Unit conversion word problem: drug dosage

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:03.580
I just received this drug
calculation problem from a

00:00:03.580 --> 00:00:06.250
nursing student, and I think
it's essential that the nursing

00:00:06.250 --> 00:00:08.890
students out there are able to
do this, just in case I'm the

00:00:08.890 --> 00:00:10.360
patient receiving the drug.

00:00:10.360 --> 00:00:10.890
So let's do it.

00:00:10.890 --> 00:00:14.450
And I think it's an interesting
unit conversion problem for

00:00:14.450 --> 00:00:17.020
pretty much anyone who wants
practice with unit conversion.

00:00:17.020 --> 00:00:19.570
So the question is that
we have a doctor.

00:00:19.570 --> 00:00:23.750
The doctor orders drug x.

00:00:27.060 --> 00:00:29.790
And this is the dosage that
the doctor's requesting.

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They're saying 5 milligrams per
pound of patient weight-- I'll

00:00:39.540 --> 00:00:46.800
just write per pound of patient
weight-- every 12 hours.

00:00:46.800 --> 00:00:48.710
This is what we're
supposed to do.

00:00:48.710 --> 00:00:58.140
But our supply of the drug--
it isn't just, you know, not

00:00:58.140 --> 00:00:59.440
just nuggets and milligrams.

00:00:59.440 --> 00:01:00.350
It's a solution.

00:01:00.350 --> 00:01:04.210
There's a certain amount of
grams for every milliliter

00:01:04.210 --> 00:01:05.340
that we have of the solution.

00:01:05.340 --> 00:01:08.190
It's dissolved in some water.

00:01:08.190 --> 00:01:11.035
So this is our
supply of drug x.

00:01:14.750 --> 00:01:18.070
We have 0.9-- I'll
write a 0 in front.

00:01:18.070 --> 00:01:20.600
My wife, who is a doctor, says
it's essential to write the

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0 in front of the decimal.

00:01:21.990 --> 00:01:29.920
We have 0.9 grams per
milliliter of solution.

00:01:29.920 --> 00:01:32.830
So if I were to take 1
milliliter out of my solution

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and give it to someone, I'm
essentially giving them

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0.9 grams of this drug.

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And the final piece of
information we're given is that

00:01:44.960 --> 00:01:51.320
the patient-- they weigh-- and
maybe we should say they mass,

00:01:51.320 --> 00:01:53.730
because kilograms is mass,
but we get the idea.

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The patient is 72.7 kilograms.

00:02:01.590 --> 00:02:03.090
So there's a couple
interesting things here.

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We have to figure out the
dosage in terms of milliliters.

00:02:06.050 --> 00:02:07.760
We have to-- oh, actually,
I didn't even tell

00:02:07.760 --> 00:02:08.580
you the question.

00:02:08.580 --> 00:02:11.070
The question is, how many
milliliters of solution

00:02:11.070 --> 00:02:15.500
do we have to give to
the patient per dose?

00:02:15.500 --> 00:02:23.420
So milliliters of
solution per dose.

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That's our question.

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And there's a couple of things.

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We have to go from
milligrams to grams.

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And then convert that
to milliliters.

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And then they tell us per
pound, but then they gave us

00:02:33.530 --> 00:02:36.480
the patient's weight, or
their mass, in kilograms.

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So we have to do some
conversion there.

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So I definitely can appreciate
how this can be a little

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daunting and maybe
confusing at times.

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So let's just do
it step by step.

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So the first interesting
thing-- and this is just

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something that you might need
to know, or you might have

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written down on paper, or you
might have a calculator that

00:02:52.250 --> 00:02:55.610
does this-- is just how to
convert kilograms to pounds.

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And it's good to know in
general, if you're converting

00:02:57.760 --> 00:03:00.570
between the metric and
the English systems.

00:03:00.570 --> 00:03:05.520
So 1 kilogram is approximately
equal to 2.2 pounds.

00:03:05.520 --> 00:03:08.520
Not exactly, but that's a
pretty good approximation.

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And 1 pound-- if you just take
1 over that-- 1 pound is

00:03:12.870 --> 00:03:18.550
approximately equal
to 0.45 kilograms.

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So we'll just put
this in a box.

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This is the only kind of
outside conversion information

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we'll need to do this problem.

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Everything else, we'll just
need a calculator, unless we

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just want to spend a lot of
time doing some arithmetic.

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So the first thing.

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Let's figure out our dosage
in terms of per kilogram.

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This is per pound, and
we really don't need to

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know every 12 hours.

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Because they're saying, how
many milliliters of solution

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do we do per dose?

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A dose is every 12 hours.

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So we just really, you know--
the every 12 hours is kind

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of extra information.

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So we want to figure out this
5 milligrams per pound.

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How do we convert that to how
many milligrams per kilogram?

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So let's do 5-- I'll write
it down here in magenta--

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5 milligrams per pound.

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And then we want to convert
this to per kilogram.

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So we can multiply this times
the number of pounds per

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kilogram-- I'll do it
in yellow-- times this

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information up here.

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Times 2.2 pounds per kilogram.

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And if you ever get confused--
you know, gee, how did Sal know

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to multiply by 2.2 instead
of dividing by 2.2?

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Which is the same thing
as multiplying by 0.5.

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You can pay very close
attention to the units.

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Notice, I wrote 2.2
pounds per kilogram.

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2.2 pounds per 1 kilogram.

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And you know this'll work out,
because you have a pound in

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the numerator and you have
a pound in the denominator.

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It's called
dimensional analysis.

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If you ever get confused with
these things-- and I think,

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once you do enough practice,
you'll find that you won't

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have to pay too much
attention to this.

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But at first, when you're
getting started, just to make

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sure you're not multiplying
or dividing by the wrong

00:05:00.360 --> 00:05:02.260
thing, just make sure the
dimensions cancel out.

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Pounds in the numerator,
pounds in the denominator.

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So let's do that.

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Pounds in the numerator, pounds
in the denominator cancel out.

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And you multiply 5 times 2.2.

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This is equal to-- let's see.

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5 times 2 is 10.

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5 times 0.2 is 1.

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So this is equal to 11.

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And then in our numerator,
we have milligrams.

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11 milligrams per kilogram.

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So we just converted our
dosage information to

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a pure metric system.

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It was actually a mix
between the metric and the

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English system before.

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Now let's see what we can do.

