KIX1002: ENGINEERING
MATHEMATICS 2
WEEK 1: VECTOR ANALYSIS
WEEK 2: ENGINEERING APPLICATIONS OF VECTOR ANALYSIS
WEEK 3: LINE INTEGRAL
WEEK 4: SURFACE INTEGRAL
WEEK 5: VOLUME INTEGRAL
WEEK 6: GAUSS’S DIVERGENCE THEOREM
WEEK 7: STOKE’S THEOREM
2018/2019
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CONTENTS
WEEK 1: VECTOR ANALYSIS .......................................................................................................... 4
1.1 INTRODUCTION .......................................................................................................................... 4
1.2 BASIC CONCEPTS ....................................................................................................................... 4
1.2.1 COMPONENTS OF A VECTOR ............................................................................................. 6
1.2.2 VECTOR ADDITION, SCALAR MULTIPLICATION ................................................................. 7
1.2.3 UNIT VECTOR..................................................................................................................... 10
1.3 DOT PRODUCT AND CROSS PRODUCT ................................................................................... 12
1.3.1 DOT PRODUCT ................................................................................................................... 12
1.3.2 CROSS PRODUCT................................................................................................................ 17
1.4 GRaDIENT, DIVERGENCE, CURL OF VECTOR Field .................................................................. 21
1.4.1 GRADIENT OF VECTOR FIELD............................................................................................ 21
1.4.2 DIRECTIONAL DERIVATIVES ............................................................................................. 22
1.4.3 DIVERGENCE OF VECTOR FIELD ....................................................................................... 25
1.4.4 CURL OF VECTOR FIELD .................................................................................................... 27
WEEK 2: ENGINEERING APPLICATIONS OF VECTOR ANALYSIS .............................................. 29
2.1 APPLICATION OF VECTOR ....................................................................................................... 29
2.2 DOT PRODUCT (WORK DONE) .................................................................................................33
2.3 TORQUE (CROSS PRODUCT) ................................................................................................... 35
2.4 TRIPLE SCALAR OR BOX PRODUCT......................................................................................... 36
2.5 INTERPRETATION OF THE DIRECTIONAL DERIVATIVES & GRADIENT ................................... 38
2.6 GRADIENTS,TANGENTS AND NORMAL TO LEVEL CURVES................................................... 41
2.7 APPLICATION OF DIVERGENCE AND CURL ............................................................................. 45
WEEK 3: LINE INTEGRALS ............................................................................................................ 50
3.1 INTRODUCTION........................................................................................................................ 50
3.2 BASIC CONCEPTS ..................................................................................................................... 52
3.3 LINE INTEGRAL: WORK DONE BY FORCE ................................................................................ 56
3.4 GREEN’S THEOREM IN THE PLANE.......................................................................................... 59
WEEK 4: SURFACE INTEGRALS .................................................................................................... 62
4.1 SURFACES FOR SURFACE INTEGRALS .................................................................................... 62
4.2 SURFACE INTEGRALS .............................................................................................................. 64
4.2.1 SURFACE INTEGRALS OF VECTOR FIELD ......................................................................... 65
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4.2.2 SURFACE INTEGRAL OF SCALAR FIELD............................................................................ 69
4.3 ORIENTATION OF SURFACE .................................................................................................... 74
4.3.3 SURFACE INTEGRALS WITHOUT REGARD TO ORIENTATION.......................................... 76
WEEK 5: VOLUME INTEGRAL ................................................................................................. 80
5.1 INTRODUCTION ................................................................................................................... 80
5.2 DEFINITION ........................................................................................................................... 80
5.3 CYLINDRICAL AND SPHERICAL COORDINATE .............................................................. 81
5.4 APPLICATION OF VOLUME INTEGRAL ........................................................................... 83
5.4.1 VOLUME ............................................................................................................................ 83
5.4.2 MASS ................................................................................................................................. 84
5.4.3 CENTRE OF MASS .............................................................................................................. 85
5.4.5 MOMENT OF INERTIA........................................................................................................ 86
5.5 ADDITIONAL EXAMPLES .................................................................................................... 88
WEEK 6: GAUSS’ DIVERGENCE THEOREM ......................................................................... 90
6.1 INTRODUCTION ................................................................................................................... 90
6.2 DIVERGENCE ........................................................................................................................ 90
6.3 DIVERGENCE THEOREM ..................................................................................................... 91
6.4 EXAMPLES ............................................................................................................................ 92
WEEK 7: STOKE’S THEOREM...................................................................................................... 100
7.1 INTRODUCTION ...................................................................................................................... 100
7.2 DIRECTION OF UNIT NORMAL VECTORS TO A SURFACEs ....................................................103
7.3 GREEN’S THEOREM ................................................................................................................. 107
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VECTOR ANALYSIS
WEEK 1: VECTOR ANALYSIS
1.1 INTRODUCTION
Forces, velocities, and various other quantities may be thought of as vectors. Vectors appear
frequently in the applications above and also in the biological and social sciences, so it is natural that
problems are modelled in 3-space. This is the space of three dimensions with the usual measurement
of distance, as given by the Pythagorean Theorem. Working in 3-space requires that we extend the
common differential calculus to vector differential calculus, that is, the calculus that deals with vector
functions and vector fields
1.2 BASIC CONCEPTS
A scalar is a quantity that is determined by its magnitude. It takes on a numerical value, i.e., a
number.
Examples of scalars are time, temperature, length, distance, speed, density, energy, and
voltage.
A vector is a quantity that has both magnitude and direction. We can say that a vector is an arrow or
a directed line segment.
For example, a velocity vector has length or magnitude, which is speed, and direction, which
indicates the direction of motion (Fig 1.1); a force vector points in the direction in which the
force acts and its length is a measure of the force’s strength.
A vector (arrow) has a tail, called its initial point, and a tip, called its terminal point. The length of the
arrow equals the distance between initial point and terminal point (Fig 1.1). This is called the length
(or magnitude) of the vector a and is denoted by |𝑎|. Another name for length is norm (or Euclidean
norm). A vector of length 1 is called a unit vector.
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Fig 1.1: The directed line segment ̅̅̅̅
𝐴𝐵 is called a vector.
Fig 1.2: The velocity vector of a particle moving along a path (a) in the plane (b) in space.
The arrowhead on the path indicates the direction of motion of the particle.
Equality of Vectors - Two vectors a and b are equal, written, if they have the same length and the
same direction as shown in Fig. 1.3.
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Fig. 1.3 (A) Equal Vectors. (B) – (D) Different Vectors
1.2.1 COMPONENTS OF A VECTOR
Let a be a given vector with initial point P: (x1, y1, z1) and terminal point Q: (x2, y2, z2). Then the three
coordinate differences
are called the components of the vector a with respect to that coordinate system, and we write
simply
a = [a1, a2, a3]. See Fig 1.4 (a). The length |𝑎| of a can now readily be expressed in terms of
components and the Pythagorean Theorem we have
A Cartesian coordinate system being given, the position vector r of a point A: (x, y, z) is the vector
with the origin (0, 0, 0) as the initial point and A as the terminal point (See Fig 1.4 (b)).
Fig 1.4(a) Components of a vector
Fig 1.4(b) Position vector r of a point A: (x, y, z)
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EXAMPLE 1.1
EXERCISES
Let u= 3𝑖 − 2𝑗 and = −2𝑖 + 5𝑗 . Find the (a) component form and (b) magnitude (length) of the
vector.
1.2.2 VECTOR ADDITION, SCALAR MULTIPLICATION
Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is
simply a real number, and is called such when we want to draw attention to its differences from
vectors. Scalars can be positive, negative, or zero and are used to “scale” a vector by multiplication
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Geometrically, place the vectors as in Fig. 1.5 (the initial point of b at the terminal point of a); then a
+ b is the vector drawn from the initial point of a to the terminal point of b. Fig. 1.5 also shows (for
the plane) that the “algebraic” way and the “geometric way” of vector addition give the same vector.
Fig 1.5 Vector Additions
Basic Properties of Vector Addition
Properties (a) and (b) are verified geometrically in Fig. 1.6 and Fig 1.7, respectively. Furthermore, -a
denotes the vector having the length |𝒂| and the direction opposite to that of a.
Fig 1.6 Commutativity of vector addition
addition
Fig 1.7 Associativity of vector
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Geometrically, if a ≠ 0 then ca with c > 0 has the direction of a and with c < 0 the direction opposite
to a. In any case, the length of ca is |𝑐𝒂| = |𝑐| |𝒂|, and ca = 0 if a = 0 or c = 0 (or both). (See Fig 1.8)
Fig 1.8 Scalar multiplication [multiplication of vectors by scalars (numbers)]
Basic Properties of Scalar Multiplication. From the definitions we obtain directly
EXAMPLE 1.2
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EXERCISES
1.2.3 UNIT VECTOR
A vector v of length 1 is called a unit vector. In this representation, i, j, k are the unit vectors in the
positive directions of the axes of a Cartesian coordinate system. The standard unit vectors are
𝑖 = ⟨1, 0, 0⟩,
𝑗 = ⟨0,1,0⟩
𝑘 = ⟨0, 0, 1⟩
Any vector can be written as a linear combination of the standard unit vectors as follows:
𝑣 = ⟨𝑣1 , 𝑣2 , 𝑣3 ⟩ = ⟨𝑣1 , 0, 0⟩ + ⟨0, 𝑣2 , 0⟩ + ⟨0, 0, 𝑣3 ⟩
= 𝑣1 ⟨1, 0, 0⟩ + 𝑣2 ⟨0, 1, 0⟩ + 𝑣3 ⟨0, 0, 1⟩
𝑣 = 𝑣1 𝑖 + 𝑣2 𝑗 + 𝑣3 𝑘
From Figure 1.9, we call the scalar (or number) v1 the i-component of the vector v, v2 the jcomponent, and v3 the k-component. In component form, the vector from P1(x1, y1, z1) to P2(x2, y2,
z2) is
⃑⃑⃑⃑⃑⃑⃑⃑
𝑃
1 𝑃2 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘
Fig. 1.9 The vector from P1 to P2 is ̅̅̅̅̅̅
𝑃1 𝑃2
Whenever v ≠ 0, its length |𝑣| is not zero and
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|
1
1
|𝑣| = 1
𝑣| =
|𝑣|
|𝑣|
That is, 𝑣⁄|𝑣| is a unit vector in the direction of v, called the direction of the nonzero vector v.
EXAMPLE 1.3
EXAMPLE 1.4
***PROVE that the length of unit vector is 1.
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1.3 DOT PRODUCT AND CROSS PRODUCT
1.3.1 DOT PRODUCT
The inner product is a kind of multiplication of two vectors, defined in such a way that the outcome is
a scalar. It involves the length of vectors and the angle between them. Indeed, another term for inner
product is scalar product, a term we shall not use here. The definition of the inner product is as follows.
Fig 1.10 Angle between vectors and value of inner product
EXAMPLE 1.5
Perpendicular (Orthogonal) Vectors (Fig 1.10)
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EXAMPLE 1.6
Angle between Vectors
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EXAMPLE 1.7
The vector projection of 𝑢 = ⃑⃑⃑⃑⃑
𝑃𝑄 onto a nonzero vector 𝑣 = ⃑⃑⃑⃑
𝑃𝑆 (Figure A) is the vector
⃑⃑⃑⃑⃑
𝑃𝑅determined by dropping a perpendicular from Q to the line PS. The notation for this vector is
If u represents a force, then 𝑝𝑟𝑜𝑗𝑣 𝑢 represents the effective force in the direction of v (Figure B).
FIGURE A The vector projection of u onto v.
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FIGURE B
If we pull on the box with force u, the effective force moving the box forward in the direction v is
the projection of u onto v.
If the angle ∅ between u and v is acute (less than 90o), 𝑝𝑟𝑜𝑗𝑣 𝑢 has length |𝑢| cos ∅ and direction
𝑣/|𝑣|(𝐅𝐢𝐠𝐮𝐫𝐞 𝐂). If ∅ is obtuse (90 < ∅ < 180), and 𝑝𝑟𝑜𝑗𝑣 𝑢 has length −|𝑢| cos ∅ and direction
−𝑣/|𝑣|. In both cases,
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EXAMPLE 1.8
EXERCISES
Find the followings
(a) 𝑣. 𝑢, |𝑣|, |𝑢|
(b) The cosine of the angle between v and u
(c) The scalar component of u in the direction of v
(d) The vector 𝑝𝑟𝑜𝑗𝑣 𝑢
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for
1) 𝑣 = 10𝑖 + 11𝑗 − 2𝑘, 𝑢 = 3𝑗 + 4𝑘,
2) 𝑣 = 2𝑖 + 10𝑗 − 11𝑘, 𝑢 = 2𝑖 + 2𝑗 + 𝑘
3) 𝑣 = −𝑖 + 𝑗, 𝑢 = √2𝑖 + √3𝑗 + 2𝑘
1.3.2 CROSS PRODUCT
Unlike the dot product, the cross product is a vector. For this reason it’s also called the vector product
of u and v, and applies only to vectors in space. The vector 𝑢 × 𝑣 is orthogonal to both u and v because
it is a scalar multiple of n (Fig 1.11).
Fig 1.11: The construction of 𝑢 × 𝑣
We start with two nonzero vectors u and v in space. If u and v are not parallel, they determine a plane.
We select a unit vector n perpendicular to the plane by the right-hand rule. This means that we choose
n to be the unit (normal) vector that points the way your right thumb points when your fingers curl
through the angle from u to v (Fig ). Then the cross product 𝑢 × 𝑣 (“u cross v”) is the vector defined
as follows.
The cross product obeys the following laws.
