Math Boot Camp

Math Boot CampThese notes cover all of the mathematical background that you will need to Ph 1b and Ph 1c.

Trigonometry

Notation

A 2D vector A may be written in Cartesian coordinates as

A = Ax x + Ay y (1)A = A1 x + A2 y (2)

A = Ax, Ay (3)A 3D vector A may be written in Cartesian coordinates as

A = Ax x + Ay y + Az z (4)A = A1 x + A2 y + A3 z (5)

A = Ax, Ay, Az (6)The norm of a 2D vector A = Ax x + Ay y is A = Ax2 + Ay2 A. The norm of a 3D vector A = Ax x + Ay y + Az z is A = Ax2 + Ay2 + Az2 A.Definitions that appear for the first time are denoted by triple equal signs.

Vectors are denoted by an overfixed arrow (Ex: A). Unit vectors (vectors with norm 1) are denoted by an overfixed hat (Ex: x). Any object without an overfixed or ^ is a scalar.The zero vector will be denoted by 0 and simply implies a vector (in the dimension we are working in) that has 0 for all of its components.

I use Mathematica notation, which means that the functions names are camel cased (i.e. a single word with the first letter of each word capitalized) and are denoted by hard brackets. For example, a function f of a variable x is written as f [x]. Therefore, Sin

2 = 1 and ArcTan[1] =

4. The benefit is that you can type this code directly into

Mathematica and evaluate it

Sin 2 1

Dot and Cross Products

Given vectors A = Ax x + Ay y + Az z and B = Bx x + By y + Bz z with angle between them, A B = Ax Bx + Ay By + Az Bz = A B Cos[] (7)

AB = Ay Bz - Az By x + (Az Bx - Ax Bz) y + Ax By - Ay Bx z = A B Sin[] n (8)

where n is a unit vector given applying the right-hand rule to A and B. A useful neumonic to remember the cross product is

AB = x y zAx Ay AzBx By Bz

(9)

Properties of the Dot Product

The dot product is: A scalar Commutative (A B = B A) Distributive (Any vectors A, B, C satisfy A B + C = A B + A C) The dot product of a vector A = Ax x + Ay y + Az z with a unit vector equals the length of A along that unit vector. For example, A x = Ax. A A = A2 If A A = 0, then A = 0

Properties of the Cross Product

The cross product is: A vector Anti-commutative (AB = -B A) For any vectors A, B and scalar c, c AB = c AB = Ac B For any vector A, A A = 0 The cross product does not obey the cancellation law (Given vectors A, B, C, if AB = AC it is not true that either A = 0 or B = C. Counter-example: A = k, B = i, C = i +k) Distributive over addition. That is, any vectors A, B, C satisfy AB + C = AB + AC Not associative. Counter-example: With A = i, B = j, C = j, then ABC = 0 - i = ABC

Calculus

The calculus portion of this bootcamp closely follows David Griffiths book Introduction to Electrodynamics Chapter 1.

Position

The position vector r is defined asr = x x + y y + z z (10)

with norm

2 Math Boot Camp.nb

r = (x2 + y2 + z2)1/2 (11)The corresponding unit vector (pointing radially outwards) equals

r = r

r= x x+y y+z zx2+y2+z21/2 (12)

We define the infinitesimal displacement vector from the point (x, y, z) to the point (x + x, y + y, z + z) as s = x x + y y + z z (13)(We could call this r, since that is what it is, but it is useful to reserve a special letter for infinitesimal displacements.)

