Math 55 LE2 Notes

M E M B E R S H I P A C A D E M I C D E V E L O P M E N T

Math 55

2nd Long Exam Notes

J.Quintos

T.Delos Santos

Preface

This handout is intended as a reviewer only and should not be substituted for a complete lecture, or used as a reference material. The goal of this reviewer is to refresh the student on the concepts and techniques in one reading. But this is more than enough to replace your own notes :)

A Topological Note

Definition (Region). A region is an open, con-nected, and non-empty set of points in Rn.

In this reviewer notes, we’ll usually denote re-gions with D, E, or R. We’ll also be working

1

Triple Integrals

From integration over regions in R2_{, we progress}

to regions in R3.

Definition (Triple Integral). Given a region D ⊆ R3

and a function f : D → R, the triple integral of f over D is Z Z Z D f = lim δ→0 l X i=0 m X j=0 n X k=0 f (x∗i, yj∗, zk∗)∆Vijk

provided the limit exists. If the limit does exist, f is said to be integrable over D.

Remark. For a more detailed development of the definition, refer to the appendix.

Other notations used for a triple integral include: Z Z Z D f (x, y, z) dV Z Z Z D f (x, y, z) dx dy dz

To compute triple integrals, we invoke again Fu-bini’s Theorem, which equates a triple integral into an iterated integral.

1.1

Applications

I’ll discuss first the typical applications of the triple integral because they’ll be used again and again in most examples.

1.1.1 Volume

V(D) =Z Z Z

D

dV (1)

1.1.2 Mass and Density

You are very familiar with the equation ρ = M

V

Where ρ, m, andV are density, mass and volume respectively. Rearranging and with equation (1), we have:

M = ρ Z Z Z

D

dV (2)

This is only applicable whenever the density is uniform/constant throughout the object.

Suppose the density varies in R3 _{space. Let ρ be}

a function of position in space. Then:

M = Z Z Z

D

ρ(x, y, z) dV (3)

1.1.3 Center of Mass

The center of mass of an object is the average position of all the mass in the object.

x = 1 M Z Z Z D x dV (4) y = 1 M Z Z Z D y dV (5) z = 1 M Z Z Z D z dV (6)

Where M is given by equation (3).

Now we move on to actually computing triple in-tegrals.

1.2

Box-like regions

For a simple case of a box region D = [a, b] × [i, j] × [m, n] the triple integral becomes:

Z Z Z D f (x, y, z) dV = Z n m Z j i Z b a f (x, y, z) dx dy dz

The right-hand side expression can be thought of as evaluating the integrals “from the inside out”, as in: Z n m " Z j i Z b a f (x, y, z) dx ! dy # dz

treating “outer variables” constant when evaluat-ing the inner integral with respect to the “inner variable”

Example. Evaluate Z Z Z

E

xyz dV

where the region E = [1, 2] × [3, 4] × [5, 6]. Solution. Using Fubini’s Theorem we have,

= Z 6 5 Z 4 3 Z 2 1 xyz dx dy dz

out-ermost, Z 2 1 xyz dx = yz (2) 2_{− (1)}2 2  =3 2yz Z 4 3 3 2yz dy = 3 2z  (4)2_{− (3)}2 2  =21 4 z Z 6 5 21 4 dz = 21 4  (6)2_{− (5)}2 2  = 231 8

Try to change the order of integration. Verify that for box-like regions, the order of integration does not matter.

1.3

General regions

Consider the region in figure 1.3.1a. If we project the shadow of the blob on the xy-plane, we get a planar region D in R2 (figure 1.3.1b).

Observe that D can be described, previously learned in double integrals, as the set of points:

D = {(x, y) : a ≤ x ≤ b and f (x) ≤ y ≤ F (x)} It means that as we move x in the direction from point a to point b, our y includes the values from f (x) to F (x). All the points (x, y) that are “touched” belong to the region D.

With similar reasoning, we can think of region E as the set of points

E = {(x, y, z) : (x, y) ∈ D and g(x, y) ≤ z ≤ G(x, y)} Think of g(x, y) as a “floor” and G(x, y) as a “ceil-ing” of the region. For any point (x, y) that be-long in region D : z goes from it’s “floor” function g(x, y), including every value in between, up to its “ceiling” function G(x, y).

x

y

x

y

z

a

b

Figure 1.3.1: A blob-like region in R3

The triple iterated integral of a function over the blob region E is then:

Z b a Z F (x) f (x) Z G(x,y) g(x,y) (some function) dz dy dz

Remark. I emphasised by typing “some func-tion” that the function to be integrated is not necessarily the same function describing the re-gion. This has been a common mistake, as noted by my instructor.