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Well, let's see if we can get
it in terms of how many

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milliliters we have to
deliver per pound.

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So once again, we want this--
well, actually, let's

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go to grams first.

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Because we have
milligrams here.

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We have grams up here.

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So let's see if we can
convert this thing to grams.

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So just like we did before,
we want a milligrams

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in the denominator.

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I'll do it in orange.

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We want a milligrams in the
denominator and we want

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a gram in the numerator.

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Why did I say that?

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Because I want this
and this to cancel.

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And I want a grams
in the numerator.

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So how many grams are
there per milligram?

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You can just think it through.

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There's 1 gram per
1,000 milligrams.

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Or 1,000 milligrams per gram.

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And you just multiply it out.

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So the milligrams cancels with
the milligrams, and then

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we get-- this is equal to
11/1,000 grams per kilogram.

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So now we have everything in
terms of grams, but we want

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it in terms of milliliters.

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The question is, how
many milliliters of

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solution per dose?

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So let me go down here on
this line right here.

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So we had this result.

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We have 11/1,000-- I won't do
the division just yet-- grams

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of drug x per kilogram.

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This is really just a re--
we've just rewritten

00:07:08.250 --> 00:07:11.105
this dosage information
in different units.

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And let's see how much solution
we need per kilogram.

00:07:14.830 --> 00:07:18.900
So I want to cancel out
the grams here and have

00:07:18.900 --> 00:07:19.920
a milliliters there.

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So to cancel out that grams,
I'm going to have to have a

00:07:21.790 --> 00:07:24.980
gram in the denominator and a
milliliter in the numerator.

00:07:29.780 --> 00:07:34.450
So in our solution, how many
grams are there per milliliter?

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Well, they told us.

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There are 0.9 grams
per milliliter.

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Or for every 1 milliliter,
there are 0.9 grams.

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Notice, I just took
the inverse of that.

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Because we want a milliliter
in the numerator, grams in

00:07:48.510 --> 00:07:50.790
the denominator, so that
these two cancel out.

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And let's do this
multiplication now.

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So our grams cancel out.

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We have milliliters
per kilogram.

00:08:00.710 --> 00:08:01.910
And then we multiply it out.

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11/1,000 times 1 over 0.9.

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So I'll just keep-- let me
just write it like this.

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So there's going to be 11/1,000
times 0.9 milliliters of

00:08:17.400 --> 00:08:19.850
our solution per kilogram.

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So we've gotten this far.

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So this is per kilogram
of patient body weight.

00:08:24.250 --> 00:08:27.080
And then finally, they tell
us how many kilograms

00:08:27.080 --> 00:08:28.030
our patient weighs.

00:08:28.030 --> 00:08:30.200
So let's do that last
multiplication, and then we can

00:08:30.200 --> 00:08:33.230
actually get our calculator out
and do it all at once.

00:08:33.230 --> 00:08:36.140
So let's multiply this times--
we want to know how many

00:08:36.140 --> 00:08:38.320
milliliters per patient.

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We want the kilograms
to cancel out.

00:08:40.290 --> 00:08:44.330
So we want kilograms
per patient.

00:08:44.330 --> 00:08:46.150
Now we're talking about
this particular patient.

00:08:46.150 --> 00:08:48.250
Not every patient is
going to be the same

00:08:48.250 --> 00:08:49.400
number of kilograms.

00:08:49.400 --> 00:08:51.720
But if we do this,
kilograms will cancel out.

00:08:51.720 --> 00:08:54.630
We'll have milliliters per
patient-- milliliters of

00:08:54.630 --> 00:08:58.370
solution per patient-- which
is exactly what we want.

00:08:58.370 --> 00:09:00.850
We want milliliters of solution
per dose per patient.

00:09:00.850 --> 00:09:03.960
But everything we've assumed
so far has been per dose.

00:09:03.960 --> 00:09:06.650
So how many kilograms
does the patient weigh?

00:09:06.650 --> 00:09:12.090
Well, there's 72.7
kilograms per patient.

00:09:12.090 --> 00:09:13.670
That's how much the
patient weighs.

00:09:13.670 --> 00:09:15.710
So we just do this final
multiplication and

00:09:15.710 --> 00:09:16.300
we'll be done.

00:09:16.300 --> 00:09:18.970
So our answer-- and as these
two things are going to cancel

00:09:18.970 --> 00:09:24.720
out-- so our final answer is
going to be 11 times 72.7

00:09:24.720 --> 00:09:28.730
divided by 100 times--
actually, 100 times 0.9 is

00:09:28.730 --> 00:09:29.800
pretty easy to figure out.

00:09:29.800 --> 00:09:31.080
That's 900.

00:09:31.080 --> 00:09:35.160
Divided by 900
milliliters per patient.

00:09:35.160 --> 00:09:37.590
Or you can just say
milliliters per dose.

00:09:37.590 --> 00:09:38.510
However you want to say it.

00:09:38.510 --> 00:09:39.470
Per dose per patient.

00:09:39.470 --> 00:09:41.720
Let's get our calculator
out and do this.

00:09:41.720 --> 00:09:54.920
So we have 11 times 72.7 is
equal to 799 divided by 900.

00:09:54.920 --> 00:09:58.080
Is equal to 0.88-- well,
we could round up.

00:09:58.080 --> 00:09:59.370
0.889.

00:09:59.370 --> 00:10:00.690
Hopefully the
doctor won't mind.

00:10:00.690 --> 00:10:04.450
So that is equal to-- I'll
write it in a nice, vibrant

00:10:04.450 --> 00:10:14.840
color-- 0.889 milliliters
of solution per dose.

00:10:14.840 --> 00:10:17.140
So this is what we're going
to give every 12 hours.

00:10:17.140 --> 00:10:19.540
If they ask, how many total
milliliters over the

00:10:19.540 --> 00:10:21.020
course of 2 days?

00:10:21.020 --> 00:10:22.550
We would have to say,
oh, there's 48 hours.

00:10:22.550 --> 00:10:23.920
We'd multiply it by 4.

00:10:23.920 --> 00:10:26.260
But that 12 hours was extra
information in this problem.

00:10:26.260 --> 00:10:32.760
But anyway, hopefully this is
useful, and it'll ensure that

00:10:32.760 --> 00:10:35.720
any nurses serving me in the
future are giving me

00:10:35.720 --> 00:10:37.040
my proper dosage.

