The choice for positive or negative sign in the formulas given in (5.17) and (5.20)depend on whether the projection of ds is in the same direction as dx or dy. This will beshown in the following example.

Example 5.2

Consider the scalar function given by

f(x, y) = 2x + y + 3

and the elliptical path of integration given in Example 5.1.

Based on the parameterized form of the path of integration, as t goes from 0 to 1,

x(t) = −3 − 4 cos (2πt)

y(t) = −2 − 2 sin (2πt)

and

dx = 8π sin (2πt) dt

dy = −4π cos (2πt) dt

ds = 4π√

4 sin2 (2πt) + cos2 (2πt)dt


Thus,

∫

C

f(x, y)dx =

∫ 1

0

(2 (−3 − 4 cos (2πt)) + (−2 − 2 sin (2πt)) + 3) (8π sin (2πt)) dt

= −8π∫

C

f(x, y)dy =

∫ 1

0

(2 (−3 − 4 cos (2πt)) + (−2 − 2 sin (2πt)) + 3) (−4π cos (2πt)) dt

= 16π∫

C

f(x, y)ds =

∫ 1

0

(2 (−3 − 4 cos (2πt)) + (−2 − 2 sin (2πt)) + 3)

×(

4π√

4 sin2 (2πt) + cos2 (2πt)

)

dt

= −96.885

When we use the explicit function formulation, e.g. y = y(x), for describing the pathof integration with

C = Cabc + Ccda

Cabc : y = yabc = −2 −

√

4 −(

x + 3

2

)2

from x = −7 to x = 1

Ccda : y = ycda = −2 +

√

4 −(

x + 3

2

)2

from x = 1 to x = −7

The integrand and differentials for the subpaths (denoted by the subscripts abc or cda) aregiven by:

f(x, y)abc = 2x + 3 +

−2 −

√

4 −(

x + 3

2

)2

f(x, y)cda = 2x + 3 +

−2 +

√

4 −(

x + 3

2

)2

(

dy

dx

)

abc

= − x + 3

2√

(1 − x) (x + 7)(

dy

dx

)

cda

= +x + 3

2√

(1 − x) (x + 7)(

ds

dx

)

abc

=√

1 + dy2abc

(

ds

dx

)

cda

= −√

1 + dy2cda


Note that we used the negative sign for ds in the subpath [cda] since the direction of ds isopposite that of dx in this region.

Using these evaluations the line integrals will be given by

∫

C

f(x, y)dx =

∫ 1

−7

f(x, y)abcdx +

∫ −7

1

f(x, y)cdadx

= −8π∫

C

f(x, y)dy =

∫ 1

−7

f(x, y)abc

(

dy

dx

)

abc

dx +

∫ −7

1

f(x, y)cda

(

dy

dx

)

cda

dx

= 16π∫

C

f(x, y)ds =

∫ 1

−7

f(x, y)abc

(

ds

dx

)

abc

dx +

∫ −7

1

f(x, y)cda

(

ds

dx

)

cda

dx

= −96.885

This example shows that using either approach, the value of line integrals will be thesame. The choice is usually determined by the tradeoffs between the complexity of theparameterization procedure and the complexity of the resulting integral.

♦♦♦

Since we will be dealing with path of integration that are either closed or not, we willuse the following notation to indicate that the line integral has a closed contour:

∮

C

fds C is a closed, sectionally smooth, nonintersecting path

The calculation procedures remain the same.

5.2 Surface Integrals

Surface integrals are specified by three components:

1. The surface of integration, S(x, y, z), which is a sectionally smooth differentiable con-tinuous surface in (x, y, z).


3. The differential area, which could either be dxdy, dydz, dxdz or dS where dS is thedifferential element of the surface of integration


Based on these three components, the surface integrals are defined as the limit of asummation of the products of the integrand F (x, y, z) and one of the differential elementalareas, ∆S, ∆x∆y, etc. as x, y and z vary along the surface of integration.

∫

S

F (x, y, z)dxdy = lim∆xi→0,n→∞

n∑

i=1

F (xi, yi, zi)∆xi∆yi (5.21)

( surface integral with respect to (x,y) plane.

along the surface of integration S )

∫

S

F (x, y, z)dS = lim∆si→0,n→∞

n∑

i=1

F (xi, yi, zi)∆Si (5.22)

( surface integral with respect to the surface domain S

along the surface of integration S )

Similar definitions apply to∫

SF (x, y, z)dydz and

∫

SF (x, y, z)dxdz.

To help us visualize an interpretation of surface integrals, we could go back to thescenario of the creature that was mining for a substance A in the three dimensional ground.This time, the scalar function f(x, y, z) could be imagined as the amount of A per unit area.Then the surface integral,

∫

Sf(x, y, z)dS, becomes the total amount of A mined by this

creature as it sweeps a surface area S out of the ground. Note that the surface S will notnecessarily be a flat plane.

5.2.1 Surface of Integration

A general description of smooth and differentiable three dimensional surface is given by theform

φ(x, y, z) = 0

where φ is a general differentiable nonlinear function. However, since the surface is necessarilytwo dimensional regardless of whether the surface is flat or curved, it is often described bythe use of two independent parameters, say u and v. This means that the surface can usuallybe described as follows:

S : (x(u, v), y(u, v), z(u, v)) as u and v vary independently in a closed domain.(5.23)

In some cases, the surface can be described by letting u = x and v = y. If this is so,then the surface can be described as an explicit function for z

S : z = z(x, y) as x and y vary independently in a close domain (5.24)

From the parameterized description of the surface, i.e. S : (x(u, v), y(u, v), z(u, v)), wecan evaluate two important variables. One variable is the direction of a vector normal to the


surface at a point (x(uo, vo), y(uo, vo), z(uo, vo)). The other variable is the differential area,dS, which is related to the product dudv (note that du and dv are not necessarily orthogonalto each other). Fortunately, these two variables can be determined at the same time.

Recall that the position vector

R = xδx + yδy + zδz (5.25)

is the vector starting at the origin and ending at the point (x, y, z). As the parameter uis varied while v is fixed, a path, say C1, is traced. Likewise, as the parameter v is variedwhile parameter u is held fixed, (x(uo, v), y(uo, v), z(uo, v)) will trace out another path, sayC2. This is shown in Figure 5.4. From the same figure, we see that the differential changein R tangential to the path C1 is given by

a = du∂R

∂u= du

(

∂x

∂uδx +

∂y

∂uδy +

∂z

∂uδz

)

(5.26)

while the differential change in R tangential to the path C2 is given by

b = dv∂R

∂v= dv

(

∂x

∂vδx +

∂y

∂vδy +

∂z

∂vδz

)

(5.27)

Figure 5.4: Graphical representation of differential area, dS, and the normal vector, N.