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To visualize Property 3, for example, notice that when the fingers of your right hand curl through the
angle from v to u, your thumb points the opposite way; the unit vector we choose in forming is the
negative of the one we choose in forming. (Fig 1.12)
Fig 1.12 : The construction of 𝑣 × 𝑢
When we apply the definition to calculate the pairwise cross products of i, j, and k, we find
|𝑼 × 𝑽| Is the Area of a Parallelogram
Because n is a unit vector, the magnitude of u x v is
This is the area of the parallelogram determined by u and v (Fig. 1.13), |𝑢|being the base of the
parallelogram and |𝑣||𝑠𝑖𝑛𝜃| the height.
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Fig 1.13 The parallelogram determined by u and v.
EXERCISES
2) 𝐹𝑖𝑛𝑑 𝑢 × 𝑣 𝑓𝑜𝑟 𝑢 = −8𝑖 − 2𝑗 − 4𝑘, 𝑣 = 2𝑖 + 2𝑗 + 𝑘
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EXAMPLE 1.9
EXAMPLE 1.10
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EXAMPLE 1.11
1.4 GRADIENT, DIVERGENCE, CURL OF VECTOR FIELD
1.4.1 GRADIENT OF VECTOR FIELD
Using scalar fields instead of vector fields is of a considerable advantage because scalar fields are
easier to use than vector fields. It is the “gradient” that allows us to obtain vector fields from scalar
fields, and thus the gradient is of great practical importance to the engineer. Gradients are useful in
several ways, notably in giving the rate of change of in any direction in space, in obtaining surface
normal vectors, and in deriving vector fields from scalar fields.
𝑔𝑟𝑎𝑑 𝑓 = ∇𝑓 = [
𝜕𝑓 𝜕𝑓 𝜕𝑓
𝜕𝑓
𝜕𝑓
𝜕𝑓
,
,
𝑖+
𝑗+
𝑘
]=
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
The notation ∇𝑓 is suggested by the differential operator ∇ (read nabla) defined by
EXAMPLE 1.12
If 𝑓(𝑥, 𝑦, 𝑧) = 2𝑦 3 + 4𝑥𝑧 + 3𝑥 , then 𝒈𝒓𝒂𝒅 𝒇 = [4𝑧 + 3, 6𝑦 2 , 4𝑥]
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EXERCISES
Find the gradient of the following function at the given point.
(a) 𝑓(𝑥, 𝑦) = ln(𝑥 2 + 𝑦 2 ) at point (1, 1)
(b) 𝑓(𝑥, 𝑦) = √2𝑥 + 3𝑦 at point (-1, 2)
1.4.2 DIRECTIONAL DERIVATIVES
From gradient we know that the partial derivatives give the rates of change of f(x, y, z) in the directions
of the three coordinate axes. It seems natural to extend this and ask for the rate of change of in an
arbitrary direction in space. This leads to the concept of directional derivative.
Fig. 1.14 Directional Derivative (Refer to above Equation)
The above equation can be derived into
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Where we can say the derivative of a differentiable function ƒ in the direction of b at P is the dot
product of b with the special vector called the gradient of ƒ at P.
The notation ∇𝑓 is read “grad ƒ” as well as “gradient of ƒ” and “del ƒ.” The symbol ∇ by itself is read
“del.” Another notation for the gradient is grad ƒ.
EXAMPLE 1.13
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EXAMPLE 1.14
EXERCISES
Find the derivative of the function at Po in the direction of u
𝑥−𝑦
(a) 𝑔(𝑥, 𝑦) = 𝑥𝑦+2 , 𝑃𝑜 (1, −1), 𝑢 = 12𝑖 + 5𝑗
(b) ℎ(𝑥, 𝑦, 𝑧) = cos 𝑥𝑦 + 𝑒 𝑦𝑧 + ln(𝑧𝑥) , 𝑃𝑜 (1, 0, 1/2), 𝑢 = 1𝑖 + 2𝑗 + 2𝑘
(c) ℎ(𝑥, 𝑦, 𝑧) = 3𝑒 𝑥 cos 𝑦𝑧 , 𝑃𝑜 (0, 0, 0), 𝑢 = 2𝑖 + 𝑗 − 2𝑘
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1.4.3 DIVERGENCE OF VECTOR FIELD
From a scalar field we can obtain a vector field by the gradient. Conversely, from a vector field we
can obtain a scalar field by the divergence or another vector field by the curl
To begin, let 𝑣(𝑥, 𝑦, 𝑧) be a differentiable vector function, where x, y, z are Cartesian coordinates, and
let 𝑣1 , 𝑣2 , 𝑣3 be the components of v. Then the function
𝑑𝑖𝑣 𝑣 =
𝜕𝑣1 𝜕𝑣2
𝜕𝑣3
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
is called the divergence of v or the divergence of the vector field defined by v. For example, if
𝑣 = [3𝑥𝑧, 2𝑥𝑦, −𝑦𝑧 2 ] = 3𝑥𝑧𝑖 + 2𝑥𝑦𝑗 − 𝑦𝑧 2 𝑘
Then
𝑑𝑖𝑣 𝑣 = 3𝑧 + 2𝑥 + 2𝑦𝑧
Another common notation for the divergence is
div v = ∇
• 𝑣=[
𝜕
= ( 𝑖+
𝜕𝑥
=
𝜕𝑣1
𝜕𝑥
+
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕𝑣2
𝜕𝑦
𝜕𝑦
,
𝜕𝑥
𝑗+
+
𝜕𝑦
,
𝜕𝑥
𝜕
𝜕𝑧
] • [𝑣1 , 𝑣2 , 𝑣3 ]
𝑘 ) • [𝑣1 𝑖 + 𝑣2 𝑗 + 𝑣3 𝑘]
𝜕𝑣3
𝜕𝑧
𝜕
with the understanding that the “product” ( ) 𝑣1 in the dot product means the partial derivative
𝜕𝑥
𝜕𝑣1
𝜕𝑥
, etc. This is a convenient notation, but nothing more. Note that 𝛁. 𝒗 means the scalar div v,
whereas 𝛁𝒇 means the vector grad f
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EXAMPLE 1.15
𝐼𝑓 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑧 𝒊 + 𝑥𝑦𝑧 𝒋 − 𝑦 2 𝒌, 𝑓𝑖𝑛𝑑 𝑑𝑖𝑣 𝑓
𝑑𝑖𝑣 𝑓 = ∇ ∙ 𝑓
𝜕
𝜕
𝜕
(𝑥𝑧) +
(𝑥𝑦𝑧) + (−𝑦 2 )
=
𝜕𝑥
𝜕𝑦
𝜕𝑧
= 𝑧 + 𝑥𝑧
*Div F is a scalar field.
Let us turn to the more immediate practical task of gaining a feel for the significance of the
divergence. Let f(x, y, z) be a twice differentiable scalar function. Then, its gradient exists
and we can differentiate once more, the first component with respect to x, the second with
respect to y, the third with respect to z, and then form the divergence,
Hence we have the basic result that the divergence of the gradient is the Laplacian
EXERCISES
1) Find divergence from the gradient, div (grad f)
(a) 𝑓 = 𝑒 𝑥𝑦𝑧
(b) 𝑓 = 𝑧 − √𝑥 2 + 𝑦 2
2) Find div v and its value at P
(a) 𝑣 = 𝑥 2 𝑖 + 4𝑦 2 + 9𝑧 2 at P (-1, 0, ½)
(b) 𝑣 = cos 𝑥𝑦𝑧 + sin 𝑥𝑦𝑧
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1.4.4 CURL OF VECTOR FIELD
Let 𝑣(𝑥, 𝑦, 𝑧) = [𝑣1 , 𝑣2 , 𝑣3 ] = 𝑣1 𝑖 + 𝑣2 𝑗 + 𝑣3 𝑘 be a differentiable vector function of the Cartesian
coordinates x, y, z. Then the curl of the vector function v or of the vector field given by v is defined by
the “symbolic” determinant
𝑖
𝑗
𝑘
𝜕
𝜕
𝛿
𝜕𝑦
𝛿𝑧
curl v = ∇ × 𝑣 = [
𝜕𝑥
𝑣1
= (
𝑣2
]
𝑣3
𝜕𝑣3 𝜕𝑣2
𝜕𝑣1 𝜕𝑣3
𝜕𝑣2 𝜕𝑣1
−
−
−
)𝑖 + (
)𝑗 + (
)𝑘
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
This is the formula when x, y, z are right-handed. If they are left-handed, the determinant has a minus
sign in front. Instead of curl v one also uses the notation rot v or rotation of v.
EXAMPLE 1.16
EXAMPLE 1.17
𝐼𝑓 𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑧 𝒊 + 𝑥𝑦𝑧 𝒋 − 𝑦 2 𝒌, 𝑓𝑖𝑛𝑑 𝑐𝑢𝑟𝑙 𝐹
𝑖
𝛿
𝑐𝑢𝑟𝑙 𝐹 = ∇ × 𝐹 = ||
𝛿𝑥
𝑥𝑧
𝑗
𝛿
𝛿𝑦
𝑥𝑦𝑧
𝑘
𝜕
|
𝜕𝑧 |
−𝑦 2
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𝜕
𝜕
𝜕
𝜕
= [ (−𝑦 2 ) −
(𝑥𝑦𝑧)] 𝒊 − [ (−𝑦 2 ) − (𝑥𝑧)] 𝒋
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕
𝜕
+ [ (𝑥𝑦𝑧) −
(𝑥𝑧)] 𝒌
𝜕𝑥
𝜕𝑦
= (−2𝑦 − 𝑥𝑦)𝒊 − (0 − 𝑥)𝒋 + (𝑦𝑧 − 0)𝒌
= −𝑦(2 + 𝑥) 𝒊 + 𝑥 𝒋 + 𝑦𝑧 𝒌
*Curl F is a vector field.
EXERCISES
Compute the curl of the following vector field
a)
𝑓(𝑥, 𝑦, 𝑧) =< 𝑒 𝑥 cos 𝑦 , 𝑒 𝑥 sin 𝑦, 0 >
b) 𝑓(𝑥, 𝑦, 𝑧) =
2𝑥𝑦
𝑖
𝑧
+ 𝑥𝑒 𝑥𝑦 𝑗 + cos(𝑥𝑦 2 )𝑘
c) 𝑓(𝑥, 𝑦, 𝑧) = (𝑥𝑦𝑧)𝑖 + (𝑥 2 + 2𝑦𝑧)𝑗 + (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑘
28
ENGINEERING APPLICATIONS OF
VECTOR ANALYSIS
WEEK 2: ENGINEERING APPLICATIONS OF VECTOR ANALYSIS
2.1 APPLICATION OF VECTOR
The word problems encountered most often with vectors are navigation problems. These
navigation problems use variables like speed and direction to form vectors for computation.
Some navigation problems ask us to find the groundspeed of an aircraft using the combined
forces of the wind and the aircraft. For these problems it is important to understand the
resultant of two forces and the components of force
Each of the three vectors in the triangle of velocities has two properties – magnitude and
direction. This means that there are a total of six components. These are the True Air Speed
(TAS) and heading (HDG) of the aircraft, the speed and direction of the wind (W/V), and the
Ground Speed (GS) and track (TR) of the path over the ground. This is shown in Figure 2.1.
Figure 2.1: Triangle of Vectors (Velocities)
29
To summarize:
Course—the direction of a line drawn on a chart representing the intended airplane path,
expressed as the angle measured from a specific reference datum clockwise from 0° through
360° to the line.
Heading—is the direction in which the nose of the airplane points during flight.
Drift angle—is the angle between heading and track.
Airspeed—is the rate of the airplane’s progress through the air.
Groundspeed—is the rate of the airplane’s in-flight progress over the ground.
Example 2.1
30
Example 2.2
31
Exercies
1) A boat leaves port on a heading of 40 with the automatic pilot set for 12 knots. On this
particular day, there is a 6-knot current with a heading of 75.
a. Sketch and label vectors to represent the intended path of the boat, the current and the
resultant path of the boat with the effects of the current.
b. Calculate the speed and heading at which the boat will actually travel due to the effects of
the current.
32
2) A plane leaves the airport on a heading 45° traveling a 400 mph. The wind is blowing at a
heading of 135° at a speed of 40 mph. What is the actual velocity of the plane?
3) Consider a 100-N weight suspended by two wires as shown in the accompanying figure.
Find the magnitudes and components of the force vectors F1 and F2.
2.2 DOT PRODUCT (WORK DONE)
After investigating the dot product, we apply it to finding the projection of one vector onto
another (as displayed in Figure 2.2) and to finding the work done by a constant force acting
through a displacement.
Fig 7.0 shows that the scalar quantity we seek is the length |𝐹| cos 𝜃 where 𝜃 is the angle
between the two vectors F and D. Then
Figure 2.2: The work done by a constant force F during a displacement D is |𝐹| cos 𝜃 , which
is the dot product F.D.
33
Forces Perpendicular to the Motion Do No Work
When an object is displaced horizontally on a flat table, the normal force n and the
gravitational force Fg do no work since cos θ = 90◦ = 0
Example 2.3
If |𝐹| = 40 𝑁 (newtons), D = |𝐷| = 3 𝑚, and 𝜃 = 60𝑜 , the work done by Fin acting from P to
Q is
Exercies
1) How much work does it take to slide a crate 20 m along a loading dock by pulling on
it with a 200 N force at an angle of 30° from the horizontal?
2) A 30 kg box is placed 10 m up a ramp that is inclined at 23◦ to the horizontal.
Calculate the work done by the force of gravity as the box slides down to the bottom
of the ramp
34
2.3 TORQUE (CROSS PRODUCT)
When we turn a bolt by applying a force F to a wrench (Figure 2.3), we produce a torque that
causes the bolt to rotate.
The torque vector points in the direction of the axis of the bolt according to the right-hand
rule (so the rotation is counter clockwise when viewed from the tip of the vector).
The magnitude of the torque depends on how far out on the wrench the force is applied and
on how much of the force is perpendicular to the wrench at the point of application.
The number we use to measure the torque’s magnitude is the product of the length of the
lever arm r and the scalar component of F perpendicular to r.
𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡𝑜𝑟𝑞𝑢𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 = |𝑟||𝐹|𝑠𝑖𝑛𝜃,
𝑜𝑟
|𝑟 × 𝐹|
If we let n be a unit vector along the axis of the bolt in the direction of the torque, then a
complete description of the torque vector is
𝑇𝑜𝑟𝑞𝑢𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 = (|𝑟||𝐹|𝑠𝑖𝑛𝜃)𝑛,
𝑜𝑟
𝑟×𝐹
Fig 2.3: The torque vector describes the tendency of the force F to drive the bolt forward.
35
EXAMPLE 2.4
Find the magnitude of the torque generated by force F at the pivot point P in Figure 2.4 is
Figure 2.4: Torque exerted by F at P
The magnitude of the torque exerted by F at P is about 56.4 ft-lb. The bar rotates
counterclockwise around P.
2.4 TRIPLE SCALAR OR BOX PRODUCT
The product is called the triple scalar product of u, v, and w (in that order). As you can see
from the formula
|(𝑢 × 𝑣) ∙ 𝑤| = |𝑢 × 𝑣||𝑤||𝑐𝑜𝑠𝜃|
the absolute value of this product is the volume of the parallelepiped (parallelogram-sided
box) determined by u, v, and w (Figure 2.5). The number |(𝑢 × 𝑣)| is the area of the base
parallelogram. The number |𝑤||𝑐𝑜𝑠𝜃|is the parallelepiped’s height. Because of this geometry,
(𝑢 × 𝑣) ∙ 𝑤 is also called the box product of u, v, and w.
36
Figure 2.5: The number |(𝑢 × 𝑣) ∙ 𝑤| = is the volume of a parallelepiped.
The triple scalar product can be evaluated as a determinant:
37
EXAMPLE 2.5:
EXERCISES
Find the volume of the parallelepiped (box) determined by u, v, and w.
2.5 INTERPRETATION OF THE DIRECTIONAL DERIVATIVES & GRADIENT
Figure 2.6: Interpretation of Directional Derivatives.
38
From the fig above, the slope of curve C at Po is (Duf)Po in which it generalizes two partial
derivatives. We can now ask for the rate of change of ƒ in any direction u, not just the
directions i and j.
For a physical interpretation of the directional derivative, suppose that T = f (x, y) is the
temperature at each point (x, y) over a region in the plane. Then f (xo, yo) is the temperature
at the point Po (xo, yo) and (Duf)Po is the instantaneous rate of change of the temperature at
Po stepping off in the direction u.
EXAMPLE 2.6:
39
EXAMPLE 2.7:
EXERCISES
Find the directions in which the functions increase and decrease most rapidly at Po. Then find
the derivatives of the functions in these directions.
(d) 𝑓(𝑥, 𝑦, 𝑧) = ln 𝑥𝑦 +ln 𝑦𝑧 +ln 𝑥𝑧, 𝑃𝑜 (1, 1, 1)
(e) 𝑔(𝑥, 𝑦, 𝑧) = 𝑥𝑒 𝑦 + 𝑧 2 , 𝑃𝑜 (1, ln 2 , 1/2)
40
(f) 𝑓(𝑥, 𝑦, 𝑧) = (𝑥⁄𝑦) − yz, 𝑃𝑜 (4, 1, 1)
2.6 GRADIENTS,TANGENTS AND NORMAL TO LEVEL CURVES
The streams flow perpendicular to the contours. The streams are following paths of steepest
descent so the waters reach the ocean as quickly as possible. Therefore, the fastest
instantaneous rate of change in a stream’s elevation above sea level has a particular direction.
In this section, you will see why this direction, called the “downhill” direction, is perpendicular
to the contours.
Figure 2.7: Contours of the hill
Based on above contours:
41
Figuure 2.8: The gradient of a differentiable function of two variables at a point is always
normal to the function’s level curve through that point.
Where it shows our observation that streams flow perpendicular to the contours in
topographical maps (see Figure 2.7). Since the downflowing stream will reach its destination
in the fastest way, it must flow in the direction of the negative gradient vectors from Property
2 for the directional derivative.
Now let us restrict our attention to the curves that pass through Po (Figure Below). All the
velocity vectors at Po are orthogonal to f at Po so the curves’ tangent lines all lie in the plane
through Po normal to f . At every point along the curve, f is orthogonal to the curve’s
velocity vector.
Figure Below: The gradient f is orthogonal to the velocity vector of every smooth curve in
the surface through Po. The velocity vectors at Po therefore lie in a common plane, which we
call the tangent plane at Po.
We now define this plane.
The tangent plane and normal line have the following equations:
42
EXAMPLE 2.8:
43
EXAMPLE 2.9:
:
EXERCISES
Find equations for the
(a) Tangent plane and (b) normal line at the point Po on the given surface.
44
2.7 APPLICATION OF D IVERGENCE AND CURL
Divergence at a given point measures the net flow out of a small box around the point, that
is, it measures what is produced (source) or consumed (sink) at a given point in space.
For example, it is used to describe the flow of gas within a domain space. A gas is compressible,
unlike a liquid, and the divergence of its velocity field measures to what extent it is expanding
or compressing at each point. Intuitively, if a gas is expanding at the point (xo, yo) the lines of
flow would diverge there (hence the name) and, since the gas would be flowing out of a small
rectangle about (xo, yo), the divergence of F at (xo, yo) would be positive. If the gas were