Derivatives

Given a function of one variable f [x], its derivative f tells us the rate of change of f , f = fx x (14)In other words, if we change x by an amount x, then f will change by an amount f , with the derivative being the proportionality factor. This has a nice geometrical interpretation as the slope of the function f [x].A function of 3 variables T[x, y, z] requires us to generalize our definition of the derivative toT = Tx x + Ty y + Tz z (15)where the notation Tx is the partial derivative of T with respect to x (which means that y and z are treated as con-stants when taking this derivative (for example, if T = 2 x2 y then Tx = 4 x y, Ty = 2 x2, and Tz = 0)). We could write this equation more compactly as T = T s (16)where T Tx x + Ty y + Tz z (17)is the gradient of T (i.e. the 3D analog of the derivative) and s = x x + y y + z z (18)is the infinitesimal displacement vector from above. Using a property of the dot product, T = T s Cos[] (19)where is the angle between the vectors T and s. So if we fix the magnitude s and rotate it around to cover all possible directions, then the maximum change in T will occur when = 0 (which implies that s points in the same direction as T). Therefore, a geometric interpretation of the gradient is that T points in the direction of maximum increase of the function T (20)and furthermore that T gives the slope of T along this maximal direction (21)Using the above expression, we also immediately see that -T is the direction of maximum decrease of the func-tion T, and that any direction perpendicular to T will result is no change to T (i.e. this plane will have a constant T value). Of course, all of these statements only hold for an infinitesimal displacement s.We can write

Math Boot Camp.nb 3

T = Tx x + Ty y + Tz z= x x + y y +z z T (22)where = x x + y y +z z (23)is the del operator. (Only this once, I will write the unit vectors on the left so that you will not misinterpret this

statement as xx which would be zero.)Note that is not a normal vector, but instead an operator that acts upon a scalar to produce a vector (so T does not mean ordinary multiplication). In face, we know of three ways that vectors can be multiplied: Multiply vector A by a scalar c: c A Take the dot product of two vectors A and B: A B Take the cross product of two vectors A and B: ABCorrespondingly, the operator can act in three ways: On a scalar function T: T (the gradient of T) On a vector A via the dot product: A (the divergence of A) On a vector A via the cross product: A (the curl of A)The gradient was discussed above, so we now focus on the other two operations.

Divergence

The divergence of a vector A = Ax x + Ay y + Az z equals A = x x + y y + z z Ax x + Ay y + Az z= Axx + Ayy + Azz (24)Note that the divergence of a vector is a scalar (and there is no such thing as the divergence of a scalar)! Geometri-cally, the divergence of a vector A represents how much A spreads out from any point. More precisely, if you treat A as describing the flow of a fluid and you put an infinitesimally small volume at a point r, then the volume will empty out in time if the divergence is positive, fill up with time if the divergence is negative, and hold the same amount of fluid if the divergence equals zero. In other words, the divergence measures the "amount of arrow" leaving an infinitesimal volume at each point in space, with larger arrows counting more than smaller arrows.

Critical point: In this course, we will often work with vectors which are themselves functions of position. For example, A = x x + y y is a different vector at each point and space (and it has a positive divergence at each point in space). We can sample different points in space to get a feel for what this looks like, as shown below. These "vectors" are actually called vector fields.

4 Math Boot Camp.nb

-1.0 -0.5 0.0 0.5 1.0-1.0-0.50.0

0.5

1.0

A = x x + y y

ExampleCalculate the divergence of the vector field A = x + y .SolutionSince this is a constant vector field, the amount of arrows entering any infinitesimal volume equals the number of arrows exiting that volume.

-1.0 -0.5 0.0 0.5 1.0-1.0-0.50.0

0.5

1.0

A = x + y

And indeed, A = x [1] + y [1] = 0, as expected. ExampleCalculate the divergence of the vector field A = x x + y2 y .Solution

Plugging into our equation, A = xx + y2y = 1 + 2 y. But lets visually check this by plotting the 3D vector field.A = x x + y2 y

-10

1

-10

1

-10

1

Math Boot Camp.nb 5

This plot - while correct - is extremely cluttered. Because A does not depend upon z, we can simply take a cut of this vector field at z = 0.