Then a vector perpendicular to both a and b will be the cross product of a and b, i.e.

N = a × b


=

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

δx δy δz

∂x/∂u ∂y/∂u ∂z/∂u

∂x/∂v ∂y/∂v ∂z/∂v

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

dudv

=

(

∂(y, z)

∂(u, v)δx +

∂(z, x)

∂(u, v)δy +

∂(x, y)

∂(u, v)δz

)

dudv (5.28)

where,

∂(a, b)

∂(c, d)=

∣

∣

∣

∣

∣

∣

∂a/∂c ∂a/∂d

∂b/∂c ∂b/∂d

∣

∣

∣

∣

∣

∣

is a shorthand notation for the Jacobian determinant.

To obtain the unit normal vector, we simply divide by the norm, i.e.

n =N

‖N‖This also removes the dependency on the factor dudv.

From the previous chapters, recall that the magnitude of cross product of the twovectors evaluates the area of the parallelogram formed by these two vectors. Thus thedifferential area dS is given by

dS = ‖N‖

=

√

(

∂(y, z)

∂(u, v)

)2

+

(

∂(z, x)

∂(u, v)

)2

+

(

∂(x, y)

∂(u, v)

)2

dudv (5.29)

If the surface of integration can be described by z = z(x, y), we let u = x and v = y.This reduces the formulas above to

N =

(

∂z

∂xδx −

∂z

∂yδy + δz

)

dxdy (5.30)

n =

(

∂z∂x

δx − ∂z∂y

δy + δz

)

√

1 +(

∂z∂x

)2+

(

∂z∂y

)2(5.31)

dS =

√

1 +

(

∂z

∂x

)2

+

(

∂z

∂y

)2

dxdy (5.32)

The choice for sign depends on the interpretation of the surface direction. For a surfacethat encloses a region of three dimensional space, the surface outward of the enclosed regionis often given a positive sign.


5.2.2 Computation of Surface Integrals

With the parameterization of the surface using the parameters u and v, the domain is aclosed two dimensional plane in the (u, v) space.

In one case, the domain is described by independent ranges of u and v. For instance,u and v could have been scaled to be between 0 and 1, i.e. domain D is given as

D : 0 ≤ u ≤ 1 ; 0 ≤ v ≤ 1 (5.33)

If the domain can be described as (5.33), then the surface integral can be computedas follows:

∫

S

f(x, y, z)dS =

∫ 1

0

∫ 1

0

g(u, v)dudv (5.34)

where

g(u, v) = f(x(u, v), y(u, v), z(u, v))

√

(

∂(y, z)

∂(u, v)

)2

+

(

∂(z, x)

∂(u, v)

)2

+

(

∂(x, y)

∂(u, v)

)2

(5.35)

and the computation proceeds to two sequential steps:

1. Step 1. Calculate h(v), where

h(v) =

∫ 1

0

g(u, v)du holding v constant while integrating

2. Step 2.∫

S

f(x, y, z)dS =

∫ 1

0

h(v)dv

In more general cases, the boundaries of u and v may be interdependent. If this is thecase, two descriptions of the domain may be used:

Du : ulower ≤ u ≤ uupper ; φ0(u) ≤ v ≤ φ1(u) (5.36)

or

Dv : vlower ≤ v ≤ vupper ; ψ0(v) ≤ u ≤ ψ1(v) (5.37)

where ulower, uupper, vlower and vupper are constants. Both domain descriptions are shown inFigure 5.5 are equally valid. The choice usually depends on which description will yield moreefficient computation.

Using the first description given in (5.36), the computation proceeds in two sequentialsteps, using g(u, v) in (5.35),


Figure 5.5: The two possible domain descriptions: (a) boundary is partitioned into twosegments such that v = φ(u), and (b) boundary is partitioned into two segments such thatu = ψ(v) .

1. Step 1. Calculate h(u), where

h(u) =

∫ φ1(u)

φ0(u)

g(u, v)dv holding u constant while integrating

2. Step 2.∫

S

f(x, y, z)dS =

∫ uupper

ulower

h(u)du

Using the second description given in (5.37), the computation also proceeds in twosequential steps:

1. Step 1. Calculate h(v), where

h(v) =

∫ ψ1(v)

ψ0(v)

g(u, v)du holding u constant while integrating

2. Step 2.∫

S

f(x, y, z)dS =

∫ vupper

vlower

h(v)dv


In the special case that the surface is given explicitly by z = z(x, y), we use thefollowing substitutions to the steps provided above,

u = x

v = y

g(u, v) = g(x, y) = f(x, y, z(x, y))

√

1 +

(

∂z

∂x

)2

+

(

∂z

∂y

)2

ulower = xlower uupper = xupper

φ1(u) = φ1(x) φ0(u) = φ0(x)

vlower = ylower vupper = yupper

ψ1(v) = ψ1(y) ψ0(v) = ψ0(y)

Example 5.3

We wish to find the surface integral of the function given by

f(x, y, z) = 2x + y − z + 3

on the surface given by the ellipse:

x2 +(y

2

)2

+ z2 = 1

We could parameterize the surface as follows:

z = cos(u)

x = sin(u) cos(v)

y = sin(u) sin(v)

with the given domain0 ≤ u ≤ 2π 0 ≤ v ≤ π

The Jacobian determinants can be found to be:

∂(x, y)

∂(u, v)= −2 sin(u) cos(u)

∂(y, z)

∂(u, v)= −2 sin2(u) cos(v)

∂(x, z)

∂(u, v)= sin2(u) cos(v)


and this gives us

g(u, v) = f(x, y, z)

√

(

∂(y, z)

∂(u, v)

)2

+

(

∂(z, x)

∂(u, v)

)2

+

(

∂(x, y)

∂(u, v)

)2

= α(u, v)β(u, v)

where

α(u, v) = 2 sin(u) cos(v) + 2 sin(u) sin(v) − cos(u) + 3

β(u.v) =√

3 cos2(v)(

(cos(u) − 1)2 (cos(u) + 1)2) + (1 + 2 cos2(u) − 3 cos4(u))

The surface integral can then be solved numerically to be

∫ π

0

∫ 2π

0

g(u, v)dudv = 64.4

As an alternative, we partition the elliptical surface into two halves. The upper halfcan be described by the function

zu =

√

1 − x2 −(

y

2

2)

while the lower half is described by

zℓ = −√

1 − x2 −(

y

2

2)

