math intro script

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Mathematical basics for Fundamentals of

Electrical Engineering 1

Prof. Roland Schmechel

Script by Lucas Bitzer

University of Duisburg-Essen

Institute for Nano Structures and Technology (NST)

Version 1.0


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Contents

1 Introduction 3

2 Linear Algebra 4

2.1 Vectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Cartesian coordinate system . . . . . . . . . . . . . . . . . . . . . 52.2.2 Polar coordinate system . . . . . . . . . . . . . . . . . . . . . . . 62.2.3 Cylindrical coordinate system . . . . . . . . . . . . . . . . . . . . 72.2.4 Spherical coordinate system . . . . . . . . . . . . . . . . . . . . . 82.2.5 A note on unit vectors in the different coordinate systems . . . . 9

2.3 The scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 The vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Vectoranalysis 18

3.1 The line Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1.1 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1.2 Why are integrals needed? . . . . . . . . . . . . . . . . . . . . . . 18

3.1.3 Conservative fields . . . . . . . . . . . . . . . . . . . . . . . . . . 193.1.4 Example: Calculation of the work needed to move a charge in a

conservative force field F . . . . . . . . . . . . . . . . . . . . . . . 203.2 The double integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2.1 Introduction of the double integral . . . . . . . . . . . . . . . . . 213.2.2 Some basic examples . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3 The triple integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3.1 Introduction of the triple integral . . . . . . . . . . . . . . . . . . 263.3.2 Example: Volume of a sphere . . . . . . . . . . . . . . . . . . . . 27


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1 Introduction

This mathematical summary shall provide the basic mathematical skills that are nec-essary to understand the content of the course Fundamentals of Electrical Engineering1 (short: FEE1), read by Prof. Roland Schmechel. The script is not a substitute formathematical lectures, but shall create a practical understanding of the matter and givethe students tools to solve the typical problems of the course.

The mathematical content is based on a script of the course Mathematische Meth-

oden der Physik, read by Prof. Karl-Heinz Lotze in Jena, Germany [1]. The authorgreatfully thanks Prof. Lotze for providing his work as basic for the script presented.Additionally, his script was created in LaTex by Simon Stutzer, who is thanked forproviding the script and his agreement to use his figures. Some illustrations were takenand modified from the book Introductory Electromagnetics by Herbert P. Neff, whichthe author recommends for compact informations about vectors and coordinate systems[2].

As this is the first version of the mathematical script, suggestions for its improvementare very welcome.

Lucas BitzerDuisburg in September 2012

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2 Linear Algebra

2.1 Vectors

A displacement of space that moves every point of a considered area in the same directionby a certain distance shall be considered a translation.

As all considered points of the area move the same distance in one direction, thetranslation of all points is known if we know the displacement of one point, no matterwhich point in space we choose. For example, imagine a box that you move one meter

to the right, without turning or rotating it. Then, every point in the box has beensubjected to the translation which is well known if the movement of one point of thebox is known.

To describe these movements mathematically we introduce the notion of a vector, forexamplea. Vectors are defined by their absolute value(or magnitude) |a| =a and theirdirection, which is given by a unit vector ea that shares the same direction as a, buthas an absolute value of 1. Vectors can correspond to geometrical or physical quantitiessuch as translation, velocity, acceleration, force, current density, electric and magneticfield strength.

Every vector can be written as the product of its absolute value|a| =a and its unitvectorea

a= a ea (2.1)Transposed, the equation is resulting in the definition of the unit vector ea, that can

be calculated for every given vector:

ea =a

a, with|ea| = 1 (2.2)

In a coordinate system, a unit vector for every dimension is needed. Let us name

these vectors in a general 3-dimensional casee1,e2 ande3. They are linear independentfrom and always perpendicular to each other. In three-dimensional space, every vectorcan be decomposed in 3 parts, which are dedicated to the 3 dimensions (unit vectors) ofthe chosen coordinate system (see chapter 2.2for the coordinate systems). In general,a vector is then given by:

a=

a1a2a3

=a10

0

+ 0a2

0

+ 00

a3

= a1 e1+a2 e2+a3 e3. (2.3)

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2.2 Coordinate systems

2.2 Coordinate systems

A coordinate system in three dimensions is defined by three independent unit vectors,called basis, spanning the space of the coordinate system in independent directions.