00:10:37.040 --> 00:10:40.490
And hopefully, the doctor even
got the right dosage to begin

00:10:40.490 --> 00:10:42.060
with, because otherwise
it's all for naught.

00:10:42.060 --> 00:10:43.360
Anyway.


Salman Khan on CNN

WEBVTT
Kind: captions
Language: en

00:00:01.124 --> 00:00:03.290
DAN SIMON (NARRATING): Sal
Khan lives in California,

00:00:03.290 --> 00:00:05.170
outside of San Francisco.

00:00:05.170 --> 00:00:08.840
Sara Shadid, halfway
around the globe in Dubai.

00:00:08.840 --> 00:00:12.420
Their lives are interconnected,
but Khan doesn't know it.

00:00:12.420 --> 00:00:14.470
This is a story
about how one man

00:00:14.470 --> 00:00:17.190
is helping to educate
the world, but has never

00:00:17.190 --> 00:00:20.110
seen any of his
students face to face.

00:00:20.110 --> 00:00:23.030
DAN SIMON: Where does
this passion come from?

00:00:23.030 --> 00:00:25.510
SALMAN KHAN: I think I've
always enjoyed teaching.

00:00:25.510 --> 00:00:27.301
DAN SIMON (NARRATING):
It's about a man who

00:00:27.301 --> 00:00:29.300
gave up a lucrative
career in Silicon Valley

00:00:29.300 --> 00:00:32.090
for what might look
like a boring desk job.

00:00:32.090 --> 00:00:34.184
SALMAN KHAN: Right now,
I'm cash flow negative.

00:00:34.184 --> 00:00:36.350
DAN SIMON (NARRATING): But
it was never about money.

00:00:36.350 --> 00:00:38.550
And with his drive
and education,

00:00:38.550 --> 00:00:40.530
Khan could have made millions.

00:00:40.530 --> 00:00:42.620
He was valedictorian
of his high school,

00:00:42.620 --> 00:00:45.690
with a perfect math
score on the SAT.

00:00:45.690 --> 00:00:49.770
Then came MIT, Bill Clinton
handing him his diploma.

00:00:49.770 --> 00:00:51.850
Next, Harvard Business School.

00:00:51.850 --> 00:00:54.600
He was lured into hedge
funds and did well.

00:00:54.600 --> 00:00:57.815
But Khan, who has a wife and
son to support, gave it up.

00:00:57.815 --> 00:01:00.190
SALMAN KHAN: A lot of people
thought I was kind of crazy.

00:01:00.190 --> 00:01:03.030
Obviously, when every
waking hour you have

00:01:03.030 --> 00:01:05.030
you would sneak into a
room and make math videos

00:01:05.030 --> 00:01:08.010
and put them on YouTube, people
kind of question what's up.

00:01:08.010 --> 00:01:09.760
DAN SIMON (NARRATING):
Here's what's up --

00:01:09.760 --> 00:01:11.910
Khan's YouTube videos.

00:01:11.910 --> 00:01:13.030
SALMAN KHAN (ON YOUTUBE): Let's
do a couple more examples,

00:01:13.030 --> 00:01:14.120
and I think you might get it.

00:01:14.120 --> 00:01:16.536
DAN SIMON (NARRATING): They've
been clicked on more than 9

00:01:16.536 --> 00:01:18.890
million times from
users around the world.

00:01:18.890 --> 00:01:22.842
The topics range from math
to chemistry to economics.

00:01:22.842 --> 00:01:24.300
SARA SHADID: From
every curriculum,

00:01:24.300 --> 00:01:26.330
they can use the videos.

00:01:26.330 --> 00:01:27.590
He's really helpful.

00:01:27.590 --> 00:01:29.256
DAN SIMON (NARRATING):
One of his users,

00:01:29.256 --> 00:01:32.250
19-year-old college sophomore
Sara Shadid in Dubai.

00:01:32.250 --> 00:01:34.420
She says the videos
made all the difference

00:01:34.420 --> 00:01:36.720
in helping her conquer calculus.

00:01:36.720 --> 00:01:38.750
SARA SHADID: Before
each and every exam,

00:01:38.750 --> 00:01:41.970
I take two days checking
all of his videos

00:01:41.970 --> 00:01:45.050
and understanding the
small details he explains.

00:01:45.050 --> 00:01:47.120
DAN SIMON: This
gigantic virtual school

00:01:47.120 --> 00:01:49.680
originates from the
smallest of places,

00:01:49.680 --> 00:01:54.130
from a tiny converted closet
inside Sal's master bedroom.

00:01:54.130 --> 00:01:57.450
This is where he's able to
reach an estimated 80,000

00:01:57.450 --> 00:01:59.442
knowledge-seeking users a month.

00:01:59.442 --> 00:02:01.650
DAN SIMON (NARRATING): It
all started a few years ago

00:02:01.650 --> 00:02:04.510
when a cousin wanted some
online math tutoring.

00:02:04.510 --> 00:02:06.290
Pretty soon, other
relatives started

00:02:06.290 --> 00:02:08.030
asking for similar help.

00:02:08.030 --> 00:02:11.160
Suddenly, people he didn't
know started watching.

00:02:11.160 --> 00:02:13.810
DAN SIMON: People must
have put in math terms,

00:02:13.810 --> 00:02:15.060
and suddenly they find you.

00:02:15.060 --> 00:02:16.450
SALMAN KHAN: Yeah, I think
that's just completely right.

00:02:16.450 --> 00:02:17.970
It's that small
subset of people who

00:02:17.970 --> 00:02:20.250
go to YouTube and do a
web search for greatest

00:02:20.250 --> 00:02:21.289
common divisor.

00:02:21.289 --> 00:02:23.580
DAN SIMON (NARRATING): People
did it, and it caught on.

00:02:23.580 --> 00:02:25.920
It's now called
the Khan Academy.

00:02:25.920 --> 00:02:29.640
The videos are short and simply
produced-- only his voice--

00:02:29.640 --> 00:02:30.530
SALMAN KHAN (ON VIDEO): So
that's your slope of the--

00:02:30.530 --> 00:02:32.738
DAN SIMON (NARRATING): With
a simple black background

00:02:32.738 --> 00:02:33.570
and graphics.

00:02:33.570 --> 00:02:35.153
SALMAN KHAN: There
are a lot of people

00:02:35.153 --> 00:02:39.040
who need help with
math, science, whatever.