The challenge is cleverly thinking of a convenient representation of the region that simplifies com-putation compared to other forms.

Note that the order of integration is not inter-changeable for a particular set of bounds describ-ing a general region. Reorderdescrib-ing the integral needs rewriting the bounds.

Example. Setup the iterated triple integral that will give the volume of a half-cylindrical region with r = 2 and l = 1, shown below. Write in the order (a) dx dy dz and (b) dz dy dx.

1 2 2 x y z Solution.

(a) The outermost variable is z, which takes val-ues 0 ≤ z ≤ 2.

Next is y. Since z and y are related by the circle z2 _{+ y}2 _{= 2}2_{, we can express}

y(z) = p4 − z2_{. As z goes from 0 to 2, y}

goes from −p4 − z2 _to p_{4 − z}2_{. So there’s}

our bounds.

The x variable doesn’t depend on the previ-ous variables, and we have 0 ≤ x ≤ 1. The iterated integral is then:

Z 2 0 Z √ 4−z2 −√4−z2 Z 1 0 1 dx dy dz

(b) The outermost variable is x which is just 0 to 1.

For y, observe that it goes from −2 to +2. Now z is a function of y alone (recall the circle), for each y value, z goes from 0 to p4 − y2_{. The integral is:}

Z 1 0 Z 2 −2 Z √ 4−y2 0 1 dz dy dx

1.4

Cylindrical coordinates

We now look at the analogue of the polar coordin-ates in R3_{. In figure (1.4.1), we see a coordinate}

O L

A

z

r

Figure 1.4.1: Cylindrical coordinate system.

system with origin at O, a longitudinal axis L and a polar axis A.

A point in 3D space can be described by its radial distance (r) from the origin; it’s angular rotation from the A-axis (θ); and its longitudinal distance (z) from the origin.

Taking z- and x-axis to be the longitudinal and polar axes respectively, the coordinate transform-ation is:

x = r cos θ y = r sin θ z = z x2_{+ y}2_{= r}s

And the “differential volume” becomes: dVrect= r dVcyl

Remark. There are cases where z-axis is not the most convenient choice for the longitudinal

axis. You should choose carefully which it should be and apply the corresponding coordinate trans-formation.

Example. Evaluate the triple integral in the pre-vious example using cylindrical coordinates. Solution. We choose the x-axis as our longitud-inal axis, and the y-axis as our polar axis. Our function is just the constant 1, so no trans-formation needed. However we must analyse what our bounds in terms of the new coordinates will be.

x : from 0 to 1 r : from 0 to 2 θ : from 0 to π The iterated integral is then:

Z 1 0 Z π 0 Z 2 0 r dr dθ dx Which evaluates to 2π

In this next example, we show how to apply the transformation Example. Transform Z 1 −1 Z √ 1−y2 0 Z √ x2_+y2 x2_+y2 xyz dz dx dy into cylindrical coordinates.

Solution. Observe the bounds of the region: −1 ≤y ≤ 1

0 ≤x ≤p1 − y2

x2+ y2≤z ≤px2_{+ y}2

The planar shadow of this region projected onto the xy-plane is a half disc of radius 1 lying on the 1st and 4th quadrant. So, −π

2 ≤ θ ≤ π 2 and

0 ≤ r ≤ 1.

As for the longitudinal component, note that x2_{+ y}2_{= r}2_{, so by algebra, r}2_{≤ z ≤ r. We now}

have the bounds.

Transforming the function:

xyz = (r cos θ)(r sin θ)z = r2_{z cos θ sin θ}

The integral is then Z π₂ −π 2 Z 1 0 Z r r2 r3_{z cos θ sin θ dz dr dθ}

1.5

Spherical coordinates

x

z

(ρ,θ,φ)

φ

θ

ρ

y

Figure 1.5.1: Spherical coordinates

A point in R3 _{can also be described by its}

• radial distance, ρ, from the origin; • polar angle, ϕ; and

• azimuthal angle, θ.

The transformation to spherical coordinates, with z axis as the zenith direction, is:

x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ z = ρ cos ϕ x2+ y2+ z2= ρ2

The corresponding “differential volume” is dVrect = ρ2sin ϕ dVsph

Remark. Some instructors use different conven-tions other than (ρ, θ, ϕ) . Be careful.