Every vector can be described as a linear combination of these vectors. We will discussfour basic coordinate systems, that are commonly used in FEE. In some coordinatesystems, angles are used to describe points in 3- or 2-dimensional space. The carefulchoice of the coordinate system can simplify a given task. The most important hint forthat choice gives the symmetry of the problem. For example, if the exercise is discussinga spherical capacitor, the spherical coordinate system will be the best coordinate systemto calculate a solution.

2.2.1 Cartesian coordinate system

The commonly known cartesian coordinate system (sometimes: rectangular coordinatesystem) is given by three axes x,y and zand their directions (see figure2.1), defined bythree unit vectors:

ex=

100

, ey =01

0

, ez =00

1

. (2.4)

Figure 2.1: Cartesian coordinate system with axes x,y and z with corresponding unitvectorsex, ey andez.

It is used for problems with cuboid geometry, for example a cuboid electrical conduc-tor. The decomposition of a vector in cartesian coordinates can be seen in figure2.2and is written as

a=

axayaz

=

ax00

+

0

ay0

+

00

az

= ax ex+ay ey+ az ez, (2.5)

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2.2.2 Polar coordinate system

where ax, ay und az are called the coordinates ofa. Compare this decomposition withthe general case in eqation2.3.

Figure 2.2: Decomposition of a vector (a) in cartesian coordinates.

Properties of cartesian coordinates (Compare figure2.2)

az = 0, ifa is in the plane spanned by x and y

The absolute value ofais given by

a= |a| =

a2x+a2y+ a

2z (2.6)

The unit vectorea corresponding to a vectorais given by

ea =a

a=

ax ex+ay ey+ az eza2x+a

2y+ a

2z

(2.7)

2.2.2 Polar coordinate system

Imagine a vector called position vector r, that is connecting the origin of a given coor-dinate system with a chosen point in space and has a direction and a length (see figure2.3). It can thus be imagined as pointer inside the coordinate system.

Now, instead of the cartesian coordinates x, y and z, that vector is used to describea chosen point in space. Let us begin with the 2-dimensional case, where a plane isdescribed by the coordinates xand y . Then, a point P in the cartesian xy-plane can bedescribed by itspolar coordinates: P(r, ). Theradius|r| =r represents the distance ofthe point of interest to the point of origin, and the angle (Phi) describes the direction,that can be computed by sin and cos .

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2.2.3 Cylindrical coordinate system

Figure 2.3: Polar coordinate system with coordinates

r and

, and position vector

r.(modified from [1])

As it can be seen in figure 2.3, the polar to cartesian coordinates transformation(x, y) (r, ) is given as

x= r cos

y= r sin (2.8)

Unit vectors:

er, e (2.9)

Instead of the cartesian unit vectorsex andey, the unit vectorser ande are used inpolar coordinates. They are giving the direction ofr and the rotational direction of.

2.2.3 Cylindrical coordinate system

One can further generalize polar coordinates by adding one more coordinate, so thatevery point can be expressed in 3-dimensional space. One way to do that is by usinga cylindrical coordinate system, which adds a cartesian z-coordinate for the height of apoint above the 2-dimensional plane of polar coordinates. Notice, that r is the distanceto the z-axis and not longer the distance to point P.

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2.2.4 Spherical coordinate system

Figure 2.4: Cylindrical coordinate system with coordinatesr, andz.

To convert cylindrical to cartesian coordinates (3-dimensional!), it holds (x , y , z )(r,,z):

x= r cos

y = r sin

z=z (2.10)

Unit vectors:

er, e, ez (2.11)

To the polar unit vectors, the cartesian unit vector ez is added.

