Vector Analysis



Vector Analysis

Chapter 1 Algebra of Vectors

I. Elementary properties of vectors

Consider a Cartesian coordinate system, in which we define unit vectors, i, j, and k having the directions of positive x, y, and z axes, respectively. Any vector v whose projections on these axes are vx, vy, and vz, can be written as

v ’ vxi + vyj + vzk or v ’ [vx, vy, vz]

The number vx, vy, and vz are called the scalar components of v in the x, y, and z directions. If, when the initial point of v coincides with the origin, the angle measured to v from the positive x, y, and z axes are denoted by α, β, and γ, respectively, there follows

vx ’ vcosα, vy ’ vcosβ, vz ’ vcosγ,

where v ’ | v | ’[pic] is the magnitude or length of v, the number cosα, cosβ, and cosγ are called the direction cosines of v. It is clearly that the direction cosines satisfy the equation

[pic]

cos2α + cos2β + cos2γ ’ 1.

Addition of vectors: If a ’ a1i + a2j + a3k, b ’ b1i + b2j + b3k, then

a + b ’ (a1 + b1)i + (a2 + b2)j + (a3 + b3)k ’ b + a

ca ’ ca1i + ca2j + ca3k

A pair of nonparallel two vectors in a plane forms a basis of the plane. That is, any vector of the plane can be uniquely determined by a linear combination of the basis. Similarly a basis for three-space is obtained from three noncoplanar vectors.

1.

Prove that the diagonals of a parallelogram bisect each other.

Solution: Let [pic], then

[pic]

[pic]

m(a + b) ’ (1 − n)a + nb ( (m + n − 1)a + (m − n)b ’ 0

Since a and b are independent, m + n − 1 ’ 0, and m − n ’ 0. We get m ’ n ’ 1/2, and point O is the midpoint of the diagonals.

2.

[pic]

If [pic].

[pic]

3.

[pic]

If [pic].

[pic]

II. Inner product (Dot product or Scalar product)

Definition: a ( b ’ abcosθ,

where θ, 0 ( θ ( π, is the angle between a and b. In components

a ( b ’ a1b1 + a2b2 + a3b3

Properties: a ( b ’ b ( a

a ( a ’ a2

Theorem: The inner product of two nonzero vectors is zero if and only if the two vectors are perpendicular.

[pic]

Projection of a vector a in the direction of a vector b is

p ’ acosθ ’ a[pic]’ a ( eb,

where eb is the unit vector in the direction of b.

4.

An orthonormal basis for 3-space is {a, b, c}, a ’ b ’ c ’ 1, any vector v in the space can be written as

v ’ (v ( a)a + (v ( b)b + (v ( c)c.

5.

Find the straight line which passes through the two points P(x1, y1, z1), Q(x2, y2, z2).

Solution: Assume any point in the plane is X(x, y, z), then

[pic] ( [(x − x1)i + (y − y1)j + (z − z1)k] ’ t[(x2 − x1)i + (y2 − y1)j + (z2 − z1)k]

[pic] is the equation of the line.

6.

Find a plane which passes through the point P(x0, y0, z0) and is normal to the vector

N ’ ai + bj + ck.

Solution: Assume any point in the plane is X(x, y, z), [pic]

(ai + bj + ck) ( [(x − x0)i + (y − y0)j + (z − z0)k] ’ 0

a(x − x0) + b(y − y0) + c(z − z0) ’ 0 is the equation of the plane.

[pic]

[pic]

7.

Prove that the triangle inscribed in a semicircle is a right triangle.

Solution: Let [pic], then

[pic]

[Exercises] 1. Find all vectors v such that (v ( v)v ’ 169v and 4i ( v ’ 3j ( v ’ k ( v.

2. Find the angle between the planes 2x − y + 2z ’ 1 and x − y ’ 2.

3. Find the scalar and vector projections of −7i + 14j + 7k along 2i + 3j − 6k.

4. Find the magnitude of the scalar component of the force vector F ’ i + 2j + 2k in the direction of the straight line with equations x ’ y ’ 2z.