-1.0 -0.5 0.0 0.5 1.0-1.0

-0.5

0.0

0.5

1.0

A = x x + y2 y + z3 z

Consider the point (0.4, 0) shown by the black square above. There are no vectors going in or out of it in the y-direction, but there are definitely larger vectors coming out of the square than going into it. This signifies that there will be positive divergence, in line with the result (1 + 2 y)x=0.4,y=0 = 1 found above. Similarly, the green square around (-0.5, -0.5) has larger vectors entering it than leaving it on the top and bottom and larger vectors leaving it than entering it on its sides. These two effects seem rather equal, and indeed we find (1 + 2 y)x=-0.5,y=-0.5 = 0 from above. Lastly, the purple square centered at (-0.3, 0.7) has divergence (1 + 2 y)x=-0.3,y=0.7 = 2.4 which agrees with the fact that larger arrows are leaving the square than entering it. This shows the geometrical description of the divergence, but be warned: you must use an infinitesimal squares for this to be correct (I only showed a large square here for convenience!) Example

Compute the divergence of the vector field A = rr2

.

SolutionBecause r = x,y,zx2+y2+z21/2 and r2 = x2 + y2 + z2,

A = xx2+y2+z23/2 , yx2+y2+z23/2 , zx2+y2+z23/2 (25)Therefore, A = x xx2+y2+z23/2 + y yx2+y2+z23/2 + z zx2+y2+z23/2 (26)Computing the first of these terms,

6 Math Boot Camp.nb

x xx2+y2+z23/2 = 1x2+y2+z23/2 - 3 x2x2+y2+z25/2= -2 x2+y2+z2x2+y2+z25/2 (27)By symmetry (i.e. by redefining x y, y z, and z x), the other two terms must bey yx2+y2+z23/2 = x2-2 y2+z2x2+y2+z25/2 (28)z zx2+y2+z23/2 = x2+y2-2 z2x2+y2+z25/2 (29)and therefore A = 0 (30)That is pretty shocking, for if we draw this vector field (taking a cut at z = 0), what do we expect for its divergence to be?

A = r

r2

But upon closer inspection, maybe this isnt so crazy. Lets consider the area between a circle of radius r1 and r2 where r1 < r2. Note that the divergence represents the amount of the arrows leaving an infinitesimal infinitesimal volume at each point in space, and not on a finitely large circle. However in the special case where the divergence equals 0 everywhere, that means that at each infinitesimal volume in space, the amount of arrows leaving that point equal the amount of arrows entering it, so in this case only the amount of arrows entering a finite volume should equal the amount of arrows leaving a finite volume. Lets check this for the donut shown above.The arrows enter the donut all along its inner perimeter, which has length 2 r1, and these arrows have magnitude 1r1

2 . Therefore, the weight of these arrows equals (2 r1) 1r12 = 2 r1 . Similarly, the arrows leaving the donut occur at the perimeter 2 r2 with magnitude 1

r22 . Therefore, the weight of these arrows equals (2 r2) 1r22 = 2 r2 . So we

expect that the divergence, which equals the weight of the arrows leaving the figure minus the weight of the arrows entering the figure, should equal 2

r2- 2

r1. Since r1 < r2, this is a negative number, and certainly not zero. So what

went wrong?

Math Boot Camp.nb 7

The problem is that we are working in 2D, rather than 3D! Indeed, in 2D the divergence is negative (in fact, it is - 1r3

), but in 3D we have to take into account the surface area of the inner and outer spheres. Therefore, the the

weight of the arrows entering the inner sphere equals 4 r12 1r12 = 4 while the weight of the arrows leaving the outer sphere equals 4 r22 1r22 = 4 , and these two contributions do indeed cancel to leave a net 0 divergence!Great, but we still have one more mystery to solve. While it may seem reasonable that the divergence of A = r

r2 is 0

at every point, there is one point at which this cannot be so: the origin. For at the origin, there are vectors eminating outwards in every direction, so putting an infinitesimal sphere at the origin must surely yield a positive divergence. In fact, as it turns out, the divergence at the origin is infinitely large, and later in the course we will quantify this infinity explicitely to determine the electric potential from a point charge.

Curl

In 2D, you can construct a del operator = x x + y y and the divergence of a 2D vector A = Ax x + Ay y equals A = Axx + Ayy . However, there is no analogy in 2D for the curl operator. In this class we will work almost exclusively in 3D.

