Engineering Mathematics 2: MTH 204 (2015)

Section : Vector Calculus

Lecturer : Dr. S. Mushayabasa

Department : Mathematics, University of Zimbabwe

Course Outline

1. Vectors in Euclidean Space

• Vector Algebra, Dot Product,

• Cross Product, Lines and Planes,

• Curvilinear Coordinates,

• Vector-Valued Functions, Arc Length

2. Functions of Several Variables

• Partial Derivatives

• Tangent Plane to the Surface

• Directional Derivatives and the Gradient

• Gradient, Divergence, curl and Laplacian

• Maxima and Minima

3. Multiple Integrals

• Double integrals

• Triple integrals

• Change of Variables in Multiple Integrals

4. Line and Surface Integrals

• Line Integrals

• Green’s Theorem

• Surface Integrals

• Divergence’s Theorem

• Stokes’ Theorem

1

References

1. K.F. Riley, M.P. Hobson and S.J. Bence. Mathematical Methods for Physicists and

Engineers. Cambridge University Press 1998.

2. D.E. Bourne and P.C. Kendall. Vector Analysis and Cartesian Tensors. Chapman and

Hall 1992.

3. T.M. Apostol. Calculus. Wiley 1975.

4. H. Anton. Calculus. Wiley 1995.

5. M.L. Boas. Mathematical Methods in the Physical Sciences. Wiley 1983.

6. E. Kreyszig. Advanced Engineering Mathematics. Wiley 1999.

7. J.E. Marsden and A.J.Tromba. Vector Calculus. Freeman 1996.

8. M.R. Spiegel. Vector Analysis. McGraw Hill 1974.

2

Chapter 1

Vectors in Euclidean Space

1.1 Definitions

A vector may be defined in essentially three different ways: geometrically, analytically and

axiomatically.

Definition 1.1.1. Geometrically, a vector is defined as a collection of equivalent line segments.

Equivalence, in this case, means that the line segments have the same magnitude/length and

are parallel. Thus, a vector is characterised by its direction and magnitude/length.

Under this definition, the algebraic operations on vectors are introduced and studied geo-

metrically, making maximum use of our geometric intuition.

Definition 1.1.2. In the analytic approach, a vector in three-dimensional space is defined as

an ordered triple of real numbers (A1, A2, A3) relative to a given coordinate system. The real

numbers A1, A2, A3 are called the components of the vector.

By far this approach is the most convenient for theoretical and computational considerations.

The axiomatic definition of a vector is left out as an exercise for you.

On the other hand, those quantities that are characterised by numerical magnitude alone,

and have nothing to do with direction, are called scalars or scalar quantities. In order to

distinguish vectors from scalars, we will use letters with an over-bar A, B, C, · · · , a, b, c, · · ·, to

denote vectors. In these typed notes I will be using boldfaced letters A,B,C, · · · ,a,b, c, · · ·.

Definition 1.1.3. For any given (three-dimensional) vector A such that

A = (A1, A2, A3),

A1 is called the first component, A2 is called the second component, A3 is called the third

component of the vector A.

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Definition 1.1.4. A vector whose components are all zero is called the zero vector and is

denoted by 0 or 0, that is, 0 = (0, 0, 0).

Definition 1.1.5. The magnitude of a vector A = (A1, A2, A3), denoted by ‖A‖, is the real

number defined by

‖A‖ =√A2

1 +A22 +A2

3 (1.1)

Definition 1.1.6. A vector whose magnitude is unit (or 1) is called a unit vector.

For any vector A, the corresponding unit vector, denoted A, is given by

A =A

‖A‖. (1.2)

From (1.2) we deduce that the vector A can be expressed as

A = ‖A‖A. (1.3)

1.2 Vector Algebra

Let us consider the Cartesian coordinate system in space, obtained by introducing three mu-

tually perpendicular axes, labelled x, y, z, with the same unit of length along the three axes.

The unit vectors in the positive x-, y- and z-directions are denoted by i, j and k, respectively.

Thus, a vector A can be expressed in the form

A = A1i +A2j +A3k. (1.4)

The numbers A1, A2 and A3 are called the orthogonal projections or components of A

in the x-, y- and z-directions, respectively.

Two vectors A = A1i +A2j +A3k and B = B1i +B2j +B3k are said to be equal if

A1 = B1, A2 = B2, A3 = B3.

Exercise: If ‖A‖ = ‖B‖, is it necessarily true that A = B?

1.2.1 Addition of vectors

Vector addition proceeds component-wise, that is

A + B = (A1 +B1)i + (A2 +B2)j + (A3 +B3)k.

It follows that addition of vectors satisfies the commutative law

A + B = B + A,

and the associative law

A + (B + C) = (A + B) + C,

for any vectors A,B,C.

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1.2.2 Scalar multiplication of vectors

λA = λ(A1)i + λ(A2)j + λ(A3)k,

where λ is a scalar.

The following theorem summarizes the basic laws of vector algebra.

Theorem 1.2.1. For any vectors u, v, w and scalars k, l we have

(a) v + w = w + v commutative Law

(b) u + (v + w) = (u + v) + w Associative Law

(c) v + 0 = v = 0 + v Additive Identity

(d) v + (v) = 0 Additive inverse

(e) k(lv) = (kl)v Associative Law

(f) k(v + w) = kv + kw Distributive Law

(g) (k + l)v = kv + lv Distributive Law

Remark 1.2.1. If k = 0, then kv = 0v = 0.

We can prove the basic laws of vector algebra in Theorem 1.2.1. Let u = (u1, u2, u3), v =

(v1, v2, v3), w = (w1, w2, w3) be vectors. Then

u + (v + w) = (u1, u2, u3) + [(v1, v2, v3) + (w1, w2, w3)]

= (u1, u2, u3) + (v1 + w1, v2 + w2, v3 + w3)

= [u1 + (v1 + w1), u2 + (v2 + w2), u3 + (v3 + w3)]

= [(u1 + v1) + w1, (u2 + v2) + w2, (u3 + v3) + w3]

= (u1 + v1, u2 + v2, u3 + v3) + (w1, w2, w3)

= (u + v) + w.

Exercise 1.2.1. Verify the basic laws of vector algebra.

1.2.3 Dot product

Definition 1.2.1. Let v = (v1, v2, v3) and w = (w1, w2, w3) be vectors. The dot product of v

and w, denoted by v ·w, is given by

v ·w = v1w1 + v2w2 + v3w3. (1.5)

The dot product can also be defined in a geometric way, which we will now develop as a

consequence of analytic definition

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Definition 1.2.2. The angle between two nonzero vectors with the same initial point is the

smallest angle between them.

Theorem 1.2.2. Let v,w be nonzero vectors, and let θ be the angle between the. Then

cos θ =v ·w‖v‖‖w‖

. (1.6)

Exercise 1.2.2. Prove theorem 1.2.2.

Example

Find the angle θ between the vectors v = (2, 1,−1) and w = (3,−4, 1).

Solution

Since v ·w = (2)(3) + (1)(−4) + (−1)(1) = 1, v| =√

6 and |w| =√

26, then

cos θ =1√

6√

26≈ 0.08,⇒ θ = 85.41o. (1.7)

Corollary 1.2.1. Two nonzero vectors v and w are perpendicular if and only if v ·w = 0.

Exercise 1.2.3.

• Are the vectors v = (−1, 5,−2) and w = (3, 1, 1) orthogonal?

From the discussion on dot product we can deduce that an alternatively description of a vector

in space is obtained by specifying its magnitude and direction. The direction can be specified

by prescribing the three direction angles α, β and γ between the vector and the positive x, y

and z directions respectively. The magnitude is determined using the Pythagorean Theorem

as

‖A‖ =√A2

1 +A22 +A2

3.

Exercise: Verify the following formulas, for a vector A = A1i +A2j +A3k:

cosα =A1

‖A‖, cosβ =

A2

‖A‖, cos γ =

A3

‖A‖,

and hence

cos2 α+ cos2 β + cos2 γ = 1.

The following theorem summarizes the basic properties of the do product.

Theorem 1.2.3. For any vectors u, v, w and scalar k, we have

(a) v ·w = w · v commutative Law

(b) kv ·w = v · k(w) Associative Law

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(c) v · 0 = 0 = 0 · v

(d) u · (v + w) = u · v + u ·w Distributive Law

(e) (u + v) ·w = u ·w + v ·w Distributive Law

(f) |v ·w| ≤ ‖v‖ ‖w‖ Cauchy-Schwarz Inequality

Proof. The proofs of parts (a)-(e) are straightforward applications of the definition of the dot

product, and have been left as an exercise. We will prove part (f). We know that

cos θ =v ·w‖v‖ ‖w‖

.

⇒ v ·w = cos θ ‖v‖ ‖w‖

Thus,

|v ·w| = | cos θ| ‖v‖ ‖w‖|v ·w| ≤ ‖v‖ ‖w‖ , since | cos θ| ≤ 1.

The dot product can be used to derive properties of the magnitude of vectors, the most

important of which is the Triangle Inequality, as given in the following theorem:

Theorem 1.2.4. For any vectors v,w, we have

(a) ‖v‖2 = v · v

(b) ‖v + w‖ ≤ ‖v‖+ ‖w‖, Triangle inequality

(c) ‖v −w‖ ≥ ‖v‖ − ‖w‖, Triangle inequality

Exercise 1.2.4.

• Prove theorem 1.2.7.

1.2.4 Cross product

The cross product (also known as the vector product) of two vectors A,B is defined by

A×B = (A2B3 −A3B2)i + (A3B1 −A1B3)j + (A1B2 −A2B1)k. (1.8)

A more convenient formula for the vector product which makes use of the notion of a deter-

minant of a matrix is as follows:

A×B =

∣∣∣∣∣∣∣∣i j k

A1 A2 A3

B1 B2 B3

∣∣∣∣∣∣∣∣ . (1.9)

Exercises: By applying the definition of a vector product, verify the following

7

i× i = 0, (b) i× j = k, (c) k× i = j,

for unit vectors i, j, k.