2.2.4 Spherical coordinate system

Another way to adapt polar coordinates to 3-dimensional space is to add a furtherdirection by another angle as coordinate. This new angle describes the angle betweenthe xy-plane and the line that connects the point Pwith the origin (see figure2.5). Wewill call that angle [Theta]. A pointPin space is then defined by three coordinates

(r,,). Therebye, the radius r is not equal to the one of the polar coordinate system.Notice that the radius in spherical coordinates is introduced as the distance from theorigin to the point of interest in the coordinate system, whereas the radius in polarcoordinates is always in a plane, parallel to the xy-plane. Therefore, we name theradius r. If the radius r is described for a certain angle , it holds:

r =r sin (2.12)

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2.2.5 A note on unit vectors in the different coordinate systems

Figure 2.5: Spherical coordinates with coordinatesr, and.

To convert spherical to cartesian coordinates (3-dimensional!), we can then use thetransformation from polar coordinates (equation2.8) with the new radius r (x , y , z ) (r,,):

x= r cos = r sin cos

y= r sin = r sin sin

z= r cos (2.13)

Unit vectors:

er, e, e (2.14)

To the polar unit vectors, the unit vector of the second angle, e is added.


In figure2.6, the unit vectors of the different coordinate systems are shown. While the

unit vectors in cartesian coordinates are constant (see eq. 2.4), they can change directionin the other so far mentioned coordinate systems. In case of cylindrical coordinates (b),only ez stays constant, while the direction ofer ande are changing in different pointsP in space. The change can be expressed by their relation to the constant cartesian unitvectors:

er = cos ex+ sin ey

e = sin ex+ cos eyez =ez (2.15)

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In spherical coordinates (c), all unit vectors change in space:

er = sin cos ex+ sin cos ey+ cos ez

e = cos cos ex+ cos sin ey sin eze = sin ex+ cos ey (2.16)

Use the mentioned Right-hand-rule in section 2.4 to check if the unit vectors arepointing in the correct direction.

Figure 2.6: Unit vectors in (a) cartesian, (b) cylindrical and (c) spherical coordinatesystem. (modified from [2])

With this result in mind, the orthogonal surfaces where each coordinate is constantare shown in figure2.7. Imagine each unit vector orthogonal to the respective surfaceswhere its dedicated coordinate is kept constant. It can be noticed, that the z=const.plane of cartesian coordinates equals that plane in cylindrical coordinates. Furthermore,= const. in cylindrical and spherical coordinates is the same, too. A unit vector cantherefore be used to describe the orientation of a plane in space, as the vector is pointingalways perpendicular to the described surface.

Other important properties of the unit vectors in coordinate systems can be found inthe examples of section2.3and2.4.

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2.3 The scalar product

Figure 2.7: Orthogonal surfaces in (a) cartesian, (b) cylindrical and (c) spherical co-ordinate system. (modified from [1])


In this section we introduce the scalar product (or dot product), a method to multiplytwo vectors. It is used for example in physics to calculate work. It is defined as:

a b= ab cos() =a1b1+a2b2+a3b3 (2.17)

Where a and b are vectors, a1, a2, a3 and b1, b2, b3 are their components and [Gamma] is the given angle between the vectorsa and b(see figure2.8).

The equation can also be read as:

cos =a

a

b

b (2.18)

It becomes clear, that the scalar product of two unit-vectors (aa

and bb

are unit-vectors)equals the cosine of the angle between these vectors.

Properties of the scalar product

It is the product of two vectors, but the result is a scalar number It is commutative, which means:

a b= b a (2.19)

It is notassociative, which means:a(bc)

= (ab)c, (2.20)

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Figure 2.8: Multiplication of two vectorsa andb by the scalar product. The vectorcis defined asc= a b, the angle is measured betweena andb.

asa(bc) gives a vector with the same direction as a, whereas (ab)c gives a vectorwith the same direction as c

The absolute value of a vector can be expressed by the scalar product. Letsassume a vector that is multiplicated by himself:

a a= a2 >0 |a| =a =

a a (2.21)

The scalar product is zero for perpendicular orientated vectors a b, which means=

2 (= 90):

a b= ab cos

2

= ab 0 = 0 (2.22)

Mathematical proof of the scalar product To justify the definition of the scalarproduct, the first step is to derive a mathematical expression for the angle between thetwo vectorsaand b.

In figure2.8,three vectors are given by the relation

c= a b.

Additionally, the trigonometrical law of cosines is defined as:

c2 =a2 +b2 2ab cos .