In either half, the domain can be described by

D : − 1 ≤ x ≤ 1 − 2√

1 − x2 ≤ y ≤ 2√

1 − x2

For the upper half,

dzu

dx=

−2x√

4 − 4x2 − y2

dzu

dy=

−y/2√

4 − 4x2 − y2

which then makes the integrand

gu(x, y) =

(

2x + y −√

1 − x2 −(

y

2

2)

+ 3

)(

1

2

√

3y2 − 16

−4 + 4x2 + y2

)


For the lower half,

dzℓ

dx=

2x√

4 − 4x2 − y2

dzℓ

dy=

y/2√

4 − 4x2 − y2

which then makes the integrand

gℓ(x, y) =

(

2x + y +

√

1 − x2 −(

y

2

2)

+ 3

)(

1

2

√

3y2 − 16

−4 + 4x2 + y2

)

Combining everything, we can calculate the surface integral via numerical integrationto be

Iu =

∫ +1

−1

∫ 2√

1−x2

−2√

1−x2

gu(x, y)dydx = 26.6

Iℓ =

∫ +1

−1

∫ 2√

1−x2

−2√

1−x2

gℓ(x, y)dydx = 37.8

∫

S

fdS = Iu + Iℓ = 64.4

which is the same value as the previous answer using the parameterized description.

♦♦♦

5.3 Volume Integrals

Volume integrals are one-dimensional extensions to surface integrals. They involve threecomponents:

1. The volume of integration, V (x, y, z), which is a closed three dimensional regionbounded by sectionally smooth differentiable surfaces.


3. The differential area, given by dxdydz


Based on these three components, the volume integrals are defined as the limit of asummation of the products of the integrand F (x, y, z) and the differential elemental volume,∆V = ∆x∆y∆z, which varies along the volume of integration.

∫

V

F (x, y, z)dxdydz = lim∆Vi→0,n→∞

n∑

i=1

F (xi, yi, zi)∆xi∆yi∆zi (5.38)

To continue the visual interpretation tool used for line and surface integrals, we go backto the scenario of the creature that was mining for a substance A in the three dimensionalground. This time, the scalar function F (x, y, z) could be imagined as the amount of A perunit volume. Then the volume integral,

∫

VF (x, y, z)dV , becomes the total amount of A

mined by this creature as it covers the region V carved out of the ground.

5.3.1 Volume of Integration

In some cases, another set of coordinates might allow for easier computation, e.g. cylindricalor spherical coordinates. In general, this set of new coordinates could be seen as parametersdefining x, y and z,

x = x(u, v, w) (5.39)

y = y(u, v, w) (5.40)

z = z(u, v, w) (5.41)

where u, v and w are the new coordinates or parameters.

To convert the differential volume dV in terms of dudvdw, we can use the same proce-dures used when obtaining dS for surface integrals. First, recall the definition of the positionvector, R,

R = xδx + yδy + zδz

By tracing changes in R as we vary u while keeping v and w constant, a curve can begenerated inside the region V . Let us denote this curve as C1. Likewise, we could trace twomore curves. Denote by C2, the curve that is generated by tracing R as we vary v whilekeeping u and w constant. Also, denote by C3, the curve that is generated by tracing Ras we vary w while keeping u and v constant. As we can see from Figure 5.6, a differentialchange along each of these curves will yield three non-coplanar vectors:

a =

(

∂R

∂u

)

du =

(

∂x

∂uδx +

∂y

∂uδy +

∂z

∂uδz

)

du (5.42)

b =

(

∂R

∂v

)

dv =

(

∂x

∂vδx +

∂y

∂vδy +

∂z

∂vδz

)

dv (5.43)

c =

(

∂R

∂w

)

dw =

(

∂x

∂wδx +

∂y

∂wδy +

∂z

∂wδz

)

dw (5.44)


Figure 5.6: Graphical representation of differential volume, dV , as function of u, v and w.

The differential volume is the volume occupied by the parallelepiped formed by thevectors a, b and c. The volume can be computed by taking the absolute value of the tripleproduct of these three vectors, i.e.

dV = |c · (a × b)| = |a · (b × c)|

=

∣

∣

∣

∣

(

∂R

∂u

)

·(

∂R

∂v× ∂R

∂w

)∣

∣

∣

∣

dudvdw

=

∣

∣

∣

∣

(

∂(x, y, z)

∂(u, v, w)

)∣

∣

∣

∣

dudvdw (5.45)

where we used the shorthand notation for the Jacobian determinant

∂(x, y, z)

∂(u, v, w)= det

∂x/∂u ∂x/∂v ∂x/∂w

∂y/∂u ∂y/∂v ∂y/∂w

∂z/∂u ∂z/∂v ∂z/∂w

(5.46)

5.3.2 Computation of Volume Integrals

Having determined the differential volume and the integrand, one needs to identify the limitsof integration in each of the variables x, y and z, or equivalently, of parameters u, v and w.

In the simpler case in which the limits are independent, i.e. if the limits of the bound-aries are given by

umin ≤ u ≤ umax (5.47)


vmin ≤ v ≤ vmax (5.48)

wmin ≤ w ≤ wmax (5.49)

the volume integral can be integrated in a nested fashion by the following

∫

V

FdV =

∫ wmax

wmin

∫ vmax

vmin

∫ umax

umin

F (x(u, v, w), y(u, v, w), z(u, v, w))

∣

∣

∣

∣

∂(x, y, z)

∂(u, v, w)

∣

∣

∣

∣

dudvdw

(5.50)

For the more general case where the variables are interdependent when describing thesurface boundaries, there are six possible descriptions that can be used based on the sequenceof dependencies. We will describe the sequence w → v → u, but the other descriptions arevery similar.

As shown in Figure 5.7, we can first identify the maximum and minimum value thatz can take, i.e.

wmin ≤ w ≤ wmax (5.51)

Figure 5.7: A nested description of volume boundaries.

Next, we take a slice of the volume by fixing w that belongs to the allowed range.This slice will be an enclosed curve for which we can identify also a maximum and minimumvalue for v,

ηmin(w) ≤ v ≤ ηmax(w) (5.52)

Finally, we can see from Figure 5.7 that the limits of v for this slice will divide the closedcurve into two segments. Each of these segments can then be described by functions of vand w (that were fixed for this slice), i.e.