Theorem 1.2.5. If the cross product v×w of two nonzero vectors v and w is also a nonzero

vector, then it is perpendicular to both v and w.

Proof. We need to show that (v ×w) · v = 0.

(v ×w) · v = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1) · (v1, v2, v3)= v2w3v1 − v3w2v1 + v3w1v2 − v1w3v2 + v1w2v3 − v2w1v3

= 0.

Theorem 1.2.6. If θ is the angle between nonzero vectors v and w, then

‖v ×w‖ = ‖v‖‖w‖ sin θ

Exercise 1.2.5.

• Prove theorem 1.2.6.

Example 1.2.1. Show that the area A of a triangle with adjacent sides v and w is given by

A =1

2‖v ×w‖.

Solution

Show that the area A of a parallelogram with adjacent sides v and w is given by

A =1

2bh =

1

2‖v‖‖w‖ sin θ

=1

2‖v ×w‖.

Exercise 1.2.6.

1. Prove that ‖A×B‖2 = ‖A‖2‖B‖2 − (A ·B)2.

2. Show that the area A of a parallelogram with adjacent sides v and w is given by

A = ‖v ×w‖.

The following theorem summarizes the basic properties of the cross product and can easily be

verified.

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Theorem 1.2.7. For any vectors u,v,w, and scalar k, we have

(a) v ×w = −w × v, Anticommutative Law

(b) u× (v + w) = u× v + u×w, Distributive Law

(c) (u + v)×w = u×w + v ×w, Distributive Law

(d) u× (v ×w) = (u ·w)v − (u · v)w

(d) (kv)×w = v × (kw) = k(v ×w), Associative Law

(e) v × 0 = 0 = 0× v

(f) v × v = 0

(g) v ×w = 0, if and only if v parallel to w.

1.2.5 Equations of lines

The position vector of a point (x, y, z) is the vector r = xi + yj + zk. Geometrically, the

position vector of a point is the directed line extending from the origin to the point.

The equation of a line is completely determined if we know any two points on the line or

if we know a point on the line and the orientation or direction of the line. The direction of

the line can be described by a vector to which the line is parallel. Suppose we wish to find

the equation of a line that passes through a given point (x0, y0, z0) and is parallel to a given

non-zero vector v = v1i + v2j + v3k. Let r denote the position vector of an arbitrary point

(x, y, z) on the line. if r0 = x0i + y0j + z0k denotes the position vector of the point (x0, y0, z0),

then the vector r − r0 is parallel to the vector v. Hence there exists a scalar t such that

r− r0 = tv. Thus the position vector r of an arbitrary point on the line is given by

r = r0 + tv, t1 ≤ t ≤ t2. (1.10)

this is a vector equation of the line. The scalar t is called the parameter; and as the parameter

ranges from t1 to t2, the vector r traces the line from one end to the other with t = 0

corresponding to the point (x0, y0, z0). Equating the corresponding components of the vectors

in (1.10) we obtain the parametric equations of the line:

x = x0 + tv1, y = y0 + tv2, z = z0 + tv3. (1.11)

If we eliminate the parameter t from the equations (1.11), we obtain the symmetric form of

the equation of the line:

x− x0v1

=y − y0v2

=z − z0v3

. (1.12)

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Example 1.2.2.

1. Find a vector equation of the line passing through the points (1,−2, 1) and (3, 1, 1).

2. Deduce the corresponding parametric equations and symmetric equations.

Solution 1.2.1.

1. We note that the line is parallel to the vector

v = (3− 1)i + (1− (−2))j = 2i + 3j.

Let r0 = i− 2j + k, then

r = r0 + tv = (1 + 2t)i + (3t− 2)j + k.

2. By equating the corresponding components of the vector equation, we obtain the para-

metric equations

x = 1 + 2t, y = −2 + 3t, z = 1.

By eliminating the parameter t, we obtain the equation in symmetric form

x− 1

2=y + 2

3, z = 1.

Exercise 1.2.7.

1. Find the parametric equations of the line that passes through the point (−1, 3,−2) and

is perpendicular to the vectors A = 3i + 4j + k and B = i + 2j.

2. Derive the symmetric form (1.12) of the equation of the line from the equation

(r− r0)× v = 0. (1.13)

3. Find the pint of intersection (if any) of the following line:

x+ 1

3=y − 2

2=z − 1

−1, and, x+ 3 =

y − 8

−3=z + 3

2.

1.2.6 Distance between a point and a line

Let L be a line in vector form r + tv for −∞ < t < ∞ and Let P be a point not on L. The

distance d from P to L is the length of the line segment from P to L which is perpendicular

to L. Now, if we pick any point, on L, say Q, and we let w be the vector from Q to P, and if

θ is the angle between w and v, then, the distance between a point and a line d is given by

d = ‖w‖ sin θ. So since ‖v ×w‖ = ‖v‖‖w‖ sin θ and v 6= 0, thus

d =‖v ×w‖‖v‖

(1.14)

Exercise 1.2.8.

• Find the distance d from point P = (1, 1, 1) to the line L = (−3, 1,−4) + t(7, 3,−2).

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1.2.7 Plane through a point, perpendicular to a vector

Let P be a plane, and suppose it contains a point P0 = (x0, y0, z0). Let n = (a, b, c) be a

nonzero vector which is perpendicular to the plane P. Such a vector is called a normal vector

(or just normal) to the plane. Now let (x, y, z) be any point in the plane P . Then the vector

r = (x − x0, y − y − y0, z − z0) lies on the plane P. Thus r ⊥ n and hence n · r = 0. And if

r = 0 then we still have n · r = 0. Therefore

n · r = (a, b, c) · (x− x0, y − y0, z − z0) = a(x− x0) + b(y − y0) + c(z − z0) = 0. (1.15)

The equation (1.15) is called the point-normal form of the plane P.

Exercise 1.2.9.

• Find the equation of the plane P containing the point (−3, 1, 3) and perpendicular to

the vector n = (2, 4, 8).

1.2.8 Curvilinear coordinates

The Cartesian coordinates of a point (x, y, z) are determined by following a family of straight

paths from the origin: first along the x− axis, then parallel to the y− axis, then parallel, to

the z−axis, so that

r = xi + yj + zk.

In curvilinear coordinate system, these paths can be curved. The two types of curvilinear coor-

dinates which we will discuss are cylindrical and spherical coordinates. Instead of referencing

a point in terms of sides of a rectangular parallelepiped, as with Cartesian coordinates, we will

think of the point as lying on cylinder or sphere. Cylindrical coordinates are often used when

there is symmetry about origin.

Let P = (x, y, z) be a point in Cartesian coordinates, and let P0 = (x, y, 0) be the projection

of P upon the xy− plane. Treating (x, y) as a point in 2 dimensional, let (r, θ) be its polar

coordinates.

Thus, the Cylindrical coordinates (r, θ, z):x = r cos θ, r =

√x2 + y2, y = r sin θ, θ = tan−1

(yx

), z = z,

where 0 ≤ θ ≤ π, if, y ≥ 0 and π < θ < 2π if y < 0.

(1.16)

For spherical coordinates, let ρ be the length of the line segment from the origin to P , and let

φ be the angle between that line segment and the positive z−axis.

Thus, the spherical coordinates coordinates (ρ, θ, φ):

x = ρ sinφ cos θ, ρ =√x2 + y2 + z2, y = ρ sinφ sin θ, θ = tan−1

(yx

),

z = ρ cosφ, φ = cos−1

(z√

x2 + y2 + z2

),

where 0 ≤ θ ≤ π, if, y ≥ 0 and π < θ < 2π if y < 0.

(1.17)

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Remark 1.2.2. Both θ and φ are measured in radians. Note that r ≥ 0.

Example 1.2.3.

Convert the point (−2,−2, 1) to

(a) cylindrical coordinates.

(b) spherical coordinates.

Solution 1.2.2.

(a)

r =√

(−2)2 + (−2)2 = 2√

2, θ = tan−1(−2

−2

)=

5π

4, since y = −2 < 0.

Therefore (r, θ, z) =

(2√

2,5π

4, 1

)(b)

ρ =√

(−2)2 + (−2)2 + 12 =√

9 = 3, φ = cos−1(

1

3

)≈ 1.23 radians.

Therefore (ρ, θ, φ) =

(3,

5π

4, 1.23

)Exercise 1.2.10.

(a) Write the equation of the cylinder x2 + y2 = 4 in cylindrical coordinates.

(b) Write the equation (x− 2)2 + (y − 1)2 + z2 = 9 in spherical coordinates.

1.2.9 Vector-valued functions

In this section we will consider vector-valued functions of one real variable. The simplest of

this type of function is exemplified by the parametric representation of a straight line. For

example, the equation of a straight line passing through the point P (−1, 2, 4) and parallel to

the direction (−3, 4, 8) is

r(t) = r0 + sv

= (−1,−2, 4) + s(−3, 4, 8)

= (−1− 3s)i + (2 + 4s)j + (4 + 8s)k

Definition 1.2.3. A vector function or vector-valued function F(t) is a mathematical

rule that associates a vector F with each real number t in some set, usually an interval t1 ≤t ≤ t2 or a collection of intervals (t1 ≤ t ≤ t2) ∪ (t3 ≤ t ≤ t4) ∪ · · · ∪ (tn−1 ≤ t ≤ tn).

12

In the Cartesian coordinate system, the vector function can be expressed in component form,

say

F(t) = f1(t)i + f2(t)j + f3(t)k (1.18)

where f1, f2 and f3 are scalar-valued functions of t and are called the components of F(t). For

example, the equation

r = r0 + tv, t1 ≤ t ≤ t2,

of a line that passes through the point r0 and is parallel to the vector v is a vector function of

the parameter t.

