MATH2420
Multiple Integrals and Vector Calculus
Prof. F.W. Nijhoff
Semester 1, 2007-8.
Course Notes
and
General Information
Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. In ordinary differential and integral calculus, you have already seen how derivatives and integrals interrelate. A derivative can be used as the opposite of an integration; it also occurs in changing variables in an integral. The same interrelation applies in multiple dimensions, but with more richness and variety.
This module starts with a discussion of different coordinate systems in two and three dimensions. The use of Cartesian, plane polar, cylindrical polar and spherical polar coordinates will run through the whole module.
The second section starts with a discussion of vector functions, which are the two- and three-dimensional equivalents of the functions of ordinary calculus. These can be used to describe curves in space. Next we look at functions of several variables: that is, functions of a vector. With these two concepts we can introduce derivatives for fully three-dimensional functions (gradient, divergence and curl).
This brings us to the halfway point of the module, and we will pause to review our new understanding before moving on to multiple-dimensional integrals. Here we extend the familiar idea of integration in one dimension to integration over an area or a volume.
Finally, with the introduction of line and surface integrals we come to the famous integral theorems of Gauss and Stokes. These encompass beautiful relations between line, surface and volume integrals and the vector derivatives studied at the start of this module.
Most real-life problems are not one-dimensional. The amount of heat stored in a piece of metal can be calculated by integrating its temperature in three dimensions; and the diffusion of dye in water is governed by differential equations based on three-dimensional derivatives. This is why a knowledge of vector calculus is essential for further study in many areas of applied mathematics.
Chapter 0
REVIEW
0.1
Calculus
Differentiation
The curvey=f(x) has a slope at pointx=agiven by thederivativeoff with respect toxata:
f′(a) = df dx a = lim ∆x→0 f(a+ ∆x)−f(a) ∆x . (1)
A few particular derivatives are:
f(x) =axn f′(x) =anxn−1 f(x) =eλx f′(x) =λeλx
f(x) =u(x)v(x) f′(x) =u′(x)v(x) +u(x)v′(x)
f(x) =u(x)/v(x) f′(x) = [u′(x)v(x)−u(x)v′(x)]/v2(x).
Integration
Integration is the opposite of differentiation:
Z b a f′(x) dx= [f(x)]b a=f(b)−f(a) or Z f′(x) dx=f(x) (2)
and it may also be seen as giving the area under a curve – so the integralRb
af(x) dxgives the area under
the curvey=f(x) betweenx=aandx=b. Some specific integrals are:
Z xndx=xn+1/(n+ 1) forn6=−1 Z x−1dx= lnx Z eλxdx=eλx/λ Z u(x)v′(x) dx= [u(x)v(x)]− Z u′(x)v(x) dx
0.2
Lines and Circles
The vector equation of a straight line in three-dimensional space is
x=a+ubwithu∈(−∞,∞) a real scalar. (3) 2
The equation of a circle of radiusr0centred about the origin isx2+y2=r20 and if the circle is centred
on the pointP(a, b) the the equation is (x−a)2+ (y−b)2=r2 0.
0.3
Trigonometry
Recall the functions sinxand cosx, with the identities:
sin2x+ cos2x = 1 (4)
tanx = sinx/cosx (5)
and the derivatives
(d/dx) sinx = cosx (6)
(d/dx) cosx = −sinx (7)
(d/dx) tanx = 1 + tan2x. (8)
There are memorable values:
x sinx cosx 0 0 1 π/2 1 0 π 0 −1 3π/2 −1 0 2π 0 1
0.4
Determinant
The determinant of a 3x3 matrixAis
|A|= a11 a12 a13 a21 a22 a23 a31 a32 a33 =a11(a22a33−a32a23)−a12(a21a33−a31a23) +a13(a21a32−a31a22) (9)
0.5
Vector Products
We consider two vectorsa= (a1, a2, a3) andb= (b1, b2, b3). Their scalar dot productis given by
a·b=a1b1+a2b2+a3b3 (10)
and theirvector cross productby
a×b= i j k a1 a2 a3 b1 b2 b3 = ([a2b3−b2a3],[a3b1−b3a1],[a1b2−a2b1]). (11)
Note that although for the scalar producta·b=b·a, for the vector product we havea×b=−b×a. Recall that
a·b=||a|| ||b||cosθ and ||a×b||=||a|| ||b||sinθ (12)
Chapter 1
COORDINATE SYSTEMS
1.1
Cartesian Coordinates
We are familiar with the rectangularCartesian coordinates x, y in a plane, or x, y, z in space. If the unit vectors in thex,y andz directions are given byi,j andk then we can write the position vectorr
of a point as
r = (x, y, z) (1.1)
= xi+yj+zk. (1.2)
These two notations will be used interchangeably in this course.
1.2
Plane Polar Coordinates
In a fixed frame of reference the rectangular coordinates (x, y) are related to the polar coordinates (r, θ) through the relations
x=rcosθ, y=rsinθ, r >0, θ∈[0,2π) (1.3) r= (x2+y2)1/2, θ= arctany x . (1.4)
1.3
Cylindrical Coordinates
In a fixed frame of reference the Cartesian coordinates (x, y, z) are related to the cylindrical coordinates (r, θ, z) through the relations:
x=rcosθ, y=rsinθ, z=z, r >0, θ∈[0,2π) (1.5) r= (x2+y2)1/2, θ= arctany x , z=z. (1.6)
1.4
Spherical Coordinates
In a fixed frame of reference the Cartesian coordinates (x, y, z) are related to the spherical coordinates (ρ, θ, φ) through the relations:
x=ρsinθcosφ, y=ρsinθsinφ, z=ρcosθ, (1.7)
ρ >0, θ∈[0, π), φ∈[0,2π) (1.8) 4
θ
φ
x
ρ
Course Notes 6
ρ= (x2+y2+z2)1/2, φ= arctany
x
, θ= arccos [z(x2+y2+z2)−1/2] (1.9)
In the above equations,θis the latitude or polar angle, andφis the longitude.
Remark: Spherical coordinates are commonly used in applications where there is a centre of symmetry. The centre of symmetry is then taken as the origin.
VECTOR CALCULUS
2.1
Vector Functions
Avector-valued functionf is avector functionwhose components aresingle-valued functions(scalar valued functions).
For example, given three single-valued functions f1(t), f2(t), f3(t) we can form the vector-valued
function
f(t) = (f1(t), f2(t), f3(t)) =f1(t)i+f2(t)j+f3(t)k (2.1)
Themagnitudeof the vector-valued functionf(t) is a scalar-valued function and is defined by
||f(t)||=
f2
1(t) +f22(t) +f32(t)
1/2
(2.2) In general, the graph of the vector function f(t) =f1(t)i+f2(t)j+f3(t)kis a curveC, in the sense
that, astvaries, the tip of the position vectorf(t) traces outC. The equations
x=f1(t), y=f2(t), z=f3(t) (2.3)
corresponding to the components off are theparametricequations ofC. If one of the components is zero, e.g. f(t) =f1(t)i+f2(t)j, thenC is said to be aplanar curve, otherwiseC is aspace curve.
2.1.1
Derivative of a vector function
Given the vector functionf(t) = (f1(t), f2(t), f3(t)) thederivativeoff is defined by f′(t) = (f′
1(t), f2′(t), f3′(t)). Properties of the derivative
• (f+g)′(t) =f′(t) +g′(t)
• (αf)′(t) =αf′(t) whereαis a constant scalar
• (uf)′(t) =u(t)f′(t) +u′(t)f(t), where uis a scalar function
• (f·g)′(t) =f(t)·g′(t) +f′(t)·g(t).
• (f×g)′(t) =f(t)×g′(t) +f′(t)×g(t).
• (f(u))′(t) =f′(u(t))u′(t).
Course Notes 8
2.1.2
Integral of a vector function
Given the vector function f(t) = (f1(t), f2(t), f3(t)), the integral of f is defined by Rabf(t) dt =
(Rb af1(t) dt, Rb a f2(t) dt, Rb a f3(t) dt).