compressing instead of expanding, the divergence would be negative.
Figure 2.9: If a gas is expanding at a point the lines of flow have positive divergence; if the
gas is compressing, the divergence is negative. Otherwise, it will get zero.
EXAMPLE 2.10:
45
Determine the divergence’s characteristic of the vector fields; 𝐹(𝑥, 𝑦) = 3𝑥 2 𝒊 − 6𝑥𝑦𝒋
*In fluid dynamics, when the velocity field of a flowing liquid always has divergence equal to
zero, as in those cases, the liquid is said to be incompressible.
If we think of the vector field as a velocity vector field of a fluid in a motion, the curl measures
the rotation. At a given point, the curl is a vector parallel to the axis of rotation of flow lines
near the point, with direction determined by the Right Hand Rule.
EXAMPLE 2.11:
Determine wherther the curl of each vector field at the origin is the zero vector or points in
the certain directions as ±𝑖, ±𝑗 , ±𝑘.
46
EXAMPLE 2.12:
The following vector fields represent the velocity of a gas flowing in the xy-plane. Find the
divergence and curl of each vector field and interpret its physical meaning. Figure displays the
vector fields.
47
Figure 2.10: Velocity fields of a gas flowing in the plane
DIVERGENCE
48
CURL
49
LINE INTEGRALS
WEEK 3: LINE INTEGRALS
3.1 INTRODUCTION
Since in evaluating line integral we need to express the curve in parametric equation as
function of t, let recognize parametric representation first.
Bodies that move in space form paths that may be represented by curves C. This and other
applications show the need for parametric representations of C with parameter t, which may
denote time or something else (see Fig. 3.1). A typical parametric representation is given by
Figure 3.1: Parametric representation of curve
Here t is the parameter and x, y, z are Cartesian coordinates, that is, the usual rectangular
coordinates. To each value 𝑡 = 𝑡𝑜 , there corresponds a point of C with position vector 𝑟(𝑡𝑜 )
whose coordinates are 𝑥(𝑡𝑜 ), 𝑦(𝑡𝑜 ), 𝑧(𝑡𝑜 ).
When we give parametric equations and a parameter interval for a curve, we say that we have
parametrized the curve. The equations and interval together constitute a parametrization of
the curve. A given curve can be represented by different sets of parametric equations.
The advantages of using parametric representation are that, the coordinates x, y, z all play an
equal role, that is, all three coordinates are dependent variables. Moreover, the parametric
representation induces an orientation on C. This means that as we increase t, we travel along
50
the curve C in a certain direction. The sense of increasing t is called the positive sense on C.
The sense of decreasing t is then called the negative sense on C.
Table 3.1: Parametric Equation for Some Basic Curves
EXAMPLE 3.1: Circle. Parametric Representation. Positive Sense
51
3.2 BASIC CONCEPTS
In a line integral, we shall integrate a given function, also called the integrand, along a curve C in
space or in the plane as shown in Figure 3.2. (Hence curve integral would be a better name but line
integral is standard.)
Figure 3.2: Oriented curves
This requires that we represent the curve C by a parametric representation
(1)
52
The curve C is called the path of integration. Look at Fig. 2.1a. The path of integration goes from A to
B. Thus A: r(a) is its initial point and B: r(b) is its terminal point. C is now oriented. The direction from
A to B, in which t increases is called the positive direction on C. We mark it by an arrow. The points A
and B may coincide, as it happens in Fig. 2.1b. Then C is called a closed path.
A line integral of a vector function F(r) over a curve C: r(t) is defined by
(2)
Where 𝑟 ′ =
𝑑𝑟
𝑑𝑡
and r(t) is the parametric representation of C as given in (2). Writing (2) in terms of
components, with dr = [dx, dy, dz] and ′ = 𝑑⁄𝑑𝑡 , we get
Note that the integrand in (2) is a scalar, not a vector, because we take the dot product. Indeed,
𝐹● 𝑟 ′ ⁄|𝑟 ′ | is the tangential component of F. Line integrals arises naturally in mechanics, where they
give the work done by a force F in a displacement along C. This will be explained in detail below. We
may thus call the line integral (2) the work integral.
53
EXAMPLE 3.2: Evaluation of a Line Integral in the Plane
Find the value of line integral when 𝐹(𝑟) = [−𝑦, −𝑥𝑦] = −𝑦𝒊 − 𝑥𝑦𝒋 and C is the circular arc
from A to B.
EXAMPLE 3.3: Line Integral in Space
The evaluation of line integral in space is practically the same as it is in the plane. To see this,
find the value of line integral when 𝐹(𝑟) = [𝑧, 𝑥, 𝑦] = 𝑧𝒊 + 𝑥𝒋 + 𝑦𝒌 and C is the helix. Given
that
𝑟(𝑡) = [cos 𝑡, sin 𝑡, 3𝑡] = cos 𝑡𝒊 + sin 𝑡𝒋 + 3𝑡𝒌
54
EXAMPLE 3.4: Evaluation of a Line Integral along the Curve, C
Evaluate ∫c F ∙ dr, where F (x, y, z) = zi + xy j – y2k along the curve C given by 𝑟(𝑡) = 𝑡 2 𝒊 + 𝑡𝒋 +
√𝑡𝒌, 0 ≤ 𝑡 ≤ 1.
𝑺𝑶𝑳 =
𝟏𝟕
𝟐𝟎
Simple general properties of the line integral (2)
55
Figure 3.3 Formula c)
where in (c) the path C is subdivided into two arcs and that have the same orientation as C (Fig. 3.3).
In (b) the orientation of C is the same in all three integrals. If the sense of integration along C is
reversed, the value of the integral is multiplied by -1.
3.3 LINE INTEGRAL: WORK DONE BY FORCE
Suppose that the vector field F = M(x, y, z)i + N(x, y, z)y + P(x, y, z)k represents a force throughout a
region in space (it might be the force of gravity or an electromagnetic force of some kind) and that
𝑟(𝑡) = 𝑔(𝑡)𝒊 + ℎ(𝑡)𝒌 + 𝑘(𝑡)𝒌,
𝑎 ≪ 𝑡 ≪ 𝑏,
is a smooth curve in the region. For a curve C in space, we define the work done by a continuous
force field F to move an object along C from a point A to another point B as follows.
In other words, the work done by a force F is the line integral of the scalar component F.T over the
smooth curve from A to B as shown in Fig 2.2. The sign of the number we calculate with this integral
depends on the direction in which the curve is traversed. If we reverse the direction of motion, then
we reverse the direction of T in Figure 2.2 and change the sign of and its integral.
56
Figure 3.4: The work done by a force F is the line
EXAMPLE 3.5
57
EXAMPLE 3.6
Find the work done by the force fields F = xi + yj +zk in moving an object along the curve C
parameterized by 𝑟(𝑡) = cos(𝜋t) i + 𝑡 2 𝑘 + sin(𝜋𝑡) 𝑘, 0 ≤ 𝑡 ≤ 1
58
Path Dependence
Take, for instance, the straight segment 𝐶1 ∶ 𝑟1 (𝑡) = [𝑡, 𝑡, 0] and the parabola 𝐶2 ∶ 𝑟2 (𝑡) =
[𝑡, 𝑡 2 , 0] as shown in Figure 3.4 with 0 ≤ t ≤ 1 and integrate F = [0, 𝑥𝑦, 0]. Then
𝐹(𝑟1 (𝑡))●𝑟1′ (𝑡) = 𝑡 2 , 𝐹(𝑟2 (𝑡))●𝑟2′ (𝑡) = 2𝑡 4
so that integration gives 1/3 and 2/5, respectively.
Figure 3.4: Proof of Theorem
3.4 GREEN’S THEOREM IN THE PLANE
Green's theorem gives the relationship between a line integral around a simple closed curve C and a
double integral over the plane region D bounded by C. It will convert line integrals of vector fields over
C into double integral over R.
59
EXAMPLE 3.7: Verification of Green Theorem
Verify Green’s Theorem for the vector field
𝐹(𝑥, 𝑦) = (𝑥 − 𝑦)𝒊 + 𝑥𝒋
And the region R bounded by the unit circle
𝐶: 𝑟(𝑡) = cos(𝑡) 𝒊 + sin(𝑡) 𝒋,
0 ≤ 𝑡 ≤ 2𝜋
Figure 3.5: The vector field has a counter clockwise circulation of 2π around the unit circle.
60
EXAMPLE 3.8
Evaluate the line integral
∮ 𝑥𝑦 𝑑𝑦 − 𝑦 2 𝑑𝑥,
Where C is the square cut from the first quadrant by the lines x = 1 and y = 1
Use Green’s theorem to compute ∮ xy dx + xy dy, over the counter clockwise rectangle with corners
(1, 1), (3, 1), (3, 2), (1, 2). The double integral is
3
2
3
∫ ∫ 𝑦 − 𝑥 𝑑𝑦𝑑𝑥 = ∫
1
1
3
2
1
3
8
− 𝑥 𝑑𝑥 = 3 − = −1
2
2
3
We could also note that ∫1 ∫1 𝑦 − 𝑥 𝑑𝑦𝑑𝑥 = 𝐴𝑟𝑒𝑎 . (𝑦 − 𝑥) = 2 . (2 − 2) = −1
We could check our answer by computing the line integral directly. We get
3
2
3
2
∮ xy dx + xy dy = ∫ 𝑡 𝑑𝑡 + ∫ 3𝑡 𝑑𝑡 − ∫ 𝑡. 2 𝑑𝑡 − ∫ 𝑡 𝑑𝑡 =
1
1
1
1
8 9 16 3
+ −
− = −1.
2 2
2
2
The double integral is a little easier than the line integral, because the region has so many sides. This
is one use of Green’s Theorem that is to simplify line integrals.
61
SURFACE INTEGRALS
WEEK 4: SURFACE INTEGRALS
4.1 SURFACES FOR SURFACE INTEGRALS
The idea that will lead to the concept of a surface integral is quite similar to that which led to a line
integral in week 2. We say briefly “surface” also for a portion of a surface, just as we said “curve” for
an arc of a curve, for simplicity. Representations of a surface S in xyz-space are
z  f x , y 
or
gx , y , z   0
(1)
For example, z   a2  x 2  y 2 or x  y  z  a  0 (z≥ 0) represents hemisphere of radius a
2
2
2
2
and centre 0.
Now for curves C in line integrals, it was more practical and gave greater flexibility to use a parametric
representation r = r(t) where a ≤ t ≤ b. This is a mapping of the interval a ≤ t ≤ b located on the t-axis,
onto the curve C in the xyz-space. It maps every t in that interval onto the point of C with position
vector r(t) (see Figure 4.1)
Figure 4.1 Parametric representations of a curve and a surface
Similarly, for surfaces S in surface integrals, it will often be more practical to use a parametric
representation. Surfaces are two-dimensional. Hence we need two parameters, which we call u and v.
Thus a parametric representation of a surface S in space is of the form
r u, v   x u, v , y u, v , z u, v   x u, v  i  y u, v  j  z u, v  k
(2)
when (u,v) varies in some region R of the uv-plane. This mapping (2) maps every point (u,v) in R onto
the point of S with position vector r(u,v) (see Figure 1 (B)).
62
Parametric Representation of a Cylinder
The circular cylinder x  y  a , -1 ≤ z ≤ 1, has radius a, height 2, and the z-axis as axis. A parametric
representation is
2
2
2
r u, v   a cos u, a sinu, v   a cos u i  a sinu j  v k
The components of r arex=a cos u, y = a sin u, z = v. The parameters u, v vary in the rectangle R: 0 ≤ u
≤ 2π, -1 ≤ v ≤ 1 in the uv-plane. The curves u = const are vertical straight lines. The curves v = const
are parallel circles. The point P in Figure 2 corresponds to u = π/3=60°, v=0.7.
Figure 4.2 Parametric representation of a cylinder
Parametric Representation of a Sphere
A sphere x  y  z  a can be represented in the form
2
2
2
2
r u,v   a cos v cos u i  a cos v sinu j  a sin v k
(3)
where the parameters u, v vary in the rectangle R in the uv-plane given by the inequalities 0 ≤ u ≤ 2π,
-π/2 ≤ v ≤ π/2. The components of r are
x  a cos v cos u ,
y  a cos v sinu ,
z  a sin v
The curves u = const and v = const are the “meridian” and “parallels” on S (as shown in Figure 4.3).
This representation is used in geography for measuring the latitude and longitude of points on the
globe. Another parametric representation of the sphere also used in mathematics is
r u, v   a cos u sin v i  a sinu sin v j  a cos v k
(3*)
where 0 ≤ u ≤ 2π, 0 ≤ u ≤ π
63
Figure 4.3 Parametric representation of a sphere
Parametric Representation of a Cone
A circular cone z  x 2  y 2 , 0 ≤ t ≤ H can be represented by
r u, v   u cos v , u sinv , u  u cos v i  u sin v j  u k
(4)
in components x=u cos v, y = u sin v, z = u. The parameters vary in the rectangle R: 0 ≤ u ≤ H, 0 ≤ v ≤
2π
4.2 SURFACE INTEGRALS
The extension of the idea of an integral to line and double integrals are not the only generalization
that can be made. We can also extend the idea to integration over a general surface, S. Two types of
such integrals occur:
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆
c
𝑆
̂ 𝑑𝑆 = ∬ 𝑭(𝒓). 𝑑𝑺
∬ 𝑭(𝒓). 𝒏
𝑆
c
Scalar Field
(5)
Vector Field
(6)
𝑆
̂ 𝑑𝑆 is the vector element of area, where 𝒏
̂ is the unit outward-drawn normal vector
Note that 𝑑𝑺 = 𝒏
to the element dS.
64
4.2.1 SURFACE INTEGRALS OF VECTOR FIELD
In general, the surface S can be described in terms of two parameters, u and v say, so that on S
r u, v   x u, v , y u, v , z u, v   x u, v  i  y u, v  j  z u, v  k
where (u,v) varies over a region R in the uv-plane. We assume S to be piecewise smooth , so that S has
a normal vector
N  ru  rv
and unit normal vector
n
1
N
N
at every point (except perhaps for some edges or cusps, as for a cube or cone). For a given vector
function F we can now define the surface integral over S by
 F. n dA   Fru, v . N u, v  du dv
S
(7)
R
Here N  N n by equation (4), and N  ru  rv is the area of the parallelogram with sides ru and rv, by
the definition of cross product. Hence
n dA  n N du dv  N du dv
(7*)
and we see that dA  N du dv is the element of area of S.
also F .n is the normal component of F. This integral arises naturally in flow problems, where it gives
the flux across S when F = ρv. The flux acrossS is the mass of fluid crossing S per unit time. Furthermore,
ρ is the density of the fluid and v the velocity vector of the flow, as illustrated by Example 4.1 below.
We may thus call the surface integral (7) the flux integral.
We can write (7) in components, using F = [F1,F2, F3], N = [N1,N2,N3], and n = [cos α, cos β, cos ϒ]. Here
α, β, ϒ are the angles between n and the coordinate axes; indeed, for the angle between n and i.
F . n dA   F cos  F cos   F cos   dA
1
S
2
3
(8)
S
  F1N1  F2N2  F3N3  du dv
R
In (8) we can write cos αdA = dydz, cos β dA=dzdx, cosϒdA = dxdy.Then (8) becomes the following the
integral for the flux:
 F . n dA   F dydz  F dzdx  F dxdy 
1
S
2
3
(9)
S
We can use this formula to evaluate surface integrals by converting them to double integrals over
regions in the coordinate planes of the xyz-coordinate system. But we must carefully take into account
the orientation of S (the choice of n).
65
The tangent vectors of all the curves on a surface S through a point P of S form a plane, called the
tangent plane of S at P (refer to Figure 4.4). Exceptions are points where S has an edge or a cusp (like
a cone), so that S cannot have a tangent plane at such a point. Furthermore, a vector perpendicular
to the tangent plane is called a normal vector of S at P.
Figure 4.4 Tangent plane and normal vector
Now since S can be given by r = r (u,v) in (2), the new idea is that we get a curve C on S by taking a pair
of differentiable functions
u = u(t),
v = v(t)
whose derivatives u’ = du/dt and v’ = dv/dt are continuous. Then C has the position vector
~r (t)  r ut ,vt  . By differentiation and the use of the chain rule, we obtain a tangent vector of C on
S
~
~r ' (t)  d r  r u' r v'
dt u
v
Hence the partial derivatives ru and rv at P are tangential to S at P. We assume that they are linearly
independent, which geometrically means that the curves u = const and v= const on S intersect at P at
a nonzero angle. Then ru and rv span the tangent plane of S at P. hence their cross product gives a
normal vector N of S at P.
N  ru  rv  0
(10)
The corresponding unit normal vector n of S at P is (Figure 4)
n
1
1
N
ru  rv
N
ru  rv
(11)
Also if S is represented by g(x,y,z)=0, then
n
1
grad g
grad g
(11*)
A surface S is called a smooth surface if its surface normal depends continuously on the point of S.S is
called piecewise smooth if it consists of finitely many smooth portions.
66
Tangent Plane and Surface Normal
If a surface S is given by (2) with continuous ru 
r
r
and rv 
satisfying (4) at every point of S, then
u
v
S has, at every point P, a unique tangent plane passing through P and spanned by ru and rv and a
unique normal whose direction depends continuously on the points of S. A normal vector is given by
(4) and the corresponding unit normal vector by (5) (see Figure 4).
Example 4.1:
Compute the flux of water through the parabolic cylinder S : y  x , 0  x  2, 0  z  3 if the
2