00:02:39.040 --> 00:02:41.920
And I think they find
these YouTube videos

00:02:41.920 --> 00:02:44.172
as kind of the ideal nuggets
to fill in their gaps.

00:02:44.172 --> 00:02:45.880
DAN SIMON (NARRATING):
These mini classes

00:02:45.880 --> 00:02:49.000
filled in Sara Shadid's gaps,
who says she and her friends

00:02:49.000 --> 00:02:52.590
refer to Khan as the man
with the colored fonts.

00:02:52.590 --> 00:02:54.860
Whatever you call
him or his teachings,

00:02:54.860 --> 00:02:57.940
the success of his site
has validated that decision

00:02:57.940 --> 00:03:00.290
to walk away from hedge funds.

00:03:00.290 --> 00:03:03.820
Dan Simon, CNN, San Francisco.


Green's theorem example 2

WEBVTT
Kind: captions
Language: en

00:00:00.570 --> 00:00:03.260
Let's say I have a path in the
xy plane that's essentially

00:00:03.260 --> 00:00:03.990
the unit circle.

00:00:06.860 --> 00:00:15.080
So this is my y-axis, this is
my x-axis, and our path is

00:00:15.080 --> 00:00:16.780
going to be the unit circle.

00:00:21.550 --> 00:00:24.070
And we're going to traverse
it just like that.

00:00:24.070 --> 00:00:25.940
We're going to traverse
it clockwise.

00:00:33.470 --> 00:00:35.460
I think you get the idea.

00:00:35.460 --> 00:00:37.530
And so its equation
is the units circle.

00:00:37.530 --> 00:00:41.600
So the equation of this is x
squared plus y squared is

00:00:41.600 --> 00:00:44.440
equal to 1; has a radius
of 1 unit circle.

00:00:44.440 --> 00:00:49.690
And what we're concerned
with is the line integral

00:00:49.690 --> 00:00:52.700
over this curve c.

00:00:52.700 --> 00:00:54.310
It's a closed curve c.

00:00:57.580 --> 00:01:11.400
It's actually going in that
direction of 2y dx minus 3x dy.

00:01:11.400 --> 00:01:14.760
So, we are probably
tempted to use Green's

00:01:14.760 --> 00:01:16.520
theorem and why not?

00:01:16.520 --> 00:01:17.200
So let's try.

00:01:17.200 --> 00:01:18.970
So this is our path.

00:01:18.970 --> 00:01:25.510
So Green's theorem tells us
that the integral of some curve

00:01:25.510 --> 00:01:33.640
f dot dr over some path where f
is equal to-- let me write

00:01:33.640 --> 00:01:35.020
it a little nit neater.

00:01:35.020 --> 00:01:46.650
Where f of x,y is equal to P
of x, y i plus Q of x, y j.

00:01:46.650 --> 00:01:53.960
That this integral is equal to
the double integral over the

00:01:53.960 --> 00:01:56.510
region-- this would be the
region under question

00:01:56.510 --> 00:01:57.780
in this example.

00:01:57.780 --> 00:02:10.140
Over the region of the partial
of Q with respect to x

00:02:10.140 --> 00:02:18.170
minus the partial of
P with respect to y.

00:02:18.170 --> 00:02:22.250
All of that dA, the
differential of area.

00:02:22.250 --> 00:02:25.110
And of course, the region
is what I just showed you.

00:02:25.110 --> 00:02:30.690
Now, you may or may not
remember-- well, there's a

00:02:30.690 --> 00:02:32.500
slight, subtle thing in
this, which would give

00:02:32.500 --> 00:02:33.590
you the wrong answer.

00:02:33.590 --> 00:02:36.370
In the last video we said that
Green's theorem applies when

00:02:36.370 --> 00:02:39.220
we're going counterclockwise.

00:02:39.220 --> 00:02:40.905
Notice, even on this little
thing on the integral I made

00:02:40.905 --> 00:02:42.410
it go counterclockwise.

00:02:42.410 --> 00:02:45.740
In our example, the
curve goes clockwise.

00:02:45.740 --> 00:02:47.430
The region is to our right.

00:02:47.430 --> 00:02:50.295
Green's theorem-- this applies
when the region is to our left.

00:02:55.620 --> 00:02:58.000
So in this situation when the
region is to our right and

00:02:58.000 --> 00:02:59.340
we're going-- so this
is counterclockwise.

00:03:04.020 --> 00:03:07.140
So in our example, where we're
going clockwise, the region is

00:03:07.140 --> 00:03:09.960
to our right, Green's theorem
is going to be the

00:03:09.960 --> 00:03:10.730
negative of this.

00:03:10.730 --> 00:03:14.970
So in our example, we're going
to have the integral of c and

00:03:14.970 --> 00:03:16.670
we're going to go in the
clockwise direction.

00:03:16.670 --> 00:03:21.910
So maybe I'll draw it like that
of f dot dr. This is going

00:03:21.910 --> 00:03:25.000
to be equal to the double
integral over the region.

00:03:25.000 --> 00:03:28.960
You could just swap these two--
the partial of P with respect

00:03:28.960 --> 00:03:35.880
to y minus the partial of
Q with respect to x da.

00:03:35.880 --> 00:03:36.650
So let's do that.

00:03:36.650 --> 00:03:39.480
So this is going to be equal
to, in this example, the

00:03:39.480 --> 00:03:41.420
integral over the region--
let's just keep it

00:03:41.420 --> 00:03:43.000
abstract for now.

00:03:43.000 --> 00:03:45.420
We could start setting the
boundaries, but let's just

00:03:45.420 --> 00:03:46.930
keep the region abstract.

00:03:46.930 --> 00:03:51.790
And what is the partial of P
with respect-- let's remember,

00:03:51.790 --> 00:03:55.650
this right here is our-- I
think we could recognize right

00:03:55.650 --> 00:03:59.010
now that if we take f dot dr
we're going to get this.

00:03:59.010 --> 00:04:00.940
The dr contributes
those components.

00:04:00.940 --> 00:04:04.330
The f contributes
these two components.

00:04:04.330 --> 00:04:05.570
So this is P of x,y.

00:04:08.370 --> 00:04:13.370
And then this is Q of x,y.

00:04:13.370 --> 00:04:13.865
And we've seen it.