[Answers] 1. v ’ ( 3i ( 4j ( 12k 2. 45( 3. −2, (−4i − 6j + 12k)/7 4. 8/3

III. Outer product (Cross product, Vector product)

Definition: The outer product of two vectors a and b is a vector

v ’ a ( b

The magnitude of v is v ’ absinθ, 0 ( θ ( π, θ is the angle between a and b. The direction of v is the direction of advance of a right-hand screw as vector a is turned into vector b, and it is perpendicular to both a and b. In components, a ’ a1i + a2j + a3k, b ’ b1i + b2j + b3k, v ’ v1i + v2j + v3k

∵i ( i ’ j ( j ’ k ( k ’ 0,

i ( j ’ k, j ( k ’ i, k ( i ’ j,

j ( i ’ −k, k ( j ’ −i, i ( k ’ −j

∴v1 ’ a2b3 − a3b2, v2 ’ a3b1 − a1b3, v3 ’ a1b2 − a2b1

or v ’[pic]

[pic]

Properties: 1. a ( b ’ −b ( a

2. (a ( b) ( c ( a ( (b ( c) in general

3. If a ( b ’ 0, (1) then one of the two vectors is zero.

(2) a and b has the same or opposite direction.

4. | a ( b | ’ absinθ is the area of the parallelogram formed by the vectors a and b, as shown.

8.

[pic]

Find the distance d of point O to the line [pic].

Solution:[pic]

9.

Find the distance d between two lines which is not at the same plane.

[pic]

Solution: Suppose the unit vectors along the two lines are e1, e2 respectively, and P, Q are points on the two lines respectively, then

d ’[pic]

10.

Find the distance of two lines L1:[pic].

Solution: (1)L1的方向向量為v1 ’ 4i − 3j − k，L2的方向向量為v2 ’ 3i − 4j − 2k

設P(4m + 11,−3m − 5,−m − 7)為L1上的點，

Q(3n − 5,−4n + 4,−2n + 6)為L2上的點，且P與Q的距離為二歪斜線的距離，則

[pic](v1 ’ 0 ( [(3n − 4m − 16)i + (−4n + 3m + 9)j + (−2n + m + 13)k] ( (4i − 3j − k) ’ 0

4(3n − 4m − 16) − 3(−4n + 3m + 9) − (−2n + m + 13) ’ 0 ( 26n − 26m ’ 104……(

[pic](v2 ’ 0 ( [(3n − 4m − 16)i + (−4n + 3m + 9)j + (−2n + m + 13)k] ( (3i − 4j − 2k) ’ 0

3(3n − 4m − 16) − 4(−4n + 3m + 9) − 2(−2n + m + 13) ’ 0 ( 26n − 29m ’ 110……(

( − (: 3m ’ −6 ( m ’ −2，代入(得n ’ 2

m, n代回得P(3, 1, −5), Q(1, −4, 2)

二歪斜線的距離為[pic]

(2)設包含L1且平行L2的平面E為a(x − 11) + b(y + 5) + c(z + 7) ’ 0，則

E的法向量 ( L1 ( (ai + bj + ck) ( (4i − 3j − k) ’ 0 ( 4a − 3b − c ’ 0

E的法向量 ( L2 ( (ai + bj + ck) ( (3i − 4j − 2k) ’ 0 ( 3a − 4b − 2c ’ 0

a: b: c ’[pic]’ 2: 5: (−7)

平面E為2(x − 11) + 5(y + 5) − 7(z + 7) ’ 0 ( 2x + 5y − 7z − 46 ’ 0

L2過點A(−5, 4, 6)，則A到平面E的距離即為兩歪斜線的距離，其值為

[pic]

(3)L1的方向向量為v1 ’ 4i − 3j − k，L2的方向向量為v2 ’ 3i − 4j − 2k

A(11, −5, −7), B(−5, 4, 6)分別在直線L1, L2上

同時與v1, v2垂直的向量為

[pic]

11.