A combination of equations2.23and2.24results in

|a b|2 = a2 +b2 2ab cos 2ab cos = a2 b2 |a b|2

(with eq. 2.6) = (a21+a22+a

23) + (b

21+b

22+b

23)

(a1 b1)2 (a2 b2)2 (a3 b3)2 ab cos = a1b1+a2b2+a3b3 = a b

which is the definition of the scalar product a b introduced in the beginning of thissection.

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2.4 The vector product

Example Scalar product of unit vectors.

For all coordinate systems, it holds that the scalar product of different unit vectorsis zero, and the scalar product for identical unit vectors is 1. This result is shown in

cartesian coordinates for the following table:

ex ey ezex 1 0 0ey 0 1 0ez 0 0 1

You can check that table by calculating the scalar product (equation 2.17) for thecartesian unit vectors (equation2.4).


Another product of vectors is thevector product(or cross product). It can be representedby the determinant of a formal 3-dimensional matrix in cartesian coordinates (compareunit vectors in eq. 2.4):

ab=

ex ey eza1 a2 a3b1 b2 b3

= (a2b3a3b2)ex+(a3b1a1b3)ey +(a1b2a2b1)ez =a2b3 a3b2a3b1 a1b3

a1b2 a2b1

(2.25)

For solving the determinant of a 3-dim matrix, please have a look at the Rule ofSarrus.

Example Vector product of two given vectors in cartesian coordinates654

12

3

=ex ey ez6 5 31 2 3

= (5 3 4 2)ex+ (4 1 6 3)ey+ (6 2 5 1)ez

= 7ex 14ey+ 7ez = 7

147

Properties of the vector product

The vector product returns a vector, the scalar product in contrast returns a scalarnumber

Right-hand rule: When taking the vector product of two vectors in a plane, youcan use the right hand rule to determine its direction. The fingers just have to bepointed in the following directions:

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It is notcommutative:a b = b a (2.26)

This can be easily shown by the right hand rule. When changing the order of thevectors, your middle finger points in the opposite direction: ab= ba, soab= b

a

It is not associative:a (b c) = (a b) c (2.27)

That can be seen in the following example. Ifc= a b, then

a (a b) = a c= bbut (a a) b= 0.

So, the result is not the samenot associative! It is distributive:

(a+ b) c= a c+ b c (2.28)

Geometric interpretation of the vector products absolute value (from[1]):

The absolute value of the scalar product is given by the area of the parallelogramspanned byaand b:

c= |c| = |a b| =ab sin (2.29)

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Parallel properties: If the vectors are aligned parallel or anti-parallel, no parallel-ogram is spanned and the vector product will be zero.

Parallel vectors: a

b, so that = 0

Anti-parallel vectors: a b, so that = a b= 0. (2.30)Note that the special case ofb= ais included in the parallel case, and results in

a a= 0 (2.31)

Example 1 A concrete problem from physics...The vector product has applications in physics, for example to calculate rotatory motionor torque. It proves helpful in the answer of the following question, which shall motivatethe introduction of the vector product.

Figure 2.9: The earth. The radius vectorr is connecting the earths origin with pointP on the surface, where the linear velocity isv. The earths rotation is given by theangular velocity . The radius r is the distance of point P to the z-axis. (modifiedfrom[1])

Question: What is the linear velocity of a point P on the earths surface resultingfrom earths rotation (see figure2.9)?

To phrase this problem mathematically, the center of the earth is set in the pointof origin of our spherical coordinate system. Then, a point Pon the earths surface isdescribed by the radius vector r, which conncects point P with the center of the earth.The earths rotation is given by a vector of angular velocity , with an absolute value of|| = = 2

T and the periode time of one complete rotationT. The direction of is the

axis of rotation, so that looking down from the tip of , we rotate counter-clockwise.The angle is given as (r, w) = = (90

latitude).

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As Pmoves on a circle of latitude with radius r, the velocity on the radius will be

v= r (2.32)

|r| =

rand

r are connected by the angle

:

r = r sin (2.33)By combining both equations, the result of the absolut value of the vector product ofand r can be obtained (compare equation2.29):

v= r sin = | r| = |v| (2.34)The velocityv is by definiton of the vector product perpendicular to and r.