Properties of the integral
• Rb a(f(t) +g(t)) dt= Rb af(t) dt+ Rb ag(t) dt • Rb a(αf)(t) dt=α Rb
af(t) dt, whereαis a constant scalar
• Rb
a(c·f)(t) dt=c(t)·
Rb
af(t) dt, wherec is a constant vector.
• Rb
a(c×f)(t) =c(t)×
Rb
a f(t) dt, wherec is a constant vector.
2.2
Curves
1. The equation ofa straight lineis parametrised by
r(t) =r0+td, t∈(−∞,∞) (2.4)
2. More generally, every vector function
r(t) =x(t)i+y(t)j+z(t)k (2.5) parametrises acurve in space(or acurve in planeif one of the componentsx(t),y(t) orz(t) is zero). It is also important to understand that a parametrised curve Cis an orientedcurve in the sense that as t increases on some interval of definitionI, the tip of the position vectorr(t) traces outC in a certain direction. For example, the unit circle parametrised by
r(t) = cos(t)i+ sin(t)j, t∈[0,2π) is traversed in the anticlockwise direction, starting at the point (1,0).
2.2.1
Tangent vector, tangent line
For a given curveC parametrised by
r(t) = (x(t), y(t), z(t)),
the derivative vector
r′(t) = (x′(t), y′(t), z′(t))
is called the tangent vector to the curve C at the point P(x(t), y(t), z(t)), and r′(t) points out in the
direction of increasingt.
For a given curveCparametrised byr(t) = (x(t), y(t), z(t)) thetangent lineat a pointtis the vector function
2.2.2
Intersecting curves
Two curves
(C1) : r1(t) =x1(t)i+x2(t)j+x3(t)k,
(C2) : r2(u) =y1(u)i+y2(u)j+y3(u)k,
intersect iff there are numbers t and u for which r1(t) = r2(u). The angle between two intersecting
curves (which, by definition, is the angle between the corresponding tangent lines) can be obtained by examining the tangent vectors at the point of intersection.
2.2.3
The unit tangent
For a given curveC parametrised byr(t) = (x(t), y(t), z(t)), the vector
T := r′(t)
||r′(t)|| (2.7)
is called theunit tangent vectorto the curveC at the point P(x(t), y(t), z(t)). The unit tangent points in the direction of increasingt along the curve and is parallel to the curve.
2.2.4
Reversing the sense of a curve
We make a distinction between the curve
r=r(t), t∈[a, b] (2.8) and the curve
R(u) =r(a+b−u), u∈[a, b]. (2.9) Both vector functions trace out the same set of points, but the order has been reversed. Whereas the first curve starts atr(a) and ends atr(b), the second curve starts atr(b) and ends at r(a).
Thus, for example, the vector function
r(t) = cos(t)i+ sin(t)j, t∈[0,2π) gives the unit circle traversed anticlockwise while the reversed curve
R(u) = cos(2π−u)i+ sin(2π−u)j, u∈[0,2π) gives the unit circle traversed clockwise.
Unit tangent
When we reverse the sense of a curve, the unit tangentT reverses direction (is multiplied by−1) because it always points in the direction of increasingtor u.
2.3
Arc Length
The length of a continuously differentiable curve (C): r=r(t),t∈[a, b] is given by
L(C) =
Z b
a
Chapter 3
FUNCTIONS OF SEVERAL
VARIABLES
3.1
Introduction
Since we live in a three-dimensional world, in applied mathematics we are interested in functions which can vary with any of the three space variablesx,y,zand also with timet. For instance, if the functionf
represents the temperature in this room, thenf depends on the location (x, y, z) at which it is measured and also on the timet when it is measured, sof is a function of the independent variablesx, y,z and
t, i.e. f(x, y, z, t).
3.2
Geometric Interpretation
For a function of two variables, f(x, y), consider (x, y) as defining a point P in thexy-plane. Let the value of f(x, y) be taken as the length P P′ drawn parallel to the z-axis (or the height of point P′
above the plane). Then asP moves in the xy-plane, P′ maps out asurfacein space whose equation is
z=f(x, y).
Example: f(x, y) = 6−2x−3y
The surfacez= 6−2x−3y, i.e. 2x+ 3y+z= 6, is a plane which intersects thex-axis wherey=z= 0, i.e.x= 3; which intersects they-axis wherex=z= 0, i.e.y= 2; which intersects thez-axis wherex=y= 0, i.e.z= 6.
Example: f(x, y) =y2−x2
In the plane x= 0, there is aminimumaty= 0; in the planey= 0, there is amaximumatx= 0. The whole surface is shaped like a horse’s saddle; and the picture shows a structure for which (0,0) is called asaddle point.
3.2.1
Plane polar coordinates
Since the variables x and y respresent a point in the plane, we can express that point in plane polar coordinates simply by substituting the definitions:
x=rcosθ y=rsinθ.
-3 -2 -1 0 1 2 3 x -3 -2 -1 0 1 2 3 y -10 -8 -6 -4 -2 0 2 4 6 8 10
Figure 3.1: The surfacez=x2−y2
-3 -2 -1 0 1 2 3 x -3 -2 -1 0 1 2 3 y 0 2 4 6 8 10 12 14 16 18
Course Notes 12
Example: f(x, y) =x2+y2
The surfacez=x2+y2may be drawn most easily by first converting into plane polar coordinates.
Substituting x=rcosθ andy =rsinθ givesz =r2. The surface is symmetric about thez-axis
and its cross-section is a parabola. [Check with the original function and the plane y= 0.] Thus the whole surface is a paraboloid (a bowl).
Another way to picture the same surface is to do as map-makers or weather forecasters do and draw
contour lines(orlevel curves) – produces by taking a section, using a planez= const. and projecting it onto thexy-plane. Forz=x2+y2 as above, the contour lines are concentric circles.
3.3
Partial Differentiation
For a functionf(x) that depends on a single variable,x, we can form theordinary derivativedf /dx. For example, iff(x) = 3x4+sin(x) then df /dx= 12x3+ cos(x). Similarly, for a functionf that depends
onseveralvariablesx,y, . . . we can differentiate with respect to each of these variables. This process is calledpartial differentiation(partial derivatives). Iff(x, y) =yx4+ sin(x) then we treatyas a constant
as we did 3 for the function ofx, and have∂f /∂x= 4yx3+ cos(x).
To find apartial derivative we hold all but one of the independent variables constant and differ-entiate with respect to that one variable using the ordinary rules for one-variable calculus.
Notation: If f depended onx, y, . . . but if onlyxis allowed to vary,the partial derivative of f with respect to xis denoted by∂f /∂xor byfx. (Distinguish carefully between df /dxtheordinary
derivative with straight d and∂f /∂xwith curly∂.) Similarly, we can denote the partial derivative with respect toy by∂f /∂yor byfy, etc.
3.3.1
Second-order partial derivatives
Forf(x, y) we can form∂f /∂xand∂f /∂y. Each of these can then be differentiated again with respect toxory to form thesecond-order derivatives
∂/∂x(∂f /∂x) denoted by∂2f /∂x2orf
xx;
∂/∂y(∂f /∂x) denoted by∂2f /∂y∂xor f
xy;
∂/∂y(∂f /∂y) denoted by∂2f /∂y2orfyy;
∂/∂x(∂f /∂y) denoted by∂2f /∂x∂yor f
yx.
fxy andfyx are calledmixed derivatives.
The concept can be extended, e.g. iff depends onx,y andzwe can form∂/∂z ∂2f /∂x2
, written as∂3f /∂z∂x2 orf
xxz, etc.
TheMixed Derivatives Theoremstates that iffxy andfyx are continuous thenfxy=fyx. Thus
to calculate a mixed derivative we can calculate in either order. [Think about calculating∂/∂x(∂f /∂y) if f(x, y) = xy+ 1/(sin(y2) +ey).] For third-order derivatives the mixed derivatives theorem gives
fxxy=fxyx=fyxx.
For a functionf(x), the ordinaryderivative df /dxgives the slope of the tangent to the curve at any pointP. For a function f(x, y), the partialderivative ∂f∂x is evaluated holding y constant, and so gives the slope of the tangent to the surface atP in a planey= const..