The curl of a vector A = Ax x + Ay y + Az z equals A = x y zx y z

Ax Ay Az= Azy - Ayz x + Axz - Azx y + Ayx - Axy z(31)

Note that the curl of a vector yields another vector. You cannot take the curl of a scalar, or a 1D or 2D vector. Therefore, when taking the curl of a vector such as A = x + y we will always assume it is a 3D vector with zero for its z component.The curl measures how much a vector field "curves around" a point. If an object in your vector field would rotate (ignore its translation and just focus on rotation) then that point has a non-zero curl. For example, considering the motion of water in a pond, a whirlpool would be a region of large curl with the direction of the curl given by the right-hand rule.

ExampleCalculate the curl of the vector field A = - y x + x y .SolutionWe expect from the shape of the vector field that there will be a curl in the +z direction.

8 Math Boot Camp.nb

-1.0 -0.5 0.0 0.5 1.0-1.0-0.50.0

0.5

1.0

A = -y x + x y

And indeed, A = Azy - Ayz x + Axz - Azx y + Ayx - Axy z = 2 z. This may perhaps seem like an obvious statement about the origin, but the result means more than that, since A = 2 z implies that placing a small object at any point will result in that object rotating at the same rate (although for any point that isnt the origin the object will also translate). ExampleCalculate the curl of the vector field A = x x + y y + z z.SolutionThis vector field points radially outward. What do we expect the curl to be?

A = x x + y y + z z

-10

1

-10

1

-10

1

From our formula, A = 0. Does this make sense? Looking at the plane z = 0,

-1.0 -0.5 0.0 0.5 1.0-1.0-0.50.0

0.5

1.0

A = x x + y y + z z

we see that although any small object would certainly be pushed radially outwards, it will not rotate. For example, placing a sphere - which in the z = 0 plane is a circle - at the point (0, 1) shows that the rotation on the points in the

Math Boot Camp.nb 9

left side of the sphere will be perfectly balanced by the rotation on the points in the right side of the sphere. Because the sphere will not rotate, its curl must be zero. This applies to a circle at any other point by the symmetry of the vector field. Line Integral

A line integral is an expression of the form v l (32)where v is a vector function, l is the infinitesimal displacement vector, and the integral is to be carried out along a prescribed path . If the path forms a closed loop (i.e. its two endpoints are the same), the line integral is often denoted as v l (33)At each point on the path we take the dot product of v (evaluated at that point) with the displacement l to the next point on the path. A familiar example of a line integral is the work done by a force F : W = F l.ExampleCalculate the line integral of the function v = y2 x + 2 x (y + 1) y from the point a = (1, 1, 0) to the point b = (2, 2, 0) along the path (1) and (2) in the figure below. What is v l for the loop that goes from a to b along (1) and then returns along (2)?

Solution

As always, l = ( x) x + ( y) y + ( z) z. Path 1 consists of two parts: part (i) is a horizontal line segment where y = z = 0. Thus l = ( x) x and y = 1 so that v l = x + 4 x y ( x) x = x and hence(i)v l = 12 x = 1 (34)On the vertical stretch (ii) of this path, x = z = 0 and x = 2 so that l = ( y) y and v l = y2 x + 4 (y + 1) y ( y) y = 4 (y + 1) y. Therefore(ii)v l = 124 (y + 1) y = 10 (35)Putting the two paths together, we find that along path (1),(1)v l = (i)v l+(ii)v l = 11 (36)On path (2), define t as a variable that will range from 1 to 2, so that x = y = t along path (2). Then x = y = t and z = 0 so that l = ( t) x + ( t) y and v l = t2 x + 2 t (t + 1) y ( t) x + ( t) y = 3 t2 + 2 t t. Thus

10 Math Boot Camp.nb

the path integral equals (2)v l = 123 t2 + 2 t t = 10 (37)For the loop that goes out through (1) and comes back through path (2), v l = (1)v l-(2)v l = 1 (38)If we considered the loop in the reverse direction (out through path (2) and back through path (1)) then we would have gotten the negative of this answer, so make sure to keep your orientation straight! Surface Integral

A surface integral is an expression of the form Sv a (39)where v is again some vector function, and a is an infinitesimal patch of area, with direction perpendicular to the surface, as shown in the figure below.