ξmin(v, w) ≤ u ≤ ξmax(v, w) (5.53)


Thus we end up with a slightly different nested integration given by

∫

V

FdV =

∫ wmax

wmin

∫ ηmax(w)

ηmin(w)

∫ ξmax(v,w)

ξmin(v,w)

F

∣

∣

∣

∣

∂(x, y, z)

∂(u, v, w)

∣

∣

∣

∣

dudvdw (5.54)

Example 5.4

Consider the volume of integration given by the ellipse

x2 +(y

2

)2

+ z2 ≤ 1

and the integrand given byF (x, y, z) = 2x + y − z + 3

Using the following parameterization:

x = u sin(v) cos(w)

y = 2u sin(v) sin(w)

z = u cos(v)

where the boundaries of the variables are given as

0 ≤ u ≤ 1

0 ≤ v ≤ 2π

0 ≤ w ≤ π

The differential volume is then given by

dV =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

det

sin(v) cos(w) u cos(v) cos(w) −u sin(v) sin(w)

2 sin(v) sin(w) 2u cos(v) sin(w) 2u sin(v) cos(w)

cos(v) −u sin(v) 0

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

dudvdw

= 2u2 |sin(v)| dudvdw

while the integrand becomes

F = 2u sin(v) (cos(w) + sin(w)) − u cos(v) + 3

Combining all the elements together, we can compute the volume integral as

∫ π

0

∫ 2π

0

∫ 1

0

(2u sin(v) (cos(w) + sin(w)) − u cos(v) + 3)(

2u2 |sin(v)| dudvdw)

= 8π


As an alternative, we could use the original variables x, y and z. Doing so, we havethe differential volume as dV = dxdydz.

The boundaries of the volume of integration can be described by

Surface boundary : −√

1 − z2 −(y

2

)2

≤ x ≤√

1 − z2 −(y

2

)2

−2√

1 − z2 y 2√

1 − z2

−1 ≤ z ≤ 1

Thus the volume integral is given by

∫ 1

−1

∫ 2√

1−z2

−2√

1−z2

∫

√1−z2−(y/2)2

−√

1−z2−(y/2)2(2x + y − z) dxdydz = 8π

which is the same answer obtained by using the parameterized description.

♦♦♦

5.4 Green’s Lemma

We begin with one of the basic formulas that consists of the relationship between a surfaceintegral on surface S and a line integral on closed curve C that is the boundary of S. Wewill state this formula in terms of parameters u and v to allow for the application of theformulas to curved surfaces in three dimensional space.

Lemma 5.1 Green’s Lemma: Let F (u, v) and G(u, v) be differentiable functions in thedomain D ∈ R2. Then

∮

C

(F (u, v)du + G(u, v)dv) =

∫

S

(

∂G

∂u− ∂F

∂v

)

dudv (5.55)

where

1. The closed curve C is the boundary of the surface of integration S, which should besectionally smooth.

2. The positive direction of contour C is consistent with the definition of the positivedirection of the vector: N = [(∂R/∂u) × (∂R/∂v)], where R is the position vector.


For the special case in which the surface of integration S lies in the (x, y)-plane,Green’s lemma is stated in terms of x and y, i.e. (5.55) is replace by a formula with u = xand v = y. In this case, the positive direction of the contour is conventionally set to becounterclockwise. Equivalently, based on the usual arrangements of the x axis and the yaxis, the positive contour direction is chosen such that the region in S is always to the leftof the contour’s path.

Proof of Green’s lemma. To prove the lemma, we make use of two possibledescriptions of the boundary as given in (5.36) and (5.37).

Recalling (5.36), the domain of the surface of integration S is given by

D : ulower ≤ u ≤ uupper ; φ0(u) ≤ v ≤ φ1(u)

where the closed contour C is the sum given by

C = C0,v − C1,v

where C0,v and C1,v are the curves described by φ0(u) and φ1(u), respectively,are positive with increasing values of u.

Applying this description to the second surface integral in (5.55),

∫

S

∂F (u, v)

∂vdudv =

∫ uupper

ulower

(

∫ φ1(u)

φ0(u)

∂F (u, v)

∂vdv

)

du

=

∫ uupper

ulower

(F (u, φ1(u)) − F (u, φ0(u))) du

= −∮

C

F (u, v)du (5.56)

Likewise, using (5.37), the domain of the surface of integration S is given by

D : vlower ≤ v ≤ vupper ; ψ0(v) ≤ u ≤ ψ1(v)

where the closed contour C is now equal to the sum given by

C = C1,u − C0,u

where C0,v and C1,v are the curves described by ψ0(u) and ψ1(u), respectively,are positive with increasing values of v.

Applying this domain description to the first surface integral in (5.55),

∫

S

∂G(u, v)

∂ududv =

∫ vupper

vlower

(

∫ ψ1(v)

ψ0(v)

∂G(u, v)

∂udv

)

du

=

∫ vupper

vlower

(G(ψ1(v), v) − G(ψ0(v), v)) du

=

∮

C

G(u, v)dv (5.57)


Combining (5.56) and (5.57), we arrive at the formula given in Green’s lemma,∮

C

G(u, v)dv +

∮

C

F (u, v)du =

∫

S

∂G(u, v)

∂ududv −

∫

S

∂F (u, v)

∂vdudv

QED

Example 5.5

We want to verify Green’s lemma for the functions F (x, y, z) and G(x, y, z) given by

F (x, y, z) = x2 + y

G(x, y, z) = 2x + 3y − z + 5

Suppose further that the surface of integration to be the top portion of the unit sphere with0 ≤ θ ≤ π/4 and 0 ≤ φ ≤ 2π as shown in Figure 5.8.

Figure 5.8: The surface of integration given by the top of a unit sphere.

One parameterization of the variables, with u = φ and v = θ, is given by

x = sin(v) cos(u)

y = sin(v) sin(u)

z = cos(v)

Substituting, we get

F (u, v) = (sin(v) cos(u))2 + sin(v) sin(u)

G(u, v) = 2 sin(v) cos(u) + 3 sin(v) sin(u) − cos(v) + 5

whose partial derivatives are given by

∂F

∂v= cos(v)

(

sin(u) + 2 sin(v) cos2(u))

∂G

∂u= sin(v) (3 cos(u) − 2 sin(u))


The surface integral in the lemma can now be evaluated to be

∫ π/4

0

∫ 2π

0

(

∂G

∂u− ∂F

∂v

)

dudv = − π

2

Since the chosen parameterization was for u = φ and u = θ, the closed contour in the(u, v) plane is shown in Figure 5.9. The positive direction is counterclockwise to yield anvector that is normal to S that is outward from the center of the sphere.