00:04:13.865 --> 00:04:16.010
I don't want to go into the
whole dot dr and take the dot

00:04:16.010 --> 00:04:16.950
product over and over again.

00:04:16.950 --> 00:04:19.120
I think you can see that
this is the dot product

00:04:19.120 --> 00:04:20.570
of two vectors.

00:04:20.570 --> 00:04:23.310
This is the x component
of f, y component of f.

00:04:23.310 --> 00:04:28.300
This is the x component of dr,
y component of dr. So let's

00:04:28.300 --> 00:04:31.290
take the partial of P
with respect to y.

00:04:31.290 --> 00:04:33.740
You take the derivative of this
with respect to y, you get 2.

00:04:33.740 --> 00:04:35.740
Derivative of 2y is just 2.

00:04:35.740 --> 00:04:40.590
So you get 2, and then,
minus the derivative of

00:04:40.590 --> 00:04:42.270
Q with respect to x.

00:04:42.270 --> 00:04:44.100
Derivative of this with
respect to x is minus 3.

00:04:44.100 --> 00:04:49.800
So we're going to get minus
3, and then all of that da.

00:04:49.800 --> 00:04:53.430
And this is equal to the
integral over the region.

00:04:53.430 --> 00:04:55.540
What's this, it's
2 minus minus 3?

00:04:55.540 --> 00:04:57.510
That's the same
thing as 2 plus 3.

00:04:57.510 --> 00:05:01.190
So it's the integral over
the region of 5 dA.

00:05:01.190 --> 00:05:04.280
5 is just a constant, so we can
take it out of the integral.

00:05:04.280 --> 00:05:07.230
So this is going to turn out
to be quite a simple problem.

00:05:07.230 --> 00:05:11.790
So this is going to be equal to
5 times the double integral

00:05:11.790 --> 00:05:14.870
over the region R dA.

00:05:14.870 --> 00:05:16.080
Now what is this thing?

00:05:16.080 --> 00:05:18.610
What is this thing right here?

00:05:18.610 --> 00:05:20.880
It looks very abstract,
but we can solve this.

00:05:20.880 --> 00:05:25.770
This is just the
area of the region.

00:05:25.770 --> 00:05:27.550
That's what that double
integral represents.

00:05:27.550 --> 00:05:29.470
You just sum up all
the little dA's.

00:05:29.470 --> 00:05:31.490
That's a dA, that's a dA.

00:05:31.490 --> 00:05:33.270
You sum up the infinite
sums of those little

00:05:33.270 --> 00:05:35.150
dA's over the region.

00:05:35.150 --> 00:05:37.580
Well, what's the area
of this unit circle?

00:05:37.580 --> 00:05:40.730
Here we just break out a little
bit of ninth grade-- actually,

00:05:40.730 --> 00:05:43.630
even earlier than that--
pre-algebra or middle

00:05:43.630 --> 00:05:44.860
school geometry.

00:05:44.860 --> 00:05:47.960
Area is equal to pi r squared.

00:05:47.960 --> 00:05:49.360
What's our radius?

00:05:49.360 --> 00:05:52.800
So unit circle,
our radius is 1.

00:05:52.800 --> 00:05:53.980
Length is 1.

00:05:53.980 --> 00:05:56.200
So the area here is pi.

00:05:56.200 --> 00:05:59.290
So this thing over here,
that whole thing is

00:05:59.290 --> 00:06:01.350
just equal to pi.

00:06:01.350 --> 00:06:06.460
So the answer to our line
integral is just 5 pi, which

00:06:06.460 --> 00:06:07.570
is pretty straightforward.

00:06:07.570 --> 00:06:10.420
I mean, we could have taken the
trouble of setting up a double

00:06:10.420 --> 00:06:12.870
integral where we take the
antiderivative with respect to

00:06:12.870 --> 00:06:16.990
y first and write y is equal to
the negative square root of 1

00:06:16.990 --> 00:06:19.990
minus x squared y is equal to
the positive square root.

00:06:19.990 --> 00:06:21.810
x goes from minus 1 to 1.

00:06:21.810 --> 00:06:26.060
But that would have been
super hairy and a huge pain.

00:06:26.060 --> 00:06:28.250
And we just have to realize,
no, this is just the area.

00:06:28.250 --> 00:06:31.000
And the other interesting thing
is I challenge you to solve

00:06:31.000 --> 00:06:34.170
the same integral without
using Green's theorem.

00:06:34.170 --> 00:06:38.310
You know, after generating a
parameterization for this

00:06:38.310 --> 00:06:41.230
curve, going in that direction,
taking the derivatives

00:06:41.230 --> 00:06:42.660
of x of t and y of t.

00:06:42.660 --> 00:06:44.880
Multiplying by the appropriate
thing and then taking the

00:06:44.880 --> 00:06:48.540
antiderivative-- way hairier
than what we just did using

00:06:48.540 --> 00:06:51.810
Green's theorem to get 5 pi.

00:06:51.810 --> 00:06:54.645
And remember, the reason why
it wasn't minus 5 pi here

00:06:54.645 --> 00:06:57.980
is because we're going in
a clockwise direction.

00:06:57.980 --> 00:07:00.040
If we were going in a
counterclockwise direction we

00:07:00.040 --> 00:07:02.330
could have applied the straight
up Green's theorem, and we

00:07:02.330 --> 00:07:04.200
would have gotten minus 5 pi.

00:07:04.200 --> 00:07:06.530
Anyway, hopefully you
found that useful.


Green's theorem example 1

WEBVTT
Kind: captions
Language: en

00:00:00.710 --> 00:00:03.070
Let's see if we can use our
knowledge of Green's theorem

00:00:03.070 --> 00:00:05.200
to solve some actual
line integrals.

00:00:05.200 --> 00:00:06.950
And actually, before I show an
example, I want to make one

00:00:06.950 --> 00:00:08.960
clarification on
Green's theorem.

00:00:08.960 --> 00:00:15.310
All of the examples that I did
is I had a region like this,

00:00:15.310 --> 00:00:18.970
and the inside of the region
was to the left of

00:00:18.970 --> 00:00:19.820
what we traversed.

00:00:19.820 --> 00:00:24.940
So all my examples I went
counterclockwise and so our

00:00:24.940 --> 00:00:28.800
region was to the left of-- if
you imagined walking along the

00:00:28.800 --> 00:00:31.120
path in that direction, it
was always to our left.