Derive the law of sines using vectors.

[pic]

Solution: c ’ b − a ( c ( c ’ c ( (b − a) ( 0 ’ c ( b − c ( a ( c ( b ’ c ( a

cbsinα ’ casin(π − β) ( [pic]

12.

Find the plane which contains three points A, B and C such that [pic] are not parallel.

Solution: Let X ’ (x, y, z), then

[pic]

where we can find the equation of the plane.

[Exercises] 1. Find a unit vector perpendicular to both i − 2j + k and −5i + 4j − 2k.

2. Find the equation of the line containing (1, 4, 3) which is perpendicular to both of the lines [pic].

3. Find parametric equations for the intersection of the planes 2x − y + z ’ −2 and x + y + z ’ 0.

4. Write the equation of the plane containing the lines [pic] and 2x’ 2−y’z.

[Answers] 1. ((j + 2k)/[pic] 2. [pic] 3. x ’ 2t, y ’ 1 + t, z ’ −1 − 3t

4. 2x + 2y + z ’ 4

IV. Multiple products

1. Scalar triple product (a ( b) ( c

Since a ( b is a vector perpendicular to the plane of a and b, having a magnitude equal to the area of the parallelogram of which a and b form coterminous sides, and the projection of c on a ( b is the altitude of the parallelepiped with a, b, and c as coterminous edges, it follows that (a ( b) ( c is equal to the volume of this parallelepiped. From the fact it follows easily that

(a ( b) ( c ’ (b ( c) ( a ’ (c ( a) ( b

Since (b ( c) ( a ’ a ( (b ( c), we get (a ( b) ( c ’ a ( (b ( c). It follows that the dot and cross can be interchanged in a scalar triple product, the notation (a b c) is frequently used to indicate the common value of the products, and in components

(a b c) ’[pic]

2. Vector triple product (a ( b) ( c

Since (a ( b) ( c is perpendicular to a ( b, which is itself perpendicular to the plane of a and b, and also perpendicular to c, it follows that (a ( b) ( c lies in the plane of a and b and perpendicular to c. Thus

(a ( b) ( c ’ ma + nb ( 0 ’ ma ( c + nb ( c

If we write n ’ λa ( c, then m ’ −λb ( c, and

(a ( b) ( c ’ λ[(a ( c)b − (b ( c)a] (1.1)

To determine λ, let u be a unit vector parallel to a, v be a second unit vector perpendicular to a such that b is in the plane of u and v, and w ’ u ( v, then

a ’ a1u, b ’ b1u + b2v, c ’ c1u + c2v + c3w

Substitute into (1.1), there follows

[a1u ( (b1u + b2v)] ( (c1u + c2v + c3w) ’ λ{[a1u ( (c1u + c2v + c3w)](b1u + b2v) − [(b1u + b2v) ( (c1u + c2v + c3w)](a1u)}

a1b2w ( (c1u + c2v + c3w) ’ λ[a1c1(b1u + b2v) − (b1c1 + b2c2)(a1u)]

a1b2c1v −a1b2c2u ’ λ(a1b2c1v − a1b2c2u)

where we can get λ ’ 1

and

(a ( b) ( c ’ (a ( c)b − (b ( c)a

Similarly

a ( (b ( c) ’ (a ( c)b − (a ( b)c

[Note] (a ( b) ( c ( a ( (b ( c)

3. (a ( b) ( (c ( d) (1.2)

Let u ’ c ( d, (1.2) becomes

(a ( b) ( u ’ a ( (b ( u) ’ a ( [b ( (c ( d)] ’ a ( [(b ( d)c − (b ( c)d]

’ (a ( c)(b ( d) − (a ( d)(b ( c)

4.(a ( b) ( (c ( d)