Figure 5.9: The closed path of integration in the (u, v) plane.

Based on Figure 5.9, the line integrals can be calculated to be

∮

C

F (u, v)du =

∫ 2π

0

F (u, 0) du +

∫ 0

2π

F(

u,π

4

)

du

= 0 − π

2

∮

C

G(u, v)dv =

∫ π/4

0

G (2π, v) dv +

∫ 0

π/4

G (0, v) dv

=

(

−3√

2

2+ 2 + 5

π

4

)

+

(

3√

2

2− 2 − 5

π

4

)

Combining all the results, we arrive at

∮

C

Fdu +

∮

C

Gdv =

∫

S

(

∂G

∂u− ∂F

∂v

)

dudv

♦♦♦


5.4.1 Surfaces with Holes

From example 5.5, the computation of∫

CG(u, v)dv resulted in zero. Two of the path seg-

ments involved the same path except that they were directed opposite of each other. Asexpected, the line integrals corresponding to both these paths canceled each other out.

We will use this observation to allow us to apply Green’s lemma to surfaces that haveholes in them. For illustration purposes, consider the surface shown in Figure 5.10b.

This surface, S1, was obtained from the original surface S after another surface S2 wasremoved. In Figure 5.10c, a continuous contour C1 can be generated to be the boundary ofS1. At some point along the outer boundary, contour C1 cuts a path to reach the inner hole.Then the contour will traverse the boundary of this hole. After reaching the point where thecontour first entered the hole, the contour C1 will retrace the same cutting path backwardsto continue tracking the outer path.

With this technique, the two line integrals that have traversed the same path segments,but in opposite directions, will cancel each other out. Thus, as shown in Figure 5.10d, theline integral along C1 will be equal to the line integral along C minus the line integral alongC2, which is the contour of the removed surface S2, i.e.

∮

C1

ωds =

∮

C

ωds −∮

C2

ωds (5.58)

From the definition of surface integrals, we know that∫

S1

ΩdS =

∫

S

ΩdS −∫

S2

ΩdS (5.59)

Applying Green’s lemma and (5.58) to (5.59),∫

S1

(

∂G

∂u− ∂F

∂v

)

dudv =

∮

C

(Fdu + Gdv) −∮

C2

(Fdu + Gdv)

=

∮

C1

(Fdu + Gdv) (5.60)

Equation (5.60) implies that the strategy of cutting through the region to connect the outercontour with the inner contours of the holes validates the applicability of the formula ofGreen lemma to surfaces with holes.

5.5 Divergence Theorem

Consider a vector field F. The divergence theorem states that the volume integral of thedivergence ∇ · F in a given volume region V can also be evaluated indirectly by taking thesurface integral of the flux of F, i.e. F · n, over the surface S that bounds the region V .


Figure 5.10: The surface of integration with holes: (a) original surface S and contour C,(b) S1 obtained by removing S2, (c) obtaining continuous contour C1 for S1, (d) due to thecancelation of line integrals in oppositely directed segments,

∮

C1=

∮

C−

∮

C2.


Theorem 5.1 Let F(x, y, z) be a vector field which is differentiable in V then∫

S

F · n dS =

∫

V

∇ · F dV (5.61)

Proof of Divergence Theorem:

In rectangular coordinates, let F be given by

F = Fx δx + Fy δy + Fz δz

The volume integral in (5.61) can be expanded to be the sum of three terms∫

V

∇ · F dV =

∫

V

∂Fx

∂xdV +

∫

V

∂Fy

∂ydV +

∫

V

∂Fz

∂zdV (5.62)

For the first term in (5.62), we can use the following description of the volumeof integration: 2

V : zmin ≤ z ≤ zmax

ηmin(z) ≤ y ≤ ηmax(z)

ξmin(y, z) ≤ x ≤ ξmax(y, z)

to obtain the following triple integral formulation

∫

V

∂Fx

∂xdV =

∫ zmax

zmax

∫ ηmax(z)

ηmin(z)

∫ ξmax(y,z)

ξmin(y,z)

∂Fx

∂xdxdydz

After performing the inner integration with respect to x, the result is a differenceof two surface integrals

∫

V

∂Fx

∂xdV =

∫ zmax

zmin

∫ ηmax(z)

ηmin(z)

Fx(ηmax(y, z), y, z) dydz

−∫ zmax

zmax

∫ ηmax(z)

ηmin(z)

Fx(ηmin(y, z), y, z) dydz (5.63)

The first surface integral in (5.63) is with respect to the surface: S1 : x =ξmax(y, z). To determine the differential area of the surface, dS1 at a point in

2This assumes that any line that is parallel to the x axis will intersect the surface boundary of region V

at two points, except at the edges of the boundary where it touches at one point. If this assumption is nottrue for V , it can always be divided into subsections for which this assumption can hold. After applying thedivergence theorem to these smaller regions, they can be added up later and the resulting sum can be shownto satisfy the divergence theorem.


the surface, we can use the position vector R of the point in surface S1. Alongthe curve in the surface, in which z is fixed, we have a tangent vector given by(∂R/∂y) dy. Likewise, along the curve in the surface, in which y is fixed, we havea tangent vector given by (∂R/∂z)dz. This is shown in Figure 5.11. By takingthe cross product of these two tangent vectors, we obtain a vector N1 which isnormal to surface S1 whose magnitude is the area of the parallelogram formedby the two tangent vectors, i.e.

N1 = dS1 n1

where n1 is the unit normal vector.

Figure 5.11: The normal vector to the surface x = ξmax(y, z) is given by N1 which has amagnitude equal to the differential surface, dS1.

Thus, with the position vector R along the surface given by

R = ξmax(y, z) δx + y δy + z δz

we have

dS1 n1 =

(

∂R

∂y× ∂R

∂z

)

dydz

=

(

∂ξmax

∂yδx + δy

)

×(

∂ξmax

∂zδx + δz

)

dydz

=

(

δx −∂ξmax

∂zδy −

∂ξmax

∂yδz

)

dydz

By taking the dot product of both sides with δx,

(n1 · δx) dS1 = dydz (5.64)


The same arguments can be used for the other surface given by x = ξmin(y, z).The difference is that, as shown in Figure 5.12, the normal vector N2 =(∂R/∂z) × (∂R/∂y), and thus

(n2 · δx) dS2 = − dydz (5.65)

Figure 5.12: The normal vector to the surface x = ξmin(y, z) is given by N2 which has amagnitude equal to the differential surface, dS2.