There are, of course, two directions perpendicular to any surface, so the sign of a surface integral is intrinsically ambiguous. If the surface is closed (forming a "balloon"), then a surface integral may be written as v a (40)By convention, the direciton "outward" is treated as positive for closed surfaces, but for closed surfaces it is arbi-trary. If v describes the flow of a fluid (mass per unit area per unit time), then Sv a represents the total mass per unit time passing through the surface - hence its alternative name, flux. ExampleCalculate the surface integral of v = 2 x z x + (x + 2) y + y z2 - 3 z over the square area shown below bounded by the points (2, 0, 0), (2, 2, 0), (2, 2, 2), and (2, 0, 2). Let the direction + x denote the positive direction.

0

1

2

3

x

0.00.5

1.01.5

2.0y

0.0

0.5

1.0

1.5

2.0

z

SolutionFor this surface, x = 2 and a = y z x so that v a = 4 z y z. Therefore, the surface integral equals

Math Boot Camp.nb 11

Sv a = 02024 z y z = 16 (41)Had we chosen the - x to denote the positive direction for the integral, we would have found the negative of this answer. Volume Integral

A volume integral is an expression of the form T (42)where T is a scalar function and dr is an infinitesimal volume element. In Cartesian coordinates, = x y z (43)For example, if T is the density of a substance (which might vary from point to point), then the volume integral would give the total mass. You can also take the volume integral of a vector functionA = x Ax + y Ay + z Az (44)where the unit vectors come outside the integrals because they are constant.

ExampleCalculate the volume integral of T = x y z2 over the prism shown below.

SolutionThis is just plain calculus, but you need to keep track of your variable ranges carefully to ensure that you cover the entire figure. You can take the integrals in any order (and doing so in multiple ways is often a great way to check your work), T = 03z2 01 y 01-yx x y z= 03z2 01 12 y (y - 1)2 y z= 03 124 z2 z= 38

(45)

which is the desired answer. ExampleMost people know that the volume of a cone with base radius r and height h is V = 1

3 r2 h. Prove this for a right

cone whose base of radius r is centered about (0, 0, 0) and whose apex is at (0, 0, h).


SolutionThe volume equals the volume integral of the constant function 1 over the entire cone. Taking slices of thickness z along the z-axis,

V = 0h-1- zh r1- zh r -1- z

h2 r2-y21/21-

z

h2 r2-y21/2 x y z

= 0h 1 - zh 2 r2 z= 13 r2 h3(46)

We could also be a bit more clever and know that the bases will be (first order) cylinders with radius 1 - zh r and

height z, and then just skip to the second step. Alternatively, we could do the calculation in Mathematica, Integrate1, {z, 0, h}, y, - 1 - z

hr, 1 - z

hr, x, - 1 - z

h2r2 - y2 , 1 - z

h2r2 - y2 , Assumptions 0 < h && 0 < r

13h r2

A very similar calculation shows that the volume of any prism with base area A and height h has volume 13

A h. If we

orient the prism so that its base lies at z = 0 and its tip at z = h, then from the second line in (46) we find that its area at height z [0, h] equals A 1 - z

h2. (One factor of 1 - z

h comes from contracting the base in the x-direction

and another from contracting the base in the y-direction.) Therefore the volume equals

V = 0h A 1 - zh 2 z= 13 A h (47)as desired.