Let u ’ c ( d,

(a ( b) ( (c ( d) ’ (a ( b) ( u ’ (a ( u)b − (b ( u)a ’ [a ( (c ( d)]b − [b ( (c ( d)]a

’ (a c d)b − (b c d)a

If we let v ’ a ( b,

(a ( b) ( (c ( d) ’ v ( (c ( d) ’ (v ( d)c − (v ( c)d ’ [(a ( b) ( d]c − [(a ( b) ( c]d

’ (a b d)c − (a b c)d

[Exercise] Derive the identity (A ( B) ( (B ( C) ( (C ( A) ’ (A B C)2

Chapter 2 Differential Calculus of Vector

I. Derivative of a vector

Definition: A vector function v(t) is said to be differentiable at a point t if the limit

v((t) ’[pic]

exists. The vector v((t) is called the derivative of v(t). In terms of components with respect to a given Cartesian coordinate system, the derivative v((t) is obtained by differentiating each component separately

v((t) ’ v(1(t)i + v(2(t)j + v(2(t)k

It follows

(u + v)( ’ u( + v(

(u ( v)( ’ u( ( v + u ( v(

(u ( v)( ’ u( ( v + u ( v(

1.

Let v(t) be a vector function, whose length is constant. Then v ( v ’ v2 ( (v ( v)( ’ 0, we get

2v ( v( ’ 0. This yields the important result: the derivative of a vector function v(t) of constant length is either the zero vector or is perpendicular to v(t).

II. Geometry of a space curve

1. Arc length

A Cartesian coordinate system being given, we may represent a curve C by a vector function

r(t) ’ x(t) i + y(t) j + z(t) k

v(t) ’ r((t) ’ x((t) i + y((t) j + z((t) k

If s represents arc length along the curve

v(t) ’ r((t) ’[pic]

The expression in parentheses is clearly a unit vector tangent to the curve at point P, we denote this unit vector by T, and

[pic]

The arc length is

[pic] (2.1)

2.

Find the tangent to the ellipse [pic].

Solution: r(t) ’ 2cost i + sint j, and P corresponds to t ’ π/4

r((t) ’ −2sint i + cost j

r((π/4) ’ −[pic]

3.

Find the arc length of a circular helix: r(t) ’ a cost i + a sint j + ct k, (c ( 0) 0 ( t ( b.

Solution: v(t) ’ −a sint i + a cost j + c k

v ( v ’ a2 + c2

s ’[pic]

4.

Find the length of the space curve: y ’ sin2πx, z ’ cos2πx, from (0, 0, 1) to (1, 0, 1).

Solution: r ’ x i + sin2πx j + cos2πx k ( dr/dx ’ i + 2πcos2πx j − 2πsin2πx k

s ’[pic]

5.

Find the length of the curve: r ’ a(1 − cosθ), 0 ( θ ( 2π, a > 0.

Solution: r ’ rer ([pic]’ asinθ er + r eθ ’ asinθ er + a(1 − cosθ) eθ

[pic]

2. Frenet formula

We have a unit tangent vector T ’[pic], and define

[pic]’ κN, κ ( 0 (2.2)

where κ is the magnitude of [pic] and is called the curvature, and N is a unit vector normal to T and is called the normal or principal normal to the curve. The reciprocal of the curvature, ρ ’[pic], is called the radius of curvature. In addition to the two perpendicular vector T and N, a third unit vector known as the binormal vector, is definede by

B ’ T ( N

then

[pic]

Since [pic] is perpendicular to B, and from the right hand side of the above equation, [pic] is perpendicular to T, [pic] can be written in the form

[pic]’ −τN (2.3)

where τ is the torsion or twisting of the curve.

[pic]’ −τN ( T + B ( κN ’ −κT + τB (2.4)

The three formulas, equations (2.2), (2.3), and (2.4) are called Frenet formulas.

In general, the position vector r is a function of t, to find κ and τ, we differentiate r with respect to t:

[pic]

6.