Returning to equation (5.63), we can now use the results in (5.64) and (5.65) toobtain,

∫

V

∂Fx

∂xdV =

∫

S1

Fx (n1 · δx) +

∫

S2

Fx (n2 · δx) =

∫

S

Fx δx · n (5.66)

Following the same procedure, we could show that the other two terms in (5.62)can be evaluated to be

∫

V

∂Fy

∂ydV =

∫

S

Fy δy · n (5.67)∫

V

∂Fz

∂zdV =

∫

S

Fz δz · n (5.68)

Adding the three equations: (5.66), (5.67) and (5.68), we end up with the diver-gence theorem, i.e.

∫

V

(

∂Fx

∂x+

∂Fy

∂y+

∂Fy

∂z

)

dV =

∫

S

(

Fx δx + Fy δy + Fz δz

)

· n dS (5.69)

QED


5.6 Stokes’ Theorem

Theorem 5.2 Let F = Fx(x, y, z) δx + Fy(x, y, z) δy + Fz(x, y, z) δz be adifferentiable vector field in a surface S, where S is a connected and sectionallysmooth surface in (x, y, z)-space bounded by the closed curve C. Then

∫

C

F · dR =

∫

S

(∇× F) · n dS (5.70)

Proof of Stokes’ Theorem:

Let the surface S be parameterized by u and v, i.e. with x = x(u, v), y = y(u, v)and z = z(u, v). Then the differentials dx, dy and dz can be described by

dx =∂x

∂udu +

∂x

∂vdv

dy =∂y

∂udu +

∂y

∂vdv

dz =∂z

∂udu +

∂z

∂vdv

The line integral in (5.70) can then be put in terms of u and v:∮

C

F · dR =

∮

C

Fxdx + Fydy + Fzdz

=

∮

C

Fx

(

∂x

∂udu +

∂x

∂vdv

)

+

∮

C

Fy

(

∂y

∂udu +

∂y

∂vdv

)

+

∮

C

Fz

(

∂z

∂udu +

∂z

∂vdv

)

=

∮

C

f(u, v)du + g(u, v)dv (5.71)

where,

f(u, v) = Fx∂x

∂u+ Fy

∂y

∂u+ Fz

∂z

∂u(5.72)

g(u, v) = Fx∂x

∂v+ Fy

∂y

∂v+ Fz

∂z

∂v(5.73)

By applying Green’s lemma, i.e. (5.55), to (5.71)∮

C

(f(u, v)du + g(u, v)dv) =

∫

S

(

∂g

∂u− ∂f

∂v

)

dudv (5.74)


Based on f(u, v) and g(u, v) given in (5.72) and (5.73), respectively, the integrandof the surface integral in (5.74) can be put in terms of the curl of F as follows:


∂g

∂u− ∂f

∂v=

(

∂Fx

∂u

∂x

∂v+ Fx

∂2x

∂v∂u+

∂Fy

∂u

∂y

∂v+ Fy

∂2y

∂v∂u

+∂Fz

∂u

∂z

∂v+ Fz

∂2z

∂v∂u

)

−(

∂Fx

∂v

∂x

∂u+ Fx

∂2x

∂u∂v+

∂Fy

∂v

∂y

∂u+ Fy

∂2y

∂u∂v

+∂Fz

∂v

∂z

∂u+ Fz

∂2z

∂u∂v

)

=∂Fx

∂x

∂x

∂u

∂x

∂v+

∂Fx

∂y

∂y

∂u

∂x

∂v+

∂Fx

∂z

∂z

∂u

∂x

∂v

+∂Fy

∂x

∂x

∂u

∂y

∂v+

∂Fy

∂y

∂y

∂u

∂y

∂v+

∂Fy

∂z

∂z

∂u

∂y

∂v

+∂Fz

∂x

∂x

∂u

∂z

∂v+

∂Fz

∂y

∂y

∂u

∂z

∂v+

∂Fz

∂z

∂z

∂u

∂z

∂v

−∂Fx

∂x

∂x

∂v

∂x

∂u− ∂Fx

∂y

∂y

∂v

∂x

∂u− ∂Fx

∂z

∂z

∂v

∂x

∂u

−∂Fy

∂x

∂x

∂v

∂y

∂u− ∂Fy

∂y

∂y

∂v

∂y

∂u− ∂Fy

∂z

∂z

∂v

∂y

∂u

−∂Fz

∂x

∂x

∂v

∂y

∂u− ∂Fz

∂y

∂y

∂v

∂y

∂u− ∂Fz

∂z

∂z

∂v

∂y

∂u

=∂Fx

∂y

∂(y, x)

∂(u, v)+

∂Fx

∂z

∂(z, x)

∂(u, v)

+∂Fy

∂x

∂(x, y)

∂(u, v)+

∂Fy

∂z

∂(z, y)

∂(u, v)

+∂Fz

∂x

∂(x, z)

∂(u, v)+

∂Fz

∂y

∂(y, z)

∂(u, v)

=

(

∂Fy

∂x− ∂Fx

∂y

)

∂(x, y)

∂(u, v)+

(

∂Fz

∂y− ∂Fy

∂z

)

∂(y, z)

∂(u, v)

+

(

∂Fx

∂z− ∂Fz

∂x

)

∂(z, x)

∂(u, v)

= (∇× F) ·(

∂(y, z)

∂(u, v)δx +

∂(z, x)

∂(u, v)δy

+∂(x, y)

∂(u, v)δz

)

(5.75)

Recall (5.28) for calculating N which is the vector normal to the surface S and


having a magnitude equal to dS,

N = n dS =

(

∂(y, z)

∂(u, v)δx +

∂(z, x)

∂(u, v)δy +

∂(x, y)

∂(u, v)δz

)

dudv (5.76)

Combining (5.71), (5.74), (5.75) and (5.76), will result in∮

C

F · dR =

∫

S

(∇× F) · ndS

which is Stokes’ theorem.QED

5.7 Path Independence of Line Integrals

Definition 5.1 Let F = Fx δx + Fy δy + Fz δz be a given vector field and R =x δx +y δy + z δz be the position vector of a point (x, y, z). Then the line integral

IC =

∫

C

F · dR (5.77)

is independent of path in a region V , if for any pair of curves, C1 and C2 insidethe region V ,

IC1 = IC2 (5.78)

where C1 and C2 are continuous and sectionally smooth curves that have the sameinitial point (xi, yi, zi) and the same end point (xf , yf , zf ).

If we leave out the specification of region V , then it is usually understood that thephrase “independent of path” refers to the whole three dimensional space as the region ofinterest.