3.4
Summary
• We form partial derivatives with respect to one variable by holding all the other variables constant and differentiating.
• The order of mixed derivatives is not important.
• Iff(x, y) represents a surface above thexyplane then∂f /∂xis the slope of a section taken in the
xdirection at a pointP and∂f /∂ythe slope of a section in they direction. Between them these two tangents define a plane which (we will see later) is the tangent planeto the surface atP.
Chapter 4
GRADIENTS
In this chapter we introduce derivatives in multiple dimensions. Although we limit ourselves to differ-entiating known functions, the real use of these derivatives is in differential equations. When we learnt calculus, there was a big gap between starting to study differentiation and solving our first differential equation. In this chapter we meet three-dimensional derivatives. Because this is a major concept we won’t get to any physical applications; but almost no physical problem can be solved without using these tools.
4.1
Gradient and Directional Derivative
For a function of three variablesf(x, y, z), the partial derivatives ∂f∂x, ∂f∂y and ∂f∂z measure the rates of change of f along x, y and z directions, respectively. We now ask how we can calculate the rate of change off in any direction in space. The answer lies in the vector
∇f = ∂f ∂x, ∂f ∂y, ∂f ∂z = ∂f ∂xi+ ∂f ∂yj+ ∂f ∂zk (4.1)
call thegradientoff. From its definition, the component of∇f alongiis (∇f·i) = ∂f∂x = rate of change alongiand similarly for theyandzdirections. But foranydirection in space we are free to temporarily call it thei-direction and carry over the above analysis. Thus in general:
For any direction in space defined by a unitvector u the rate of change off along u is given by (∇f ·u) and is called thedirectional derivativeoff alongu.
4.1.1
Two further properties of the gradient
We look at cases whereuis parallel or perpendicular to∇f. The changedf in f due to a change in the positionP bydr=dru is given by change inf = (rate of change with distance)×(distance), i.e.
df = (∇f·u)dr=∇f·(udr) =∇f ·dr=||∇f|| ||dr||cos(θ) (4.2)
whereθis the angle between the vectorsdrand∇f. From this equation it can be seen that the direction
drfor whichdf is amaximumis obviously that for which cos(θ) = 1, orθ= 0, i.e. the direction of∇f. Thus
Property 1. At any point,∇f points in the direction in whichf is increasing most rapidly and its magnitude||∇f||gives this maximum rate of change.
Again from eq. (4.2),df= 0 corresponds toθ=π/2, when∇f anddrare perpendicular. Butdf = 0 means thatf has not changed - so the displacementdr is along the surfacef = const.. Thus
Property 2. At any point,∇f is perpendicular to the surfacef = const. through that point. 14
4.2
Linear Approximations (Tangents)
Motivation: Many functions arising in applications are difficult to deal with. We thus need ways of
approximatingsuch functions by others which are easier to handle. The most useful approximations are polynomials. We consider first the single variable case,f(x), which will guide us into the treatment of the two-variable case.
4.2.1
One-variable case (tangent line)
The tangent to the curvey =f(x) atAwhere x=ahas the slopef′(a) and therefore has the equation
y=f′(a)x+ const.. It passes through A, i.e. forx=a, y=f(a), so const. =f(a)−f′(a)a. Thus the
equation of the tangent line (the lineparallel to the curve) is
y=f(a) + (x−a)f′(a). (4.3)
For points close toAthe tangent gives a close approximation to the curve. The approximation
f(x)≈f(a) + (x−a)f′(a) (4.4)
is called thelinear approximationto f(x) nearx=a.
4.2.2
Two-variables case (tangent plane)
Above we saw that, for a function of one variable, approximating its curve by a tangent line gave alinear approximation. For a function of two variables,f(x, y), the corresponding linear approximation arises when we approximate a surface by its tangent plane. Suppose the surface is z =f(x, y). The tangent plane atAis perpendicular to the normal atA, i.e. is perpendicular to the gradient vector atA. Now the surface isf(x, y)−z= 0, orF(x, y, z) = const. = 0, whose gradient is∇F = ∂f∂xi+∂f∂yj−k. Thus the tangent planer·n= const. atAis
x ∂f ∂x A +y ∂f ∂y A −z= const. (4.5)
This plane passes throughA(a, b) so the constant in eq. (4.5) has the valuea∂f∂x
A+b
∂f
∂y
A−f(a, b)
and thus (4.5) becomes
z=f(a, b) + (x−a) ∂f ∂x A + (y−b) ∂f ∂y A . (4.6)
Thus the linear approximationto the surface close toA(a, b) is given by the tangent plane (4.6), so
f(x, y)≈f(a, b) + (x−a) ∂f ∂x A + (y−b) ∂f ∂y A (4.7) and we note that we can rewrite this using the gradient∇f as
Course Notes 16
4.3
The Chain Rule
For a function of one variable,f(x), ifxalso depends on another variable t, then f depends on t and the chain rule gives
df dt = df dx dx dt (4.9)
We now generalise this result for a function of several variables. Forf(x, y), supposexandy depend on
t. Let t increase by ∆t and so xand y increase by ∆xand ∆y. Then from our earlier work on linear approximations we obtain f(t+ ∆t)≈f(t) + ∆x ∂f ∂x + ∆y ∂f ∂y (4.10) so if we rearrange and let ∆t→0 we obtain thechain rulefor a function of two variables, which is,
df dt = ∂f ∂x dx dt + ∂f ∂y dy dt. (4.11)
Notethat f depends on xand y [so partial derivatives∂f /∂x, ∂f /∂y] whilst xand y depend on the single variable t [so ordinary derivatives dx/dt, dy/dt]. Thus f depends on t and has the derivative df /dtgiven by the chain rule (4.11).
The chain rule extends directly to functions of three or more variables.
4.3.1
Extended chain rule
Forf(x, y) suppose that xandy depend ontwovariablessandt. Then changing eithersort changes
xandy, so changesf, i.e. producing ∂f∂s and ∂f∂t according to theextended chain rule
∂f ∂s = ∂f ∂x ∂x ∂s+ ∂f ∂y ∂y ∂s, ∂f ∂t = ∂f ∂x ∂x ∂t + ∂f ∂y ∂y ∂t. (4.12)
4.4
The vector differential operator
∇
(grad)
Thevector differential operator∇is defined formally by setting
∇= ∂ ∂x, ∂ ∂y, ∂ ∂z . (4.13)
By ‘formally’ we mean that this is not an ordinary vector but its ‘components’ are differentiation symbols. As the term ‘operator’ suggests,∇ is thought of as something that ‘operates’ on things. What sort of things? Scalar functions and vector functions.
Suppose thatf(x, y, z) is a scalar function. Then∇operates onf as follows:
∇f = ∂ ∂x, ∂ ∂y, ∂ ∂z f = ∂f ∂x, ∂f ∂y, ∂f ∂z . (4.14)
This is thegradientoff, which we have just discussed.
Question How does∇operate on vector functions?
4.4.1
Divergence
Ifq(x, y, z) = (q1(x, y, z), q2(x, y, z), q3(x, y, z)) is a vector function, then by definition
div(q) =∇ ·q= ∂q1 ∂x + ∂q2 ∂y + ∂q3 ∂z. (4.15)
This “product”,∇ ·q, defined in imitation of the ordinary dot product, is a scalar called thedivergence
ofq.
Physical Meaning
Consider a fluid whose velocity everywhere is given by the vector functionu(x, y, z) = (u1, u2, u3). Then
imagine we have a cubic volume [a, a+ ∆x]×[b, b+ ∆y]×[c, c+ ∆z] and we want to know how much fluid is flowing out of it per unit time.
-6 x y z a a+ ∆x c c+ ∆z
The fluid flowing through one face is the velocity normal to the face multiplied by the area of that face. So for the face on whichx=a, u1 gives the velocity pointing in and the face has area ∆y∆z, so
the flow out is
−∆y∆zu1(x=a).
Looking at the face on which x=a+ ∆x, the area of the face is still ∆y∆z but the velocity out is u1
(since the inside of the cube is on the other side of this face). The contribution from this face is ∆y∆zu1(x=a+ ∆x).