Given a space curve r(t) ’ a cost i + a sint j + ct k, find (1)curvature (2)torsion (3)unit tangent, normal, and binormal vectors.

[pic]

Substituting into a, we have N ’ −costi − sintj

[pic]

7.

Show that for a curve y ’ y(x) in the x-y plane, κ(x) ’[pic]

Solution: r ’ x i + y j

v ’ i + y( j ’ v T, ( v ’[pic]

a ’ y(j ’[pic]T + κv2N

v ( a ’ κv3 B ( y(k ’ κv3 B ( κ ’[pic]

[Exercise] The position vector of a particle P is r(t) ’ t i + t2 j + 2t3/3 k. Find (1) the velocity, acceleration, and speed of P, (2) the tangential and centripetal components of acceleration, (3) the curvature of the curve C traversed by P, and (4) the minimum radius of curvature of C.

[Answer] (1) v ’ i + 2t j + 2t2 k, a ’ 2j + 4t k, v ’ 1 + 2t2 (2) at ’ 4t, an ’ 2 (3) κ ’ 2/(1 + 2t2)2 (4) ρmin ’ 1/2

III. Gradient

Definition: For a given scalar function f(x, y, z) the gradient of f is the vector function

grad f ’ (f ’[pic]

where ( ’[pic], read del, is a differential operator. Then

[pic]

where T is the unit tangent vector to the curve described by r(s). Since (f ( T ’ | (f | cosθ, where θ is the angle between (f and T, then [pic]’ | (f | cosθ; so the component of (f in the direction of T is the directional derivative of f in that direction. It readily follows from this result that (f is in the direction in which [pic] has its maximum values; this maximum value is equal to | (f |.

Consider a surface S in space given by f(x, y, z) ’ constant. As long as we move along this surface on any curve C, f has constant value and consequently [pic]’ 0. Thus

[pic]( (f ( T

where r(s) is the position vector describing the curve C. Since (f is completely determined by f and C is any curve on the surface, (f must be normal to the surface.

8.

Find a unit normal vector n of the cone of revolution z2 ’ 4(x2 + y2) at point P(1, 0, 2).

Solution: The cone is the level surface f ’ 0 of f(x, y, z) ’ 4(x2 + y2) − z2, Thus

(f ’ 8x i + 8y j − 2z k

and at P

(f|(1, 0, 2)’ 8i − 4k

n ’[pic]

9.

Find the directional derivative of f(x, y, z) ’ 4x2y + 3y2 + 2xz2 at the point (1, −1, 0) in the direction of the vector b ’ −3i + 4j.

Solution: (f ’ (8xy + 2z2) i + (4x2 + 6y) j + 4xz k

(f|(1, −1, 0)’ −8i − 2j

eb ’[pic]

(f ( eb ’ 16/5

[Exercise] The temperature at any point in space is given by T ’ xy + yz + zx.

(1) Find the direction cosines of the direction in which the temperature changes most rapidly with distance from the point (1, 1, 1), and determine the maximum rate of change.

(2) Find the derivative of T in the direction of the vector 3i − 4k at the point (1, 1, 1).

[Answer] (1)((i + j + k)/[pic] (2)−2/5

IV. Divergence of a vector field

Definition: Let v(x, y, z) be a differentiable vector function and v ’ v1i + v2j + v3k. Then the function

div v ’ (( v ’[pic]

is called the divergence of v.

For a small rectangular solid with edges (x, (y, and (z. In the y direction the outflow of v through the right-hand face is v2(x, y + (y, z)(x(z, through the left-hand face is −v2(x, y, z) (x(z, the total outflow in the y direction is the sum

[v2(x, y + (y, z) − v2(x, y, z)](x(z

The difference in these values of v2 is

[pic](y

Thus the net outflow from the two faces is

[pic](x(y(z

Similarly, the two faces in the x and z direction contribute respectively

[pic](x(y(z and [pic](x(y(z

After we divide by the volume (x(y(z, we have

(( v ’[pic]

as the net outflow per unit volume.