00:00:31.120 --> 00:00:33.910
And that's the situation which
Green's theorem would apply.

00:00:33.910 --> 00:00:36.880
So if you were to take a line
integral along this path, a

00:00:36.880 --> 00:00:39.010
closed line integral, maybe we
could even specify

00:00:39.010 --> 00:00:39.520
it like that.

00:00:39.520 --> 00:00:42.900
You'll see that in some
textbooks-- along the curve c

00:00:42.900 --> 00:00:50.620
of f dot dr. This is what
equals the double integral over

00:00:50.620 --> 00:00:58.680
this region r of the partial of
q with respect to x minus

00:00:58.680 --> 00:01:03.670
the partial of p with
respect to y d area.

00:01:03.670 --> 00:01:06.520
And just as reminder, now this
q and p are coming from the

00:01:06.520 --> 00:01:09.430
components of f in
the situation.

00:01:09.430 --> 00:01:16.600
f would be written as f of xy
is equal to P of xy time the i

00:01:16.600 --> 00:01:22.290
component plus Q of xy
times the j component.

00:01:22.290 --> 00:01:26.010
So this is a situation where
the inside of the region is

00:01:26.010 --> 00:01:29.180
to the left of the direction
that we're taking the path.

00:01:29.180 --> 00:01:31.200
If it was in the reverse,
then we would put a

00:01:31.200 --> 00:01:32.550
minus sign right here.

00:01:32.550 --> 00:01:35.110
If this arrow went the other
way, we'd put a minus sign and

00:01:35.110 --> 00:01:37.730
we can do that because we know
that when we're taking line

00:01:37.730 --> 00:01:39.540
integrals through vector
fields, if we were reverse

00:01:39.540 --> 00:01:41.500
the direction it becomes
the negative of that.

00:01:41.500 --> 00:01:44.370
We showed that, I think,
4 or 5 videos ago.

00:01:44.370 --> 00:01:46.320
With that said, it was
convenient to write

00:01:46.320 --> 00:01:47.590
Green's theorem up here.

00:01:47.590 --> 00:01:50.060
Let's actually solve a problem.

00:01:50.060 --> 00:01:54.820
Let's say I have the line
integral and let's say

00:01:54.820 --> 00:01:57.140
we're over a curve.

00:01:57.140 --> 00:01:58.845
I'll define the
curve in a second.

00:01:58.845 --> 00:02:01.540
But let's say that the integral
we're trying to solve is x

00:02:01.540 --> 00:02:13.230
squared minus y squared
dx plus 2xy dy.

00:02:13.230 --> 00:02:16.870
And then our curve-- they're
giving us the boundary.

00:02:16.870 --> 00:02:18.430
The boundary is the region.

00:02:18.430 --> 00:02:22.420
I'll do it in a
different color.

00:02:22.420 --> 00:02:37.820
So the curve is boundary of the
region given by all of the

00:02:37.820 --> 00:02:43.710
points x,y such that x is a
greater than or equal to 0,

00:02:43.710 --> 00:02:46.590
less than or equal to 1.

00:02:46.590 --> 00:02:53.210
And then y is greater than
or equal to 2x squared and

00:02:53.210 --> 00:02:55.700
less than or equal to 2x.

00:02:55.700 --> 00:02:59.520
So let's draw this region that
we're dealing with right now.

00:02:59.520 --> 00:03:06.450
So let me draw my x-axis
or my y-axis, sorry.

00:03:06.450 --> 00:03:12.210
My y-axis and then my
x-axis right there.

00:03:12.210 --> 00:03:16.360
And let's see. x goes from
0 to 1, so if we make--

00:03:16.360 --> 00:03:17.910
that's obviously 0.

00:03:17.910 --> 00:03:21.950
Let's say that that is x
is equal to 1, so that's

00:03:21.950 --> 00:03:23.600
all the x values.

00:03:23.600 --> 00:03:27.640
And y varies, it's above
2x squared and below 2x.

00:03:27.640 --> 00:03:30.120
Normally, if you get to large
enough numbers, 2x squared is

00:03:30.120 --> 00:03:32.530
larger, but if you're below 1
this is actually going to

00:03:32.530 --> 00:03:33.660
be smaller than that.

00:03:33.660 --> 00:03:38.880
So the upper boundary is
2x, so there's 1 comma 2.

00:03:38.880 --> 00:03:43.240
This is a line y is equal to
2x, so that is the line y is--

00:03:43.240 --> 00:03:45.770
let me draw a straighter
line than that.

00:03:45.770 --> 00:03:49.270
The line y is equal to 2x
looks something like that.

00:03:49.270 --> 00:03:53.060
That right there is
y is equal to 2x.

00:03:53.060 --> 00:03:54.710
Maybe I'll do that in yellow.

00:03:54.710 --> 00:03:58.520
And then the bottom curve right
here is y is going to be

00:03:58.520 --> 00:04:00.110
greater than 2x squared.

00:04:00.110 --> 00:04:02.250
It might look
something like this.

00:04:02.250 --> 00:04:04.830
And of course, the region that
they're talking about is this,

00:04:04.830 --> 00:04:07.620
but we're saying that the curve
is the boundary of this region

00:04:07.620 --> 00:04:10.200
and we're going to go in I
counterclockwise direction.

00:04:10.200 --> 00:04:11.300
I have to specify that.

00:04:17.410 --> 00:04:20.400
So our curve, we could start at
any point really, but we're

00:04:20.400 --> 00:04:23.900
going to go like that.

00:04:23.900 --> 00:04:26.740
Then get to that point and then
come back down along that

00:04:26.740 --> 00:04:29.950
top curve just like that.

00:04:29.950 --> 00:04:33.090
And so this met the condition
that the inside of the region

00:04:33.090 --> 00:04:35.340
is always going to be our left,
so we can just do the straight

00:04:35.340 --> 00:04:38.660
up Green's theorem, we don't
have to do the negative of it.

00:04:38.660 --> 00:04:42.140
And let's define our region.

00:04:42.140 --> 00:04:43.530
Let's just do our region.

00:04:43.530 --> 00:04:46.820
This integral right here is
going to go-- I'll just do it

00:04:46.820 --> 00:05:01.940
the way-- y varies from y is
equal to 2x squared to

00:05:01.940 --> 00:05:05.290
2y is equal to 2x.