Example 2 Unit vectors

Unit vectors in a coordinate system are always perpendicular orientated, as they haveto be independent from each other (compare section 2.2 and figure2.6). Therefore, thevector product of unit vectors is either one for different unit vectors, or zero for equalunit vectors.

Expressed in cartesian coordinates:

ex ex=ey ey =ez ez = 0ex

ey =ez

ey ez =exez ex = ey. (2.35)

Summed up in tabular form we get:

ex ey ezex 0 ez -eyey -ez 0 exez ey ex 0

Here, the vector in the column on the left is the one that is the first vector in the vector

product. Please compare these results with the ones for unit vector scalar products insection2.3.

These results are the same for all the other coordinate systems (see figure 2.6. Incylindrical coordinates:

er er =e e=ez ez = 0er e=eze ez =erez

er =e. (2.36)

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And in spherical coordinates:

er er =e e = e e = 0er e=ee e=ere er =e. (2.37)

Simple proof With the properties of the cartesian unit vectors (see equation 2.35),

the vector product (eq. 2.25) can be shown. Inserting the vectorsa andb in cartesiancoordinates (eq. 2.5):

c= a b= (a1ex+a2ey+a3ez) (b1ex+b2ey+ b3ez) (2.38)

Then, the distributivity (see eq. 2.28) of the vector product is used:

c=a1b1(ex ex) +a1b2(ex ey) +a1b3(ex ez)+a2b1(ey ex) +a2b2(ey ey) +a2b3(ey ez)+a3b1(ez ex) +a3b2(ez ey) +a3b3(ez ez)

with the table for the unit vector vector products, this results in the definition of thevector product:

c= (a2b3

a3b2)ex+ (a3b1

a1b3)ey+ (a1b2

a2b1)ez = ex ey eza1 a2 a3b1 b2 b3 (2.39)

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3 Vectoranalysis

In the following chapter, one- two- and three-dimensional integrals that are importantfor solving exercises in FEE1 will be discussed in the different coordinate systems.

3.1 The line Integral

3.1.1 Vector fields

Most electrical quantities have a direction and an absolute value, as for example theelectric field E, that is acting on a charge in a defined direction. Therefore, thesequantities are given as vectors (compare section 2.1). To describe such a quantityin space and in a coordinate system, every point P in space (for example ( x , y , z ) in

cartesian coordinates) has his own vector of the quantity, and the function E(x , y , z ) isthen called vector fieldor in that specific case electrical field.

3.1.2 Why are integrals needed?

Imagine a point charge that is moved along an arbitrary pathway in three-dimensional

cartesian coordinate space (see figure3.1(a)). This point charge can be subject to forces,for example an electrical force

Fel(x , y , z ) (3.1)

which has a dedicated vector for each point (x , y , z ) in space.Question: What amount of work is needed to move a point charge Q in the electrical

field from point a to point b on a given path or line (see figure 3.1(a))?The integral is needed to give an answer: The amount of work to move the charge

along a defined pathway fromato bcan be calculated by the line integral of the electricalforce Fel along that way:

W=

ba

Felds (3.2)

Wheredscan be thought as an elementary or infinitesimal length of the total pathways. The integral sums up these infinitesimally small parts multiplied with the functionterm. In general, the result of integration can be different for different pathways asshown in figure3.1(a)and3.1(b),even if the points aand bstay the same.

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3.1.3 Conservative fields

(a) Movement on random pathway (b) Movement on straight pathway

Figure 3.1: Line of movement between two points a and b in a vector field

cartesian cylindrical sphericalds dx ex, dy ey, dz ez dr er, rd e, dz ez dr er, r sin de, rd e

... ...

...

Table 3.1: Possible simple line elements for the line integration on simple pathways indifferent coordinate systems (compare figure3.2).