velocity vector is v  F  3z , 6 , 6 xz , speed being measured in meters/sec. (Generally, F=ρv, but
2
3
water has density ρ = 1 g/cm = 1 ton/m3)
Solution:
Writing x = u and z = v, we have y  x  u . Hence a representation of S is
2
S:

r  u, u2 , v
2

0  x  2,
0  v  3
By differentiation and by the definition of the cross product,
N  ru  rv  1, 2u, 0 0, 0, 1  2u, - 1, 0
On S, writing simply F(S) for F[r(u,v)], we have F(S) = [ 3v2, 6, 6uv]. Hence F(s).N = 6uv2-6. By integration
we thus get from (6) the flux
 F.n
S

dA   6uv  6 dudv   3u v  6u
3 2
3
2
0 0
2 2
0


2

dv
u 0


  12v 2  12dv  4v 3  12v v 0  108  36  72 m3 / sec
3 2
3
0 0
67
Or 72,000 liters/sec. Note that the y-component of F is positive (equal to 6), so that in Figure above,
the flow goes from left to right.
Example 4.2:
Compute the flux through the surface S, of the cylinder 𝑥 2 + 𝑦 2 = 16 in the first octant between z=0
and z=5 if the velocity vector ∬𝑆 𝑭 . 𝑑𝑺 where 𝑭 = [𝑧, 𝑥, −3𝑦 2 𝑧].
Solution:
Writing x=u and z=v, we have 𝑦 = √16 − 𝑢2 (from 𝑥 2 + 𝑦 2 = 16)
𝒓(𝑢, 𝑣) = [𝑢, √16 − 𝑢2 , 𝑣]
𝜕𝑟
𝒓𝒖 = 𝜕𝑢 = [1, −
𝒓𝒗 =
𝜕𝑟
𝜕𝑣
𝑢
, 0]
√16−𝑢2
𝑖̃
∴ 𝑵 = 𝒓𝒖 × 𝒓𝒗 = |1
= [0, 0,1]
0
𝑘̃
𝑢
0| = [− √16−𝑢2 , −1,0]
𝑗̃
−
𝑢
√16−𝑢2
0
1
𝑭[𝒓(𝑢, 𝑣)] = [𝑣, 𝑢, −3(16 − 𝑢2 )𝑣]
−𝑢𝑣
𝑭[𝒓(𝑢, 𝑣)]. 𝑵 =
√16 − 𝑢2
5
0
∬ 𝑭. 𝑵𝑑𝐴 = ∫ ∫
0
𝑆
4
−𝑢
−𝑢𝑣
√16 − 𝑢2
− 𝑢 𝑑𝑢𝑑𝑣
Sustitute 𝑎 = 16 − 𝑢2
𝑑𝑎 = −2𝑢 𝑑𝑢 ∴ 𝑑𝑢 = −
5
4
1
𝑑𝑎
2𝑢
5
4
4
4
𝑣
1
∴ ∫ (𝑣 [∫
𝑑𝑢] − ∫ 𝑢 𝑑𝑢) = ∫ [ ∫
𝑑𝑢 − ∫ 𝑢 𝑑𝑢] 𝑑𝑣
2 √16 − 𝑢2
√16 − 𝑢2
0
−𝑢
0
0
5
= ∫ 𝑣 [√16 −
0
𝑢2 ]
5
0
0
0
4
𝑢2
𝑑𝑣 − ∫ [ ] 𝑑𝑣
2 0
0
4
0
68
5
5
= ∫0 −4𝑣 𝑑𝑣 − ∫0 8 𝑑𝑣 = - 90
4.2.2 SURFACE INTEGRAL OF SCALAR FIELD
The surface S can be specified by a scalar point function C(r)= c, where c is a constant. Curves may be
drawn on that surface, and in particular if we fix the value of one of the two parameters u and v then
we obtain two families of curves. On one, 𝐶𝑢 (𝒓(𝑢, 𝑣0 )) the value of u varies while v is fixed, and on
the other, , 𝐶𝑣 (𝒓(𝑢0 , 𝑣)), the values of v varies while u is fixed, as shown in Figure 4.5. Then as
indicated on Figure 5, the vector element of area dS is given by:
Figure 4.5 Parametric curves on a surface
𝑑𝑺 =
𝜕𝒓
𝜕𝒓
𝜕𝒓 𝜕𝒓
𝑑𝑢 ×
𝑑𝑣 =
×
𝑑𝑢𝑑𝑣
𝜕𝑢
𝜕𝑣
𝜕𝑢 𝜕𝑣
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕𝑥 𝜕𝑦 𝜕𝑧
= ( , , ) × ( , , ) 𝑑𝑢 𝑑𝑣 = (𝐽1 𝒊 + 𝐽2 𝒊 + 𝐽3 𝒊)𝑑𝑢 𝑑𝑣
𝜕𝑢 𝜕𝑢 𝜕𝑢
𝜕𝑣 𝜕𝑣 𝜕𝑣
where
𝜕𝑦 𝜕𝑧
𝜕𝑦 𝜕𝑧
𝜕𝑧 𝜕𝑥
𝐽1 = 𝜕𝑢 𝜕𝑣 − 𝜕𝑣 𝜕𝑢,
𝜕𝑧 𝜕𝑥
𝐽2 = 𝜕𝑢 𝜕𝑣 − 𝜕𝑣 𝜕𝑢,
𝜕𝑥 𝜕𝑦
𝜕𝑥 𝜕𝑦
𝐽3 = 𝜕𝑢 𝜕𝑣 − 𝜕𝑣 𝜕𝑢
(12)
Hence
∬ 𝑭(𝑟). 𝑑𝑺 = ∬(𝑃𝐽1 + 𝑄𝐽2 + 𝑅𝐽3 ) 𝑑𝑢𝑑𝑣
𝑆
𝐴
∬ 𝑓(𝑥, 𝑦, 𝑧). 𝑑𝑆 = ∬ 𝑓(𝑢, 𝑣)√(𝐽1 2 + 𝐽2 2 + 𝐽3 2 ) 𝑑𝑢𝑑𝑣
𝑆
𝐴
where F(r)=(P,Q,R) and A is the region of (u,v) plane corresponding to S. Here, of course, the terms in
the integrands have to be expressed in terms of u and v.
In particular, u and v can be chosen as any two of x, y and z. for example, if z=z(x,y) describes a surface
as in Figure 6 then
69
𝒓 = (𝑥, 𝑦, 𝑧(𝑥, 𝑦))
with x and y as independent variables. This gives
𝜕𝑧
𝐽1 = − 𝜕𝑥
𝜕𝑧
𝐽2 = − 𝜕𝑦
𝐽3 = 1
and
∬ 𝑭(𝑟). 𝑑𝑺 = ∬ (−𝑃
𝑆
𝐴
𝜕𝑧
𝜕𝑧
−𝑄
+ 𝑅) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
----------------(13)
𝜕𝑧 2
𝜕𝑧 2
∬ 𝑓(𝑥, 𝑦, 𝑧). 𝑑𝑆 = ∬ 𝑓(𝑥, 𝑦, 𝑧(𝑥, 𝑦))√(1 + ( ) + ( ) ) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝑆
𝐴
----------------(14)
Figure 4.6 A surface described by z=z(x,y)
Example 4.3:
Evaluate the surface integral
∬(𝑥 + 𝑦 + 𝑧) 𝑑𝑆
𝑆
where S is the portion of the sphere 𝑥 2 + 𝑦 2 +𝑧 2 = 1 that lies in the first quadrant.
70
(a) Surface S (b) quadrant of a circle in the (x,y) plane
Solution:
The surface S is illustrated in the above figure. Taking
𝑧 = √1 − 𝑥 2 − 𝑦 2
we have
𝜕𝑧
−𝑥
=
𝜕𝑥 √1 − 𝑥 2 − 𝑦 2
𝜕𝑧
−𝑦
=
𝜕𝑦 √1 − 𝑥 2 − 𝑦 2
𝜕𝑧 2
𝜕𝑧 2
𝑥 2 +𝑦2 +(1−𝑥 2 −𝑦2 )
(1−𝑥 2 −𝑦 2 )
giving √[1 + (𝜕𝑥) + (𝜕𝑦) ] = √
∴ ∬(𝑥 + 𝑡 + 𝑧)𝑑𝑆 = ∬[𝑥 + 𝑦 + (1 − 𝑥 2 − 𝑦 2 )]
𝑆
𝐴
=
1
√1−𝑥 2 −𝑦 2
1
√1 − 𝑥 2 − 𝑦 2
𝑑𝑥𝑑𝑦
where A is the quadrant of a circle in the (x,y) plane illustrated in Figure (b) (refer to the above Figure).
Thus,
1
√1−𝑥 2
∬(𝑥 + 𝑦 + 𝑧) 𝑑𝑆 = ∫ 𝑑𝑥 ∫
𝑆
0
0
[
𝑥
√1 − 𝑥 2 − 𝑦 2
+
𝑦
√1 − 𝑥 2 − 𝑦 2
+ 1] 𝑑𝑦
1
√1−𝑥 2
𝑦
= ∫ [𝑥 𝑠𝑖𝑛−1 (
) − √1 − 𝑥 2 − 𝑦 2 + 𝑦]
𝑑𝑥
√1 − 𝑥 2
0
0
1
1
𝜋
𝜋
3
= ∫ [ 𝑥 + 2√1 − 𝑥 2 ] 𝑑𝑥 = [ 𝑥 2 + 𝑥√1 − 𝑥 2 + 𝑠𝑖𝑛−1 𝑥] = 𝜋
2
4
4
0
0
An alternative approach to evaluating the surface integral in this example is to evaluate it directly over
the surface of the sphere using spherical polar coordinates. As illustrated in the Figure (c), on the
surface of a sphere of radius a we have,
𝑥 = 𝑎 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜙, 𝑦 = 𝑎𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜙, 𝑧 = 𝑎 𝑐𝑜𝑠𝜃,
𝑑𝑆 = 𝑎2 𝑠𝑖𝑛 𝜃𝑑𝜃𝑑𝜙
71
(c) Surface element in spherical polar coordinates
The radius a=1, so that
𝜋/2 𝜋/2
∬(𝑥 + 𝑦 + 𝑧)𝑑𝑆 = ∫ ∫ (𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜙 + 𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜙 + 𝑐𝑜𝑠𝜃)𝑠𝑖𝑛𝜃𝑑𝜃𝑑𝜙
𝑆
0
0
𝜋/2
1
1
1
3
= ∫ [ 𝜋𝑐𝑜𝑠𝜙 + 𝜋𝑠𝑖𝑛𝜙 + ] 𝑑𝜙 = 𝜋
4
4
2
4
0
In similar manner, when evaluating surface integral over the surface of a cylinder of radius a, we have
as illustrated
(d) Surface element in cylindrical polar coordinates
𝑥 = 𝑎 cos 𝜙 , 𝑦 = 𝑎 sin 𝜙,
𝑧 = 𝑧,
𝑑𝑆 = 𝑎𝑑𝑧𝑑𝜙
Example 4.4:
Calculate the surface integral ∬𝑆 (𝑥 + 𝑦 + 𝑧) 𝑑𝑆 where S is the portion of the plane 𝑥 + 2𝑦 + 4𝑧 =
4 lying in the first octant (x 0, y  0, z  0)
Solution:
Rewrite linear equation:
𝑧=
𝜕𝑧
𝜕𝑥
4 − 𝑥 − 2𝑦
𝑥 𝑦
=1− −
4
4 2
1
𝜕𝑧
𝜕𝑦
= −4
1
= −2
𝜕𝑧 2
𝜕𝑧 2
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ 𝑓(𝑥, 𝑦, 𝑧(𝑥, 𝑦))√1 + ( ) + ( ) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝑆
𝐴
72
𝑥 𝑦
1 2
1 2
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ (𝑥 + 𝑦 + 1 − − ) √1 + (− ) + (− ) 𝑑𝑥𝑑𝑦
4 2
4
2
𝑆
𝐴
3𝑥 𝑦
√21
= ∬ ( + + 1)
𝑑𝑥𝑑𝑦
4 2
4
2 4−2𝑦
2
4−2𝑦
3𝑥 𝑦
3𝑥 2
√21
√21
=
∫ ∫ ( + + 1) 𝑑𝑥𝑑𝑦 =
∫(
+ 2𝑦𝑥 + 4𝑥)
4
4 2
16
2
0
0
0
𝑑𝑦
0
2
2
√21
√21
=
∫ 3(16 − 16𝑦 + 4𝑦 2 )2 + 16𝑦 − 8𝑦 2 + 32 − 16𝑦 𝑑𝑦 =
∫(80 − 48𝑦 + 4𝑦 2 ) 𝑑𝑦
32
32
0
2
0
2
𝑦3
8
7√21
√21
√21
√21
=
∫ 20 − 12𝑦 + 𝑦 2 𝑑𝑦 =
[20𝑦 − 6𝑦 2 + ] =
(40 − 24 + ) =
4
4
3 0
4
3
3
0
Example 4.5:
Evaluate the surface integral ∬𝑆 (𝑧 2 ) 𝑑𝑆, where S is the total area of the cone √𝑥 2 + 𝑦 2 ≤ 𝑧 ≤ 2
S1 : surface of the cone
S2 : base of the cone
73
𝑆1 = ∬ 𝑧 2 𝑑𝑆1 = √2 ∬ 𝑥 2 + 𝑦 2 𝑑𝑥𝑑𝑦
𝑆1
Change to polar coordinate
2𝜋 2
= √2 ∫ ∫ 𝑟 2 𝑟𝑑𝑟𝑑𝜃 = 8√2 𝜋
0
0
S2 – base of the cone at z=2
dS2 = area of the base =r2=(22)=4
𝑆2 = ∬ 22 𝑑𝑆2 = 4 ∬ 𝑑𝑆2
𝑆2
𝑆2
= 4 ∬ 𝑑𝑆2 = 4(4𝜋) = 16𝜋
𝑆2
𝑇𝑜𝑡𝑎𝑙 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 = 𝑆1 + 𝑆2 = 8√2𝜋 + 16𝜋
4.3 ORIENTATION OF SURFACE
From (7) and (8), we see that the value of the integral depends on the choice of the unit normal vector
n. We express this by saying that such an integral is an integral over an oriented surface S, that is, over
a surface S on which we have chosen one of the two possible unit normal vectors in a continuous
fashion.
If we change the orientation of S, this means that we replace n with –n. Then each component of n
in (4) is multiplied by -1, so that we have
Change of Orientation in a Surface Integral
The replacement of n by –n (hence of N by –N) corresponds to the multiplication of the integral in (6)
and (7) by -1
Example 4.6:


r  v, v 2 , u , 0 ≤ v ≤ 2, 0 ≤ u ≤ 3. Then
In Example 4.1, we now represent S by ~
~
N  ~ru  ~rv  0,0,1 1,2v,0   2v,1,0


~


~
~
2
For F  3z ,6,6 xz we now get F( S )  3u ,6,6uv . Hence F( S ) . N  6u v  6 and integration
2
2
gives the old result times -1.
74
3 2
3
~
~
2
2
 FS  . N dv du    6u v  6 dv du   - 12u  12 du  72

R

0 0


0
Orientation of Smooth Surfaces
A smooth surface S is called orientable if the positive normal direction, when given at an arbitrary
point P0 of S, can be continued in a unique and continuous way to the entire surface. In many practical
applications, the surfaces are smooth and thus orientable
Orientation of Piecewise Smooth Surfaces
For a smooth orientable surface S with boundary curve C we may associate with each of the two
possible orientations of S an orientation of C, as shown in Figure 5(a). Then a piecewise smooth surface
is called orientable if we can orient each smooth piece of S so that along each curve C* which is
common boundary of two pieces S1 and S2 the positive direction of C* relative to S1 is opposite to the
direction of C* relative to S2. See Figure 5(b) for two adjacent pieces; note the arrows along C*.
Figure 5 Orientation of a Surface
75
4.3.3 SURFACE INTEGRALS WITHOUT REGARD TO ORIENTATION
Another type of surface integral is
 Gr  dA   Gr u, v Nu, v
S
(9)
du dv
R
Here dA = |N| du dv = |ru x rv| du dv is the element of area of the surface S represented by (2) and
we disregard the orientation.
The mean value theorem for surface integrals, which states that if R in (9) is simply connected and
G(r) is continuous in a domain containing R, then there is point (u0, v0) in R such that
 Gr  dA  Gr u , v A
o
 A  Area of S 
o
(10)
S
As for applications, if G(r) is the mass density of S, then (9) is the total mass of S. If G = 1, then (9)
gives the area A(S) of S,
AS    dA   ru  rv du dv
S
R
Representations z = f (x,y). If a surface S is given by z = f (x,y), then setting u = x , v = y, r = [u,v,f]
gives
N  ru  rv  1,
0,
fu  0,
1,
fu    fu , - fv ,
1  1  fu2  fv2
and, since fu=fx, fv = fy, formula (9) becomes
 Gr dA   Gx, y, f x, y 
S
R*
2
 f   f 
1       dxdy
 x   y 
2
(11)
Here R* is the projection of S into the xy-plane (Figure above) and the normal vector N on S points
up. If it points down, the integral on the right is preceded by a minus sign. From (11) with G = 1 we
obtain for the area A(S) of S: z = f (x.y) the formula
76
2
 f   f 
1       dxdy
 x   y 
2
AS   
R*
where R* is the projection of S into the xy-plane, as before.
Example: 4.7
x 2  y 2 over the region bounded by x  y  1
2
Find the area of the surface z 
2
Solution:
z = f(x,y)
2
 f   f 
1       dxdy
 x   y 
2
AS   
R*
f
f
 f   f 
So we now find
and
and determine 1       which is
y
x
 x   y 
2
2

f x, y   x 2  y 2
f 1 2
 x  y2
y 2


1/ 2

1 / 2

f 1 2
 x  y2
x 2


1 / 2
2x 
x
x  y2
2
y
2y 
x  y2
2
2
x2  y 2
 f   f 
 1        1  2
 2
x  y2
 x   y 
2
AS   
R*
2
 f   f 
1       dxdy  2  dxdy
 x   y 
R
2
But R is bounded by x  y  1 , i.e. a circle, centre the origin and radius 1. area  
2
2
AS   2  dxdy  2
R
Example 4.8:
Find the area of the surface S of the paraboloid z  x 2  y 2 cut off by the cone z  2 x 2  y 2
77
We can find the point of intersection A by considering the y-z plane i.e. put x=0. Coordinates of A are
A(2,4).
The projection of the surface S on the x-y plane is
x2  y2  4
AS   
R*
2
 f   f 
1       dxdy
 x   y 
2
For this we use the equation of the surface S. The
information from the projection R on the x-y plane will
later provide the limits of the two stages of integration.
For the time being, then
AS    1  4 x 2  4y 2 dxdy
R*
using Cartesian coordinates, we could integrate with respect to
y from y=0 to y  4  x and then with respect to x from x =0
2
to x=2. Finally we should multiply by four to cover all four
quadrants
i.e. AS   4
2
x 2 y  4  x
 
x 0
1  4 x 2  4 y 2 dydx
y 0
but how do we carry out the actual integration? It becomes a lot easier if we use polar coordinates.
The same integral in polar coordinates is
AS  
 2 r 2
 

1  4r 2 rdrd 
0 r 0
78
1/2
 2 r 2
AS     1  4r 
2
 0 r 0
2
2
3/2 
1
rdrd     1  4r 2   d
12
0
0
2
1
173 / 2  1d  5.7577 20  36.18