Let us first consider two paths C1,AB and C2,AB in region V that does not intersecteach other except for the start and end points. If the line integrals are independent of pathin V ,

∫

C1,AB

F · dR =

∫

C2,AB

F · dR (5.79)

which could be combined into one integral,∫

C1,AB

F · dR −∫

C2,AB

F · dR = 0

∫

C1,AB−C2,AB

F · dR =

∮

C

F · dR = 0 (5.80)


where C = C1,AB − C2,AB is a simple closed path as shown in Figure 5.13. With any choiceof C1,AB and C2,AB (so far assumed to be nonintersecting at midpath), by reversing the pathdirection of C2,AB, a closed path C is generated. Thus path independence guarantees thatthe line integral using the closed path will have to be zero.

Figure 5.13: Two paths with the same start point A and end point B.

Alternatively, we could have rearranged (5.79) to be∫

C2,AB−C1. AB

F · dR = 0 =

∮

C′

F · dR (5.81)

where C ′ = C2,AB − C1,AB is also a simple closed path but in the opposite direction of C.

Now consider the situation where path C1,AB and C2,AB might intersect each othersomewhere between the end points, say at point D. Path C1,AB can then be partitioned tobe the sum of two subpaths: C1,AD and C1,DB. Likewise, path C2,AB can also be partitionedto be the sum of two subpaths: C2,AD and C2,DB. This is shown in Figure 5.14.

Equation (5.79) can now be expanded to be∫

C1,AB

F · dR −∫

C2,AB

F · dR = 0

∫

C1,AD

F · dR +

∫

C1,DB

F · dR −∫

C2,AD

F · dR −∫

C2,DB

F · dR = 0


Figure 5.14: Two paths with the same start point A and end point B plus an intersectionat point D.

∫

C1,AD

F · dR −∫

C2,AD

F · dR +

∫

C1,DB

F · dR −∫

C2,DB

F · dR = 0

∫

C1,AD−C2,AD

F · dR +

∫

C1,DB−C2,DB

F · dR = 0

∮

CAD

F · dR −∮

CDB

F · dR = 0 (5.82)

where CAD is the closed path formed by adding path C1,AD and reverse of path C2,AD.Similarly, CDB is the closed path formed by adding path C1,DB and reverse of path C2,DB.Since the definition of path independence applies to the subpaths, the two closed paths willeach generate a zero line integral also.

Thus, the conditions for path independence is equivalent to having the line integral ofany closed paths in region V all be zero, including whichever direction the closed path takes.

Having established this equivalence, we can use Stokes’ theorem to offer one more toolto help determine path independence. Recall that Stokes’ theorem relates the line integralin a closed path to a surface integral involving the curl of vector field F:

∮

C

F · dR =

∫

S

n · (∇× F) (5.83)

If ∇×F = 0 for all point in region V , then Stokes’ theorem suggests that this condition willalso guarantee path independence, since for any surface S inside region V a zero curl impliesa zero line integral on the left hand side of (5.83).

One more detail is still needed, however. Stokes’ theorem requires that the integrandin the surface integral be bounded. This means that in the chosen region V , no singularitiesof the integrand can be present. In relation to the closed path integral, this kind of regionis formally referred to as a simply connected region.


Definition 5.2 A region V is simply connected if any simple closed path in the region canbe continuously deformed to a single point. Otherwise, the region is multiply connected.

A set of example regions are shown in Figure 5.15. The first case is shown in Fig-ure 5.15(a) where the region is a solid rectangular box. In this case, we see that any closedpath contained in the region can be deformed continuously into a point inside the box. Thesecond case is shown in Figure 5.15(b). For this case, a cylindrical subregion has been re-moved from the center. Even though we could find some closed paths that could deformto a point, the existence of at least one closed path that will not continuously deform to apoint is sufficient to have the region be classified as multiply connected. The third case isshown in Figure 5.15(c). In this case, the rectangular box region has a spherical subregionremoved. However, unlike case (b), any closed path inside the region of case (c) can stillbe continuously deformed to a point. Thus this third case is also considered to be a simplyconnected region.

Figure 5.15: Examples of simply and multiply connected regions. (a) Solid rectangularregion: simply connected (b) Solid rectangular region with a cylindrical subregion removedfrom the center: multiply connected (c) Solid rectangular region with a spherical subregionremoved from the center: simply connected.


We can now summarize the preceding discussion in the following theorem:

Theorem 5.3 Let F be a vector field whose curl is equal to zero inside a simply connectedregion V . Then the integral,

∫

CF · dR, is independent of path inside the region V .

Example 5.6

Consider the following vector fields:

H = y δx + x δy + 2z δz

G = x2 + y2 + z2 δx + 2xy δy + (y − z) δz

F = − y + 2

(x − 1)2 + (y + 2)2δx +

x − 1

(x − 1)2 + (y + 2)2δy + 2 δz

and the following paths with the parameter t ranging from 0 to 1:

Path C1 : x = 3 + cos(3πt)

y = 3 + sin(3πt)

z = 2t

Path C2 : x = −12t2 + 10t + 4

y = 3 + 4t(1 − t)

z = 2 − 2(1 − t)2

Path C3 : x = −6t2 + 5t + 1

y = 2t2 − 5t

z = −6t2 + 7t

Path C4 : x = 10t2 − 11t + 1

y = 2t2 − 5t

z = 10t2 − 9t

Figure 5.16 shows paths C1 and C2. Note that C1 has a helical form while C2 isa simpler three dimensional curve. Figure 5.17 shows paths C3 and C4 together.Included in the figure is a line described by: (x, y, z) = (1,−2, z). This lineincludes the singular point of vector field F. Thus the line is only relevant whenconsidering the F.


Figure 5.16: Paths C1 and C2.

Table 5.1: Line integrals based on different vector fields and paths.

PathsStart: (4,3,0) Start: (1,0,0)

Vector Fields End: (2,3,2) End: (0,-3,1) CurlC1 C2 C3 C4

H -2 -2 1 1 0G -36.062 -45.2 -4.4 -11.333 −δx + 2xδy

F 4.343 4.343 -1.927 4.356 0

We can tabulate the results of calculating the various type of line integrals of∫

CH ·dR,

∫

CG ·dR and

∫

CF ·dR for different paths. This is shown in Table 5.1

together with curl of the respective vector fields.