3.1.3 Conservative fields

A field is named a conservative field, if the result of the integration stays the samefor every possible pathway. Or in other words: The line integration overevery closedpathwayin a conservative field Fis always zero:

F ds= 0 (3.3)

Fortunately, the vector fields treated in FEE1 are only conservative fields. The path-way of integrationdsin eq. 3.2can therefore be chosen as simple as possible, for example

as a straight line between the two points (see figure3.1(b)). In most cases, it is chosenalong the axes directions of the coordinate system, and the coordinate system has to bechosen carefully to fit the given geometry of the problem. To integrate alonge these axes,the basic line elements ds= ds es are needed, and listed in table 3.1. The direction ofds is given by the unit vector of the respective coordinate of integration (for example:dx = dx ex). Note that for simplicity reasons, the elements for polar coordinates arenot listed, as they equal those for cylindrical coordinates. To get a graphical impressionof these elements, see figure3.2.

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3.1.4 Example: Calculation of the work needed to move a charge in a conservative

force field F

Figure 3.2: Simple line elements for line integration in (a) cartesian, (b) cylindrical

and (c) spherical coordinates (modified from[2])

3.1.4 Example: Calculation of the work needed to move a charge

in a conservative force field F

This example was taken from [2]: Given is the force field F(x , y , z ) = (2y, 2x, z), as wellas the startpoint a= (0, 0, 1) and the endpoint b= (2, 4, 1) of the movement.

The work is calculated as defined in equation3.2:

W=

b

a

F ds=

b

a

2y2xz ds (3.4)

As the force field is conservative (see equation3.3), the result of integration is not de-pending on the pathway of integration. We calculate in cartesian coordinates, thereforethe line elements can be extracted from table3.1. With them, the path of integrationcan be separated in three parts, related to the respective axis:

W =Wx+Wy+ Wz

=

x1

x0

F dsx+

y1

y0

F dsy+

z1

z0

F dsz

=

x1x0

2y2xz

dx0

0

+ y1y0

2y2xz

0dy

0

+ z1z0

2y2xz

00

dz

Solving the scalar product of the vectors and inserting the respective integration limits

gives:

W =

20

2ydx+

40

2xdy+

11

zdz (3.5)

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3.2 The double integral

Figure 3.3:Different pathways: straight line, parabolic line and separated pathway alongthe cartesian axes. (modified from [2])

Notice that the integral is now split up in three basic integrations. Furthermore, thez-integration is zero because of identical upper and lower limit. Imagine the integrationsas integrating along the specific coordinate axis, see figure 3.3. For the first integral,the integration is along the x-axis, therefore all other coordinates are set constant, andit holds: x= x, y= 0, z= 1. For the second integral, the integration is along the y -axisand it holds: x= 2, y= y, z=z. It follows:

W =

20

0dx+

40

2 2dy+ 0 =4

0

4dy= 4 4 = 16(J). (3.6)

As the field is a conservative field, this result is true for all possible pathways (someare given in figure3.3). This calculation principle can be transfered to cylindrical andspherical coordinates.

Task: Calculate the work between the two points in cylindrical coordinates, use thegeometry of that system with its line elements and a pathway that is going radially andcircular along the cylindrical axes (see figure3.4). You should get the same solution as

in cartesian coordinates.

3.2 The double integral

3.2.1 Introduction of the double integral

In the following, cartesian coordinates are used to derive the double integral. It is alreadyknown that the definite integral from a to b measures the area between a function f(x),the x-axis and the two parallels to the y-axis at the points aand b, as it can be seen infigure3.5.

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3.2.1 Introduction of the double integral

Figure 3.4: Prefered pathway in cylindrical coordinates. (modified from [2])

Figure 3.5: The value of a one-dimensional integration can be seen als the area Abetween functionf(x) andx-axis. (from [1])

The integral can be seen as base height, where the heightf(x) is dependent fromthe value on the x-axis.

A= b

a

f(x) dx (3.7)

In another possible form of the integral, the height f(x) itself is already the resultof an integration. This means that f(x) is a summation of path elements in directiony:

f(x) =

f(x)0

dy=y

f(x)

0

(3.8)

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3.2.2 Some basic examples

cartesian cylindrical sphericalds dx ex, dy ey, dz ez dr er, rd e, dz ez dr er, r sin de, rd ed A dxdy ez, rdrd ez, r sin ddr e,

dxdz ey, drdz e, rddr e,

dydz ex rddz er r2sindd er...

... ...

Table 3.2: Line elementsds and possible area elementsd A. Area elements are multi-plications of two line elements in the respective coordinate system. Therefore, threedifferent basic area elements are possible in every coordinate system.