Definition 1.2.4. Let f(t) be a vector-valued function, let a be a real number and let c be a

vector. Then limt→a f(t) = c, if limt→a ‖f(t)− c‖ = 0. If f(t) = (f1(t), f2(t), f3(t)), then

limt→a

f(t) =(

limt→a

f1(t), limt→a

f2(t), limt→a

f2(t))

(1.19)

provided that all three limits on the right side exist.

The above definition shows that continuity and derivative of vector-valued functions can also

be defined in terms of its component functions. Thus,

f ′(a) = limh→0

f(a+ h)− f(a)

h= (f1(a), f2(a), f3(a)), (1.20)

if the component derivatives exists.

Example 1.2.4.

Let f(t) =

(e−t,

2t− 3

4t− 1,3t2 + 1

t2 − t

). Then

limt→∞

f(t) =

(limt→∞

e−t, limt→∞

2t− 3

4t+ 1, limt→∞

3t2 + 1

t2 − t

)=

(0,

1

2, 3

).

Definition 1.2.5. A scalar function is a real-valued function. Note that if u(t) is a scalar

function and f(t) is a vector-valued function, then their product, defined by (uf)(t) for all t is

a vector-valued function (since the product of scalar with a vector is a vector).

The following properties of derivatives of vector-valued functions are are summarized in the

following theorem.

Theorem 1.2.8. Let f(t) and g(t) be differentiable vector-valued functions, let u(t) be a dif-

ferentiable scalar function, let k be a scalar, and let c be a constant vector.. Then

(a)d

dt(c) = 0

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(b)d

dt(kf) = k

df

dt

(c)d

dt(f ± g) =

df

dt± dg

dt

(d)d

dt(f · g) =

df

dt· g + f · dg

dt

(e)d

dt(f × g) =

df

dt× g + f × dg

dt

Example 1.2.5.

Let f(t) = (cos t, sin t, t). Determine the equation of the tangent L to the curve at f(2π).

Solution 1.2.3.

L = f(2π) + sf ′(2π) = (1, 0, 2π) + s(0, 1, 1)

In parametric form this can be written as x = 1, y = s, z = 2π + s for −∞ < s <∞.

1.2.10 Velocity and Acceleration

Let r(t) represent the position vector of a particle P in space where t is the time. The

path described by P will be some curve. The velocity and acceleration of this particle are,

respectively given by:

v(t) =dr

dt

a(t) =dv

dt=d2r

dt2

Exercise 1.2.11.

(a) Determine the velocity and acceleration of a particle P moving along the helical path

r(t) = (a cosωt, a sinωt, bt) for, t ∈[0,

4π

ω

]when t = 2π

ω .

(b) Determine the distance travelled by the particle when t =4π

ω.

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1.2.11 Arc Length

Let r(t) = (x(t), y(t), z(t)) be the position vector of an object moving in a three dimensional

space. Since ‖v(t)‖ is the speed of the object at time t, it seems natural to define the distance

s travelled by the object from time t = a to t = b as the definite integral, that is:

s =

∫ b

a‖v‖dt =

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt

=

∫ b

a

√x′(t)2 + y′(t)2 + z′(t)2dt.

Example 1.2.6.

• Find the length L of the helix f(t) = (cos t, sin t, t) from t = 0 to t = 2π.

Solution 1.2.4.

L =

∫ 2π

0

√x′(t)2 + y′(t)2 + z′(t)2dt =

∫ 2π

0

√(− sin t)2 + (cos t)2 + 12dt

= 2√

2π.

In some situations the arc length parametrization can be useful, because the arc length arises

from the shape of the curve and does not depend on a particular coordinate system. The idea

is to replace the parameter t, for any given smooth parametrization f(t) defined on [a, b], by

the parameter s given by

s = s(t) =

∫ t

0‖f ′(u)‖du (1.21)

In terms of motion along a curve, s is the distance travelled along the curve after time t has

elapsed.

Example 1.2.7.

• Parametrize the helix f(t) = (cos t, sin t, t) for t ∈ [0, 2π], by arc length.

Solution 1.2.5.

s =

∫ t

0‖f ′(u)‖du =

∫ t

0

√2du =

√2t, for allt ∈ [0, 2π].

So we can solve for t in terms of s : t =s√2. Thus,

f(s) =

(cos

s√2, sin

s√2,s√2

), for all s ∈ [0, 2

√2π].

Remark: Note that ‖f ′(s)‖ = 1.

15

Theorem 1.2.9. Suppose that r = r(t), θ = θ(t) and z = z are cylindrical coordinates of a

curve f(t), for t ∈ [a, b]. Then the arc length L of the curve over [a, b] is

L =√r′(t)2 + r(t)2θ′(t)2 + z′(t)2dt. (1.22)

Exercise 1.2.12.

• Prove theorem 1.2.9.

Example 1.2.8.

Find the arc length L of the curve whose cylindrical coordinates are (r, θ, z) = (et, t, et) for

t ∈ [0, 1]

Solution 1.2.6.

Observe that r′(t) = et, θ′(t) = 1 and z′(t) = et

L =√r′(t)2 + r(t)2θ′(t)2 + z′(t)2dt =

∫ 1

0

√e2t + e2t(1) + e2tdt

=√

3

∫ 1

0etdt

=√

3(e− 1).

1.2.12 Velocity and Acceleration in Polar Coordinates

When a particle P (r, θ) moves along a curve in the polar coordinate plane, we express its

position, velocity, and acceleration in terms of the moving unit vectors

ur = cos θi + sin θj, uθ = − sin θi + cos θj. (1.23)

The vector ur points along the position vector ~OP , so r = rur. The vector uθ, orthogonal to

ur, points in the direction of increasing θ. (see Figure 1.1).

Figure 1.1: Position of a particle moving in space

16

It can easily be verified that

durdθ

= − sin θi + cos θj = uθ,

duθdθ

= − cos θi− sin θj = −ur,

With r = rur as a position function, we can express the velocity v = r as

v =d

dt[rur] =

dr

dtur + r

durdt

= rur + rur (1.24)

Differentiating ur and uθ with respect to time t (apply the Chain Rule) to get

durdt

=durdθ

dθ

dt= uθθ

duθdt

= −urθ

Therefore, the velocity is given by

v = rur + rθuθ

Figure 1.2: Particle position, velocity and acceleration

Exercise 1.2.13.

(a) Show that the acceleration of a Particle P (r, θ) is space is given by

a = (r − rθ2)ur + (rθ + 2rθ)uθ.

(b) A charged particle moves along the curve

r =1

1 + 2 cos θwith

dθ

dt=

1

r2.

By differentiating the equation r = rur. Show that

dr

dt= 2 sin θur +

1

ruθ.

17

Chapter 2

Functions of Several Variables

2.1 Partial derivatives

If f is a function of two variables x and y, suppose we let only x vary while keeping y fixed,

say y = b, where b is a constant. Then we are really considering a function of a single variable

x, namely g(x) = f(x, b). If g has a derivative at a, then we call it the partial derivative of f

with respect to x at (a, b) and we denote it by fx(a, b). Thus,

fx(a, b) =∂f

∂x(a, b) = lim

h→0

f(a+ h, b)− f(a, b)

h(2.1)

Similarly, the partial derivative of f with respect to y at (a, b), denoted by fy(a, b), is obtained

by keeping x fixed (x = a) and finding the ordinary derivative at b. Thus,

fy(a, b) =∂f

∂y(a, b) = lim

h→0

f(a, b+ h)− f(a, b)

h(2.2)

Example 2.1.1.

If f(x, y) = x3 + x2y3 − 2y2, find (a) fx(2, 1) and (b) fy(2, 1).

Solution 2.1.1.

(a) fx(x, y) = 3x2 + 2xy3 ⇒ fx(2, 1) = 3(2)2 + 2(2)(1)3 = 16.

(b) fy(x, y) = 3x2 − 4y2 ⇒ fy(2, 1) = 3(2)2(1)2 − 4(1) = 8.

2.1.1 Higher order derivatives

If f is function of two variables, then its partial derivatives fx and fy are also functions of two

variables, so we can consider their partial derivatives (fx)x, (fx)y, (fy)x and (fy)y, which are

called the second partial derivatives of f. If z = f(x, y), we have

(fx)x = fxx =∂

∂x

(∂f

∂x

)=∂2f

∂x2=∂2z

∂x2

18

(fx)y = fxy =∂

∂y

(∂f

∂x

)=

∂2f

∂x∂y=

∂2z

∂x∂y

(fy)x = fyx =∂

∂x

(∂f

∂y

)=

∂2f

∂y∂x=

∂2z

∂y∂x

(fy)y = fyy =∂

∂y

(∂f

∂y

)=∂2f

∂y2=∂2z

∂y2.

Example 2.1.2. Find the second partial derivatives of

f(x, y) = x3 + x2y3 − 2y2.

Solution 2.1.2.

fx = 3x2 + 2xy3 ⇒ fxx = 6x+ 2y3

fy = 3x2y2 − 4y ⇒ fyy = 6x2y − 4

fxy = 6xy2 = fyx (2.3)

Exercise 2.1.1.

Show that the function u(x, y) = ex sin y is a solution of the Laplace’s equation.

∂2u

∂x2+∂2u

∂y2= 0.

2.1.2 Chain rule

Definition 2.1.1. Suppose that z = f(x, y) is a differentiable function of x and y, where

x = g(t) and y = h(t) are both differentiable funcions of t. Then z is a differentiable function

of t. Then z is a differentiable function of t and

dz

dt=∂f

∂x

∂x

∂t+∂f

∂y

∂y

∂t. (2.4)

Example 2.1.3. If z = x2y + 3xy4, where x = sin t and y = cos t. Find ∂z∂t

Solution 2.1.3.

∂z

∂t=

∂z

∂x

∂x

∂t+∂z

∂y

∂y

∂t= (2xy + 3y4)(2 cos 2t) + (x2 + 12xy3)(− sin t).

Exercise 2.1.2.