As expected, the line integrals for H are path independent because the curl iszero and the whole three dimensional space is simply connected. For G, since thecurl is not zero, the line integrals depend on the path. Finally, for F, since thereexists a simply connected region that covers C1 and C2, the two line integrals areexpected to be equal. However, as we increase the region in order to contain pathsC3 and C4, the allowed region (after removal of regions containing singularities)are multiply connected. Thus, the theorem states that we expect that pathindependence is no longer guaranteed even though the curl along paths C3 andC4 are still zero.

♦♦♦


Figure 5.17: Paths C3 and C4. (Also included here is the line (1,−2, z).

5.8 Leibnitz Derivative Formula

Theorem 5.4 Given a function F (α, x) that is differentiable in x and α, then

d

dα

∫ h(α)

g(α)

F (α, x) dx =

∫ h(α)

g(α)

∂F (α, x)

∂αdx

+F (α, h(α))∂h(α)

∂α− F (α, g(α))

∂g(α)

∂α(5.84)

Proof:

Using the definition of a derivative:

d

dα

(

∫ h(α)

g(α)

F (α, x) dx

)

= lim∆α → 0

1

∆α

(

∫ h(α+∆α)

g(α+∆α)

F (α + ∆α, x)dx

−∫ h(α)

g(α)

F (α, x)dx

)

(5.85)

The first integral in the left hand side of (5.85) can be divided into three parts,

∫ h(α+∆α)

g(α+∆α)

F (α + ∆α, x) dx =

∫ h(α+∆α)

h(α)

F (α + ∆α, x) dx


+

∫ h(α)

g(α)

F (α + ∆α, x) dx

+

∫ g(α)

g(α+∆α)

F (α + ∆α, x) dx (5.86)

We will now approximate the first integral in the left hand side of (5.86) usingthe trapezoidal rule:

∫ h(α+∆α)

h(α)

F (α + ∆α, x) dx ≈ 1

2

[

F(

α + ∆α, h(α+∆α)

)

+ F(

α + ∆α, h(α)

)] (

h(α+∆α) − h(α)

)

(5.87)

Likewise, we can also approximate the third integral in the left hand side of (5.86)as

∫ g(α)

g(α+∆α)

F (α + ∆α, x) dx ≈ 1

2

[

F(

α + ∆α, g(α+∆α)

)

+ F(

α + ∆α, g(α)

)] (

g(α) − g(α+∆α)

)

(5.88)

Substituting (5.86) (5.87) and (5.88) into (5.86),

d

dα

∫ h(α)

g(α)

F (α, x) dx = lim∆α→0

[

∫ h(α)

g(α))

(

F (α + ∆α, x) − F (α, x)

∆α

)

dx

+F

(

α + ∆α, h(α+∆α)

)

+ F(

α + ∆α, h(α)

)

2∆α

(

h(α+∆α) − h(α)

)

+F

(

α + ∆α, g(α+∆α)

)

+ F(

α + ∆α, g(α)

)

2∆α

(

g(α) − g(α+∆α)

)

=

∫ h(α)

g(α))

∂

∂αF (α, x) dx + F (α, h(α))

dh

dα− F (α, g(α))

dg

dα(5.89)

QED

Theorem 5.5 Given a function f(x, y, z, α) that is differentiable in an openregion containing three dimensional region V (α) and where both functionf(x, y, z, α) and region V (α) are both differentiable with respect to the parameterα. Then,

d

dα

∫

V (α)

f(x, y, z, α)dV =

∫

V (α)

∂f

∂αdV +

∫

S(α)

f(x, y, z, α) n · ∂R

∂αdS (5.90)

where S(α) is the bounding surface of volumetric region V (α), n(x, y, z) is theunit outward normal vector at the point (x, y, z) on surface S, and R = xδx +yδy + zδz is the position vector.


Proof:

From the definition of the derivative,

d

dα

∫

V (α)

f(x, y, z, α)dV = lim∆α→0

[

1

∆α

∫

V (α+∆α)

f(x, y, z, α + ∆α)dV

−∫

V (α)

f(x, y, z, α)dV

]

(5.91)

By adding and subtracting the term∫

V (α)f(x, y, z, α + ∆α)dV in the right hand

side,

d

dα

∫

V (α)

f(x, y, z, α)dV = lim∆α→0

1

∆α

[∫

V (α)


−∫

V (α)

f(x, y, z, α)dV

]

+ lim∆α→0

1

∆α

[∫

V (α+∆α)


−∫

V (α)


]

=

∫

V (α)

∂f

∂αdV

+ lim∆α→0

1

∆α

[∫

V (α+∆α)


−∫

V (α)


]

(5.92)

The last group of terms in the right hand side (5.92) is the difference of twovolume integrals involving the same integrand. We can combine these integralsby changing the volume of integration to be the region between V (α + ∆α) andV (α).

∫

V (α+∆α)

f(x, y, z, α + ∆α)dV −∫

V (α)

f(x, y, z, α + ∆α)dV =

∫

V (α+∆α)−V (α)

f(x, y, z, α + ∆α)dV (5.93)

We could approximate the differential volume in (5.93) as the parallelepipedformed by the vectors (∂R/∂u)du, (∂R/∂v)dv and (∂R/∂α)dα, where u and vare parameters used to describe surface S(α). This is shown in Figure 5.18.


Figure 5.18: Graphical representation of differential volume emanating from points in S(α)towards S(α + ∆α).

Recall that(

∂R

∂udu

)

×(

∂R

∂vdv

)

= n dS

which then gives a differential volume attached to S(α)

dV |(x,y,z)∈V (α+∆α)−V (α) =∂R

∂α·(

∂R

∂u× ∂R

∂u

)

dαdudv

=∂R

∂α· n dα dS (5.94)

The volume integral for points bounded between the surfaces of V (α) and V (α+∆α) can now be approximated as follows:

∫

V(α+∆α)−V(α)


≈∫

S(α)

f(x, y, z, α + ∆α)∂R

∂α· n ∆α dS (5.95)

Substituting (5.95) into (5.93) and then to (5.92),

d

dα

∫

V (α)

f(x, y, z, α)dV =

∫

V (α)

∂f

∂αdV

+ lim∆α→0

1

∆α

∫

S(α)

f(x, y, z, α + ∆α)∂R

∂α· n∆αdS

=

∫

V (α)

∂f

∂αdV


+

∫

S(α)

f(x, y, z, α)∂R

∂α· n dS (5.96)

which is the Leibnitz rule for differentiation of volume integrals.

QED

chapter 4 vectors and tensors - michigan tech it …pages.mtu.edu/~tbco/cm5100/chapter4_5.pdf ·...

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