00:05:05.290 --> 00:05:09.380
And so maybe we'll integrate
with respect to y first.

00:05:09.380 --> 00:05:11.245
And then x, I'll
do the outside.

00:05:13.850 --> 00:05:15.780
The boundary of x
goes from 0 to 1.

00:05:15.780 --> 00:05:19.940
So they set us up well to
do an indefinite integral.

00:05:19.940 --> 00:05:21.930
Now we just have to figure
out what goes over

00:05:21.930 --> 00:05:23.490
here-- Green's theorem.

00:05:26.800 --> 00:05:29.660
Our f would look like
this in this situation.

00:05:29.660 --> 00:05:37.300
f is f of xy is going to be
equal to x squared minus

00:05:37.300 --> 00:05:42.940
y squared i plus 2xy j.

00:05:42.940 --> 00:05:44.190
We've seen this in
multiple videos.

00:05:44.190 --> 00:05:47.090
You take the dot product of
this with dr, you're going to

00:05:47.090 --> 00:05:49.250
get this thing right here.

00:05:49.250 --> 00:05:54.690
So this expression right
here is our P of x y.

00:05:54.690 --> 00:05:58.070
And this expression right
here is our Q of xy.

00:06:00.640 --> 00:06:03.060
So inside of here we're just
going to apply Green's

00:06:03.060 --> 00:06:05.320
theorem straight up.

00:06:05.320 --> 00:06:12.080
So the partial of Q with
respect to x-- so take

00:06:12.080 --> 00:06:14.240
the derivative of this
with respect to x.

00:06:14.240 --> 00:06:15.390
We're just going to
end up with the 2y.

00:06:18.050 --> 00:06:20.270
And then from that we're going
to subtract the partial

00:06:20.270 --> 00:06:22.750
of P with respect to y.

00:06:22.750 --> 00:06:25.300
So if you take the derivative
of this with respect to y that

00:06:25.300 --> 00:06:30.260
becomes 0 and then, here you
have-- the derivative with

00:06:30.260 --> 00:06:32.900
respect to y here is minus 2y.

00:06:38.790 --> 00:06:39.900
Just like that.

00:06:39.900 --> 00:06:45.360
And so this simplifies
to 2y minus minus 2y.

00:06:45.360 --> 00:06:48.470
That's 2y plus plus 2y.

00:06:48.470 --> 00:06:49.950
I'm just subtracting
a negative.

00:06:49.950 --> 00:06:52.190
Or this inside, and just
to save space, this

00:06:52.190 --> 00:06:54.890
inside-- that's just 4y.

00:06:54.890 --> 00:06:56.320
I don't want to have to
rewrite the boundaries.

00:06:59.050 --> 00:07:02.940
That right there is
the same thing as 4y.

00:07:02.940 --> 00:07:08.310
The partial of Q with respect
to y, 2y minus the partial

00:07:08.310 --> 00:07:09.860
of P with respect to y.

00:07:09.860 --> 00:07:10.680
Which is minus 2y.

00:07:10.680 --> 00:07:12.510
You subtract a negative
you get a positive.

00:07:12.510 --> 00:07:13.500
You have 4y.

00:07:13.500 --> 00:07:16.780
Let's take the antiderivative
of the inside with respect

00:07:16.780 --> 00:07:21.350
to y and we're going
to get 2y squared.

00:07:21.350 --> 00:07:23.410
Let me do it a
little bit lower.

00:07:23.410 --> 00:07:29.620
We're going to get 2y squared
if you take the partial with

00:07:29.620 --> 00:07:32.100
respect to y you're
going to get 4y.

00:07:32.100 --> 00:07:42.000
We're going to evaluate that
from y is equal to 2x squared

00:07:42.000 --> 00:07:44.550
to y is equal to 2x.

00:07:44.550 --> 00:07:46.150
And then of course, we
still have the outside

00:07:46.150 --> 00:07:47.632
integral still there.

00:07:47.632 --> 00:07:51.520
x goes from 0 to 1 dx.

00:07:51.520 --> 00:07:56.000
This thing is going to be equal
to the integral from 0 to 1 and

00:07:56.000 --> 00:07:58.750
then we evaluate
it first at 2x.

00:07:58.750 --> 00:08:04.100
So you put a 2x in here,
2x squared is 4x squared.

00:08:04.100 --> 00:08:07.960
2 squared, x squared, so
4x squared times 2 is

00:08:07.960 --> 00:08:11.500
going to be 8x squared.

00:08:11.500 --> 00:08:14.790
Minus-- put this guy in there.

00:08:14.790 --> 00:08:19.600
2x squared squared is
2x to the fourth.

00:08:19.600 --> 00:08:26.130
4x to the fourth times
2 is 8x to the fourth.

00:08:26.130 --> 00:08:27.210
Did I do that right?

00:08:27.210 --> 00:08:30.350
2x squared-- going to
put it there for y,

00:08:30.350 --> 00:08:31.690
substitute y with it.

00:08:31.690 --> 00:08:34.760
That squared is 4
X to the fourth.

00:08:34.760 --> 00:08:36.430
Times 2 is 8x x to the fourth.

00:08:36.430 --> 00:08:37.040
Very good.

00:08:37.040 --> 00:08:37.770
All right.

00:08:37.770 --> 00:08:38.550
Now dx.

00:08:38.550 --> 00:08:40.850
Now this is just a
straightforward

00:08:40.850 --> 00:08:42.600
one-dimensional integral.

00:08:42.600 --> 00:08:45.400
This is going to be equal
to-- I'll just do it here.

00:08:45.400 --> 00:08:47.720
This is going to be equal to
the antiderivative of 8x

00:08:47.720 --> 00:08:50.485
squared is 8/3 x to the third.

00:08:53.920 --> 00:08:56.740
And then the antiderivative
of 8x to the fourth is

00:08:56.740 --> 00:09:03.950
minus 8/5 x to the fifth.

00:09:03.950 --> 00:09:07.503
Then we're going to have to
evaluate that from 0 to 1, or

00:09:07.503 --> 00:09:09.590
we can put a little line there.

00:09:09.590 --> 00:09:12.195
When you put 1 in there
you get-- I'll do it

00:09:12.195 --> 00:09:13.450
in a different color.

00:09:13.450 --> 00:09:16.890
We get 8/5 times 1 to
the third, which is 8/5

00:09:16.890 --> 00:09:22.550
minus 8/5 minus 8/5.