(a) Using the chain rule, determine

(i)∂w

∂u, (ii)

∂w

∂v

given that w = x2y + y2x and x = u cos v and y = u sin v.

(b) If w = f(u) and u = αx+ βy where α and β are constants, show that

α∂w

∂y= β

∂w

∂x.

19

(c) The pressure P , volume V and temperature T of a mole of an ideal gas are related by

the equation PV = 8.31T. Find the rate at which the pressure is changing when the

temperature is 300 kelvins and increasing at a rate of 0.1 kelvins per second and the

volume is 100 litres and increasing at a rate of 0.2 litres per second.

2.2 Tangent Planes

Definition 2.2.1. Suppose f has continuous partial derivatives. An equation of the tangent

plane to the surface z = f(x, y) at the point P (x0, y0, z0) is

z − z0 = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0). (2.5)

Example 2.2.1.

Find the tangent plane to the elliptic paraboloid z = 2x2 + y2 at the point (1, 1, 3).

Solution 2.2.1.

Let f(x, y) = 2x2 + y2. Then fx(x, y) = 4x ⇒ fx(1, 1) = 4 and fy(x, y) = 2y ⇒ fy(1, 1) = 2.

Thus, the equation of the tangent plane at (1, 1) is

z − 3 = 4(x− 1) + 2(y − 1)

z = 4x+ 2y − 3.

Exercise 2.2.1.

• Find the equation of the tangent plane to the surface z = x2 + y2 at the point (1, 2, 5).

Remark: If the surface S is defined implicitly by an equation of the form f(x, y, z) = 0 then

the tangent plane to the surface at a point (a, b, c) is given by

fx(a, b, c)(x− a) + fy(a, b, c)(y − b) + fz(a, b, c)(z − c) = 0 (2.6)

Exercise 2.2.2.

Find the equation of the tangent to the surface x2 + y2 + z2 = 9 at the point (2, 2,−1).

2.2.1 Tangent planes to parametric surfaces

Here we use vector functions to discuss more general surfaces, called parametric surfaces. In

much same way that we described a space curve by a vector function r(t) of a single parameter

t, we can describe a surface by a vector function r(u, v) of two parameters u and v. We suppose

that

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (2.7)

20

is a vector-valued function defined in the uv− plane. So x, y and z, the component functions

of r, are functions of two variables u and v such that

x = x(u, v), y = y(u, v), z = z(u, v). (2.8)

Example 2.2.2.

Find the parametric representation for the surface

r(u, v) = 2 cosui + vj + 2 sinuk.

Solution 2.2.2.

The parametric representation for this surface are

x = 2 cosu, y = v, z = 2 sinu

Example 2.2.3.

Find a parametric representation for the cylinder

x2 + y2 = 4, 0 ≤ z ≤ 1.

Solution 2.2.3.

The cylinder has a simple representation r = 2 in cylindrical coordinates, so we choose as

parameters θ and z in cylindirical coordinates. Thus, the parametric equations of the cylinder

are

x = 2 cos θ, y = 2 sin θ, z = z.

Exercise 2.2.3.

Find a parametric representation of the sphere

x2 + y2 + z2 = a2.

We now find the tangent plane to a parametric surface S traced out by a vector function

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (2.9)

at a point P0 with position vector r(u0, v0). If we keep u constant by putting u = u0, then

r(u0, v) becomes a vector function of the single parameter v and as we have discussed earlier

we can make use of partial derivatives to obtain the tangent vector, that is

rv =∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k (2.10)

Similarly, if we keep v constant by putting v = v0 we get

ru =∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k (2.11)

If ru × rv is not 0, then the surface S is called smooth (that is it has no corners). For smooth

surface, the tangent plane is the plane that contains the tangent vectors ru and rv and the

vector ru × rv is a normal vector to the tangent plane.

21

Example 2.2.4.

Find the tangent plane to the surface with parametric equations

x = u2, y = v2, z = u+ 2v

at the point (1, 1, 3).

Solution 2.2.4.

ru =∂x

∂ui +

∂y

∂uj +

∂z

∂uk = 2ui + k

rv =∂x

∂vi +

∂y

v∂vj +

∂z

∂vk = 2uj + 2k

ru × rv =

∣∣∣∣∣∣∣∣i j k

2u 0 1

0 2v 2

∣∣∣∣∣∣∣∣ = −2vi− 4uj + 4uvk

Notice that the point (1, 1, 3) corresponds to the parameter values u = 1, and v = 1 so the

normal vector there is

−2i− 4j + 4k

Therefore, an equation of the tangent plane at (1, 1, 1) is

−2(x− 1)− 4(y − 1) + 4(z − 3) = 0⇒ x+ 2y − 2z + 3 = 0

2.3 Directional derivatives and the gradient vector

The directional derivative of a function provides the rate of change of that function (with

respect to distance) any specified direction–not just the direction of the axes as generated by

the partial derivatives.

Definition 2.3.1. If f is a differentiable function of x and y, then f has a directional derivative

in the direction of any unit vector u = (a, b, c) and

Duf(x, y, z) = fx(x, y, z)a+ fy(x, y, z)b+ fz(x, y, z)c. (2.12)

From (2.12) we can write

Du = fx(x, y, z)a+ fy(x, y, z)b+ fz(x, y, z)c

= (fx(x, y, z), fy(x, y, z)), fz(x, y, z) · (a, b, c)= (fx(x, y, z), fy(x, y, z), fz(x, y, z)) · u= ∇f(x, y, z) · u

where ∇f(x, y) = ∂f∂x i + ∂f

∂y j + ∂f∂zk.

22

Definition 2.3.2. If f is a function of two variable x and y, then the gradient of f is the

vector function ∇f defined by

∇f(x, y, z) =∂f

∂xi +

∂f

∂yj +

∂f

∂zk.

Thus,

Du = ∇f(x, y, z) · u.

Example 2.3.1.

If f(x, y, z) = x sin yz

(a) find the gradient of f and

(b) find the direction derivative of f at (1, 3, 0) in the direction of v = i + 2j− k.

Solution 2.3.1.

(a) The gradient of f is

∇f(x, y, z) = (fx(x, y, z), fy(x, y, z), fz(x.y.z)) = (sin yz, xz cos yz, xy cos yz)

(b) At (1, 3, 0) we have ∇f(1, 3, 0) = (0, 0, 3). The unit vector in the direction of v = i+2j−k

is

u =v

‖v‖=

i + 2j− k√6

.

Therefore, the directional derivatives is

Duf(1, 3, 0) = ∇f(1, 3, 0) · u = 3k ·(

i + 2j− k√6

)= −

√3

2.

Exercise 2.3.1.

If f(x, y, z) = x2 + y2 + z2, determine the directional of f in the direction of the vector

v = i− j + 2k at the point (2,−1, 3).

Theorem 2.3.1. Let f(x, y, z) be a continuously differentiable real-valued function, with ∇f 6=0, Then

(a) The value of f(x, y) increases the fastest in the direction of ∇f .

(b) The value of f(x, y) decreases the fastest in the direction of −∇f.

Example 2.3.2.

23

Suppose that the temperature at a point (x, y, z) in space is given by

T (x, y, z) =80

1 + x2 + 2y2 + 3z2

where T is measured in degrees Celsius and x, y, z in meters. In what direction does the

temperature increase fastest at the point (1, 1,−2) ? What is the maximum rate of increase?

Solution 2.3.2.

The gradient of T is

∇T =∂T

∂xi +

∂T

∂yj +

∂T

∂zk

= − 160

(1 + x2 + 2y2 + 3z2)2(−xi− 2yj− 3zk)

At point (1, 1, 2) the gradient vector is

∇T (1, 1,−2) =160

256(−i− 2j + 6k) =

5

8(−i− 2j + 6k)

The maximum rate of increase is

|∇T (1, 1,−2)| = 5

8| − i− 2j + 6k| = 5

√41

8.

2.4 Vector Fields

Definition 2.4.1. A vector field F is a rule associating with each point (x, y, z) in a region

(domain) D.

2.4.1 Examples of vector fields

1. V(x, y, z) = x2yi− 2yz3j + x2zk.

2. ∇f , where f is a scalar field.

3. The instantaneous velocity of a fluid at every point of a region in a river.

Any vector field may be written in terms of its components:

F(x, y, z) = F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k. (2.13)

There are two basic concepts that arise in connection with the spatial rate of change of a vector

field F, namely the divergence of F and the curl of F.

Definition 2.4.2. The divergence of a vector field

F = F1i + F2j + F3k

is a scalar field, denoted div F, defined by

div F =

(∂

∂xi +

∂

∂yj +

∂

∂zk

)·(F1i + F2j + F3k

). (2.14)

24

We observe that in terms of the del operator ∇

divF = ∇ · F.

Thus,

divF = ∇ · F =∂F1

∂x+∂F2

∂y+∂F3

∂z.

Note that ∇ · F 6= F · ∇. In fact the operator F · ∇ is meaningless alone.

Example 2.4.1.

Find the divergence of the vector field

F = xeyi + exyj + sin (yz)k.

Solution 2.4.1.

div F = ∇ · F =

(i∂

∂x+ j

∂

∂x+ k

∂

∂x

)· (xeyi + exyj + sin (yz)k)

= ey + xexy + y cos (yz).

Definition 2.4.3. A vector field whose divergence is zero in a certain region is said to be

solenoidal or non-divergent in that region.

Remark: Note that, the divergence of a vector field is a scalar field that tells us, at each point

in the region of definition, the extend to which the field diverges or explodes from that point.

Definition 2.4.4. The curl of a vector field

F = F1i + F2j + F3k,

denoted curl F, is the vector field defined by

curl F =

(∂

∂xi +

∂

∂yj +

∂

∂zk

)×(F1i + F2j + F3k

)(2.15)

Again we observe that in terms of the del operator ∇

curl F = ∇× F.