Similarly, the other 4 contributions are
−∆x∆zu2(y=b) ∆x∆zu2(y=b+ ∆y) −∆x∆yu3(z=c) ∆x∆yu3(z=c+ ∆z).
We add together the six contributions to find
Rate of flow out = ∆y∆z(u1(x=a+ ∆x)−u1(x=a))
+ ∆x∆z(u2(y=b+ ∆y)−u2(y=b))
+ ∆x∆y(u3(z=c+ ∆z)−u3(z=c))
= ∆x∆y∆z[(u1(x=a+ ∆x)−u1(x=a))/∆x
+ (u2(y=b+ ∆y)−u2(y =b))/∆y
+ (u3(z=c+ ∆z)−u3(z=c))/∆z]
and if we let the ∆xetc. terms tend to zero we have
Rate of flow out = volume×(∂u1/∂x+∂u2/∂y+∂u3/∂z)
= volume×div(u).
Thus the divergence represents the flow out of the volume, per unit time, per unit volume. This is where the name comes from – it is the rate at which the fluid is diverging.
Course Notes 18
4.4.2
Curl
For the vectorq=q1i+q2j+q3k, we also have
curl(q) =∇ ×q = det i j k ∂ ∂x ∂ ∂y ∂ ∂z q1 q2 q3 = ∂q3 ∂y − ∂q2 ∂z, ∂q1 ∂z − ∂q3 ∂x, ∂q2 ∂x − ∂q1 ∂y . (4.16)
This second “product”,∇ ×q, defined in imitation of the ordinary cross product, is a vector called the
curlofq.
Example
Calculate∇ ×q for (from above)q(x, y, z) = (x−y, x+y, z).
Solution ∇ ×q= i j k ∂/∂x ∂/∂y ∂/∂z x−y x+y z =i(0) +j(0) +k(1 + 1) = 2k. Physical Meaning
Consider a solid body which is rotating about thez-axis at rateω, so that each point (x, y, z) inside it has velocityu= (−ωy, ωx,0). We can express this as u= Ω×r, wherer= (x, y, z) and Ω = (0,0, ω) is called the rotation vector. Then the curl of the velocity field(the vector field, i.e. three-dimensional vector functionu, defining the velocity) is
∇ ×u = i j k ∂/∂x ∂/∂y ∂/∂z −ωy, ωx 0 = k(ω+ω) = 2ωk= (0,0,2ω).
We have shown that for a rotation, the curl is related to the rotation vector. This is true in general – so if we had a fluid velocity which was partly rotating and partly moving in some other way, the curl would extract the rotation part (thecirculation).
4.4.3
Grad and div in polar coordinates
The forms of grad and div in polar coordinates are not examinable in this module, but you may find them useful in later years. The forms for curl are too unpleasant to be quoted here; do not be fooled into carrying out a simple cross product with the gradient “vector”.
Plane polar coordinates
The unit vector in ther direction iser; the unit vector in the θdirection iseθ. The values of these two vectors vary according to location.
We can represent any vectorv in terms of these two unit vectors:
Then we have ∇f = ∂f ∂rer+ 1 r ∂f ∂θeθ ∇ ·v = 1 r ∂(rvr) ∂r + 1 r ∂vθ ∂θ
Cylindrical polar coordinates
In a similar fashion to plane polar coordinates, we obtain
∇f = ∂f ∂rer+ 1 r ∂f ∂θeθ+ ∂f ∂zez ∇ ·v = 1 r ∂(rvr) ∂r + 1 r ∂vθ ∂θ + ∂vz ∂z
Spherical polar coordinates
Here we denote the unit vector in the ρdirection aseρ and similar for θ andφ, and a vector v can be
written
v=vρeρ+vθeθ+vφeφ.
Grad and div are given by:
∇f = ∂f ∂ρeρ+ 1 ρ ∂f ∂θeθ+ 1 ρsinθ ∂f ∂φeφ ∇ ·v = 1 ρ2 ∂(ρ2v ρ) ∂ρ + 1 ρsinθ ∂(sinθvθ) ∂θ + 1 ρsinθ ∂vφ ∂φ
4.4.4
Basic Identities
For vectors we have thata×a= 0. Is it true that∇ × ∇= 0? Define (∇ × ∇)f by (∇ × ∇)f =∇×(∇f).
Theorem (The curl of a gradient is zero)
Iff(x, y, z) is a scalar function then
∇ ×(∇f) = 0. Proof ∇ ×(∇f) = i j k ∂/∂x ∂/∂y ∂/∂z fx fy fz = i(fyz−fzy) +j(fzx−fxz) +k(fxy−fyx)
and by the mixed derivatives theorem,∇ ×(∇f) = 0.
For vectors we have thata·(a×c) = 0. The analogous operator formula∇ ·(∇ ×q) = 0 is also valid.
Theorem (The divergence of a curl is zero)
Ifq(x, y, z) is a vector function then
Course Notes 20 Proof ∇ ·(∇ ×q) = ∇ · i j k ∂/∂x ∂/∂y ∂/∂z q1 q2 q3 = ∇ · ∂q 2 ∂z − ∂q3 ∂y, ∂q3 ∂x − ∂q1 ∂z, ∂q1 ∂y − ∂q2 ∂x = ∂ 2q 2 ∂x∂z− ∂2q 3 ∂x∂y + ∂2q 3 ∂y∂x− ∂2q 1 ∂y∂z+ ∂2q 1 ∂z∂y− ∂2q 2 ∂z∂x = 0.
The following identities are left for you as an exercise.
Theorem Iff(x, y, z) is a scalar function andq(x, y, z) is a vector function then show that: (i) ∇ ·(f q) = (∇f)·q+f(∇ ·q)
(ii) ∇ ×(f q) = (∇f)×q+f(∇ ×q).
4.4.5
The Laplacian
From the operator∇ we can construct other operators. The most important of these is theLaplacian
(or theLaplace operator)∇2=∇ · ∇which operates on scalar functionsf(x, y, z) according to the rule: ∇2f =∇ ·(∇f) = ∂ 2f ∂x2 + ∂2f ∂y2 + ∂2f ∂z2. (4.17)
The forms for the Laplacian in different coordinate systems are more complex (cf. the final example of§4.3.1). They are given by:
∇2f = 1 r ∂ ∂r r∂f ∂r + 1 r2 ∂2f
∂θ2 in plane polar coordinates
∇2f = 1 r ∂ ∂r r∂f ∂r + 1 r2 ∂2f ∂θ2 + ∂2f
∂z2 in cylindrical polar coordinates
∇2f = 1 ρ2 ∂ ∂ρ ρ2∂f ∂ρ + 1 ρ2sinθ ∂ ∂θ sinθ∂f ∂θ + 1 ρ2sin2θ ∂2f
∂φ2 in spherical polar coordinates
4.5
Jacobian
TheJacobian matrix of the transformation x=x(s, t), y =y(s, t) is the matrix which relates the two sets of partial derivatives:
∂f /∂s ∂f /∂t = ∂x/∂s ∂y/∂s ∂x/∂t ∂y/∂t ∂f /∂x ∂f /∂y . (4.18)
TheJacobianof the transformation is the determinant
∂(x, y) ∂(r, θ) = ∂x ∂r ∂y ∂θ − ∂x ∂θ ∂y ∂r (4.19)
of theJacobian matrix.
Question: Ifx=x(s, t) andy=y(s, t), can we solve forsandtin terms ofxandy?
4.5.1
Jacobians of the standard coordinate transformations
The following Jacobians may be quoted as standard results:
∂(x, y)
∂(r, θ) = rfor plane polar coordinates
∂(x, y, z)
∂(r, θ, z) = rfor cylindrical polar coordinates
∂(x, y, z)
∂(ρ, θ, φ) = ρ
Chapter 5
DOUBLE AND TRIPLE
INTEGRALS
5.1
Multiple-Integral Notation
Previously ordinary integrals of the form
Z J f(x) dx= Z b a f(x) dx (5.1)
whereJ = [a, b] is an interval on the real line, have been studied. Here we studydouble integrals
Z Z
Ω
f(x, y) dxdy (5.2)
where Ω is some region in thexy-plane, and a little later we will studytriple integrals
Z Z Z
T
f(x, y, z) dxdydz (5.3)
whereT is a solid (volume) in thexyz-space.