10.

If F ’ x i + y2z j + xz3 k, find (( F.

Solution: (( F ’ 1 + 2yz + 3xz2

11.

Find (( r, given r ’ xi + yj + zk.

Solution: (( r ’ 1 + 1 + 1 ’ 3

V. Curl of a vector field

Definition: Let v(x, y, z) ’ v1i + v2j + v3k be a differentiable vector function. Then the function

curl v ’ ( ( v ’[pic]

is called the curl of v or the rotation of v.

12.

Let ω ’ ω1i + ω2j + ω3k represent the angular velocity of a rigid body, the velocity field is v ’ ω ( r ’ (ω1i + ω2j + ω3k) ( (xi + yj + zk). The curl of v is

( ( v ’[pic]’ 2ω

Hence, the curl of the velocity field in the case of the rotation of a rigid body has the direction of the axis of rotation, and its magnitude equals twice the angular speed ω of the rotation.

VI. Vector identities

((φ1φ2) ’ φ1(φ2 + φ2(φ1

( ( (φF) ’ (φ ( F + φ( ( F

( ( (φF) ’ (φ ( F + φ( ( F

(f(u) ’[pic](u

( ( (F ( G) ’ G ( (( ( F) − F ( (( ( G)

( ( (F ( G) ’ (G ( ()F − (F ( ()G + (( ( G)F − (( ( F)G

((F ( G) ’ (F ( ()G + (G ( ()F + F ( (( ( G) + G ( (( ( F)

( ( (( ( F) ’ ((( ( F) − (2F

( ( r ’ 3

( ( r ’ 0

( ( ((φ) ’ 0

( ( (( ( F) ’ 0

( ( ((φ1 ( (φ2) ’ 0

Tensor notations:

a ( b ’ ai bi

a ( b ’ εijk aj bk

(φ ’ (i φ

(2φ ’ ( ( ((φ) ’ (i (i φ

( ( F ’ (i Fi

( ( F ’ εijk (j Fk

where

εijk ’[pic]

and

εijkεips ’ δjp δks − δjs δkp

where

δij ’[pic]

13.

Prove (A ( B) ( C ’ (A ( C)B − (B ( C)A.

Solution: (A ( B) ( C ’ εijk (A ( B)j Ck ’ εijk εjlm Al Bm Ck

’ εjki εjlm Al Bm Ck ’ (δkl δim − δkm δil) Al Bm Ck

’ Ak Bi Ck − Ai Bk Ck

’ (A ( C)B − (B ( C)A

14.

Prove ( ( (F ( G) ’ (G ( ()F − (F ( ()G + (( ( G)F − (( ( F)G.

Solution: ( ( (F ( G) ’ εijk (j (F ( G)k ’ εijk εklm (j (Fl Gm)

’ εkij εklm (j (Fl Gm) ’ (δil δjm − δim δjl) (j (Fl Gm)

’ (j (Fi Gj) − (j (Fj Gi)

’ (j Fi Gj + Fi (j Gj − (j Fj Gi − Fj (j Gi

’ (G ( ()F + (( ( G)F − (( ( F)G − (F ( ()G

15.

Prove ( ( ((φ) ’ 0.

Solution: ( ( ((φ) ’ εijk (j ((φ)k ’ εijk (j (k φ

Since (j (k φ ’ (k (j φ and εijk ’ −εikj

( εijk (j (k φ ’ 0

16.

Prove (v ( ()v ’[pic](|v|2 − v ( (( ( v).

Solution: v ( (( ( v) ’εijk vj (( ( v)k ’ εijk vj (εklm (l vm)

’ εkij εklm vj (l vm ’ (δil δjm − δim δjl)vj (l vm

’ vj (i vj − vj (j vi ’[pic](i (vj vj) − vj (j vi

’[pic]((v ( v) − (v ( ()v

’[pic](|v|2 − (v ( ()v

( (v( ()v ’ [pic](|v|2 − v ( (( ( v)

[Exercise] Show that ((v ( v) ’ 2v ( (v + 2v ( (( ( v).

