Integration Along a Curve:

Kicking it up a notch

Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000

Motivation: Why do we want to integrate a function along a curve?

A Real Mathematician’s Answer:

Because we can. That’s why!

No, but really...

Motivation: A Massachusetts Dilemma

In Boston, we freeze during the wintah.

School is often cancelled due to the hazids of snow and ice.

During a snowball fight, we notice the ice coating the telephone wiahs

My buddy Maak the physics major says, “I bet you 10 bucks you can’t figure out the total mass of the ice on that wiah!”

You’re on.

But first I need to develop the theory of line integrals.

The thickness of the ice varies as one moves along the wire

The confused person

(an annoying recurring character)

Setting up the Problem…...

r = radius of wire

r

R = radius of wire + ice coating

R

Recall the Ole Physics formula:

Mass = density * volume

Density of ice = 0.92 g/cm3

Area of a cross section of ice = R- r2) cm2

Linear density f of ice on wire = 0.92 * R- r2) g/cm

So...

The total mass of ice on wire = total accumulation of the linear density function along the wire

Parameterize the wire using a continuously differentiable (i.e. smooth) function

2],[: Rba ))(),(()( tytxt

x(t)

y(t)

Real-valued, continuously differentiable functions

Now we can at least approximate the total mass of the ice along the wire by these 4 easy steps:

1. Partition2. Sample

3. Scale

4. Sum

1. Partition

Partition the arc into n subarcs

How? By partitioning [a,b] into n subintervals, we induce a partition of into n subarcs

x

y

t

tx(t),y(t))

…

… …

… t1

t1

t1

ti

ti

ti

a = t0

a = t0

t0

1

tn = b

tn = b

tnn

ti-1

ti-1

ti-1

i

2. SampleRecall f is the density function defined on [a,b]), our “frozen wire”.

On each subarc i, choose a point ixi*,yi*) and sample f at those points.

x

y

t

a = t0

tx(t),y(t))

a = t0 tn = b

tn = b

ti-1

ti-1

ti

ti

t1

t1

…

… …

…

t0

t1

tn

ti-1

ti

n

i

0x0*,y0

*)

i-1xi-1*,yi-

1*)

nxn*,yn*)

3. Scale

x

y

t

tx(t),y(t))

a = t0 tn = b ti-1 ti t1 … …

t0

t1

tn

ti-1 ti

0

n-1

i-

1

Now we scale those sampled values f(ifxi*,yi*) by the length of the subarc denoted by si

si

4. SumNow we sum those scaled sampled values to get what looks like a Riemann sum.

i

n

iii syxf

)**,(1

But we want a way to actually calculate this mutha.

So we need to bring it down to the case we know: the one variable case.

It seems so pointless.

Should I just give him the money right now?

No Way!!!!!

We gotta show him up!

Math majors, represent!

Let’s do this.

(The Encouragement Slide)

How, do you ask? By relating everything to t.

fxi*,yi*) = fti*)) for some ti*in [ti-1, ti ]

)()( 1 iii tts

almostsi

yi

xi

Notice that for si small, a continuous curve looks locally linear

(t i-1)

(t i)

ttttttt )(')(')()(

Since is continuously differentiable,

So when t is small,

t

tttt

t

)()()(' lim

0

Since fis path integrable (continuity on [a,b]) is sufficient for this), we may sample f((t)) and partition however we choose:

Let ti = (b-a)/n = t

Let ti* = ti-1

So we have

tttfsyxf i

n

iii

n

iii

*)(')*)(()**,(11

which is a Riemann sum of the one variable real-valued function f((t))|| ’(t)) ||

So letting L(P, f) and U(P, f) be our respective lower and upper Riemann sums...

b

a

dtttffds )('))((

)')(,(inf)')(,(sup)('))(( fPUfPLdtttfPP

b

a

where

Thus we define the path integral of f along the curve

Note: If f (x,y)=1 on [a,b]

dttdsb

a )('

= length of the curve

So to answer Maak’s challenge…

b

a

b

a

dttdtttds )2,1())tf((t,)('))(f( R)f(r, 2

the total mass of the ice is ………....

r = Radius of wiah = 10cm

R = Radius of wiah + ice

b

a

dttt )41()(t * 0.92 224

(t) = (t,t2) parametrization of wire (parabola)

f(r,R) = linear density = 0.92*(R2-r2)

NOW SHOW ME THE MONEY!!!!

Always Get A Prenuptual-Dr. Jock Rader

Words of Wisdom