Coordinate Systems

2D Polar Coordinates

A generic 2D vector in Cartesian coordinates is

A = Ax i+ Ay j (48)which in polar coordinates equals

A = r Cos[] i+ r Sin[] j (49)


where

r = Ax2 + Ay2 (50) = ArcTan AyAx (51)

When doing 2D integrals in Cartesian coordinates, we are familiar with the area element x y. We use this when computing an integral f [x, y] x y. But where does x y come from? If we consider a generic point (x, y) and infinitesimally displace it in the x-direction and y-direction by x and y, respectively, we find the points (x + x, y), (x, y + y), and (x + x, y + y). What is the area of the region encompassed by these points? Simply x y!The situation is slightly more complicated for polar coordinates. Consider a generic point (r, ) and the infinitesi-mally displaced points found when we vary the first coordinate by r and the second coordinate by . Thus we have four points (r, ), (r + r, ), (r, + ), and (r + r, + ). As shown in the figure below, the element of this infinitesimal area equals ( r) (r ) = r r . Thus,

polar coordinate area element = r r (52)

What does this imply? If we integrate f [r, ] over a region in polar coordinates, then the integral must take the form f [r, ] r r (53)ExampleCompute the area of a region R enclosed by the curve r[] and the rays = a and = b where 0 < b- a 2 .


SolutionTo compute an area, we set f [r, ] = 1 in equation (53) and find the areaab0r[]r r = 12 abr[]2 (54)which is a famous formula for area in polar coordinates. ExampleFind the area of an ellipse with semi-major axis a and semi-minor axis b.

SolutionOrienting our ellipse as in the figure above, the equation of an ellipse is

x2

a2+ y2

b2= 1 (55)

In polar coordinates (x = r Cos[], y = r Sin[]) this becomes r2 Cos[]2

a2+ r2 Sin[]2

b2= 1 (56)

Method 1: Integrate using Polar CoordinatesAs shown in the previous Example, the area of the ellipse equals 1

2 02 r[]2 where r[] traces the boundary of the ellipse. From equation (56),

r[]2 = 1

Cos[]2a2

+ Sin[]2b2

= a2 b2b2 Cos[]2+a2 Sin[]2 (57)

Integrating over the top-half of the ellipse and multiplying by 2 (for the bottom half), the area of the ellipse equals0 a2 b2b2 Cos[]2+a2 Sin[]2 = a b ArcTan a Sin[]b Cos[] =0= = a b- 0 = a b (58)You must be very careful when taking the bounds on this integral, because of the branch cuts (i.e. the discontinu-ities) in ArcTan. When taking the bounds on this integral you must take the modulus of the ArcTan function so that it becomes continuous. Moral of the story: check your answer with Mathematica (but note that even with Mathemat-ica it is easy to get mixed up about the ArcTan branch cuts!)


Integrate a2 b2b2 Cos[]2 + a2 Sin[]2 , {, 0, }, Assumptions 0 < b < a

a b Method 2: Integrate using Cartesian CoordinatesUsing the equation of an ellipse

x2

a2+ y2

b2= 1 (59)

we can integrate over the top-half of the ellipse in Cartesian coordinates (multiplying by 2 to account for the bottom half)

2 -aa b1 - x2a2 1/2 = b x 1 - x2a2 1/2 + a ArcSin xa x=-ax=a = a b 2 - - 2 = a b (60)This integral is nasty, so you probably wont tackle the problem this way without Mathematica at your back. However, it is nice to check that this method yields the same answer.

Method 3: A Clever TrickBy far the easiest way to find the area of an ellipse is to note that the equation

x2

a2+ y2

b2= 1 (61)

becomes a circle if we stretch our coordinate system in the y-direction using the transformation y = ba

y,

x2

a2+ y2

a2= 1 (62)

This is the equation of a circle of radius a, and therefore of area a2. An infinitesimal unit of area in our normal coordinates, x y, is related to the infinitesimal unit of area in the new coordinates, x y = b

a x y. In other

words, any area that we measure in the x-y coordinate system will be ab

bigger than the corresponding area in the x-y

coordinate system. Therefore, the area of an ellipse in the x-y coordinate systems equals ba a2 = a b.

3D Spherical Coordinates

A generic 3D vector in Cartesian coordinates is

A = Ax i+ Ay j + Az k (63)which in spherical polar coordinates equals

A = r Sin[] Cos[] i+ r Sin[] Sin[] j + r Cos[] k (64)where [0, 2 ) is the polar angle and [0, ] is the azimuthal angle.