5.2
Double Integrals
5.2.1
Properties
(1) Area property Z Z Ω dxdy= Area of Ω.In particular if Ω is the rectangle Ω = [a, b]×[c, d] thenR R
Ωdxdy= (b−a)(d−c). (2) Linearity Z Z Ω [αf(x, y) +βg(x, y)] dxdy =α Z Z Ω f(x, y) dxdy+β Z Z Ω g(x, y) dxdy (5.4) where αandβ are constants.
(3) Additivity
If Ω is broken up into a finite number of nonoverlapping basic regions Ω1,. . . ,Ωn, then
Z Z Ω f(x, y) dxdy= Z Z Ω1 f(x, y) dxdy+. . .+ Z Z Ωn f(x, y) dxdy. (5.5)
5.2.2
Geometric Interpretation
The double integral over Ω gives the volume of the solidTwhose upper boundary is the surfacez=f(x, y) and whose lower boundary is the region Ω in thexy-plane:
Z Z
Ω
f(x, y) dxdy= volume of T. (5.6)
5.2.3
The Evaluation of Double Integrals by Repeated Integrals
If an ordinary integral Rb
af(x) dx proves difficult to evaluate, it is not because of the interval [a, b]
but because of the integrandf. Difficulty in evaluating a double integral R R
Ωf(x, y) dxdy can come
from two sources: from the integrandf or from the domain Ω. Even such a simple looking integral as
R R
Ω1 dxdy is difficult to evaluate if Ω is complicated.
In this section we introduce a technique for evaluating double integrals over domains that have special shapes. The key idea is that double integrals over such special domains can be reduced to a pair of ordinary integrals.
Horizontally simple domain
a b
Theprojectionof the domain Ω onto thex-axis is a closed interval [a, b] and Ω consists of all points (x, y) with
a≤x≤b, andφ1(x)≤ y≤φ2(x). (5.7) Then Z Z Ω f(x, y) dxdy= Z b a Z φ2(x) φ1(x) f(x, y) dy ! dx. (5.8) Here we first calculate Rφ2(x)
φ1(x)f(x, y) dy by integrating f(x, y) with respect to y from y = φ1(x) to
y=φ2(x). The resulting expression is a function ofxalone, which we then integrate with respect tox
fromx=atox=b.
Example
Evaluate R R
Ω(x4−2y) dxdy, where the domain Ω consists of all points (x, y) with −1 ≤ x≤ 1
and−x2≤y≤x2. Solution Z Z Ω (x4−2y) dxdy = Z x=1 x=−1 Z y=x2 y=−x2 (x4−2y) dydx= Z x=1 x=−1 [x4y−y2]yy==x2 −x2dx = Z x=1 x=−1 2x6dx= [2x7/7]xx=1=−1= 4/7
Course Notes 24
Vertically simple domain
c
d Theprojectionof the domain Ω onto they-axis is a closed interval [c, d]
and Ω consists of all points (x, y) with
c≤y≤d, andψ1(y)≤ x≤ψ2(y). (5.9) Then Z Z Ω f(x, y) dxdy= Z d c Z ψ2(y) ψ1(y) f(x, y) dx ! dy. (5.10) Here we first calculate Rψ2(y)
ψ1(y) f(x, y) dx by integrating f(x, y) with respect to x from x = ψ1(y) to
x=ψ2(y). The resulting expression is a function ofy alone, which we then integrate with respect toy
fromy=c toy=d.
The integrals in the right-hand sides of formulae (5.8) and (5.10) are calledrepeated integrals.
Remark 1
Sometimes a domain can be expressed both as a horizontally simple domain: a≤x≤b, φ1(x)≤y ≤ φ2(x), and as a vertically simple domain: c≤y≤d,ψ1(y)≤x≤ψ2(y). Then
Z Z Ω f(x, y) dxdy = Z b a Z φ2(x) φ1(x) f(x, y) dy ! dx= Z d c Z ψ2(y) ψ1(y) f(x, y) dx ! dy. (5.11) Therefore we can, at least in theory, perform the integration in either order. However, there are situations where one order is preferable over the other.
Remark 2
Finally, if Ω, the domain of integration, is neither horizontally nor vertically simple, then it is usually possible to break it up into a finite number of domains, say Ω1, . . . , Ωn, each of which is either horizontally
or vertically simple. Then we can use the additivity property given by eq. (5.5).
5.2.4
Evaluating Double Integrals Using
Polar Coordinates
Let Ω be a domain formed with all points (x, y) that have polar coordinates (r, θ) in the set
Γ :α≤θ≤β, ρ1(θ)≤r≤ρ2(θ) (5.12) whereβ ≤α+ 2π. Then Z Z Ω f(x, y) dxdy = Z Z Γ f(rcos(θ), rsin(θ))rdrdθ = Z β α Z ρ2(θ) ρ1(θ) f(rcos(θ), rsin(θ))rdrdθ. (5.13)
5.3
Triple Integrals
5.3.1
Properties
(1) Volume property Z Z Z T dxdydz= Volume ofT.In particular if T is the box T= [a, b]×[c, d]×[e, f] thenR R R
T dxdydz= (b−a)(d−c)(f−e). (2) Linearity Z Z Z T [αf(x, y, z) +βg(x, y, z)] dxdydz =α Z Z Z T f(x, y, z) dxdydz+β Z Z Z T g(x, y, z) dxdydz (5.14) where αandβ are constants.
(3) Additivity IfT is broken up into a finite number of nonoverlapping basic regionsT1,. . . ,Tn, then
Z Z Z T f(x, y, z) dxdydz = Z Z Z T1 f(x, y, z) dxdydz+. . .+ Z Z Z Tn f(x, y, z) dxdydz. (5.15)
5.3.2
The Evaluation of Triple Integrals by Repeated Integrals
LetT be a solid whose projection onto thexy-plane is labelled Ωxy. Then the solid T is the set of all
points (x, y, z) satisfying
(x, y)∈Ωxy, χ1(x, y)≤z≤χ2(x, y). (5.16)
The triple integral overT can be evaluated by setting
Z Z Z T f(x, y, z) dxdydz= Z Z Ωxy Z χ2(x,y) χ1(x,y) f(x, y, z) dz ! dxdy. (5.17) In eq. (5.17) we can evaluate the integration with respect tozfirst and then evaluate the double integral over the domain Ωxy as studied for double integrals. In particular if Ωxy is horizontally simple, say
a≤x≤b, φ1(x)≤ y≤φ2(x). (5.18)
then the solidT itself is the set of all points (x, y, z) such that
a≤x≤b, φ1(x)≤ y≤φ2(x), χ1(x, y)≤z≤χ2(x, y) (5.19)
and the triple integral overT can be expressed by three ordinary integrals as:
Z Z Z T f(x, y, z) dxdydz= Z b a " Z φ2(x) φ1(x) Z χ2(x,y) χ1(x,y) f(x, y, z) dz ! dy # dx. (5.20) Here we first integrate withz[fromz=χ1(x, y) toz=χ2(x, y)], then with respect toy[fromy=φ1(x)
Course Notes 26
There is nothing special about this order of integration. Other orders of integration are possible and in some cases more convenient. Suppose for example that the projection of T onto the xz-plane is a domain Ωxz of the form
z1≤z≤z2, φ1(z)≤x≤φ2(z). (5.21)
IfT is the set of all (x, y, z) with
z1≤z≤z2, φ1(z)≤ x≤φ2(z), ψ1(x, z)≤y≤ψ2(x, z) (5.22) then Z Z Z T f(x, y, z) dxdydz= Z z2 z1 " Z φ2(z) φ1(z) Z ψ2(x,z) ψ1(x,z) f(x, y, z) dy ! dx # dz. (5.23) In this case we integrate first with respect toy, then with respect to x, and finally with respect to z. Still four other orders of integration are possible.