The double integral in cartesian coordinates can then be defined as:

A=

x=bx=a

y=f(x)y=0

dy dx= x=b

x=a

y=f(x)y=0

dy dx (3.9)

To put the upper equation in words: If we want to determine the area of a certain2-dimensional figure, we fill it with infinitely small area-segments. These segments arespanned by two infinitely small line segments. We encountered those already with theline integral. All the while we have to keep in mind the limits of the area we want todetermine (which appear here as integration limits).

In a more general form independent of the coordinate system, the area of a certaingeometry is then given by:

A=

1dA (3.10)

where dA is the infinitesimal area element, that is spanned by two line elements of therespective coordinate system. An area element d A is defined by an absolute value (sizeof the area) and a unit vector eA that is perpendicular orientated to the surface ofthe area. In every coordinate system, three different combinations of line elements arepossible to form a basic area element (see table 3.2).

3.2.2 Some basic examplesThe following examples illustrate that the choice of a coordinate system can make thetask of integration easier or harder, depending on the geometry we have to deal with.

A rectangular area in cartesian coordinates The area A of a rectangle shall becalculated by a double integration (see figure3.6).

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Figure 3.6: Use of double integraton to calculate the rectangular areaA (from[1])

A=

by=0

ax=0

dxdy =

by=0

a dy=a b (3.11)

The result is not surprising, as it is the area of a rectangle with the sidelengths a andb. If we start the integration with the other coordinate:

A=

ax=0

by=0

dydx=

ax=0

b dx= b a (3.12)

the result is identical. Even

A=

a0

dx

b

0

dy

= a b (3.13)

is not changing the result.The change of order of integration was possible because for both variables ( x and y)

the integration limits are constants. We could even write the integration as the productof two integrations. If the limits itself contain variables, the order of integration isimportant.

This example illustrates that cartesian coordinates are well suited for rectan-gular geometries.

A circular area in cartesian coordinates As can be seen in figure3.7, a circle can bedescribed by: x2 + y2 =r2. If we integrate in the limits from

r < x < r, then a single

integral, together with the function f(x) is obtaining the asked area A. This equationresults in two solutions for y = f(x) = r2 x2, where every solution is the functionfor a half circle. Therefore, the area A can be expressed as 2 times the area of half acircle.

A = 2rrf(x)dx= 2

rr

r2 x2 dx= 4 r0

r2 x2 dx

= 4 12

x

r2 x2 +r2 arcsin xr

r0

= 2r2 arcsin 1 = 2r2 2

=r2 (3.14)

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The result is the well known area of a circle. Another way is to integrate with a doubleintegral: integration limits shall now ber x r andr2 x2 y r2 x2,where the y-limits depends on the x-value:

A= rx=r

r2x2y=

r2x2

dydx (3.15)

Figure 3.7: Use of double integration to find the areaA of a circle. (from [1])

It is now important to perform the integration in ydirection first, as the result includes

the integration-variable x. That results in

A= 2 rx=r

r2 x2 dx (3.16)

and can be solved according to equation 3.14 that was derived geometrically withfigure3.7.

A circular area in polar coordinates The comparison of the two previous examplesillustrates, that the area of a rectangle is easier to determine, since the integration vari-ables are independent from one another and the integration limits are simply constants.

The reason for this is the shared symmetry of the rectangular dxdy-elements in cartesiancoordinates and the rectangular area itself.To make our life easier we now want to define infinitely small area elements that are

better suited to the geometry of the circle. Looking back to section 2.2, the assumptioncan be made that polar coordinates (or cylindrical coordinates with z=const.) are theright way to achieve this easy integration (see figure 3.8). For that integration, the lineelements in polar coordinates are needed (see table 3.1).

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3.3 The triple integral

Figure 3.8: Circle in polar coordinates. Compare the line elements to figure3.2(b) -they are the same. (from [1])

We deduce from the figure the infinitesimal area element dA (the red, filled area),that is a multiplication of the elementary line elements:

dA= dr rd = rdrd (3.17)

What is the part of the circular ring between the radius r and r+dr and limited bythe angle d, and in principle well known (see table 3.2). To distinct the integration,the radius r = R shall be the outer radius of the circle, and r shall be the coordinate.Now we can try to solve the integral by the help of this element.