Example. Find the volume of the region bounded above by the cone ϕ = π

4 and below by

the sphere ρ = 4 cos ϕ. Refer to the figure below. (sample problem from [3])

x

y

z

4

(a) ϕ = π/4 and ρ = 4 cos ϕ

y

z

φ

O

P

(b) Projected view on the zy-plane

Figure 1.5.2: Cone and sphere

First we need to setup the bounds. A slice of the

region of interest is the shaded region in figure 1.5.2b.

Observe that our region of interest starts at ϕ = π/4 (∠zOP ) and ends with ϕ = π/2 (∠zOy). At each ϕ, ρ goes from 0 at the origin to the sphere ρ = 4 cos ϕ.

As for θ, we fully revolve this slice about z to get our desired region, so 0 ≤ θ ≤ 2π

Our iterated integral is then Z 2π 0 Z π₂ π 4 Z 4 cos ρ 0 ρ2_{sin ϕ dρ dϕ dθ} Which evaluates to 8π 3

1.6

Change of variables* (optional)

Back in Math 54, you were taught the u-substitution technique:

Z

f (g(x))g0(x) dx = Z

f (u) du

Where u = g(x). This can be extended to multiple integrals. You may have noticed that when we changed from rectangular to cylindrical or spher-ical, there were extra terms we considered. We begin the change in variables for multiple integ-rals by defining the Jacobian for a particular case.

Definition (Jacobian). Suppose we have the transformation x = g(u, v) and y = h(u, v). Then the Jacobian of the transformation is the determ-inant JT =      xu xv yu yv     

Where the subscripts denote partial differenti-ation. You can extend this to 3 or more variables, just take note of the row and column pattern.

Theorem 1 (Change of variables for Double In-tegrals). Suppose that we want to integrate f (x, y) over the regionD.

Under the transformationx = g(u, v), y = h(u, v) the region becomes E and the integral becomes, Z Z D f (x, y) dA = Z Z E f (g(u, v), h(u, v))|JT| du dv

We extend this theorem to the triple integral in the following example:

Example. Verify with the change of variable theorem the transformation from rectangular to spherical coordinates.

Solution. We compute the Jacobian of the trans-formation JT=       

(ρ sin ϕ cos θ)ρ (ρ sin ϕ cos θ)θ (ρ sin ϕ cos θ)ϕ

(ρ sin ϕ sin θ)ρ (ρ sin ϕ sin θ)θ (ρ sin ϕ sin θ)ϕ

(ρ cos ϕ)ρ (ρ cos ϕ)θ (ρ cos ϕ)ϕ

       =       

sin ϕ cos θ −ρ sin ϕ sin θ ρ cos ϕ cos θ sin ϕ sin θ ρ sin ϕ cos θ ρ cos ϕ sin θ

cos ϕ 0 −ρ sin ϕ       

with some handwavery math magic: JT = −ρ2sin ϕ

Take the absolute value, and there you go.

2

Vector Calculus

In this section we discuss vector fields, two new types of integral, and important theorems about those new types of integral.

2.1

Vector fields

Definition(Vector field). A vector field in two or three dimensional space is a function ~F that

out-puts a vector at each point in the domain ((x, y) or (x, y, z)) given by ~F (x, y) or ~F (x, y, z)

Notation for vector fields follow that of vector function.

• Angled bracket notation ~

F = hP, Qi ~

G = hP, Q, Ri • Unit vector notation

~

F = P ˆi+Qˆj ~

G = P ˆi+Qˆj+R ˆk

Where in all cases, P, Q, R are scalar functions of x, y or x, y, z.

Some examples of vector fields IRL are shown in figure 2.1.1.

(a) Magnetic field lines

(b) Surface wind over the Ph.

Visual representations do not include all of the vectors in a vector field though. Otherwise, you’d just end up covering the whole image and see nothing. In reality, every point in the plane (or space) has a vector.

2.2

Divergence and Curl

Definition(Del operator). We let the symbol ~∇ denote the Del operator, such that:

~ ∇ = ∂ ∂x, ∂ ∂y, ∂ ∂z 

We then introduce two new objects, the diver-gence and curl.