5.3.3
Evaluating Triple Integrals Using Cylindrical Coordinates
LetT be a solid whose projection onto thexy-plane is labelled Ωxy. Then the solid T is the set of all
points (x, y, z) satisfying
(x, y)∈Ωxy, χ1(x, y)≤z≤χ2(x, y). (5.24)
The domain Ωxy has polar coordinates in some set Ωrθ and then the solidT in cylindrical coordinates
is some solidS satisfying
(r, θ)∈Ωrθ, χ1(rcos(θ), rsin(θ))≤z≤χ2(rcos(θ), rsin(θ)). (5.25)
Then Z Z Z T f(x, y, z) dxdydz= Z Z Ωxy Z χ2(x,y) χ1(x,y) f(x, y, z) dz ! dxdy = Z Z Ωrθ Z χ2(rcos(θ),rsin(θ)) χ1(rcos(θ),rsin(θ)) f(rcos(θ), rsin(θ), z) dz ! rdrdθ= Z Z Z S f(rcos(θ), rsin(θ), z)rdrdθdz. (5.26)
5.3.4
Evaluating Triple Integrals Using Spherical Coordinates
LetT be a solid in xyz-space with spherical coordinates in the solidS ofρθφ-space. Then
Z Z Z
T
f(x, y, z) dxdydz=
Z Z Z
S
f(ρsinθcosφ, ρsinθsinφ, ρcosθ)ρ2sinθdρdθdφ. (5.27)
5.4
Jacobians and changing variables in multiple integration
During the course of the last few sections you have met several formulae for changing variables in multiple integration: to polar coordinates, to cylindrical coordinates, to spherical coordinates. The purpose of this section is to bring some unity to that material and provide a general description for other changes of variable.
5.4.1
Change of variables for double integrals
Consider the change of variables x = x(u, v) and y = y(u, v), which maps the points (u, v) of some domain Γ into the points (x, y) of some other domain Ω. Then
The area of Ω = Z Z Γ ∂(x, y) ∂(u, v) dudv. (5.28)
Suppose now that we want to integrate some functionf(x, y) over Ω. If this proves difficult to do directly, then we can change variables (x, y) to (u, v) and try to integrate over Γ instead. Then
Z Z Ω f(x, y) dxdy= Z Z Γ f(x(u, v), y(u, v)) ∂(x, y) ∂(u, v) dudv. (5.29)
5.4.2
Change of variables for triple integrals
Consider the change of variables x =x(u, v, w), y = y(u, v, w), z =z(u, v, w) which maps the points (u, v, w) of some solid S into the points (x, y, z) of some other solidT. Then
The volume ofT = Z Z Z S ∂(x, y, z) ∂(u, v, w) dudvdw. (5.30) Suppose now that we want to integrate some function f(x, y, z) over T. If this proves difficult to do directly, then we can change variables (x, y, z) to (u, v, w) and try to integrate overS instead. Then
Z Z Z T f(x, y, z) dxdydz= Z Z Z S f(x(u, v, w), y(u, v, w), z(u, v, w)) ∂(x, y, z) ∂(u, v, w) dudvdw. (5.31)
Referring back to equations (5.26) and (5.27), and the Jacobians given at the end of §4.5, we can verify that this formula is correct for a change from Cartesian to cylindrical coordinates (Jacobian isr) and for a change from Cartesian to spherical coordinates (Jacobian isρ2sinθ).
Chapter 6
LINE INTEGRALS AND
SURFACE INTEGRALS
In this chapter we will study integration along curves and integration along surfaces. At the heart of this subject lie three great theorems: Green’s theorem,Gauss’s theorem(commonly known as thedivergence theorem) andStokes’s theorem. All of these are ultimately based on thefundamental theorem of integral calculus, and all can be cast in the same general form: An integral over a region S = An integral over the boundary of S.
6.1
Line integrals
Leth(x, y, z) = (h1(x, y, z), h2(x, y, z), h3(x, y, z)) be a vector function that is continuous over a smooth
curveC parametrised byC:r(u) = (x(u), y(u), z(u)) withu∈[a, b]. Theline integralofhoverC is the number Z C h(r)·dr= Z b a [h(r(u))·r′(u)] du. (6.1)
Although we stated this definition in terms of three-dimensional vectorial functionsh(x, y, z) and curves in space r(u) = (x(u), y(u), z(u)), it also includes the two-dimensional case: h(x, y) and plane curves
r(u) = (x(u), y(u)).
If the curveCis not smooth but is made up of a finite number of adjoining smooth piecesC1, . . . , Cn,
i.e. it is piecewise smooth, then we define the integral overC as the sum of the integrals overCi for
i= 1, . . . , n, that isR
C=
R
C1+· · ·+
R
Cn. All polygonal paths are piecewise smooth.
When we integrate over a parametrised curve, we integrate in the direction determined by the parametrisation. If we integrate in the opposite direction, our answer is altered by a factor of −1, that isR
−C=−
R
C.
6.1.1
Another notation for line integrals
Ifh(x, y, z) = (h1(x, y, z), h2(x, y, z), h3(x, y, z)) then the line integral over a curveC can be written as
Z C h(r)·dr= Z C {h1(x, y, z) dx+h2(x, y, z) dy+h3(x, y, z) dz}. (6.2) 28
6.2
The Fundamental Theorem for Line Integrals
In general, if we integrate a vector function hfrom one point to another, the value of the line integral depends on the path chosen. There is, however, an important exception. If the vector function h is a gradient, i.e. there exists a scalar function f such that h= ∇f, then the value of the line integral depends only on the endpoints of the path and not on the path itself. The details are spelled out in the following theorem.
Theorem
Let C, parametrised by r = r(u) with u ∈ [a, b], be a piecewise smooth curve that begins at
α=r(a) and ends atβ=r(b). Then if the vector functionhis a gradient, i.e.h=∇f, we have
Z C h(r)·dr= Z C ∇f(r)·dr=f(β)−f(α). (6.3)
NOTE: It is important to see that this result is an extension of the fundamental theorem of integral calculus: Rb
a f′(x) dx=f(b)−f(a).
Corollary
If the curveC is closed, i.e. α=β, then f(α) =f(β) andR
C∇f(r)·dr= 0.
6.3
Line integrals with respect to arc length
Suppose thatf is a scalar function continuous on a piecewise smooth curveC parametrised byr=r(u) withu∈[a, b]. Ifs(u) is the length of the curve from the tip ofr(a) to the tip ofr(u), then, as we have seen in section 2.3,s′(u) =||r′(u)||. The integral off overC with respect to arc lengths is defined by
setting Z C f(r) ds= Z b a f(r(u))s′(u) du. (6.4)
6.4
Green’s Theorem
IfP(x, y) andQ(x, y) are scalar functions defined over a domain Ω with piecewise smooth closed boundary
C, then Z Z Ω ∂Q ∂x(x, y)− ∂P ∂y(x, y) dxdy= I C P(x, y) dx+Q(x, y) dy (6.5) where the integral on the right is a line integral overC taken in the anticlockwise direction.
Remark As indicated, the symbol H
is used to denote the line integral over a simple closed curveC
taken in the anticlockwise direction.
6.5
Parametrised Surfaces; Surface Area
We have seen that a space curve C can be parametrised by a vector functionr=r(u) whereuranges over some intervalI of theu-axis. In an analogous manner, we can parametrise a surfaceS in space by a vector functionr=r(u, v) where (u, v) ranges over some domain Ω of theuv-plane.
Course Notes 30
Example (The graph of a function)
The graph of a function y = f(x), x ∈ [a, b] can be parametrised by setting r(u) = (u, f(u)),
u∈[a, b].
Similarly, the graph of a functionz=f(x, y), (x, y)∈Ω can be parametrised by settingr(u, v) = (u, v, f(u, v)), (u, v)∈Ω.
Example (A plane)
If two vectorsaandb are not parallel, then the set of all combinationsua+vb generates a plane
P0 that passes through the origin. We can parametrise this plane by settingr(u, v) =ua+vb,u, v real numbers.
The planeP that is parallel toP0 and passes through the tip of a vectorccan be parametrised by
settingr(u, v) =ua+vb+c,u,vreal numbers.