Chapter 3 Vector Integral Calculus

I. Line integrals

1. Definition

A line integral of a vector function F(r) over a curve C is

[pic]

In terms of components

[pic]

If x, y, and z are function of t, we have

[pic]

1.

Find the value of the line integral [pic], when F ’ −yi + xyj and C is the circular arc as shown from A to B.

Solution: We may represent C by

r ’ costi + sintj 0 ( t ( π/2

and

F ’ −sinti + costsintj

[pic]

2.

Find the value of the line integral [pic], when f ’ x2 + x3y and C is the line y ’ 2x from (0, 0) to

(2, 4).

Solution: We may represent C by r ’ xi + yj ( dr ’ dxi + dyj

ds ’ | dr | ’[pic]

and

f ’ x2 + x3y ’ x2 + x3(2x) ’ x2 + 2x4

[pic]

[Exercise] Evaluate the line integral [pic], with F ’ 5zi + xyj + x2zk from point A(0, 0, 0) to point B(1, 1, 1) along

(1) C: the straight line r ’ ti + tj + tk.

(2) C: the parabolic arc r ’ ti + tj + t2k.

[Solution](1) F ’ 5ti + t2j + t3k, dr ’ (i + j + k)dt

[pic]’ 37/12

(2) F ’ 5t2i + t2j + t4k, dr ’ (i + j + 2tk)dt

[pic]’7/3

2. Conservative fields

A vector field F is said to be conservative if there can be found some scalar φ such that

F ’ (φ, i.e., [pic]. Then φ is called a potential function or simply potential for F, and

F ( dr ’ (φ ( dr ’[pic]’ dφ

The line integral from A to B along a curve C is

[pic]’ φ(B) − φ(A)

This shows that the value of line integral is simply the difference of the value of φ at the two endpoints of C and is independent of the path C. Hence the line integral around a closed curve of a conservative field is zero.

Since

( ( ((φ) ’ 0 ( ( ( F ’ 0

There follows a fact that if and only if ( ( F ’ 0, the vector field F is conservative.

3.

Find the line integral of [pic], with F ’ 2xi + 2yj + 4zk from point A(0, 0, 0) to point B(2, 2, 2) along C: the straight line r ’ ti + tj + tk.

Solution: ∵ ( ( F ’ 0 ∴the line integral is independent of the path, and the potential is

[pic]

φ ’ x2 + y2 + 2z2

and

[pic]’ φ(2, 2, 2) − φ(0, 0, 0) ’ 16

[pic]

II. Surface integrals

Definition: the flux of F through the surface S to be the surface integral

[pic],

where n is the unit normal vector to S.

From the figure shown, ds | cosγ | ’ dxdy, γ is the angle between the normal of ds and z-axis.

n ( k ’ cosγ

[pic]

Similarly, if α and β are the angle between the normal of ds and x and y axes respectively,

[pic]

[pic]

When the surface S is parametrized by r(u,v)

[pic]

If F ’ n, the surface integral is area of the surface itself.

4.

[pic]

Given F ’ x2i + 3y2k and S is the portion of the plane x + y + z ’ 1 in the first octant, evaluate [pic].

Solution: Let f ’ x + y + z − 1 the unit normal vector n of S is

[pic]

5.

Calculate the surface integral of the vector function F ’ xi + yj over the portion of the surface of the unit sphere S: x2 + y2 + z2 ’ 1 above the xy-plane, z ( 0.

Solution: (1)Let f ’ x2 + y2 + z2, the unit normal vector of S is

[pic]

Let x ’ rcosθ, y ’ rsinθ, we have

[pic]

(2)Let x ’ sinθcosφ, y ’ sinθsinφ, z ’ cosθ, then r ’ sinθcosφi + sinθsinφj +cosθk,

F ’ sinθcosφi + sinθsinφj, and

[pic]

III. Volume integral

Definition: the volume integral of a function f over the volume V is

[pic]

In Cartesian coordinate system, dv ’ dxdydz. If f ’ 1, the volume integral is volume of V.