Definition (Divergence). The divergence of a vector field

~

F = hP (x, y, z, ), Q(x, y, z), R(x, y, z)i is a scalar expression given by

div( ~F ) = ~∇ · ~F Equivalently, div( ~F ) = ∂ ∂x, ∂ ∂y, ∂ ∂z  · hP, Q, Ri = ∂P ∂x + ∂Q ∂y + ∂R ∂z = Px+ Qy+ Rz

Definition(Curl). The curl of a vector field ~ F = hP (x, y, z, ), Q(x, y, z), R(x, y, z)i is a vector given by curl( ~F ) = ~∇ × ~F Equivalently curl( ~F ) = ∂ ∂x, ∂ ∂y, ∂ ∂z  × hP, Q, Ri = h(Ry− Qz), (Pz− Rx), (Qx− Py)i = (Ry− Qz)ˆi+(Pz− Rx)ˆj+(Qx− Py) ˆk

When a point in (x, y, z) is specified, the diver-gence and curl expressions can be evaluated to give scalar and vector values, respectively. A simplification for 2-D vector fields can be easily derived for

~

F = hP (x, y, z, ), Q(x, y, z), R(x, y, z)i by setting R = 0, and noting that P and Q will be functions of x and y only. Verify the following:

div( ~F ) = Px+ Qy

curl( ~F ) = h0, 0, Qx− Pyi

2.3

Conservative fields

Definition(Conservative field). A vector field ~F is said to be conservative if there exists a scalar function φ such that:

~

∇φ = ~F = hP, Q, Ri

The scalar function φ is called the potential func-tionof ~F .

An easier method utilises a theorem, the proof of which will not be shown here.

Theorem 2. ~F is conservative if and only if curl( ~F ) = 0

Remark 2.1. For 2D vector fields, this simply means that

Qx= Py

as shown in the previous subsection.

Upon showing that the vector field is conservative, we can apply the definition of gradient in the x term. Considering only 2D:

dφ dx = P Z dφ = Z P dx

Because of integration, this initial potential func-tion φ has a constant of integrafunc-tion which may be a function of y. To check, differentiate φ with respect to y and compare this to Q.

The usefulness of conservative vector fields and the potential function will be apparent later on.

2.4

Line Integrals

Definition(Line Integral). We define the line in-tegral of f over a curve C as:

Z C f (x, y, z) ds = lim ∆t→0 n X i=1 f (x∗i(t), y ∗ i(t), z ∗ i(t))∆is

where ∆is is the arc length of a segment of C,

and ∆t is the interval between subsequent points on C.

In a 2D sense, the line integral is the area of the “curtain” that connects the points of curve C plotted in the xy-plane to points in z defined as z = f (x, y).

2.4.1 Scalar Fields

The evaluation of the line integral is simple when the curve, the function, and the limits of integ-ration are given in parametric form, or can be

integral can be expressed as:

Z C f (x, y, z) ds = Z t=b t=a f (x(t), y(t), z(t))ds dtdt where ds dt = || ~R 0_{(t)|| (7)} ~ R(t) = hx(t), y(t), z(t)i (8)

Note that when the function is unity, the line in-tegral simplifies to the formula of the arc length of the curve from a to b.

2.4.2 Vector Fields

The result of an evaluated integral is usually a number. To force this on vector fields such as ~F = hP (x, y, z, ), Q(x, y, z), R(x, y, z)i, the line integral of vector fields is given by:

Z C ~ F · d ~R = Z C ~ F · ˆTds = Z C ~ F · d ~R dt ds dt ds = Z C ~ F · d ~R dt ds dt ds dtdt = Z t=b t=a ~ F ·d ~R dt dt

By using the dot product, we mean to get the component of the vector field ~F along the curve, signified by ˆT. We multiply this component to a distance dt, then integrate over a curve. Doesn’t this look familiar? This is the concept of work for any vector field. This is why the Line Integral is also called the Work Integral.

when the curve, the function, and the limits of integration are given in parametric form, or can be transformed to parametric form easily.

2.4.3 Fundamental Theorem for Line

Integ-rals

Theorem 3. LetC be a piecewise, smooth para-metric curve lying in an open region D having initial and terminal points (x0, y0) and (x1, y1),

respectively. If ~F (x, y) = hM (x, y), N (x, y)i is a conservative vector field on D, where M and N are continuous onD, then for any potential func-tionφ of ~F on D: Z C ~ F · d ~R = φ(x, y)     (x1,y1) (x0,y0)

This theorem provides us another means of cal-culating the line integral if we can show that the function ~F is conservative (that is curl( ~F ) = 0) and we can successfully find the potential func-tion φ.

2.4.4 Path independence

By a line integral independent of the path, we mean an integral Z C ~ F · d ~R satisfying Z C ~ F · d ~R = Z C0 ~ F · d ~R

for any other curve C0having the same initial and terminal points as C.