Example (A sphere)
The sphere of radius acentred at the origin can be parametrised by setting
r(u, v) = (asin(u) cos(v), asin(u) sin(v), acos(u)), (u, v)∈[0, π]×[0,2π). (6.6)
6.5.1
The fundamental vector product
LetS be a surface parametrised byr=r(u, v), (u, v)∈Ω. The cross product
N=r′
u×r′v (6.7)
is called thefundamental vector productof the surfaceS.
The vectorN(u, v) is perpendicular to the surfaceS at the point with position vectorr(u, v) and, if different from zero, can be taken as the normal to the surfaceS at that point.
Example
For the plane r(u, v) =ua+vb+c, the vectora×bis normal to the plane.
Example
The fundamental vector product for a sphere is parallel to the radius vector r(u, v). (Using the parametrisation given above,N =asin(u)r.)
6.5.2
The area of a parametrised surface
The area of a surfaceS parametrised byr=r(u, v), (u, v)∈Ω, is given by Area ofS=
Z Z
S
||N(u, v)||dudv. (6.8)
Example (The surface area of a sphere)
Using the parametrisation given by equation (6.6), we had N =asin(u)rso ||N||=a2sin(u) and
the area is Z Z S ||N(u, v)||dudv=a2 Z 2π v=0 Z π u=0
sin(u) dudv= 2πa2[−cos(v)]π0 = 4πa2. Example (The area of a plane domain)
A plane domain may be parametrised as r = (u, v,0) for (u, v) ∈ Ω. Then r′
u = (1,0,0) and
r′
v = (0,1,0) and so the fundamental vector product isN = (0,0,1) which has magnitude 1.
Z Z
Ω
6.5.3
The area of a surface
z
=
f
(x, y
)
Let the surfaceS be the graph of the functionz=f(x, y) with (x, y)∈Ω. Then Area of S= Z Z Ω " ∂f ∂x 2 + ∂f ∂y 2 + 1 #1/2 dxdy. (6.9)
In this case the parametrisation of S is r(u, v) = (u, v, f(u, v)), (u, v)∈ Ω and so N = (−fx,−fy,1).
The unit vectorn=N /||N||is called theupper unit normal.
6.6
Surface Integrals
LetH(x, y, z) be a scalar function, continuous over a surfaceS parametrised byr=r(u, v), (u, v)∈Ω. Thesurface integralofH overS is the number
Z Z S H(x, y, z) dσ= Z Z Ω H(r(u, v))||N(u, v)||dudv. (6.10) TakingH ≡1 and referring back to eq. (6.8) we get
Z Z
S
dσ= Area ofS. (6.11)
6.6.1
Flux of a vector function
Letq(x, y, z) be a vector function that is continuous over a smooth surfaceSparametrised byr=r(u, v), (u, v)∈Ω. ThefluxofqacrossS in the direction of the unit normalnto the surfaceS is the number
Z Z
S
q·ndσ (6.12)
which can be calculated as
Z Z S q·ndσ= Z Z Ω q(r(u, v))·n||N||dudv= Z Z Ω q(r(u, v))·Ndudv. (6.13) Proposition
IfS is the graph of a functionz=f(x, y) with (x, y)∈Ω andnis the upper unit normal, then the flux of the vector functionq= (q1(x, y, z), q2(x, y, z), q3(x, y, z)) acrossS in the direction ofnis
Z Z S q·ndσ= Z Z Ω (−q1fx−q2fy+q3) dxdy. (6.14) Proof
We can parametrise the surface by r = (u, v, f(u, v)) with (u, v) ∈ Ω. Then the fundamental vector product is N =r′ u×r′v= (1,0, fu)×(0,1, fv) = (−fu,−fv,1) and we have Z Z S q·ndσ = Z Z Ω (q·N) dudv = Z Z Ω (−q1fu−q2fv+q3) dudv= Z Z Ω (−q1fx−q2fy+q3) dxdy.
Course Notes 32
6.7
The Divergence (Gauss) Theorem
Recall that if P(x, y) and Q(x, y) are scalar functions defined over a domain Ω with piecewise smooth closed boundaryC, then Green’s theorem (section 6.4) allowed us to express a double integral over Ω as a line integral overC:
Z Z Ω ∂Q ∂x(x, y)− ∂P ∂y(x, y) dxdy= I C P(x, y) dx+Q(x, y) dy. (6.15) This formula can be rewritten in vector terms (usingq= (Q,−P)) to give thedivergence theorem in two dimensions as follows:
The divergence theorem in two dimensions
Let Ω be a two-dimensional domain bounded by a piecewise smooth closed curveC. Then for any (continuously differentiable) vector function q(x, y) we have that
Z Z Ω (∇ ·q) dxdy= I C (q·n) ds (6.16) wherenis the outer unit normal and the integral on the right is taken with respect to arc length. We can now give the three-dimensional analogue of the divergence (Gauss) theorem.
The divergence theorem in three dimensions
LetT be a three-dimensional solid bounded by a piecewise smooth closed surfaceS. Then for any (continuously differentiable) vector function q(x, y, z) we have that
Z Z Z T (∇ ·q) dxdydz= Z Z S (q·n) dσ (6.17) where nis the outer unit normal.
6.7.1
Divergence as outward flux per unit volume
In eq. (6.17), the right-hand side R R
S(q·n) dσ represents theq acrossS in the direction of n. In this
sense, from eq. (6.17) we can say thatthe divergence is the outward flux per unit volume, as we discussed in section 4.4.1.
Points (x, y, z)∈T for which
• ∇ ·q(x, y, z)<0 are calledsinks.
• ∇ ·q(x, y, z)>0 are calledsources.
• If∇ ·q(x, y, z)≡0 thenqis calledsolenoidal.
6.8
Stokes’s Theorem
We return to Green’s theorem (section 6.4):
Z Z Ω ∂Q ∂x(x, y)− ∂P ∂y(x, y) dxdy= I C P(x, y) dx+Q(x, y) dy. (6.18)
and this time settingq= (P, Q, R) a vector function, we have (∇ ×q)·k=det i j k ∂ ∂x ∂ ∂y ∂ ∂z P Q R ·k= ∂Q ∂x − ∂P ∂y. (6.19)
Thus in vector terms Green’s theorem can be written as
Z Z Ω [(∇ ×q)·k] dxdy = I C q(r)·dr. (6.20) Since any plane can be coordinatised as thexy-plane, this result can be phrased in the following theorem
Stokes’s theorem
Let S be a smooth surface with smooth bounding curveC. Then for any (continuously differen-tiable) vectorial function q(x, y, z) we have
Z Z S [(∇ ×q)·n] dσ= I C q(r)·dr (6.21) where nis a unit normal that varies continuously on S, and the line integral H
C is taken in the
positive sense with respect to n.
6.8.1
The normal component of
∇×
q
as circulation per unit area; Irrotational
flow
Interpret the vector function q(x, y, z) as the velocity of a fluid. In eq. (6.21), the right-hand side line integralH
Cq(r)·dris called thecirculationof qaround the curveC. In this sense, from eq. (6.21), we
can say that∇ ×qin the directionnis the circulation of qper unit area, which relates to the rotation of the fluid as discussed in section 4.4.2.
If∇ ×q≡0 then there is no circulation andq is calledirrotational, i.e. the fluid has no rotational tendency.
Chapter 7
EXAMPLE SHEETS
There are 6 example sheets, five of which are to be handed in on the timetable below:
Sheet 1 Friday 5th October
Sheet 2 Friday 19th October
Sheet 3 Friday 2nd November
Sheet 4 Friday 16th November
Sheet 5 Friday 30th November
Sheet 6 Not handed in.
GENERAL INFORMATION
Lecturer: Prof. F. W. Nijhoff Department of Applied Mathematics Room 9.20a, telephone ext. 35120 e-mail: [email protected]
Lectures and example classes There will be 33 hours total of lectures and classes during 11 weeks.