Unfortunately, as seen above, physicists take the horrible convention where is the polar angle in 2D radial coordi-nates but is the polar angle in 3D spherical coordinates. Therefore, to regain polar coordinates from spherical


coordinates you must set = 2

and then rename = . (Mathematicians, on the other hand, tend to switch the definitions of and in spherical coordinates to avoid this issue!)We can convert from Cartesian coordinates to spherical coordinates using

r = Ax2 + Ay2 + Az2 (65) = ArcTan AyAx (66) = ArcTan Ax2+Ay21/2

Az = ArcCos AzAx2+Ay2+Az21/2 (67)

(I encourage you to verify that both of these expressions for are equal!)In 3D Cartesian coordinates, we know that the volume element equals x y z. In spherical coordinates, we can compute the volume element by considering a generic point (r, , ) and looking at the volume element found by infinitesimally displacing its three coordinates by r, , and . As shown in the diagram below, the volume of this element equals ( r) (r ) (r Sin[] ) = r2 Sin[] r . Thus,

spherical coordinate volume element = r2 Sin[] r (68)

r d

We can also define at each point in space spherical coordinate unit vectors r, , and where r points radially outward, points in the x-y plane along the polar angle, and points in the azimuthal direction. In certain situa-tions, this can simplify the descriptions of systems. (For example, suppose that you had a vector associated with each point in space that pointed radially outward with magnitude proportional to its distance from the origin. This

vector field could be described by x i + y j + z k using Cartesian coordinates or by the more intuitive r r using

spherical coordinates.) However, note that r, , and actually depend upon the location of the base point (more precisely, they depend on and ). For example, when = 0 we find r = k but when = we find r = -k.


Notes:

- The unit vectors (i, j, k, r, , ) all have length 1

- Within any coordinate system, the unit vectors are all orthogonal to each other

- We always work in right-handed coordinate systems. In the language of the cross product, i j = k and r = .

3D Cylindrical Coordinates

Another 3D coordinate system that is occasionally convenient is cylindrical coordinates which has a radial distance , a polar angle , and a height component z . Essentially, cylindrical coordinates extend 2D polar coordinates to 3D by adding the Euclidean height z. A vector A = Ax i + Ay j + Az k in Cartesian coordinates would be written in cylindrical coordinates as

A = Cos[] i+ Sin[] j + z k (69)where [0, 2 ) is the polar angle.

We can convert from Cartesian coordinates to cylindrical coordinates using


= Ax2 + Ay2 (70) = ArcTan AyAx (71)

z = Az (72)As we did for spherical coordinates, we can find the volume element in cylindrical coordinates by considering a generic point (, , z) and looking at the volume element found by infinitesimally displacing its three coordinates by , , and z. As shown in the diagram below, the volume of this element equals ( ) ( ) ( z) = z. Thus,

cylindrical coordinate volume element = z (73)(Note that sometimes people use r instead of when using cylindrical coordinates, as in the diagram below.)

Cylindrical coordinates are useful in 3D problems with rotational symmetries about one axis, such as in this next problem. (In problems with radial symmetry, then spherical coordinates are typically the easiest to work with.)

ExampleIn the "Volume Integral" section above, we computed the volume of a cone with base radius r and height h using Cartesian coordinates. Repeat the problem using cylindrical coordinates and recover the answer V = 1

3 r2 h.

SolutionTaking the volume integral of the constant function 1 times the volume element z,

V = 0h02 01- zh r z= 0h02 12 1 - zh 2 r2 z= 0h 1 - zh 2 r2 z= 13 r2 h3(74)

Note that in cylindrical coordinates, this integral is significantly easier to both setup and evaluate. That is the


reason why we use different coordinate systems: cylindrical coordinates let you utilize the symmetry of the problem.

Of course, we could have chosen to go wild and choose the z-axis to lie off the axis of the cone; the integral would still yield the same answer, but the calculation would be significantly tougher. Many problems in physics are ultimately solved by approaching the problem from the right perspective.


Date post:	10-Mar-2016
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Math Boot Camp

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