12 0
79
VOLUME INTEGRAL
WEEK 5: VOLUME INTEGRAL
5.1 INTRODUCTION
In the previous note, integral of a function is defined over a curved surface in three dimensions. This
concept can be extended to define the integral of a function of three variables through a region 𝑇 of
three-dimensional space by the limit.
5.2 DEFINITION
The general mathematical form of volume integral can be expressed as
𝑛
lim
∑ 𝑓(𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 )∆𝑉𝑖 = ∭ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉
𝑛→∞
all ∆𝑉𝑖 →0 𝑖=1
𝑇
where ∆𝑉𝑖 (𝑖 = 1, … , 𝑛) is a partition 𝑇 into 𝑛 elements of volume, and (𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 ) is a point in ∆𝑉𝑖 as
illustrated in Figure 5.1.
Figure 5.1: Partition of region 𝑇 into volume elements ∆𝑉𝑖
Figure 5.2: The volume integral in terms of rectangular Cartesian coordinates.
80
In terms of rectangular Cartesian coordinates the triple integral can, as illustrated in Figure 5.2, be
written as
𝑧=𝑎 𝑦=𝑔2 (𝑦) 𝑧=ℎ2 (𝑥,𝑦)
𝑉= ∫
∫
∫
(1)
𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑧=𝑏 𝑦=𝑔1 (𝑦) 𝑧=ℎ1 (𝑥,𝑦)
Note that there are six different orders in which the integration in (1) can be carried out.
5.3 CYLINDRICAL AND SPHERICAL COORDINATE
It is possible to use coordinate system other than Cartesian to obtain volume integral. Recall
cylindrical and spherical polar coordinates system which can be represented as
𝑧
𝑃(𝑟, 𝜙, 𝑧)
𝑧
𝜙
𝑟
𝑦
𝑥
Figure 5.3: Cylindrical coordinate system where 𝑥 = 𝑟 cos 𝜙, 𝑦 = 𝑟 sin 𝜙, 𝑧 = 𝑧.
𝑧
𝑃(𝑟 , 𝜃, 𝜙)
𝜃
𝑟
𝜙
𝑦
𝑥
Figure 5.4: Cylindrical coordinate system where 𝑥 = 𝑟 sin 𝜃 cos 𝜙, 𝑦 = 𝑟 sin 𝜃 sin 𝜙, 𝑧 = 𝑟 cos 𝜃.
To obtain the volume integral using other coordinate system, first we need to transform the integral
expression from the Cartesian coordinate system to the other system.
81
The expression for the element of volume 𝑑𝑉 = 𝑑𝑥 𝑑𝑦 𝑑𝑧 under the transformation
𝑥 = 𝑥(𝑢, 𝑣, 𝑤), 𝑦 = 𝑦(𝑢, 𝑣, 𝑤), 𝑧 = 𝑧 (𝑢, 𝑣, 𝑤) may be obtained using the Jacobian
𝜕𝑥
| 𝜕𝑢
𝜕(𝑥, 𝑦, 𝑧)
𝜕𝑥
𝐽=
=
𝜕(𝑢, 𝑣, 𝑤)
| 𝜕𝑣
𝜕𝑥
𝜕𝑤
𝜕𝑦
𝜕𝑢
𝜕𝑦
𝜕𝑣
𝜕𝑦
𝜕𝑤
𝜕𝑧
𝜕𝑢 |
𝜕𝑧
𝜕𝑣 |
𝜕𝑧
𝜕𝑤
as 𝑑𝑉 = 𝑑𝑥 𝑑𝑦 𝑑𝑧 = |𝐽| 𝑑𝑢 𝑑𝑣 𝑑𝑤
For example, in the case of cylindrical polar coordinates
𝑥 = 𝑟 cos 𝜙, 𝑦 = 𝑟 sin 𝜙, 𝑧 = 𝑧
cos 𝜙
𝐽 = |−𝑟 sin 𝜙
0
sin 𝜙
𝑟 cos 𝜙
0
0
0| = 𝑟
1
so that 𝑑𝑉 = 𝑟 𝑑𝑟 𝑑𝜙 𝑑𝑧
As illustrated in Figure 4.5.
Figure 5.5: Volume element in cylindrical polar coordinates.
Similarly, for spherical polar coordinates (𝑟, 𝜃, 𝜙)
𝑥 = 𝑟 sin 𝜃 cos 𝜙, 𝑦 = 𝑟 sin 𝜃 sin 𝜙, 𝑧 = 𝑟 cos 𝜃
sin 𝜃 cos 𝜙
𝑟
𝐽 = | cos 𝜃 cos 𝜙
−𝑟 sin 𝜃 sin 𝜙
sin 𝜃 sin 𝜙
𝑟 cos 𝜃 cos 𝜙
𝑟 sin 𝜃 cos 𝜙
cos 𝜃
−𝑟 sin 𝜃 | = 𝑟 2 sin 𝜃
0
so that 𝑑𝑉 = 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
A result illustrated in Figure 5.6.
82
Figure 5.6: Volume element in spherical polar coordinates.
5.4 APPLICATION OF VOLUME INTEGRAL
Volume or triple integral has important real world application which can be applied in various
scientific and engineering matters. In this section, we will discuss some of the application examples.
5.4.1 VOLUME
The most obvious usage of volume integral is to find the volume of a space confined within 3
dimensional boundary.
Example 5.1
Find the volume of the tetrahedron defined by 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 and 𝑥 + 𝑦 + 𝑧 ≤ 1.
Solution
Figure 5.7: Tetrahedron for Example 1.
The tetrahedron is shown in Figure 5.7. Its volume is
𝑥=1 𝑦=1−𝑥 𝑧=1−𝑥−𝑦
𝑉= ∫
∫
∫
𝑥=0 𝑦=0
𝑧=0
1 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑥=1 𝑦=1−𝑥
= ∫
∫ (1 − 𝑥 − 𝑦) 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0
83
𝑥=1
= ∫
𝑥=0
1
1
(1 − 𝑥)2 𝑑𝑥 = .
2
6
∎
5.4.2 MASS
A mass of a matter bounded by 𝑉 with desity of 𝜌(𝑥, 𝑦, 𝑧) at point (𝑥, 𝑦, 𝑧) can be calculated as 𝑀 =
∭ 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥 (or ∫𝑉 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑉 (with appropriate limits)).
Example 5.2
Figure 5.8
A water reservoir shown in Figure 4.8 has the width of 𝑥 = 100m, length of 𝑦 = 400m, and the
depth of the reservoir is given by 𝑧 = 40 − 𝑦/10m. The density of the water can be approximated
by 𝜌(𝑧) = 𝑎 − 𝑏 × 𝑧 where a = 998 kgm−3 and b = 0.05 kgm−4 i.e. at the surface (𝑧 = 0) the
water has density 998 kgm−3 (corresponding to a temperature of 20◦C while 40m down i.e. z = −40,
the water has a density of 1000kgm−3 (corresponding to the lower temperature of 4℃. Find the
total mass of water in the reservoir.
Solution
The mass of water in the reservoir is given by the integral of the function 𝜌(𝑧) = 𝑎 − 𝑏 × 𝑧. For
each value of 𝑥 and 𝑦, the limits on 𝑧 will be from 𝑦/10 − 40 (bottom) to 0 (top). Limits on 𝑦 will be
0 to 400m while the limits of 𝑥 will be 0 to 100m. The mass of water is therefore given by the
integral
100 400
𝑀= ∫
∫
𝑥=0 𝑦=0
0
∫ (𝑎 − 𝑏𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑦
𝑧= −40
10
This can be solved as follow
100 400
𝑀= ∫
𝑏 2 0
∫ [𝑎𝑧 − 𝑧 ] 𝑦
𝑑𝑦 𝑑𝑥
2
𝑧= −40
𝑥=0 𝑦=0
10
84
100 400
= ∫
2
𝑦
𝑏 𝑦
∫ 0 − 𝑎 ( − 40) + ( − 40) 𝑑𝑦 𝑑𝑥
10
2 10
𝑥=0 𝑦=0
100 400
= ∫
∫ 40𝑎 −
𝑥=0 𝑦=0
𝑎
𝑏 2
𝑦+
𝑦 − 4𝑏𝑦 + 800𝑏 𝑑𝑦 𝑑𝑥
10
200
100
400
𝑎 2
𝑏 3
2
= ∫ [40𝑎𝑦 − 𝑦 +
𝑦 − 2𝑏𝑦 + 800𝑏𝑦]
𝑑𝑥
10
600
𝑦=0
𝑥=0
100
= ∫ 16000𝑎 − 8000𝑎 +
𝑥=0
100
= ∫ 8000𝑎 +
𝑥=0
= 8 × 105 𝑎 +
320000
𝑏 − 320000𝑏 + 320000𝑏 𝑑𝑥
3
100
320000
320000
𝑏 𝑑𝑥 = [8000𝑎𝑥 +
𝑏𝑥]
3
3
𝑥=0
3.2
0.16
× 107 𝑏 = 7.984 × 108 +
× 107
3
3
= 7.989 × 108 kg
So the mass of water in the reservoir is 7.989 × 108 kg.
∎
5.4.3 CENTRE OF MASS
The expressions for the center of mass (𝑥̅ , 𝑦̅, 𝑧̅) of a solid of density 𝜌(𝑥, 𝑦, 𝑧) are given below
𝑥̅ =
∫ 𝜌(𝑥, 𝑦, 𝑧)𝑥 𝑑𝑉
∫ 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑉
𝑦̅ =
∫ 𝜌(𝑥, 𝑦, 𝑧)𝑦 𝑑𝑉
∫ 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑉
𝑧̅ =
∫ 𝜌(𝑥, 𝑦, 𝑧)𝑧 𝑑𝑉
∫ 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑉
If 𝜌 does not vary with position, these simplify to
𝑥̅ =
∫ 𝑥 𝑑𝑉
∫ 𝑑𝑉
𝑦̅ =
∫ 𝑦 𝑑𝑉
∫ 𝑑𝑉
𝑧̅ =
∫ 𝑧 𝑑𝑉
∫ 𝑑𝑉
Example 5.3
A uniform tetrahedron is enclosed by the planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 𝑥 + 𝑦 + 𝑧 = 4. Find
the volume and the position of the center of mass.
85
Solution
The volume integral of tetrahedron
4 4−𝑥 4−𝑥−𝑦
𝑉= ∫ ∫
∫
𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0 𝑧=0
4 4−𝑥
= ∫ ∫
4 4−𝑥
[𝑧]4−𝑥−𝑦
𝑑𝑦 𝑑𝑥
𝑧=0
= ∫ ∫ (4 − 𝑥 − 𝑦) 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0
𝑥=0 𝑦=0
4
1 2 4−𝑥
= ∫ [4𝑦 − 𝑥𝑦 − 𝑦 ]
𝑑𝑥
2
𝑦=0
𝑥=0
4
1
= ∫ [8 − 4𝑥 + 𝑥 2 ] 𝑑𝑥
2
𝑥=0
4
1
64 32
= [8𝑥 − 2𝑥 2 + 𝑥 3 ]
= 32 − 32 +
=
.
6
6
3
𝑥=0
Then, we calculate ∫ 𝑥 𝑑𝑉 as follows
4 4−𝑥 4−𝑥−𝑦
∫ 𝑥 𝑑𝑉 = ∫ ∫
∫
𝑥 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0 𝑧=0
4 4−𝑥
= ∫ ∫
4 4−𝑥
[𝑥𝑧]4−𝑥−𝑦
𝑑𝑦 𝑑𝑥
𝑧=0
𝑥=0 𝑦=0
= ∫ ∫ [(4 − 𝑥 − 𝑦) − 0] 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0
4
4−𝑥
1
= ∫ [4𝑥𝑦 − 𝑥 𝑦 − 𝑥𝑦 2 ]
𝑑𝑥
2
𝑦=0
2
𝑥=0
4
1
= ∫ [8𝑥 − 4𝑥 2 + 𝑥 3 ] 𝑑𝑥
2
𝑥=0
4
4
1
256
32
= [4𝑥 2 − 𝑥 3 + 𝑥 4 ] = 64 −
+ 32 =
3
8
3
3
0
Thus, 𝑥̅ =
∫ 𝑥𝑑𝑉
∫ 𝑑𝑉
32/3
= 32/3 = 1.
By symmetry (or by evaluating relevant integrals), it can be shown that 𝑦̅ = 𝑧̅ = 1 i.e. the center
of mass is at (1, 1, 1).
∎
5.4.5 MOMENT OF INERTIA
The moment of inertia 𝐼 of a small particle of mass 𝑚 is defined as
𝐼 = Mass × Distance2 or 𝐼 = 𝑚𝑑2
where 𝑑 is the perpendicular distance from the particle to the axis.
86
To find the Moment of Inertia of a larger object, it is necessary to carry out a volume integration
over all such particles. The distance of a particle at (𝑥, 𝑦, 𝑧) from the z-axis is given by √𝑥 2 + 𝑦 2 so
the moment of inertia of an object about the 𝑧-axis is given by
𝐼𝑧 = ∫ 𝜌(𝑥, 𝑦, 𝑧) (𝑥 2 + 𝑦 2 )𝑑𝑉
𝑉
Similarly, the Moments of Inertia about the 𝑥- and 𝑦-axes are given by
𝐼𝑥 = ∫ 𝜌(𝑥, 𝑦, 𝑧) (𝑧 2 + 𝑦 2 ) 𝑑𝑉
𝑉
𝐼𝑦 = ∫ 𝜌(𝑥, 𝑦, 𝑧) (𝑥 2 + 𝑧 2 ) 𝑑𝑉
𝑉
Example 5.4
Find the moment of inertia of a uniform sphere of mass M and radius a about a diameter.
Solution
A sphere of radius 𝑎 has volume 4𝜋𝑎3 /3, so that its density is 3𝑀/4𝜋𝑎3 . Then the moment of
inertia of the sphere about the 𝑧 axis is
𝐼=
3𝑀
∭(𝑥 2 + 𝑦 2 ) 𝑑𝑥 𝑑𝑦 𝑑𝑧
4𝜋𝑎3
𝑉
In this example it is natural to use spherical polar coordinates (recall that 𝑥 = 𝑟 sin 𝜃 cos 𝜙, 𝑦 =
𝑟 sin 𝜃 sin 𝜙, 𝑧 = 𝑟 cos 𝜃 and 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 ), so that
𝐼=
3𝑀
∭(𝑟 2 sin2 𝜃 cos2 𝜙 + 𝑟 2 sin2 𝜃 sin2 𝜙)𝑟 2 sin 𝜃 𝑑𝜙 𝑑𝜃 𝑑𝑟
4𝜋𝑎3
𝑉
=
3𝑀
∭(𝑟 2 sin2 𝜃) 𝑟 2 sin 𝜃 𝑑𝜙 𝑑𝜃 𝑑𝑟
4𝜋𝑎3
𝑉
𝑎
𝜋
2𝜋
3𝑀
=
∫ ∫ ∫ 𝑟 4 sin3 𝜃 𝑑𝜙 𝑑𝜃 𝑑𝑟
4𝜋𝑎3
𝑟=0 𝜃=0 𝜙=0
𝑎
𝜋
2𝜋
𝑟=0
𝜃=0
𝜙=0
3𝑀
=
∫ 𝑟 4 𝑑𝑟 ∫ sin3 𝜃 𝑑𝜃 ∫ 𝑑𝜙
4𝜋𝑎3
=
𝜋
3𝑀 1 5 𝑎
1
3
[𝜙]2𝜋
𝑟
cos
3𝜃
−
cos
𝜃]
[
]
[
𝜙=0
4𝜋𝑎3 5
4
𝑟=0 12
𝜃=0
87
=
3𝑀 8
( 𝜋𝑎5 )
4𝜋𝑎3 15
2
= 𝑀𝑎2 .
5
∎
5.5 ADDITIONAL EXAMPLES
Example 5.5
Use spherical polar coordinates to evaluate
∭ 𝑥(𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑉
where 𝑉 is the region in the first octant lying within the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1.
Solution
𝜋/2 𝜋/2
2
2
∭ 𝑥(𝑥 + 𝑦 + 𝑧
2)
𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫
𝑉
1
∫(𝑟 sin 𝜃 cos 𝜙)(𝑟 2 ) 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
∫
𝜙=0 𝜃=0 𝑟=0
𝜋/2
𝜋/2
1
= ∫ cos 𝜙 𝑑𝜙 ∫ sin2 𝜃 𝑑𝜃 ∫ 𝑟 5 𝑑𝑟
𝜙=0
𝜃=0
𝑟=0
𝜋
1
2
1
𝑟6
[sin
(θ
=
𝜙]𝜙=0 [
+ sin 𝜙 cos 𝜙)]
[ ]
2
𝜃=0 6 𝑟=0
𝜋
2
1 𝜋
1
𝜋
=1×( ⋅ )×( ) =
.
2 2
6
24
∎
Example 5.6
Show that
1
𝑧
∫ ∫
𝑥+𝑧
∫ (𝑥 + 𝑦 + 𝑧) 𝑑𝑦 𝑑𝑥 𝑑𝑧 = 0
𝑧=−1 𝑥=0 𝑦=𝑥−𝑧
Solution
1
𝑧
∫ ∫
𝑥+𝑧
1
𝑧
∫ (𝑥 + 𝑦 + 𝑧) 𝑑𝑦 𝑑𝑥 𝑑𝑧 = ∫ ∫ [𝑥𝑦 +
𝑧=−1 𝑥=0 𝑦=𝑥−𝑧
𝑧=−1 𝑥=0
1
𝑥+𝑧
𝑦2
+ 𝑧𝑦]
𝑑𝑥 𝑑𝑧
2
𝑦=𝑥−𝑧
𝑧
= ∫ ∫ 2𝑧(𝑥 + 𝑧) + 2𝑥𝑧 𝑑𝑥 𝑑𝑧
𝑧=−1 𝑥=0
88
1
𝑧
3𝑥 2 𝑧
= ∫ [
+ 2𝑧 2 𝑥]
𝑑𝑧
2
𝑥=0
𝑧=−1
1
= ∫ 3𝑧 3 𝑑𝑧 = 0.
∎
𝑧=−1
89
GAUSS’S DIVERGENCE THEOREM
WEEK 6: GAUSS’ DIVERGENCE THEOREM
6.1 INTRODUCTION
In the same way that Green’s theorem relates surface and line integrals, Gauss’s theorem relates
surface and volume integrals. Among the application of divergence theorem is in electromagnetism,
gravity, fluid dynamics and quantum mechanics.
6.2 DIVERGENCE
Before we go into divergence theorem, first we need to recall the definition of divergence. A way of
defining the divergence of a vector field 𝑭(𝒓) at the point 𝒓 is to enclose the point in an elementary
volume 𝛥𝑉 and find the flow or flux out of 𝛥𝑉 per unit volume. Thus
flow out of ∆𝑉
∆𝑉→0
∆𝑉
div 𝑭 = ∇ ⋅ 𝑭 = lim
𝛥𝑉
Figure 6.1: Flow of fluid or vector field towards and away from volume 𝛥𝑉
A non-zero divergence at a point in a fluid measures the rate, per unit volume, at which the fluid is
flowing away from or towards that point. That implies that either the density of the fluid is changing
at the point or there is a source or sink of fluid there
In the case of a non-material vector field, for example temperature gradient in heat transfer, a