6.

Find the volume integral of f(x, y, z) ’ x + yz over the box bounded by the coordinate planes, x ’ 1, y ’ 2, and z ’ 1 + x.

[pic]

[pic]

7.

[pic]

Find the volume of the region of space above the xy plane and beneath the plane z ’ 2 + x + y, bounded by the planes y ’ 0, x ’ 0, and the surface y ’ 1 − x2.

[pic]

IV. Divergence theorem (Gauss theorem)

Divergence theorem: Let V be a closed bounded region in space whose boundary is a piecewise smooth oriented surface S. Let F be a vector function that is continuous first partial derivative in some domain containing V. Then

[pic]

8.

[pic]

Solution: (1) Let f ’ x2 + y2 + z

[pic]

[Exercises]

1. Evaluate [pic], S is the lateral surface of the portion of the cylinder x2 + y2 ’ 1 for which 0 ( z ( 1.

[Solution] (1) Let f ’ x2 + y2 − 1, then n ’ xi + yj

[pic]

2. Evaluate[pic], with S: x2 + y2 + z2 ’ 4.

[Solution] (1)n ’ (xi + yj + zk)/2

[pic]

(2)( ( (7xi − zk) ’ 7 − 1 ’ 6

[pic]

V. Stokes theorem

Stokes theorem: Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let F be a continuous vector function that has continuous first partial derivative in a domain in space containing S. Then

[pic]

[Note] The positive direction along C is defined as the direction along which an observer, traveling on the positive side of S, would proceed in keeping the enclosed area to his left.

9.

Find the surface integral [pic], with F ’ yi + zj + xk and S: the paraboloid z ’ 1 − (x2 + y2), z ( 0.

[pic]

[pic]

10.

Evaluate [pic], where C is the circle x2 + y2 ’ 4, z ’ −3, oriented counterclockwise as seen by a person standing at the origin, and F ’ yi + xz3j − zy3k.

Solution:

(1) Let x ’ 2cosθ, y ’ 2sinθ

F ( dr ’ (2sinθi − 54cosθj + 24sin3θk) ( (−2sinθi + 2cosθj)dθ

’ (−4sin2θ − 108cos2θ)dθ

[pic]’ −112π

(2) As surface S bounded by C we can take the plane circular disk x2 + y2 ( 4 in the plane z ’ −3. Then n ’ k, and ( ( F| z’−3 ’ (9y2 − 27x)i − 28k, (( ( F) ( n ’ −28

[pic]’ −28(π ( 22) ’ −112π

VI. Green’s theorem

If F ’ F1i + F2j is a vector function that is continuously differentiable in a domain in the xy-plane containing a simply connected bounded close region S whose boundary C is a piecewise smooth simple closed curve. Then the Stokes theorem can be reduced to

[pic]

which is the Green’s theorem.

11.

Evaluate [pic], C is a closed curve from (0, 0) to (1, 1) along y ’ x2 and back to (0, 0) along y ’ x.

Solution:

[pic]

(1) [pic]

(2) [pic]

12.

Evaluate the line integral [pic]F(dr counterclockwise around the boundary C of the region R, where F ’ xsinyi − ysinxj, R is the rectangle: 0 ( x ( π, 0 ( y ( π/2.

Solution:

(1) On C1: y ’ 0, dy ’ 0, F ’ 0([pic];

On C2: x ’ π, dx ’ 0, F ’ πsinyi ([pic];

On C3: y ’ π/2, dy ’ 0, F ’ xi −[pic];

On C4: x ’ 0, dx ’ 0, F ’ 0 ([pic].

[pic].

[pic]

-----------------------

[pic]

[pic]

[pic]

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download