A =r=Rr=0

2=0

r drd=r=R

r=0 r dr

2

=0 d

= r

2

2

2 = R2 (3.18)

The effect of the polar coordinates is very similar to the cartesian coordinates and therectangular area. The solution is the same as in eq. 3.14(except for the notationr = R),but the calculation is simplified.

3.3 The triple integral

3.3.1 Introduction of the triple integral

After having treated 1- and 2-dimensional elements, we want to have a look at the

3-dimensional space. As introduction of the triple integral, the example of measuringthe volume of a sphere in case of cartesian coordinates is used. Analogously to thecase of areas, the height z(x, y) above the xy-plane can be interpreted as the resultof an integration, with the function of a half sphere z(x, y) =

R2 x2 y2 used as

integration limit:

z(x, y) =

R2x2y2

R2x2y2dz. (3.19)

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3.3.2 Example: Volume of a sphere

cartesian cylindrical sphericalds dx ex, dy ey, dz ez dr er, rd e, dz ez dr er, r sin de, rd ed A dxdy ez, rdrd ez, r sin ddr e,

dxdz ey, drdz e, rddr e,

dydz ex rddz er r2sindd erdV dxdydz rdrddz rd r sin

r

ddr

Table 3.3: Summary: infinitesimal line elements ds, area elements d A and volumeelementsdV in the respective coordinate system.

The volume of the sphere is then (without a factor 2, that once again is the result ofsymmetry)

V =

Rx=R

R2x2

y=R2x2

R2x2y2z=

R2x2y2

dzdydx (3.20)

with R as radius of the sphere.If the integration is performed, it is necessary to integrate in the right order, as the

limits are depending on other variables.

V =

R

x=R R2x2

y=R2x22

R2 x2 y2 dydx

= 2 Rx=R2

12yR2 x2 y2 + (R2 x2) arcsin y

R2x2R2x20

dx

= 4Rx=0

(R2 x2)2

dx= 2

R2x 13

x3R

0= 4

3R3 (3.21)

The triple integral can be interpreted similarly to the double integral case. To deter-mine the volume of a certain body, we fill it with infinitely small volume elementsdV (incartesian case: dV = dxdydz) which are a multiplication of all three independent lineelements in the respective coordinate system - see table 3.3for other volume elements.

These elements are then summed up in the limits of the body. Therfore, the generaldefinition of a triple integral(volume) is given as:

V =

dV (3.22)


In cartesian coordinates, the problem of the circle (compare section3.2.2)repeats itselfin case of a sphere: The integration limits can vary and are not independent. Lookingback at how this problem was solved in the case of the circle, the solution to this problemlies once again in the right coordinate system. By choosing spherical coordinates, theintegration can be simplified (see figure3.9).

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Figure 3.9: Volume element in spherical coordinates. (modified from [1])

The volume element dV can be derived from the figure. For the base, it holds:

rd r sin d (3.23)Together with the height dr this results in the volume element given in table 3.3.

dV =r2dr sin d d (3.24)

Then, the volume of the sphere can be calculated, starting from the definition inequation3.22:

V =

dV =

R0

0

a d

20

r2dr sin d d=R

0

r2 dr

0

sin d

20

d (3.25)

The integration limits are chosen in a way that every point of the sphere appears onlyonce in the integration. So the volume of the sphere is

V = R3

3

(

cos )

0 2 =

4

3

R3 (3.26)

This calculation was a lot easier than in cartesian coordinates (compare equation 3.21)as they are not suited to the spherical geometry.

Another concrete image of spherical coordinates is that they can be seen as polarcoordinates

x= (r sin )cos , y= (r sin )sin (3.27)

with a radius r = r sin , completed by a third angle that defines the height:

z=r cos . (3.28)

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Bibliography

[1] Karl-Heinz Lotze and Simon Stutzer. Vorlesungsskript: Mathematische methodender physik. 2010.

[2] Herbert P. Neff. Introductory Electromagnetics. John Wiley & Sons, 1991.

math intro script

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