The importance of this concept is its ability to simplify a complex path without a parametric form into a simple path whose parametric form is easily obtainable (e.g. lines). To show that a line integral is not independent of the path, simply show that the line integral yields different answers in different paths.

To show that a line integral is independent of the path, we utilise the theorem, whose proof will not be shown here.

Theorem 4. IfM (x, y) and N (x, y) are continu-ous on open-connected region D, then the follow-ing are equivalent statements:

1. ~F (x, y) = hM (x, y), N (x, y)i is conservative on D.

2. Z

C

~

F ·d ~R is independent of the path to and from any point on D.

3. Z

C

~

F · d ~R = 0 for any piecewise, smooth, closed curveC in D.

2.5

Green’s Theorem

As stated above, if the curve is closed and the vector field conservative, then the line integral is simply zero. Green’s Theorem provides for the method to calculate the line integral over a closed curve with a vector field which is not conservative – and in doing so, relates the line integral to a double integral.

Theorem 5(Green’s Theorem). If ~

F (x, y) = hP (x, y), Q(x, y)i

is a smooth vector field on both the closed curve C and the region R it encloses, then:

I C ~ F · ~dR = Z Z R  ∂Q ∂x − ∂P ∂y  dA

Notice that when the function is conservative,  ∂Q ∂x − ∂P ∂y  = 0

and the line integral is zero, as stated in the pre-ceding subsection.

ging the order of integration (e.g. from Dx dy to Dx dy) or changing coordinate axes (e.g. from rectangular to polar). Review on these.

Green’s theorem has important application in sur-veying. What this video (watch me!) to be amazed. See how Green’s theorem applies here.

2.6

Surface integrals

Definition(Surface Integral). Z Z σ f (x, y, z) dS = lim ∆P →0 n X j=1 m X i=1 f (x∗ij(u, v), y ∗ ij(u, v), z ∗ ij(u, v))∆iS

where ∆iS is the infinitesimal area element, and

∆P is the interval between subsequent points. This doesn’t really give us anything substantial calculation-wise.

2.6.1 Scalar fields

If we express the infinitesimal area element using a double integral, we see that for a region, function, and limits parametrized in u and v:

dS = || ~Ru× ~Rv|| dA

and we can write:

Z Z σ f dS = Zv1 v0 Z u1 u0

f (x(u, v), y(u, v), z(u, v))|| ~Ru× ~Rv|| du dv

where:

σ : ~R(u, v) = hx(u, v), y(u, v, ), z(u, v)i

dS =q1 + [gx]2+ [gy]2dA

and we can write:

Z Z σ f dS = Z y1 y0 Z x1 x0 f (x, y, g(x, y)) q 1 + [gx]2+ [gy]2dA where: σ : z = g(x, y) 2.6.2 Vector fields

Similar to the extension of Line Integrals to Vector Fields, the Surface Integral will be calculated as

follows: _{Z Z}

σ

~ F · ˆNdS

where ˆNis an oriented unit normal to σ.

In taking the component of the vector field along the normal, we’re taking the flux of the vector field as it passes through the surface. That is why the Surface Integral is also known as the Flux Integral. It is important to note that the unit normal may be defined in two ways for every surface - one way is the opposite of the other.

In parametric form, the surface σ is given by ~

R(u, v) = hx(u, v), y(u, v, ), z(u, v)i, such that: ˆ

N= R~u× ~Rv || ~Ru× ~Rv||

dS = || ~Ru× ~Rv|| dA

where ˆN is defined as the positive orientation. The negative orientation is the negative of ˆN.

Plugging in the two equalities to the definition, using positive orientation:

Z Z σ ~ F · ˆNdS = Z Z D ~ F · ( ~Ru× ~Rv) dA

In rectangular coordinates, the surface σ is given by z = g(x, y), such that:

ˆ

N= h−gx, −gy, 1i p1 + [gx]2+ [gy]2

dS =q1 + [gx]2+ [gy]2dA

where ˆN is defined as the upward unit normal. The downward unit normal is the negative of ˆN. Plugging in the two equalities to the definition, using upward unit normal:

Z Z σ ~ F · ˆNdS = Z Z D ~ F · h−gx, −gy, 1i dA

The Surface Integral is just a complex exercise of plug-and-play. Be wary of the distinction between scalar and vector field, parametric and rectangular coordinates, positive and negative orientation, or upward and downward unit normal.

References

[1] Wikipedia, the Free Encyclopedia. [2] Paul Dawkins. Calculus III, 2012. [3] Joe Erickson. Calculus 3 Notes.