Hence,

curl F = ∇× F =

∣∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣∣∣=

(∂F3

∂y− ∂F2

∂z

)i +

(∂F1

∂z− ∂F3

∂x

)j +

(∂F2

∂x− ∂F1

∂y

)k

25

Example 2.4.2. Find curl F, if

F = xyzi + x2y2z2i + y2z3k.

Solution 2.4.2.

curl F =

∣∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

xyz x2y2z2 y2z3

∣∣∣∣∣∣∣∣ = (2yz3 − 2x2y2z)i + zyj + (2xy2z2 − xz)k.

Definition 2.4.5. A vector field with the property that its curl is identically zero is said to be

conservative.

If a vector field F is conservative in a domain D, then there can be found a scalar field φ

defined in D such that

F = ∇φ,

and φ is called a scalar potential function of F.

Exercise 2.4.1. Show that F = 2xyi + (x2 + 1)j + 6z2k is conservative, and hence find its

corresponding scalar potential φ.

The curl of a vector field is a vector field that gives us, at each point, an indication of how the

field rotates, or swirls, from that point.

2.4.2 The Laplacian

Definition 2.4.6. The Laplacian of a scalar field f , denoted by ∇2f , is defined by

∇2f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2. (2.16)

The Laplacian of f is also defined as the divergence of the gradient of f :

∇2f = div (grad f) = ∇ · (∇f) = ∇2f.

If f is a scalar field, then ∇2f(x, y, z) is a number that tells us something about the behaviour

of the scalar field in the vicinity of (x, y, z).

Definition 2.4.7. If φ is a scalar field such that

∇2φ = 0, (2.17)

then φ is said to be harmonic.

The partial differential equation (3.11) is called Laplace equation. It means that the average

value of f in any neighborhood of (x, y, z) will be exactly equal to the value of f .

Exercise 2.4.2.

26

Let r(x, y, z) = xi + yj + zk. Find

(a) the gradient of ‖r‖2.

(b) the divergence of r

(c) the curl of r

(d) the Laplacian of ‖r‖2

2.5 Maxima and Minima

The gradient can be used to find extreme points of real-valued functions of several variables,

that is, points where the function has a local maximum or local minimum.

Definition 2.5.1. A function of two variables has a local maximum at (a, b) if f(x, y) ≤ f(a, b)

when (x, y) is near (a, b). [This means that f(x, y) ≤ f(a, b) for all points (x, y) is some disk

center (a, b).] The number f(a, b) is called a local maximum value. If f(x, y) ≥ f(a, b) when

(x, y) is near (a, b), then f(a, b) is a local minimum value.

A point (a, b) is called a critical point (or stationary point) of f if fx(a, b) = 0 and fy(a, b) = 0

or if one of these partial derivatives does not exists.

Example 2.5.1.

Find the critical points of f(x, y) = x2 + y2 − 2x− 6y + 14.

Solution 2.5.1.

We have

fx(x, y) = 2x− 2, fy(x, y) = 2y − 6

These partial derivatives are equal to 0 when x = 1 and y = 3, so the only critical point is

(1, 3).

Definition 2.5.2. Suppose the second partial derivatives of f are continuous on a disk with

center (a, b), and suppose that fx(a, b) = 0 and fy(a, b) = 0 [that is, (a, b) is a critical point of

f ]. Let

D = D(a, b) =

∣∣∣∣∣ fxx fxy

fyx fyy

∣∣∣∣∣ = fxx(a, b)fyy(a, b)− [fxy(a, b)]2

(a) If D > 0 and fxx(a, b) > 0, then f(a, b) is a local minimum.

(b) If D > 0 and fxx(a, b) < 0, then f(a, b) is a local maximum.

27

(c) If D < 0, then f has neither a local minimum nor a local maximum at (a, b), but a saddle

point.

(d) if D = 0, then the test fails.

Example 2.5.2.

Find all local maxima and minima of f(x, y) = x2 + xy + y2 − 3x.

Solution 2.5.2.

fx = 2x+ y − 3, fy = x+ 2y,

then the critical points (x, y) are the common solutions of the equations

2x+ y = 3

x+ 2y = 0

which has the unique solution (x, y) = (2,−1). So (2,−1) is the only critical point. Now we

need to determine the nature of this critical point.

fxx = 2, fyy = 2, fxy = 1

Therefore D = fxx(2,−1)fyy(2,−1)− [fxy(2,−1)]2 = (2)(2)− (1)2 = 3 > 0. Since fxx(2,−1) >

0 it follows that (2,−1) is a local minimum.

Exercise 2.5.1.

Find all local maxima and minima of f(x, y) = xy − x3 − y2.

28

Chapter 3

Multiple integrals

3.1 Double Integrals

Suppose that f is a function of two variables that is continuous on the rectangleR = [a, b]×[c, d],

that is R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}. We use the notation∫ da f(x, y)dy to mean that x

is held fixed and f(x, y) is integrated with respect to y from y = c to y = d. This procedure is

called partial integration with respect to y. Now∫ dc f(x, y)dy is a number that depends of the

value of x, so it defines a function of x :

A(x) =

∫ d

cf(x, y)dy

If we now integrate the function A with respect to x from x = a to x = b, we get∫ b

aA(x)dx =

∫ b

a

[∫ d

cf(x, y)dy

]dx (3.1)

The integral on the right side of equation (3.1) is called an iterated integral.

Example 3.1.1.

Evaluate ∫ 2

1

∫ 3

0x2ydxdy =

∫ 2

1

[∫ 3

0x2ydx

]dy =

∫ 2

1

[x3

3y

]x=3

x=0

dy =27

2.

Theorem 3.1.1. If f is continuous on the rectangle R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}, then∫ ∫Rf(x, y)dA =

∫ b

a

∫ d

cf(x, y)dydx =

∫ d

c

∫ b

af(x, y)dxdy (3.2)

More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a

finite number of smooth curves, and the iterated integrals exists.

29

3.1.1 Double integrals over general regions

In the previous section, we learned how to integrate functions of two variables over rectangular

regions. In this section, we will consider integration over more general plane regions.

Definition 3.1.1.

There are two types of plane regions.

• A region D in the xy− plane is said to be of type I if it lies between the graphs of two

continuous functions of x. That is,

D = {(x, y)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}

where g1 and g2 are continuous on [a; b].

• A region D in the xy− plane is said to be of type II if it lies between the graphs of two

continuous functions of y. That is,

D = {(x, y)|c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}

where h1 and h2 are continuous on [c; d].

Theorem 3.1.2. Suppose that f is a continuous function over some region D.

1. If D is a type I region, then∫ ∫Df(x, y)dA =

∫ b

a

∫ g2(x)

g1(x)f(x, y)dydx

2. If D is a type II region, then∫ ∫Df(x, y)dA =

∫ d

c

∫ h2(y)

h1(y)f(x, y)dxdy

Example 3.1.2.

Evaluate

∫ ∫D

(x + 2y)dA, where D is the region bounded by the parabolas y = 2x2 and

y = 1 + x2.

Solution 3.1.1.

The parabolas intersect when 2x2 = 1 + x2, that is x2 = 1⇒ x = ±1. Thus,

D = {(x, y)| − 1 ≤ x ≤ 1, 2x2 ≤ y ≤ 1 + x2}

Clearly, this is a type I region, thus∫ ∫D

(x+ 2y)dA =

∫ 1

−1

∫ 1+x2

2x2(x+ 2y)dydx

30

=

∫ 1

−1[xy + y2]1+x

2

2x2dx

=

∫ 1

−1(−3x4 − x3 + 2x2 + x+ 1)dx

=

[−3

x5

5− x4

4+ 2

x3

3+x2

2+ x

]1−1

=32

15.

Example 3.1.3.

Find the volume of the solid that lies under the elliptic paraboloid z = 3x2 + y2 and above the

region D bounded by y = x and x = y2 − y.

Solution 3.1.2.

To find the points of intersection we consider y2 − y = y ⇒ y = 0 and y = 2. Thus the region

D is type II which can be described as

D = {(x, y)|0 ≤ y ≤ 2, y2 − y ≤ x ≤ y}

Thus, ∫ ∫D

(3x2 + xy2)dxdy =

∫ 2

0

∫ y

y2−ydxdy

=

∫ 2

0(x3 + xy2)|y

y2−ydy

=

∫ 2

0(y6 + y4 − 2y3)dy

=

[1

7y7 +

1

5y5 − 1

2y4]20

=794

35.

Example 3.1.4.

Evaluate ∫ 1

0

∫ 1

xsin(y2)dydx

If we try to evaluate the integral in its present form, then we must first evaluate∫

sin(y2)dy.

But it’s impossible to do so in finite terms. So we must change the order of integration.

The region of integration can be expressed as

D = {(x, y)|0 ≤ x ≤ 1, x ≤ y ≤ 1}

31

By sketching this region, we find that an alternative representation of D is

D ={

(x, y)|0 ≤ y ≤ 1, 0 ≤ x ≤ y}

Thus, the double integral can be evaluated as follows∫ 1

0

∫ 1

xsin(y2)dydx =

∫ 1

0

∫ y

0sin(y2)dxdy

=

∫ 1

0

[x sin(y2)

]x=yx=0

dy

=

∫ 1

0y sin(y2)dy

=

[−1

2cos(y2)

]10

=1

2(1− cos 1).

Exercise 3.1.1.

(a) Find the volume of the solid that lies under the paraboloid z = x2 + y2 and above the

region D in the xy− plane bounded by the line y = 2x and the parabola y = x2.

(b) Evaluate

∫ ∫D

(x+ y)dA, D is bounded by y =√x and y = x2.

(c) Evaluate

∫ 1

0

∫ 3

3yex

2dxdy.

Properties of double integrals

Suppose that f and g are integrable functions over some region D. Then,

1. ∫ ∫D

[f(x, y)± g(x, y)]dA =

∫ ∫Df(x, y)dA±

∫ ∫Dg(x, y)dA

2. ∫ ∫Dcf(x, y)dA = c

∫ ∫Df(x, y)dA

3. If f(x, y) ≤ g(x, y) for all (x, y) ∈ D, then∫ ∫Df(x, y)dA ≤

∫ ∫Dg(x, y)dA

4. If D = D1 ∪D2 where D1 and D2 do not overlap, then∫ ∫Df(x, y)dA =

∫ ∫D1

f(x, y)dA+

∫ ∫D2

f(x, y)dA.