nonzero divergence indicates a point of generation or absorption.
When the divergence is everywhere zero, the flow entering any element of the space is exactly
balanced by the outflow. This implies that the lines of flow of the field 𝑭(𝒓) where 𝑑iv 𝑭 = 0 must
either form closed curves or finish at boundaries or extend to infinity. Vectors satisfying this
condition are sometimes termed solenoidal.
90
We can summarize that:
i.
ii.
Positive divergence = Source
Negative divergence = Sink
Divergence of a vector field expression 𝒗 can be represented as:
div 𝒗 = 𝜵. 𝒗
= (𝒊 ·
=
𝜕
𝜕
𝜕
+𝒋·
+ 𝒌 · ) . (𝒊 · 𝑣1 + 𝒋 · 𝑣2 + 𝒌 · 𝑣3 )
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑣1 𝜕𝑣2 𝜕𝑣3
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
where 𝑣1 , 𝑣2 , 𝑣3 is 𝑥, 𝑦, 𝑧 component of vector 𝒗, respectively.
6.3 DIVERGENCE THEOREM
Consider the closed volume 𝑉 with surface area 𝑆 shown in Figure 6.1. The surface integral ∬𝑆 𝑭 ⋅ 𝑑𝑺
may be interpreted as the flow of a liquid with velocity field 𝑭(𝒓) out of the volume 𝑉. Previously, we
discussed that the divergence of 𝑭 could be expressed as
div 𝑭 = 𝛻 ⋅ 𝑭
flow out of ∆𝑉
∆𝑉→0
∆𝑉
= lim
In terms of differentials, this may be written as
div 𝑭 𝑑𝑉 = flow out of ∆𝑉
Consider now a partition of the volume 𝑉 given by 𝛥𝑉𝑖 (𝑖 = 1, . . . , 𝑛). Then the total flow out of
𝑉 is the sum of the flows out of each 𝛥𝑉𝑖 . That is,
𝑉
∯ 𝑭 ⋅ 𝑑𝑺 = lim ∑(flow out of ∆𝑉𝑖 )
𝑛→∞
𝑆
𝑖=1
𝑉
= lim ∑(div 𝑭 ∆𝑉𝑖 )
𝑛→∞
𝑖=1
giving
∯ 𝑭 ⋅ 𝑑𝑺 = ∭ div 𝑭 𝑑𝑉
𝑆
𝑉
This result is known as the divergence theorem or Gauss’s theorem. It enables us to convert surface
integrals into volume integrals, and often simplifies their evaluation.
91
6.4 EXAMPLES
Example 6.1
Verify the divergence theorem on the unit cube 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1 for the following
vector fields.
𝐹 = (𝑥, 𝑦, 𝑧) = 𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌
Solution 6.1
𝒏6 = 𝒌
𝒏1 = −𝒊
𝒏4 = 𝒋
𝒏3 = −𝒋
𝒏2 = 𝒊
𝒏5 = −𝒌
Figure 5.2: Cube example
Figure 5.2 illustrate the surface of the cube. To find the results using ∯𝑆 𝑭 ⋅ 𝑑𝑺, surface integral
need to be done in each of the cube surface. Since cube has 6 surfaces, the equation can be
expanded as follows
∯ 𝑭 ⋅ 𝑑𝑺 = ∬ 𝑭 ⋅ 𝒏1 𝑑𝑆1 + ∬ 𝑭 ⋅ 𝒏2 𝑑𝑆2 + ∬ 𝑭 ⋅ 𝒏3 𝑑𝑆3 + ∬ 𝑭 ⋅ 𝒏4 𝑑𝑆4 + ∬ 𝑭 ⋅ 𝒏5 𝑑𝑆5 + ∬ 𝑭 ⋅ 𝒏6 𝑑𝑆6
𝑆
𝑆1
𝑆2
𝑆3
𝑆4
𝑆5
𝑆6
Then, we solve the integral for each surface:
i.
For surface 𝑆1 , the surface boundary is 𝑥 = 0, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1, and outgoing
1
1
normal to the surface 𝑆1 is 𝒏1 =– 𝒊. Thus, ∬𝑆 𝑭 ⋅ 𝒏1 𝑑𝑆1 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅
1
1
ii.
1
−𝒊 𝑑𝑦 𝑑𝑧 = ∫0 ∫0 0 𝑑𝑦 𝑑𝑧 = 0
For surface 𝑆2 , the surface boundary is 𝑥 = 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1, and outgoing normal
1
1
to the surface 𝑆2 is 𝒏2 = 𝒊. Thus, ∬𝑆 𝑭 ⋅ 𝒏2 𝑑𝑆2 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅ 𝒊 𝑑𝑦 𝑑𝑧 =
1
1
iii.
1
∫0 ∫0 1 𝑑𝑦 𝑑𝑧 = 1.
For surface 𝑆3 , the surface boundary is 0 ≤ 𝑥 ≤ 1, 𝑦 = 0, 0 ≤ 𝑧 ≤ 1, and outgoing normal
1
1
to the surface 𝑆3 is 𝒏3 = −𝒋. Thus, ∬𝑆 𝑭 ⋅ 𝒏3 𝑑𝑆3 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅ −𝒋 𝑑𝑥 𝑑𝑧 =
3
1 1
∫0 ∫0 0
𝑑𝑥 𝑑𝑧 = 0.
92
iv.
For surface 𝑆4 , the surface boundary is 0 ≤ 𝑥 ≤ 1, 𝑦 = 1, 0 ≤ 𝑧 ≤ 1, and outgoing normal
1
1
to the surface 𝑆4 is 𝒏4 = 𝒋. Thus, ∬𝑆 𝑭 ⋅ 𝒏4 𝑑𝑆4 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅ 𝒋 𝑑𝑥 𝑑𝑧 =
4
1
v.
1
∫0 ∫0 𝑥 𝑑𝑥 𝑑𝑧 = 1/2.
For surface 𝑆5 , the surface boundary 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 0, and outgoing normal
1
1
to the surface 𝑆5 is 𝒏5 =– 𝒌. Thus, ∬𝑆 𝑭 ⋅ 𝒏5 𝑑𝑆5 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅ −𝒌 𝑑𝑦 𝑑𝑧 =
5
1
vi.
1
∫0 ∫0 0 𝑑𝑥 𝑑𝑦 = 0
For surface 𝑆6 , the surface boundary 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 1, and outgoing normal to
1
1
the surface 𝑆6 is 𝒏6 = 𝒌. Thus, ∬𝑆 𝑭 ⋅ 𝒏6 𝑑𝑆6 = ∫0 ∫0 (𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) ⋅ 𝒌 𝑑𝑥 𝑑𝑦 =
6
1
1
1
∫0 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦 = 4
Therefore,
1
1 7
∯ 𝑭 ⋅ 𝑑𝑺 = 0 + 1 + 0 + + 0 + =
2
4 4
𝑆
Next, we solve the surface integral using ∭𝑉 div 𝑭 𝑑𝑉. This method is less tedious than using
∯𝑆 𝑭 ⋅ 𝑑𝑺 because by using the volume integration, we do not have to perform the integration for
every surface, which is shown as follow
1 1 1
∭ div 𝑭 𝑑𝑉 = ∫ ∫ ∫ div(𝑥𝒊 + 𝑥𝑦𝒋 + 𝑥𝑦𝑧𝒌) 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑉
0 0 0
1 1 1
= ∫ ∫ ∫ 1 + 𝑥 + 𝑥𝑦 𝑑𝑧 𝑑𝑦 𝑑𝑥
0 0 0
1 1
1 1
= ∫ ∫[𝑧 + 𝑥𝑧 +
𝑥𝑦𝑧]1𝑧=0 𝑑𝑦
𝑑𝑥 = ∫ ∫ 1 + 𝑥 + 𝑥𝑦 𝑑𝑦 𝑑𝑥
0 0
0 0
1
1
1
𝑥𝑦 2
𝑥
= ∫ [𝑦 + 𝑥𝑦 +
]
𝑑𝑥 = ∫ 1 + 𝑥 + 𝑑𝑥
2 𝑦=0
2
0
0
1
𝑥2 𝑥2
7
= [𝑥 + + ]
= .
2
4 𝑥=0 4
This result confirms that in this example, ∯𝑆 𝑭 ⋅ 𝑑𝑺 = ∭𝑉 div 𝑭 𝑑𝑉 .
∎
Example 6.2
A vector field 𝑭(𝒓) is given by
93
𝑭(𝒓) = 𝑥 3 𝑦𝒊 + 𝑥 2 𝑦 2 𝒋 + 𝑥 2 𝑦𝑧𝒌
Find ∬𝑆 𝑭 ⋅ 𝑑𝑺, where S is the surface of the region in the first octant for which 𝑥 + 𝑦 + 𝑧 ≤ 1.
Solution 6.2
Figure 5.3: Tetrahedron for Example 1
We begin by sketching the region enclosed by the tetrahedron surface, as shown in Figure 5.3. It is
clear that evaluating the surface integral directly will be rather clumsy, involving four separate
integrals (one over each of the four surfaces).
It is simpler in this case to transform it into a volume integral using the divergence theorem,
∯𝑆 𝑭 ⋅ 𝑑𝑺 = ∭𝑉 div 𝑭 𝑑𝑉, where
div 𝑭 = 3𝑥 2 𝑦 + 2𝑥 2 𝑦 + 𝑥 2 𝑦 = 6𝑥 2 𝑦
and we obtain
1 1−𝑥 1−𝑥−𝑦
∯ 𝑭 ⋅ 𝑑𝑺 = ∫ ∫
∫
𝑆
0
0 0
6𝑥 2 𝑦 𝑑𝑧 𝑑𝑦 𝑑𝑥
1 1−𝑥
1−𝑥−𝑦
= ∫ ∫ [6𝑥 2 𝑦𝑧]𝑧=0 𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0
1 1−𝑥
= ∫ ∫ 6𝑥 2 𝑦(1 − 𝑥 − 𝑦)𝑑𝑦 𝑑𝑥
𝑥=0 𝑦=0
1
= ∫ [𝑥 2 𝑦 2 (3 − 3𝑥 − 2𝑦)]1−𝑥
𝑦=0 𝑑𝑥
𝑥=0
1
= ∫ 𝑥 2 − 3 𝑥 3 + 3𝑥 4 − 𝑥 5 𝑑𝑥
0
94
1
𝑥3 3 𝑥4 3 𝑥5 𝑥6
1
=[ −
+
− ]
=
.∎
3
4
5
6 𝑥=0 60
Example 6.3
Verify the divergence theorem
∯ 𝑭 ⋅ 𝑑𝑺 = ∭ div 𝑭 𝑑𝑉
𝑆
𝑉
when 𝑭 = 2𝑥𝑧𝒊 + 𝑦𝑧𝒋 + 𝑧 2 𝒌 and 𝑉 is the volume enclosed by the upper hemisphere 𝑥 2 + 𝑦 2 +
𝑧 2 = 𝑎2 , 𝑧 ≥ 0. In this example, it is convenient to use spherical coordinate to solve the integral.
Solution 6.3
In this example we are required to proof that divergence theorem is valid in the given example.
Hence, we need to obtain the result using both ∯𝑆 𝑭 ⋅ 𝑑𝑺 and ∭𝑉 div 𝑭 𝑑𝑉.
Figure 5.4: Hemisphere example.
The enclosed hemisphere surface that are given in the example can be illustrated as in Figure 5.4.
Note that since the theorem relates to a closed volume, the surface 𝑆 consists of the flat circular
base in the (𝑥, 𝑦) plane as well as the hemispherical surface. In this case
div 𝑭 = 2𝑧 + 𝑧 + 2𝑧
= 5𝑧
so that the volume integral is readily evaluated as
𝑎
∭ 5𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫ 5𝑧𝜋𝑟 2 𝑑𝑧
𝑉
0
𝑎
= ∫ 5𝜋𝑧(𝑎2 − 𝑧 2 ) 𝑑𝑧
0
5
= 𝜋𝑎4
4
95
Next, we obtain the result using the surface integral, as follows
∯ 𝑭 ⋅ 𝑑𝑺 =
𝑆
∬
𝑭 ⋅ 𝒏1 𝑑𝑆 +
circular base
∬
𝑭 ⋅ 𝒏2 𝑑𝑆
hemisphere
The unit normal to the base is clearly 𝒏1 = −𝒌, so
𝑭 ⋅ 𝒏1 = −𝑧 2
giving
∬
𝑭 ⋅ 𝒏1 𝑑𝑆 = 0
circular base
since 𝑧 = 0 on this surface.
The hemispherical surface is given by
𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 − 𝑎2
=0
so the outward unit normal 𝒏2 is
𝒏2 =
=
∇𝑓
|∇𝑓|
2𝑥𝒊 + 2𝑦𝒋 + 2𝑧𝒌
2√𝑥 2 + 𝑦 2 + 𝑧 2
Since 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 on the surface
𝒏2 =
𝑥
𝑦
𝑧
𝒊+ 𝒋+ 𝒌
𝑎
𝑎
𝑎
Giving
𝑭 ⋅ 𝒏2 =
2𝑥 2 𝑧 𝑦 2 𝑧 𝑧 3
+
+
𝑎
𝑎
𝑎
=
𝑥2𝑧 𝑧 2
+ (𝑥 + 𝑦 2 + 𝑧 2 )
𝑎
𝑎
Hence
∬
hemisphere
𝑭 ⋅ 𝒏2 𝑑𝑆 =
∬
hemisphere
𝑧 2
(𝑥 + 𝑎2 ) 𝑑𝑆
𝑎
since 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 on the surface. Transforming to spherical polar coordinates,
96
𝑥 = 𝑎 sin 𝜃 cos 𝜙, 𝑧 = 𝑎 cos 𝜃, 𝑑𝑆 = 𝑎2 sin 𝜃 𝑑𝜃 𝑑𝜙, the surface integral becomes
2𝜋 𝜋/2
4
𝑭 ⋅ 𝒏2 𝑑𝑆 = 𝑎 ∫ ∫ (sin 𝜃 cos 𝜙 + sin3 𝜃 cos 𝜃 cos2 𝜙)𝑑𝜃 𝑑𝜙
∬
hemisphere
0
0
2𝜋
𝜋/2
1 2
1 4
2
= 𝑎 ∫ [ sin 𝜃 + sin 𝜃 cos 𝜙]
𝑑𝜙
2
4
0
4
0
2𝜋
1 1
5
= 𝑎 ∫ [ + cos2 𝜙] 𝑑𝜙 = 𝜋𝑎4
2 4
4
4
0
Thus confirming that in this example,
∯ 𝑭 ⋅ 𝑑𝑺 = ∭ div 𝑭 𝑑𝑉 .
𝑆
∎
𝑉
Example 6.4
A field 𝑭 = 𝑥 2 𝒊 + 𝑦 2 𝒋 + 𝑧 2 𝒌 is enclosed within a solid vertical cylinder of 𝑥 2 + 𝑦 2 = 1 and the
planes 𝑧 = 1, 𝑧 = −1. Use divergence theorem to find ∯𝑆 𝑭 ⋅ 𝑑𝑺 where S is the surface of the
cylinder.
Solution 6.4
Divergence theorem shows that ∯𝑆 𝑭 ⋅ 𝑑𝑺 = ∭𝑉 div 𝑭 𝑑𝑉. Hence, it will be easier to solve
∯𝑆 𝑭 ⋅ 𝑑𝑺 using ∭𝑉 div 𝑭 𝑑𝑉 to obtain the answer. Since this is cylindrical surface, it is also easier
to use cylindrical coordinate system (recall that 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃, 𝑧 = 𝑧, 𝑑𝑥 𝑑𝑦 𝑑𝑧 =
𝑟 𝑑𝑟 𝑑𝜃 𝑑𝑧). Hence, we can solve ∯𝑆 𝑭 ⋅ 𝑑𝑺 as follows
∯ 𝑭 ⋅ 𝑑𝑺 = ∭ div (𝑥 2 𝒊 + 𝑦 2 𝒋 + 𝑧 2 𝒌) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑆
𝑉
2𝜋
1
1
= ∫ ∫ ∫ (2𝑥 + 2𝑦 + 2𝑧 ) 𝑟 𝑑𝑧 𝑑𝜃 𝑑𝑟
𝜃=0 𝑟=0 𝑧=−1
2𝜋
1
1
= ∫ ∫ ∫ (2𝑟 cos 𝜃 + 2𝑟 sin 𝜃 + 2𝑧)𝑟 𝑑𝑧 𝑑𝜃 𝑑𝑟
𝜃=0 𝑟=0 𝑧=−1
97
2𝜋
1
1
= ∫ ∫ ∫ 2𝑟 2 cos 𝜃 + 2𝑟 2 sin 𝜃 + 2𝑟𝑧 𝑑𝑧 𝑑𝜃 𝑑𝑟
𝜃=0 𝑟=0 𝑧=−1
2𝜋
1
= ∫ ∫[2𝑟 2 cos 𝜃 𝑧 + 2𝑟 2 sin 𝜃 𝑧 + 𝑟𝑧 2 ]1𝑧=−1 𝑑𝜃 𝑑𝑟
𝜃=0 𝑟=0
2𝜋
1
= ∫ ∫ 4𝑟 2 (cos 𝜃 + sin 𝜃 ) 𝑑𝑟 𝑑𝜃
𝜃=0 𝑟=0
2𝜋
1
4
= ∫ [ 𝑟 3 (cos 𝜃 + sin 𝜃 )]
𝑑𝜃
3
𝑟=0
𝜃=0
2𝜋
= ∫
𝜃=0
2𝜋
= ∫
𝜃=0
4
(cos 𝜃 + sin 𝜃 ) 𝑑𝜃
3
4
(cos 𝜃 + sin 𝜃 ) 𝑑𝜃
3
4
= [sin 𝜃 − cos 𝜃]2𝜋
𝜃=0
3
= 0.
∎
Example 6.5
Find ∯𝑆 𝑭 ⋅ 𝑑𝑺 for 𝑭 = 4𝑥𝒊 − 2𝑦 2 𝒋 + 𝑧 2 𝒌 over the region bounded by 𝑥 2 + 𝑦 2 = 4, 𝑧 = 0 and
𝑧 = 3.
Solution 6.5
∯ 𝑭 ⋅ 𝑑𝑺 = ∭ div 𝑭 𝑑𝑉
𝑆
𝑉
= ∭ div (4𝑥𝒊 − 2𝑦 2 𝒋 + 𝑧 2 𝒌 ) 𝑑𝑉
𝑉
= ∭(4 − 4𝑦 + 2𝑧 ) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑉
2𝜋
2
3
= ∫ ∫ ∫ (4 − 4𝑟 sin 𝜙 + 2𝑧)𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜙
𝜙=0 𝑟=0 𝑧=0
98
2𝜋
2
= ∫ ∫(21 − 12𝑟 sin 𝜙)𝑟 𝑑𝑟 𝑑𝜙
𝜙=0 𝑟=0
2𝜋
= ∫ (42 − 32 sin 𝜙) 𝑑𝜙
𝜙=0
= 84𝜋.
∎
99
STOKE’S THEOREM
WEEK 7: STOKE’S THEOREM
7.1 INTRODUCTION
From previous topic, we have learnt that double integrals over a region in the plane can be
transformed into line integrals over the boundary curve of the region and conversely. We shall now
see that more generally, surface integrals over a surface S with boundary curve C can be transformed
into line integrals over C and conversely.
Stoke’s theorem is the ‘curl analogue’ of the divergence theorem and related the integral of curl of a
vector field over an open surface S to the line integral of the vector field around the perimeter C
bounding the surface.
If F is a vector filed existing over surface S and around its boundary, closed curve c, then
 curl F.dS   F.dr
s
c
This means that we can express a surface integral in terms of a line integral round the boundary curve.
Example 7.1:
A hemisphere S is defined by x  y  z  4 (z ≥ 0). A vector field F=2yi-xj+xzk exists over the
2
2
2
surface and around its boundary c.