• Lectures on Tuesdays 3-4pm in RSLT 8;
• Classes on Mondays 4-5pm in RSLT 6;
• Lectures on Fridays 1-2pm in RSLT 6.
webpage The MATH2420 webpage conatining most of the material can be found under the URL:
http://www.maths.leeds.ac.uk/%7Efrank/math2420.html
or follow the link under the Lecturer’s staff page:
http://www.maths.leeds.ac.uk/ frank
and click onTeaching. This page will contain most of the material, but be aware that some ma-terial may be updated during the term.
Example sheets Every two weeks you are expected to hand in the solutions to an example sheet. Your work will be marked to monitor progress on the first 5 sheets, but worked solutions to the last sheet (on the last section of the course) will be handed out to help your revision over the Christmas vacation. The completed sheets will be handed in at the Friday lectures, on the dates given on page 34. The schedule will be as follows:
Expl. sheet Handout date Due date
# 1 25/9 5/10 # 2 5/10 19/10 # 3 19/10 2/11 # 4 2/11 16/11 # 5 16/11 30/11 # 6 27/11 not marked
Marking The exercises will be marked by postgraduate students (the top mark being 5 for each sheet). They count for up to 15% of your course marks.
General Information 36
What I expect from you
• Attend lectures and examples classes
• Take notes during lectures (this document is only a summary)
• Attempt the examples on your own
• Hand in example sheets on time
• Ask questions as they occur to you
• Turn off your mobile phone!
• be considerate to your fellow students (do not hold disruptive conversations during lectures)
Booklist
1. M. R. Spiegel, Theory and Problems of Vector Analysis, McGraw-Hill. 2. E. Kreyszig, Advanced Engineering Mathematics, Wiley.
3. P. V. O’Neil, Advanced Engineering Mathematics, PWS-Kent Publishing. 4. A. C. Bajpai, Advanced Engineering Mathematics, Wiley, 1977.
5. C. R. Wylie and L. C. Barrett, Advanced Engineering Mathematics, McGraw-Hill. 6. P. C. Matthews, Vector Calculus, 1998.
GREEK ALPHABET
Letter name Lower case Capital
Alpha α A Beta β B Gamma γ Γ Delta δ ∆ Epsilon ǫ E Zeta ζ Z Eta η H Theta θ Θ Iota ι I Kappa κ K Lambda λ Λ Mu µ M Nu ν N Xi ξ Ξ Omicron o O Pi π Π Rho ρ P Sigma σ Σ Tau τ T Upsilon υ Υ Phi φ Φ Chi χ X Psi ψ Ψ Omega ω Ω 37
Appendix B
∇
IN POLAR COORDINATES
You need not memorise these. In each coordinate system we use the unit vectors pointing along the directions of increasing coordinates: thuser is the unit vector in ther direction and so on. Any vector
can be written in terms of a set of unit vectors, e.g. in spherical polars
v=vρeρ+vθeθ+vφeφ.
Plane polar coordinates
∇f = ∂f ∂rer+ 1 r ∂f ∂θeθ ∇ ·v = 1 r ∂(rvr) ∂r + 1 r ∂vθ ∂θ ∇2f = 1 r ∂ ∂r r∂f ∂r + 1 r2 ∂2f ∂θ2 Cylindrical polar coordinates
∇f = ∂f ∂rer+ 1 r ∂f ∂θeθ+ ∂f ∂zez ∇ ·v = 1 r ∂(rvr) ∂r + 1 r ∂vθ ∂θ + ∂vz ∂z ∇2f = 1 r ∂ ∂r r∂f ∂r + 1 r2 ∂2f ∂θ2 + ∂2f ∂z2 Spherical polar coordinates
∇f = ∂f ∂ρeρ+ 1 ρ ∂f ∂θeθ+ 1 ρsinθ ∂f ∂φeφ ∇ ·v = 1 ρ2 ∂(ρ2v ρ) ∂ρ + 1 ρsinθ ∂(sinθvθ) ∂θ + 1 ρsinθ ∂vφ ∂φ ∇2f = 1 ρ2 ∂ ∂ρ ρ2∂f ∂ρ + 1 ρ2sinθ ∂ ∂θ sinθ∂f ∂θ + 1 ρ2sin2θ ∂2f ∂φ2. 38
0 REVIEW 2
0.1 Calculus . . . 2
0.2 Lines and Circles . . . 2
0.3 Trigonometry . . . 3
0.4 Determinant . . . 3
0.5 Vector Products . . . 3
1 COORDINATE SYSTEMS 4 1.1 Cartesian Coordinates . . . 4
1.2 Plane Polar Coordinates . . . 4
1.3 Cylindrical Coordinates . . . 4
1.4 Spherical Coordinates . . . 4
2 VECTOR CALCULUS 7 2.1 Vector Functions . . . 7
2.1.1 Derivative of a vector function . . . 7
2.1.2 Integral of a vector function . . . 8
2.2 Curves . . . 8
2.2.1 Tangent vector, tangent line . . . 8
2.2.2 Intersecting curves . . . 9
2.2.3 The unit tangent . . . 9
2.2.4 Reversing the sense of a curve . . . 9
2.3 Arc Length . . . 9
3 FUNCTIONS OF SEVERAL VARIABLES 10 3.1 Introduction . . . 10
3.2 Geometric Interpretation . . . 10
3.2.1 Plane polar coordinates . . . 10
3.3 Partial Differentiation . . . 12
3.3.1 Second-order partial derivatives . . . 12
3.4 Summary . . . 12
4 GRADIENTS 14 4.1 Gradient and Directional Derivative . . . 14
4.1.1 Two further properties of the gradient . . . 14
4.2 Linear Approximations (Tangents) . . . 15
4.2.1 One-variable case (tangent line) . . . 15
4.2.2 Two-variables case (tangent plane) . . . 15
General Information 40
4.3 The Chain Rule . . . 16
4.3.1 Extended chain rule . . . 16
4.4 The vector differential operator∇ (grad) . . . 16
4.4.1 Divergence . . . 17
4.4.2 Curl . . . 18
4.4.3 Grad and div in polar coordinates . . . 18
4.4.4 Basic Identities . . . 19
4.4.5 The Laplacian . . . 20
4.5 Jacobian . . . 20
4.5.1 Jacobians of the standard coordinate transformations . . . 21
5 DOUBLE AND TRIPLE INTEGRALS 22 5.1 Multiple-Integral Notation . . . 22
5.2 Double Integrals . . . 22
5.2.1 Properties . . . 22
5.2.2 Geometric Interpretation . . . 23
5.2.3 The Evaluation of Double Integrals by Repeated Integrals . . . 23
5.2.4 Evaluating Double Integrals Using Polar Coordinates . . . 24
5.3 Triple Integrals . . . 25
5.3.1 Properties . . . 25
5.3.2 The Evaluation of Triple Integrals by Repeated Integrals . . . 25
5.3.3 Evaluating Triple Integrals Using Cylindrical Coordinates . . . 26
5.3.4 Evaluating Triple Integrals Using Spherical Coordinates . . . 26
5.4 Jacobians and changing variables in multiple integration . . . 26
5.4.1 Change of variables for double integrals . . . 27
5.4.2 Change of variables for triple integrals . . . 27
6 LINE INTEGRALS AND SURFACE INTEGRALS 28 6.1 Line integrals . . . 28
6.1.1 Another notation for line integrals . . . 28
6.2 The Fundamental Theorem for Line Integrals . . . 29
6.3 Line integrals with respect to arc length . . . 29
6.4 Green’s Theorem . . . 29
6.5 Parametrised Surfaces; Surface Area . . . 29
6.5.1 The fundamental vector product . . . 30
6.5.2 The area of a parametrised surface . . . 30
6.5.3 The area of a surfacez=f(x, y) . . . 31
6.6 Surface Integrals . . . 31
6.6.1 Flux of a vector function . . . 31
6.7 The Divergence (Gauss) Theorem . . . 32
6.7.1 Divergence as outward flux per unit volume . . . 32
6.8 Stokes’s Theorem . . . 32
6.8.1 The normal component of∇ ×qas circulation per unit area; Irrotational flow . . . 33
7 EXAMPLE SHEETS 34
8 GENERAL INFORMATION 35