32

3.2 Double Integrals in Polar Coordinates

Suppose we want to evaluate the double integral∫ ∫

D f(x, y)dA, where D is the unit disk

x2+y2 ≤ 1. The description of D in terms of rectangular coordinates is somewhat complicated:

D = {(x, y)| − 1 ≤ x ≤ 1,−√

1− x2 ≤ y ≤√

1− x2}

However, D is easily described using polar coordinates

D = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}

This double integral can be easily evaluated using polar coordinates

Theorem 3.2.1.

If f is continuous on the polar region

D = {(r, θ)|α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}

then ∫ ∫Df(x, y)dA =

∫ β

α

∫ h2(θ)

h1(θ)f(r cos θ, r sin θ)rdrdθ

Note: Be careful not to forget the additional factor r in the integrand! The additional factor

comes from the fact that the area of an infinitesimal polar rectangle is dA = rdrdθ.

Example 3.2.1.

Evaluate

∫ ∫D

(3x+ 4y2)dA, where D is the region in the upper half-plane bounded by circles

x2 + y2 = 1 and x2 + y2 = 4.

Solution 3.2.1.

We have D = {(r, θ)|1 ≤ r ≤ 2, 0 ≤ θ ≤ π} so that

∫ ∫D

(3x+ 4y2)dA =

∫ π

0

∫ 2

1(3r cos θ + 4r2 sin2 θ)rdrdθ

=

∫ π

0

∫ 2

1(3r2 cos θ + 4r3 sin2 θ)drdθ

=

∫ π

0[r3 cos θ + r2 sin2 θ]r=2

r=1dθ

=

∫ π

0[7 cos θ + 15 sin2 θ)dθ

=

∫ π

0

[7 cos θ +

15

2(1− cos2θ)

]dθ

=

[7 sin θ +

15

2θ − 15

4sin 2θ

]π0

=15

2π.

33

Example 3.2.2.

Find the volume of the solid bounded by the paraboloid z = 16 − 3x2 − 3y2 and the plane

z = 4.

Solution 3.2.2.

The intersection of these surfaces is given by

16− 3x2 − 3y2 = 4⇒ x2 + y2 = 4.

In polar coordinates we D = {(r, θ)|0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}. Thus,

V =

∫ ∫D

(16− 3x2 − 3y2 − 4)dA =

∫ 2π

0

∫ 2

0(12− 3r2)rdrdθ

= 2π

∫ 2

0(12r − 3r3)dr

= 2π

[6r2 − 3

4r4]20

= 24π.

Exercise 3.2.1.

(a) Evaluate

∫ ∫DxydA, where D is the portion of the annular region 4 ≤ x2 + y2 ≤ 25 that

lies in first quadrant.

(b) Find the volume of the solid bounded by the plane z = 0 and the paraboloid z = 1−x2−y2.

(c) Evaluate

∫ ∫De−x

2−y2dA, where D is the region bounded by the semicircle x =√

4− y2

and the y− axis.

3.3 Triple integrals

Theorem 3.3.1. If f is continuous on the rectangular box B = [a, b]× [c, d]× [r, s], then∫ ∫ ∫Bf(x, y, z)dV =

∫ s

r

∫ d

c

∫ b

af(x, y, z)dxdydz (3.3)

Example 3.3.1.

Evaluate ∫ 3

0

∫ 2

0

∫ 1

0(xy + z)dxdydz =

∫ 3

0

∫ 2

0

([1

2x2y + xz

]x=1

x=0

)dydz

=

∫ 3

0

∫ 2

0

(1

2y + z

)dydz

34

=

∫ 3

0

([1

4y2 + yz

]y=2

y=0

)dz

=

∫ 3

0(1 + 2z)dz.

= 12

Definition: There are three types of solid regions

• A solid region E is said to be of type I if

E = {(x, y, z)|(x, y) ∈ D, g1(x, y) ≤ z ≤ g2(x, y)}

where D is the projection of E onto the xy-plane.

• A solid region E is said to be of type II if

E = {(x, y, z)|(y, z) ∈ D, g1(y, z) ≤ z ≤ g2(y, z)}

where D is the projection of E onto the yz-plane.

• A solid region E is said to be of type III if

E = {(x, y, z)|(y, z) ∈ D, g1(x, z) ≤ z ≤ g2(x, z)}

where D is the projection of E onto the xz-plane.

Theorem: Suppose that f is a continuous function over some region E.

1. If E is a type I region, then∫ ∫ ∫Ef(x, y, z)dV =

∫ ∫D

[∫ g2(x,y)

g1(x,y)f(x, y, z)

]dA (3.4)

2. If E is a type II region, then∫ ∫ ∫Ef(x, y, z)dV =

∫ ∫D

[∫ g2(y,z)

g1(y,z)f(x, y, z)

]dA (3.5)

3. If E is a type III region, then∫ ∫ ∫Ef(x, y, z)dV =

∫ ∫D

[∫ g2(x,z)

g1(x,z)f(x, y, z)

]dA (3.6)

In particular, if E is a type I region and its projection D onto the xy-plane is a type I region,

then ∫ ∫ ∫Ef(x, y, z)dV =

∫ b

a

∫ h2(x)

h(x)

∫ g2(x,y)

g1(x,y)f(x, yz)dzdydx (3.7)

35

Example 3.3.2.

Evaluate

∫ ∫ ∫EzdV , where E is the solid region bounded by the planes x = 0, y = 0, z = 0

and x+ y + z = 1.

Solution 3.3.1.

From the boundaries we have a = 0, b = 1, h2(x) = 1− x, h1(x) = 0, g2(x, y) = 1− x− y and

g2(x, y) = 0 so that

E = {(x, y, z)|0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x, 0 ≤ z ≤ 1− x− y}

∫ ∫ ∫EzdV =

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0zdzdydx =

∫ 1

0

∫ 1−x

0

[z2

2

]z=1−x−y

z=0

dydx

=1

2

∫ 1

0

∫ 1−x

0(1− x− y)2dydx

=1

2

∫ 1

0

[−(1− x− y)3

3

]y=1−x

y=0

dx

=1

6

∫ 1

0(1− x)3dx

=1

6

[−(1− x)4

4

]10

=1

24.

3.3.1 Triple integrals in Cylindrical and Spherical Coordinates

Some regions in space are easier to express in terms of cylindrical or spherical coordinates.

Triple integrals over these regions are easier to evaluate by converting to cylindrical or spherical

coordinates.

Theorem: (Triple Integrals in Cylindrical Coordinates)

Suppose that f is continuous function on a type I region

E = {(x, y, z)|(x, y) ∈ D, h1(x, y) ≤ z ≤ h2(x, y)},

where the projection D onto the xy- plane is given in polar coordinates

D = {(r, θ)|α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ)}.

Then ∫ ∫ ∫Ef(x, y, z)dV =

∫ β

α

∫ g2(θ)

g1(θ)

∫ h2(r cos θ,r sin θ)

h1(r cos θ,r sin θ)f(r cos θ, r sin θ, z)rdzdrdθ.

36

Example 3.3.3.

Evaluate

∫ ∫ ∫EydV, where E is the solid that lies between the cylinder x2 + y2 = 1 and

x2 + y2 = 4, above the xy− plane, and below the plane z = x.

Solution 3.3.2.

Using cylindrical coordinates, the region of integration is

E = {(r, θ, z)|0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2, 0 ≤ z ≤ r cos θ},

Then ∫ ∫ ∫E

=

∫ 2π

0

∫ 2

1

∫ r cos θ

0(r sin θ)rdzdrdθ

=

∫ 2π

0

∫ 2

1r3 sin θ cos θdrdθ

=1

4

∫ 2π

r4 sin θ cos θ|21dθ

=15

4

∫ 2π

0sin θ cos θdθ

=15

8sin2 θ|2π0 = 0

Example 3.3.4.

Evaluate

∫ ∫ ∫E

√y2 + z2dV , where E is the region bounded by the paraboloid x = y2 + z2

and the plane x = 9.

Solution 3.3.3.

The solid region E and its projection D onto the yz− plane. The region E can be expressed

as

E = {(x, y, z)|(y, z) ∈ D, y2 + z2 ≤ x ≤ 9}

where D is the disk y2 + z2 ≤ 9. In polar coordinates

D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3}

∫ ∫ ∫E

√y2 + z2dV =

∫ ∫D

[∫ 9

y2+z2

√y2 + z2dx

]dA

=

∫ ∫D

(9− y2 − z2)√y2 + z2dA

37

=

∫ 2π

0

∫ 3

0r2(9− r2)drdθ

= 2π

∫ 3

0(9r2 − r4)dr

= 2π

[9r3 − 1

5r4]30

=1944π

5.

Exercise 3.3.1.

Evaluate ∫ 2

−2

∫ √4−x2−√4−x2

∫ 2

√x2+y2

(x2 + y2)dzdydx.

Theorem: (Triple Integrals in Spherical Coordinates)

Suppose that f is continuous function on the region

E = {(ρ, θ, φ)|a ≤ ρ ≤ b, α ≤ θ ≤ β, γ ≤ φ ≤ δ}.

Then∫ ∫ ∫Ef(x, y, z)dV =

∫ δ

γ

∫ β

α

∫ b

af(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)ρ2 sinφdρdθdφ.

Example 3.3.5.

Evaluate

∫ ∫ ∫Ee(x

2+y2+z2)32 dV , where E is the unit ball.

Solution 3.3.4.