Verify Stoke’s theorem, that curl F.dS  F.dr
s
c
S : x2  y2  z2  4  0
F  2yi  xj  xzk
c is the circle x 2  y 2  4
(a)
 F.dr   2yi  xj  xzk . idx  jdy  kdz 
c
c
  2ydx  xdy  xzdz 
c
Converting to polar coordinates
x= 2cosθ;
y= 2sinθ;
z=0
dx= -2sinθ dθ;
dy=2cosθdθ;
limits θ= 0 to 2π
Making the substitution and completing the integral
100
2π
 F.dr   4 sinθ 2sinθsi   2cosθcosθ2dθ
c
0
 4  2sin2θ  cos 2θ  dθ
2π
0
2
2
 -4  1  sin   d  - 2  3 - cos2  d
2
0
(1)
0
2
sin2 

 -23  12
2  0


(b) Now we determine curl F.dS
s
 curl F.dS   curl

F. n
dS
s
i

curl F 
x
2y
n
Now
j

y
x
k

 i0  0   jz  0   k  1  2   zj  3k
z
xz
S
2xi  2yj  2zk
xi  yj  zk


2
2
2
S
2
4 x  4y  4 z


Then curl F. n dS 
s
 xi  yj  zk 
dS
2

  zj  3k .
s

1
 yz  3z  dS
2 s
Expressing this in spherical polar coordinates and integrating, because
x  2 sin cos  ;

  curl F.n dS 
s
y  2sinsin ;
z  2cos ;
dS  4sindd
1
 2 sin sin2 cos  6 cos 4 sindd
2 
s
2 / 2
 4 
 2 sin  cos sin  3 sin cos dd
2
0 0
2
 /2
 2 sin3  sin 3 sin2  
 4  

d
3
2  0
0
(2)
2
3
2
 4   sin   dφ  -12
3
2
0
So we have from our two results (1) and (2)
101
 curl F.dS  F. dr
s
c
Example 7.2:
̃, where S is the upper half of the sphere
Verify the Stoke’s Theorem for 𝐹 = (2𝑥 − 𝑦)𝒊̃ − 𝑦𝑧 2 𝒋̃ − 𝑦 2 𝑧𝒌
𝑥 2 + 𝑦 2 + 𝑧 2 = 1 and C is its boundary.
Hemispherical surface and boundary for Example 5
Solution:
The surface and boundary involved are illustrated in the above figure. We are required to show that
∮ 𝑭. 𝑑𝑟 = ∬ 𝑐𝑢𝑟𝑙 𝑭. 𝑑𝑺
𝑐
𝑆
Since C is a circle of unit radius in the (x,y) plane, to evaluate ∮𝑐 𝑭. 𝑑𝑟, we take
𝑥 = cos 𝜙,
𝑦 = sin 𝜙,
so that
𝑟 = cos 𝜙𝒊 + 𝑠𝑖𝑛𝜙𝒋
giving
𝑑𝑟 = −𝑠𝑖𝑛𝜙𝑑𝜙 𝑖 + 𝑐𝑜𝑠𝜙𝑑𝜙 𝑗
Also, on the boundary C, z=0, so that
𝐹 = (2𝑥 − 𝑦)𝒊 = (2 𝑐𝑜𝑠𝜙 − 𝑠𝑖𝑛𝜙)𝒊
Thus,
2𝜋
∮ 𝑭. 𝑑𝑟 = ∬(2 𝑐𝑜𝑠𝜙 − 𝑠𝑖𝑛𝜙)𝒊 . (−𝑠𝑖𝑛𝜙 𝒊 + 𝑐𝑜𝑠𝜙 𝒋)𝑑𝜙
𝑐
0
2𝜋
2𝜋
0
0
1
= ∫ (−2𝑠𝑖𝑛𝜙 𝑐𝑜𝑠𝜙 + 𝑠𝑖𝑛2 𝜙)𝑑𝜙 = ∫ [−𝑠𝑖𝑛2𝜙 + (1 + 𝑐𝑜𝑠2𝜙)] 𝑑𝜙 = 𝜋
2
102
𝒊
𝜕
𝑐𝑢𝑟𝑙 𝐹 = ||
𝜕𝑥
2𝑥 − 𝑦
𝒋
𝜕
𝜕𝑦
−𝑦𝑧 2
𝒌
𝜕
|=𝒌
𝜕𝑧 |
−𝑦 2 𝑧
The unit outward-drawn normal at a point (x,y,z) on the hemisphere is given by (xi+yj+zk) since 𝑥 2 +
𝑦 2 + 𝑧 2 = 1. Thus
∬ 𝑐𝑢𝑟𝑙 𝑭. 𝑑𝑺 = ∬ 𝒌. (𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌)𝑑𝑆
𝑆
𝑆
2𝜋 𝜋/2
𝜋/2
1
= ∬ 𝑧 𝑑𝑆 = ∫ ∫ 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙 = 2𝜋 [ 𝑠𝑖𝑛2 𝜃]
=𝜋
2
0
𝑆
0
0
Hence ∮𝑐 𝑭. 𝑑𝑟 = ∬𝑆 𝑐𝑢𝑟𝑙 𝑭. 𝑑𝑺 and Stoke’s Theorem is verified.
7.2 DIRECTION OF UNIT NORMAL VECTORS TO A SURFACES
When we were dealing with the divergence theorem, the normal vectors were drawn in a direction
outward from the enclosed region. With an open surface as we now have, there is in fact no inward
or outward direction. With any general surface, a normal vector can be drawn in either of two opposite
directions. To avoid confusion, a convention must therefore be agreed upon and the established rule
is as follows.

A unit normal n is drawn perpendicular to the surface S at any point in the direction indicated by
applying the right-handed screw sense to the direction of integration round the boundary c. Having
noted that point, we can now deal with the next example.
Example 7.3:
A surface consists of five sections formed by the planes x=0, x=1, y=0, y=3, z=2 in the first octant. If
the vector field F = yi+z2j+xyk exists over the surface and around its boundary, verify Stoke’s theorem.
103
If we progress round the boundary along c1, c2, c3, c4 in an anti-clockwise manner, the normal to the
surfaces will be as shown.


We have to verify that curl F.dS  F.dr
s
c

(a) We will start off by finding F.dr
c
(1) Along c1:
y=0; z=0; dy=0; dz=0
  F.dr   0  0  0  0
c1
(2) Along c2:
x=1; z=0; dx=0; dz=0
  F.dr   0  0  0  0
c2
(3) Along c3:
y=3; z=0; dy=0; dz=0
0
  F.dr   3dx  0  0   3x 10  3
c3
(4) Along c4:
1
x=0; z=0; dx=0; dz=0
0
  F.dr   0  0  0   0
c3
1
  F.dr  0  0  3  0  3
c
 F.dr  -3
c

(b) Now we have to find curl F.dS
s
First we need an expression for curl F.
F  yi  z 2 j  xyk
104
i

curl F    F 
x
y
j

y
z2
k

z
xy
 i(x  2z)  j(y  0)  k (0  1)  (x  2z)i  yj  k



Then for each section, we obtain curl F.dS  curl F. n
s
dS

(1) S1 (top) n  k

 curl F. n dS   (x  2z)i  yj. (k) dS
S1
S1
 (1)dS  (area of S )  3
1
S1
Then, likewise

(2) S2 (right-hand end): n  j

  curl F. n dS   (x  2z)i  yj. ( j) dS
S2
S2
  (y) dS
S2
But y=3 for this section

  curl F. n dS    3 dS  - 32  6
S2
S2

(3) S3 (right-hand end): n   j

 curl F. n dS   (x  2z)i  yj. ( j) dS
S3
S3
  (y) dS
S3
But y=0 over S3

  curl F. n dS  0
S2

(4) S4 (front): n  i

  curl F. n dS   (x  2z)i  yj. (i) dS
S4
S4
  (x  2z) dS
S4
105
but x=1 over S4

3 2
3
0 0
0


  curl F. n dS   (1  2z)dzdy   z  z 2 0dy
S4
2
3
   2 dy   2y 30  6
0

(5) S5 (back): n  i with x=0 over S5

Similar working to the above gives
 curl F. n dS  12
S5
Finally, collecting the five results together gives

  curl F. n dS  3  6  0  6  12  3
S4
So, referring back to our result for section (a) we see that
 curl F.dS   F.dr
s
c
Example 7.4:
A surface S consists of that part of the cylinder x 2  y 2  9 between z=0 and z=4 for y≥0 and the two
semicircles of radius 3 in the planes z=0 and z=4. If F  zi  xyj  xzk , evaluate curl F.dS over the

s
surface.
The surface S consists of three sections
(a) The curved surface of the cylinder
(b) The top and bottom semicircles
We could therefore evaluate
 curl F.dS
s
Over each of these separately.
However, we know by Stokes’ theorem that
F  zi  xyj  xzk
 F. dr   zi  xyj  xzk . idx  jdy  kdz 
c
c
106
  zdx  xydy  xzdz 
c
Now we can work through this easily enough, taking c1, c2, c3, c4 in turn, and summing the results which
gives
 curl F.dS   F.dr   zdx  xydy  xzdz 
s
c
c
(1) C1: y=0; z=0; dy=0; dz=0
 F. dr   0  0  0  0
c1
c1
(2) C2: x=-3; y=0; dx=0; dy=0
4
 3z 2 
 F. dr  c 0  0  3zdz    2   24
c2
0
2
(3) C3: y=-; z=4; dy=0; dz=0
3
 F. dr   4dx  0  0   4dx  24
c3
c3
3
(4) C4: x=3; y=0; dx=0; dy=0
0
 3z2 
 F. dr  c 0  0  3zdz    2   24
c4
4
4
7.3 GREEN’S THEOREM
Green’s theorem enables an integral over a plane area to be expressed in terms of a line integral round
its boundary curve. Let P and Q be two functions of x and y that are, along with their first partial
derivatives, finite and continuous inside and on the boundary c of a region R in the x-y plane.
If the first partial derivatives are continuous within the region
and on the boundary, then Green’s theorem states that
 P
Q 
   y  x dxdy   Pdx  Qdy 
R
c
That is, a double integral over the plane region R can be transformed into a line integral over the
boundary c of the region – and the action is reversible.
107
If P and Q are two single-valued functions of x and y, continuous
over a plane surface S, and c is its boundary curve, then
 Q
P 
 Pdx  Qdy     x  y dxdy
c
S
where the line integral is taken round c in an anticlockwise manner. In vector terms, this becomes:
S is a two-dimensional space enclosed by a simple closed curve c.
dS  dxdy

dS  n dS  k dx dy
If F  Pi  Qj where P=P(x,y) and Q=Q(x,y), then
i

curl F 
x
P
j

y
Q
But in the x-y plane,

k

Q  
P   Q P 

 i 0 
 
  j 0    k 
z 
z  
z   x y 
0
 Q P 
Q P

 0 .  curl F  k
 
z z
 x y 



So curl F.dS  curl F. n dS and in the x-y plane, n  k
 Q P 
 Q P 
  curl F.dS   k 
 .k dS   
 dxdy
x y 
x y 
S
s 
S 
 Q P 
  curl F.dS   
 dxdy
x y 
S
S 
 curl
Now by Stoke’s theorem
s

and, in this case F.dr 
c
(1)
F.dS   F.dr
c
 Pi  Qj. idx  jdy  kdz 
c
  Pdx  Qdy 
c
108
  F.dr   Pdx  Qdy
c
c
(2)
Therefore from (1) and (2)
Stoke’s
 curl F. dS   F. dr
theorem
S
 Q
in
two
dimensions
becomes
Green’s
theorem
c
P 
  x  y dxdy   Pdx  Qdy
S
c
Example 7.5:
Verify Green’s theorem for the integral
 x
2

 y 2 dx  x  2y dy taken round the boundary curve
c
c defined by
0 x 2
0 x 2
0 y 2
y 0
x2  y2  4
x 0
Green’s theorem:
 Q
P 
  x  y  dx dy   P dx  Q dy 
S

c

In this case x  y dx  x  2y dy  P dx  Q dy
2
2
P  x 2  y 2 and Q  x  2y
We now take c1, c2, c3 in turn.
(1) c1: y=0; dy=0
2
 x3  8
  P dx  Q dy    x dx    
 3 0 3
c1
0
2
2
(2) c2:
x 2  y 2  4 y 2  4  x 2 y  4 - x 2 
1/2
x  2y  x  24  x 2 
1
1 / 2
dy  4  x 2   2 x dx   x 2 dx
4x
2

 x
  P dx  Q dy    4  x  2 4  x 2 
2
 4x
c2
c2 


x2 
  4  2 x 
 dx
4  x2 
c2 
1/2



 dx


109
4  x 2  2 cos 
Putting x = 2 sin θ

Limits: x=2  
;
2
dx  2cosd
x=0 , θ=0

4 sin2  
4

4
sin


2 cosd

2 cos 
 /2 
0
  P dx  Q dy  
c2
0
1
sin2 

 4 2 sin  sin2    

2
2   / 2

 
 
 4   2  1      4
4 
 
Finally
(3) c3: x=0;
dx=0
 
0
  P dx  Q dy    2y dy  y2 2  4
c3
0
2
Therefore, collecting our three partial results
8
16
 P dx  Q dy   3    4  4    3
c
That is one part done. Now we have to evaluate
 Q
P 
  x  y  dx dy
s
P  x2  y2 
Q  x  2y 
 Q
P
 2y
y
Q
1
x
P 
  x  y  dx dy   1  2y  dy dx
s
S
It will be more convenient to work in polar coordinates, so we make the substitutions
x  r cos ;
y  r sin ;
 Q P 
  
  dx dy 
x y 
s 
 /2
dS  dxdy  r dr d
  1  2r sin  r dr d
2
0
0
 /2


0
2
 r 2 2r 3

 2  3 sin  d

0
110
 /2


0
 /2
16
 16



2  3 sin d  2  3 cos 
0
 
16
3
So we have established once again that
 Q
P 
 P dx  Q dy     x  y  dx dy
c
S
111