Here, we note that E = {(x, y, z)|x2 + y2 + z2 ≤ 1}. Therefore, the boundary of E is a sphere

whose coordinates are

E{(ρ, θ, φ)|0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}

Thus ∫ ∫ ∫Ee(x

2+y2+z2)32 dV =

∫ π

0

∫ 2π

0

∫ 1

0e(ρ

2) 32 ρ2 sinφdρdθdφ

=

∫ π

0sinφdφ

∫ 2π

0dθ

∫ 1

0ρ2eρ

3dρ

= [− cosφ]π0 (2π)

[1

3eρ

3

]10

=4π

3(e− 1).

38

Exercise 3.3.2.

1. Evaluate

∫ ∫ ∫E

(x2 + y2 + z2)dV , where E is the portion of the unit ball in the first

octant.

2. Evaluate

∫ 3

−3

∫ √9−x2−√9−x2

∫ √9−x2−y2

0z√x2 + y2 + z2dzdydx by converting to spherical co-

ordinates.

3.3.2 Change of variable in Multiple Integrals

A change of variables can be useful when evaluating double or triple integrals. For instance,

changing from Cartesian coordinates to polar coordinates is often useful. A change of variables

can be described by a transformation.

Definition: A transformation T from the uv- plane to the xy- plane is a function which maps

points (u, v) to points (x, y) according to the rule:

T (u, v) = (x(u, v), y(u, v)) = (x, y).

A transformation T maps a region S in the uv− plane to a region R in the xy-plane called the

image of S. That is,

R = {T (u, v)|(u, v) ∈ S}.

We will usually work with C1 transformations which are transformations whose component

functions x(u, v) and y(u, v) have continuous first-order partial derivatives.

Example 3.3.6.

Find the image of the set S = {(u, v)|0 ≤ u ≤ 2, 0 ≤ v ≤ 1} under the transformation T

defined by the equations x = u− 2v and y = 2u− v.

Solution

The rectangular region S = [0, 2]× [0, 1] in the uv-plane is mapped to the parallelogram in the

xy-plane with vertices (0, 0), (2, 4), (−2,−1) and (0, 3).

Definition: The Jacobian of the transformation T defined by x = x(u, v) and y = y(u, v) is

∂(x, y)

∂(u, v)=

∣∣∣∣∣ ∂x∂u

∂x∂v

∂y∂u

∂y∂v

∣∣∣∣∣ =∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u.

Example 3.3.7.

Find the Jacobian of the transformation defined by x = r cos θ and y = r sin θ.

Solution The Jacobian of the transformation is

∂(x, y)

∂(r, θ)=

∣∣∣∣∣ cos θ −r sin θ

sin θ r cos θ

∣∣∣∣∣ = r(cos2 θ + sin2 θ) = r

39

Theorem 3.3.2.

Change of variables in a Double Integral: Suppose that T is a one-to-one C1 transformation

that maps a region S in the uv-plane to a region R in the xy-plane and whose Jacobian is

nonzero. Suppose that f is continuous on R and that R and S are type I and type II regions.

Then ∫ ∫Rf(x, y)dA =

∫ ∫Sf(x(u, v), y(u, v))

∣∣∣ ∂(x,y)∂(u,v)

∣∣∣ dudv (3.8)

Example 3.3.8.

Use the change of variables x = 2u and y = 3v to evaluate

∫ ∫Rx2dA, where R is the region

bounded by the ellipse 9x2 + 4y2 = 36.

solution

The given transformation maps the ellipse to a circle:

9x2 + 4y2 = 36⇒ x2

4+y2

9= 1⇒ u2 + v2 = 1.

The Jacobian of this transformation is

∂(x, y)

∂(u, v)=

∣∣∣∣∣ 2 0

0 3

∣∣∣∣∣ = 6.

By the previous Theorem we have∫ ∫Rx2dA = 6

∫ ∫4u2dA

= 24

∫ 2π

0

∫ 1

0r3 cos2 θdrdθ

= 6

∫ 2π

0cos2 θdθ

= 3

∫ π

0(1 + cos 2θ0dθ

= 3

[θ +

1

2sin 2θ

]2π0

= 6π.

Exercise 3.3.3.

1. Make an appropriate change of variables to evaluate

∫ ∫R

5 cos

(9y − xy + x

)dA, where R

is the trapezoidal region with vertices (2, 0), (5, 0), (0, 5) and (0, 2).

2. Use the transformation defined by x = uv and y = v to evaluate

∫ ∫R xydA, where R is the

region in the first quadrant bounded by the lines y = x and y = 3x and the hyperbolas

xy = 1 and xy = 3.

40

There is a similar change of variables formula for triple integrals. Consider a C1 transformation

T that maps a region S in uvw-plane to a region R in xyz-space according to the rule:

T (u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w)) = (x, y, x)

The Jacobian of this transformation is

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∣∣∂x∂u

∂x∂v

∂x∂w

∂y∂u

∂y∂v

∂y∂w

∂z∂u

∂z∂v

∂z∂w

∣∣∣∣∣∣∣∣Theorem 3.3.3.

Change of variables in Triple Integral: Suppose that T is a one-to-one C1 transformation that

maps a region S in the uvw-space to a region R in the xyz-space and whose Jacobian is

nonzero. Suppose that f is continuous on R and that R and S are type I, II, or III regions.

Then∫ ∫ ∫Rf(x, y, z)dxdydz =

∫ ∫ ∫Sf(x(u, v, w), y(u, v, w), z(u, v, w))

∣∣∣ ∂(x,y,z)∂(u,v,w)

∣∣∣ dudvdw

41

Chapter 4

Line and Surface Integrals

4.1 Line integrals of scalar fields

There are two ways to generalize the integral to functions of two or more variables. One way

is the line integral, which we shall consider in this section. The other is the multiple integral,

which was studied in the previous. In this section we define an integral that is similar to a

single integral except that instead of integrating over an interval [a, b], we integrate over a

curve C.

Definition: Consider a smooth curve C given by parametric equations

x = x(t), y = y(t), a ≤ t ≤ b

or equivalently, by the vector equation r(t) = x(t)i + y(t)j. Then, the line integral of f

along C is

∫cf(x, y)ds =

∫ b

af(x(t), y(t))

√(dx

dt

)2

+

(dy

dt

)2

dt. (4.1)

Remark In a special case where is the line segment that joins (a, 0) to (b, 0), using x as the

parameter, we can write the parametric equations as follows: x = x, y = 0, a ≤ x ≤ b.

Equation (4.1) becomes ∫cf(x, y)ds =

∫ b

af(x, 0)dx

and so the line integral reduces to an ordinary single integral in this case.

Example 4.1.1.

Find the line integral of

F(x, y) = sinx cos yi + exyj

along

42

(a) The horizontal line C1 : 0 ≤ x ≤ π, y =π

3.

(b) The vertical line C2 : 0 ≤ y ≤ 1, x = 2.

Solution

(a) ∫C1

F · dS =

∫ π

0sinx cos

(π3

)dx =

[−1

2cosx

]π0

= 1.

(b)

∫C2

F · dS =

∫ 1

0e2ydy =

1

2(e2 − 1).

Example 4.1.2.

(a) Evaluate

∫C

(2 + x2y)ds, where C is the upper half of the unit circle x2 + y2 = 1.

(b) Evaluate

∫c2xds, where C consists of the arc C1 of the parabola y = x2 from (0, 0) to

(1, 1) followed by the vertical line segment C2 from (1, 1) to (1, 2).

Solution

(a) We first parametrize the curve C as: x = cos t, y = sin t, 0 ≤ t ≤ π.

∫C

(2 + x2y)ds =

∫ π

0(2 + cos2 t sin t)

√(dx

dt

)2

+

(dy

dt

)2

dt

=

∫ π

0(2 + cos2 t sin t)

√sin2 t+ cos2 tdt

=

∫ π

0(2 + cos2 t sin t)dt

=

[2t− 1

3cos3 t

]π0

= 2π +2

3

(b) Since the graph is a function of x we can choose x as the parameter and the parametric

equations for C1 are

x = x, y = x2, 0 ≤ x ≤ 1

Therefore ∫C1

2xds =

∫ 1

02x

√(dx

dx

)2

+

(dy

dx

)2

dx

43

=

∫ 1

02x√

1 + 4x2dx

=5√

5− 1

6.

On C2 we choose y as the parameter, so the equations of C2 are

x = 1, y = y, 1 ≤ y ≤ 2

and ∫C2

2xds =

∫ 2

12(1)

√(dx

dy

)2

+

(dy

dy

)2

dy

=

∫ 2

12dy = 2.

Therefore

∫C

2xds =

∫C1

2xds+

∫C2

2xds =5√

5− 1

6+ 2.

Theorem Suppose that C is piecewise-sooth curve; that is, C is a union of finite number of

smooth curves C1, C2, · · · , Cn, The we define the integral of f along C as the sum of the

integrals of f along each of the smooth pieces of C:∫cf(x, y)ds =

∫C1

f(x, y)ds+

∫C2

f(x, y)ds+ · · ·+∫Cn

f(x, y)ds.

Properties of line integrals

1.

∫CP (x, y)dx+Q(x, y)dy =

∫CP (x, y)dx+

∫CQ(x, y)dy

2.

∫ (a2,b2)

(a1,b1)P (x, y)dx+Q(x, y)dy = −

∫ (a1,b1)

(a2,b2)P (x, y)dx+Q(x, y)dy

3.

∫ (a2,b2)

(a1,b1)P (x, y)dx+Qdy =

∫ (a3,b3)

(a1,b1)P (x, y)dx+Qdy +

∫ (a2,b2)

(a3,b3)P (x, y)dx+Qdy

where (a3, b3) is another point on C.

Exercise 4.1.1.

1.

∫C

(x− y)2dx along the parabola y = x− x2/4 from (0, 0) to (4, 0).

2.

∫C

(x− y)2dx around a circle of radius a traced counter-clockwise.

3.

∫C

(x2 + y2)dy around the ellipsex2

a2+y2

b2= 1 traced counter-clockwise.

4.

∫C

(xz + yz + xy)dy around the intersection of the cylinder x2 + y2 = 1 and the plane

y + z = 1 in the clockwise direction as viewed from the origin.

5. Evaluate

∫cy2dx+xdy, where (a) C = C1 is the line segment from (−5,−3) to (0,2) and

(b) C = C2 is the arc of the parabola x = 4− y2 from (−5,−3) to (0,2).

44

4.2 Line Integrals in Space

Suppose that C is a smooth space curve given by the parametric equations

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

or by a vector equation r(t) = x(t)i + y(t)j + z(t)k. If f is a function of three variables that

is continuous on some region containing C, the we define the line integral of f along C (with

respect to arc length) as :

∫cf(x, y, z)ds =

∫ b

af(x(t), y(t), z(t))

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt. (4.2)

Exercise 4.2.1.

Evaluate

∫Cy sin zds, where C is the circular helix given by the equations

x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.

4.3 Line Integrals of vector fields

In a vector filed the line integral can be thought in the notion of work done by a constant

force. If the vector field f(x, y) represents the force moving an object along a curve C, then

the work W done by this force is

W =

∫C

f ·Tds =

∫C

f · dr (4.3)

where T =r′(t)

‖r(t)‖is the unit vector to C.

Example 4.3.1.

Find the work done by the force field F(x, y) = x2i − xyj in moving a particle along the

quarter-circle r(t) = cos ti + sin tj, 0 ≤ t ≤ π2 .

Solution Since x = cos t and y = sin t, and r′(t) = − sin ti + cos tj and therefore the work done

is ∫C

F · dr =

∫ π2

0F(r(t)) · r′(t)dt =

∫ π2

0(cos2 ti− cos t sin tj) · (− sin ti + cos tj)dt

=

∫ π2

0(−2 cos2 t sin t)dt.

= −2

3.

Exercise 4.3.1.

45

Evaluate

∫C

F · dr, where F(x, y, z) = xyi + yzj + zxk and C is the twisted cubic given by

x = t, y = t2, z = t3, 0 ≤ t ≤ 1.

Theorem 4.3.1.∫cF · dr is independent of the path D if and only if

∫C

F · dr = 0 for every closed path C in D.

Theorem 4.3.2.

If F(x, y) = P (x, y)i +Q(x, y)j is a conservative vector field, where P and Q have continuous

first-order partial derivatives on a domain D the throughout D we have

∂P

∂y=∂Q

∂x

Example 4.3.2.

Determine whether or not the vector field

F(x, y) = (x− y)i + (x− 2)j

is conservative.

Solution Let P (x, y) = x− y and Q(x, y) = x− 2. Then

∂P

∂y= −1,

∂Q

∂x= 1

Therefore F is not conservative.

Remark If F is conservative then there exists a (potential) function f such that F = ∇f.

Exercise 4.3.2.

If F(x, y) = (3 + 2xy)i + (x2 − 3y2)j, find its function f such that F = ∇f.

4.4 Green’s Theorem

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. In stating Green’s Theorem

we use the convention that the positive orientation of simple closed curve C which refers to a

single counter-clockwise traversal of C.

Green’ Theorem: Let C be a positively oriented, piecewise-smooth, simple closed curve in the

plane and let D be the region bounded by C. If P and Q have continuous partial derivatives

on an open region that contains D, then∫CPdx+Qdy =

∫ ∫D

(∂Q

∂x− ∂P

∂y

)dA.

46

Example 4.4.1.

Evaluate

∮C

(3y − esinx)dx + (7x +√y4 + 1)dy, where C is the circle x2 + y2 = 9 oriented

counterclockwie.

Solution

The region D bounded by C is the disk x2 + y2 ≤ 9, so we need to change to polar coordinates

after applying Green’s Theorem:∮(3y − esinx)dx+ (7x+

√y4 + 1)dy =

∫ ∫ [∂

∂x

(7x+

√y4 + 1

)− ∂

∂y

(3y − esinx

)]dA

=

∫ 2π

0

∫ 3

0(7− 3)rdrdθ

= 36π.

Exercise 4.4.1.

1. Evaluate

∮Cy2dx+ 3xydy, where C is the boundary of the semi-annular region D in the

upper half-plane between the circles x2 + y2 = 1 and x2 + y2 = 4.

2. Evaluate

∮Cx4dx+xydy, where C is the triangular curve consisting of the line segments

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

3. Evaluate by Green’s Theorem the line integral∮C

y

x+ 1dx+ 2xydy

where C is the closed curve of the region bounded by y = x2 and y = x.

4. Verify Green’s theorem in the plane for∮C

(x3 − x2y)dx+ xy2dy

where C is the boundary of the region enclosed by circles x2 + y2 = 1 and x2 + y2 = 16.

4.5 Surface Integrals

In this section, we will use double integrals to compute the area of a surface defined by the

equation z = f(x, y).

Theorem: (Surface Area) Suppose that f has continuous first-order partial derivatives fx and

fy. Then the area of the surface defined by z = f(x, y), (x, y) ∈ D is

A =

∫ ∫D

√[fx(x, y)]2 + [fy(x, y)]2 + 1dA.

47

Example 4.5.1.

Find the area of the part of the hyperbolic paraboloid z = y2 − x2 that lies above the annular

region 1 ≤ x2 + y2 ≤ 4.

Solution The annular region D can be described in polar coordinates by

D = {(r, θ)|0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2}.

Note that f(x, y) = y2 − x2, and fx = −2x, fy = 2y. Thus

A =

∫ ∫ √(−2x)2 + (2y)2 + 1dA

=

∫ 2π

0

∫ 2

1

(√4r2 + 1

)rdrdθ

=π

6(5√

5− 1).

Exercise 4.5.1.

Find the area of the part of the surface z = x + y2 that lies above the triangle with vertices

(0,0), (1,1) and (0,1).

4.6 Divergence Theorem

In section 4.4 we presented the Green’s Theorem in a vector form as∫C

F · dr =

∫ ∫D

divF(x, y)dA

where C is the positively oriented boundary curve of the plane region D. If were are to extend

this theorem to a three dimensional fields we have

∫ ∫S

F · dr =

∫ ∫ ∫V

divF(x, y, z)dV

=

∫ ∫ ∫V∇ · FdV (4.4)

Equation (4.4) is the divergence theorem or Green’s theorem in space. It states that the sur-

face integral of the normal component of a vector F taken over a closed surface is equal to the

integral of the divergence of F taken over the volume enclosed by the surface.

Example: Evaluate

∫ ∫S

F · dS where F(x, y, z) = xyi + (y2 + exz2)j + sin(xy)k, and S is the

surface of the region E bounded by the parabolic cylinder z = 1 − x2 and the planes z = 0,

48

y = 0 and y + z = 2.

Solution

divF =∂

∂x(xy) +

∂

∂y(y2 + exz

2) +

∂

∂z(sinxy) = 3y.

Now we use Divergence Theorem to transform the given surface integral into a triple integral.

The easiest way to evaluate the triple is to express E as a type III region, that is:

E = {(x, y, z)| − 1 ≤ x ≤ 1, 0 ≤ z ≤ 1− x2, 0 ≤ y ≤ 2− z}

Thus ∫ ∫S

F · dS =

∫ ∫ ∫3ydV

= 3

∫ 1

−1

∫ 1−x2

0

∫ 2−z

0ydydzdx

= 3

∫ 1

−1

∫ 1−x2

0

(2− z)2

2dzdx

=3

2

∫ 1

−1

[−(2− z)3

3

]1−x20

dx

= −1

2

∫ 1

−1[(x2 + 1)3 − 8]dx

=184

35.

4.7 Stokes Theorem

Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem. Whereas

Green’s Theorem relates a double integral over a plane region D to a line integral around its

plane boundary curve, Stokes’ Theorem relates a surface integral over a surface S to a line

integral around the boundary curve of S (which is a space).

Stokes’ Theorem

Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-

smooth boundary curve C with positive orientation. Let F be a vector field whose components

have continuous partial derivatives on an open region <3 that contains S. Then∫cF · dr =

∫ ∫S

curlF · dS

=

∫ ∫S

(∇× F) · ndS

Example 4.7.1.

49

Use Stokes’ Theorem to evaluate

∫cF · dr, where F(x, y, z) = 3yi − xzj + yz2k and S is the

surface of the paraboloid 2z = x2 + y2 bounded by z = 2 and C is its boundary.

Solution

The boundary C of S is a circle with equations x2+y2 = 4, z = 2 and the parametric equations

for this case are x = 2 cos t, y = 2 sin, z = 2 with 0 ≤ t ≤ 2π. Then

Example 4.7.2.

Evaluate

∫C

F · dr, where F(x, y, z) = −y2i +xj + z2k and C is the curve of intersection of the

plane y + z = 2 and the cylinder x2 + y2 = 1.

Solution 4.7.1.

curlF =

∣∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

−y2 x z2

∣∣∣∣∣∣∣∣ = (1 + 2y)k.

Thus, ∫C

F · dr =

∫ ∫S

curl F · dS =

∫ ∫D

(1 + 2y)dA.

The projection D of S on the xy-plane is the disk x2 + y2 ≤ 1 and z = g(x, y) = 2− y. Hence

we can introduce parametric equations x = r cos θ, y = r sin θ, with 0 ≤ θ ≤ 2π, and 0 ≤ r ≤ 1

so that ∫ ∫D

(1 + 2y)dA =

∫ 2π

0

∫ 1

0(1 + 2r sin θ)rdrdθ

=

∫ 2π

0

[r2

2+ 2

r3

3sin θ

]10

dθ

=

∫ 2π

0

(1

2+

2

3sin θ

)dθ

=

[1

2θ − 2

3cos θ

]2π0

= π.

Exercise 4.7.1.

Use Stokes’ theorem, to evaluate∮C F · dr around the ellipse C : x2 + y2 = 1, z = y, where

F = xi + (x+ y)j + (x+ y + z)k.

THE END

50