00:09:22.550 --> 00:09:25.335
And then we're going to have
minus-- when you put 0 in

00:09:25.335 --> 00:09:27.490
here you're just going
to get a bunch of 0's.

00:09:27.490 --> 00:09:29.170
Oh sorry, I made a mistake.

00:09:29.170 --> 00:09:31.210
It would have been a blunder.

00:09:31.210 --> 00:09:31.760
It's 8/3.

00:09:34.310 --> 00:09:38.640
8/3 times 1 to the third minus
8/5 times 1 to the fifth,

00:09:38.640 --> 00:09:40.780
so that's minus 8/5.

00:09:40.780 --> 00:09:44.705
And then, when you subtract the
0, you evaluate 0 here. you're

00:09:44.705 --> 00:09:45.850
just going to get a bunch of
0's so you don't have

00:09:45.850 --> 00:09:46.670
to do anything else.

00:09:46.670 --> 00:09:49.470
So now we just have to
subtract these two fractions.

00:09:49.470 --> 00:09:52.990
So let's get a common
denominator of 15.

00:09:52.990 --> 00:09:55.540
8/3 is the same thing if
we multiply the numerator

00:09:55.540 --> 00:09:57.690
and denominator by 5.

00:09:57.690 --> 00:10:00.820
That is 40/15.

00:10:00.820 --> 00:10:03.020
And then if we multiply this
numerator and denominator by

00:10:03.020 --> 00:10:06.320
3, that's going to be 24/15.

00:10:06.320 --> 00:10:15.710
So minus 24/15 and we get
it being equal to 16/15.

00:10:15.710 --> 00:10:22.390
And so using Green's theorem we
were able to find the answer

00:10:22.390 --> 00:10:24.100
to this integral up here.

00:10:24.100 --> 00:10:26.780
It's equal to 16/15.

00:10:26.780 --> 00:10:27.960
Hopefully you found
that useful.

00:10:27.960 --> 00:10:30.790
I'll do one more example
in the next video.


Green's theorem proof (part 2)

WEBVTT
Kind: captions
Language: en

00:00:00.550 --> 00:00:04.770
Let's say we have the same path
that we had in the last video.

00:00:04.770 --> 00:00:09.710
Draw my y-axis,
that is my x-axis.

00:00:09.710 --> 00:00:12.630
Let's say the path looks
like this, it looks

00:00:12.630 --> 00:00:13.570
something like this.

00:00:13.570 --> 00:00:16.080
It's the same one we
had in the last video.

00:00:16.080 --> 00:00:18.270
Might not look exactly like
it, let me see what I

00:00:18.270 --> 00:00:19.430
did in the last video.

00:00:19.430 --> 00:00:22.330
It looked like that in the
last video, but close enough.

00:00:22.330 --> 00:00:24.270
Let's say we're dealing
with the exact same

00:00:24.270 --> 00:00:25.450
curve as the last video.

00:00:25.450 --> 00:00:27.840
We could call that curve c.

00:00:27.840 --> 00:00:32.850
Now last video, we dealt with
a vector field that only had

00:00:32.850 --> 00:00:34.240
vectors in the i direction.

00:00:34.240 --> 00:00:37.590
Let's build with another vector
field that only has vectors in

00:00:37.590 --> 00:00:39.680
j direction, or the
vertical direction.

00:00:39.680 --> 00:00:46.810
So let's say that q, the vector
field q of xy, say it's equal

00:00:46.810 --> 00:00:54.430
to capital Q of xy times j, and
we are going to concern

00:00:54.430 --> 00:00:58.120
ourselves we're going to
concern ourselves with though

00:00:58.120 --> 00:01:09.310
closed line integral around the
path c of q dot dr. And

00:01:09.310 --> 00:01:11.760
we've seen it already.

00:01:11.760 --> 00:01:20.240
dr can be rewritten as dx
times i, plus dy times j.

00:01:20.240 --> 00:01:23.940
So if we were to take the dot
product of these two, this

00:01:23.940 --> 00:01:26.300
line integral is going to
be the exact same thing.

00:01:26.300 --> 00:01:29.830
This is going to be the same
thing as the closed line

00:01:29.830 --> 00:01:34.090
integral over c of q dot dr.
Well, Q only has a j-component.

00:01:34.090 --> 00:01:34.830
So if you take [? its ?]

00:01:34.830 --> 00:01:38.780
0i, so 0 times dxs is 0,
and then you're going

00:01:38.780 --> 00:01:41.190
to have Q xy times dy.

00:01:41.190 --> 00:01:44.810
They had no i-component, so
this is just going to be Q,

00:01:44.810 --> 00:01:51.780
I'll switch to that same color
again, Q of x and y times dy.

00:01:51.780 --> 00:01:52.900
That's the dot product.

00:01:52.900 --> 00:01:56.960
There was no i-component,
that's why we lose the dx.

00:01:56.960 --> 00:02:00.030
Now let's see if there's any
way that we can solve this line

00:02:00.030 --> 00:02:02.710
integral without having to
resort to a third parameter, t.

00:02:02.710 --> 00:02:03.780
Just like we did in
the last video.

00:02:03.780 --> 00:02:05.875
Actually, it will be almost
identical, we're just dealing

00:02:05.875 --> 00:02:08.080
with y's now instead of x's.

00:02:08.080 --> 00:02:09.920
So what we can do is, we
could say, well, what's our

00:02:09.920 --> 00:02:12.070
minimum y and our maximum y?

00:02:12.070 --> 00:02:15.120
So our minimum y, let's
say it's right here.

00:02:15.120 --> 00:02:17.840
The minimum y,
let's call that a.

00:02:17.840 --> 00:02:22.906
Let's say our maximum y that we
attain is right over there.

00:02:22.906 --> 00:02:24.880
Let's call that b.

00:02:24.880 --> 00:02:26.960
Oh, and just like in the last,
I forgot to tell you the

00:02:26.960 --> 00:02:28.250
direction of the curve.

00:02:28.250 --> 00:02:30.270
But this is the same path as
last time, so we're going in a

00:02:30.270 --> 00:02:32.070
counterclockwise direction.

00:02:32.070 --> 00:02:36.500
The exact same curve,
exact same path.

00:02:36.500 --> 00:02:39.180
So we're going in
that direction.

00:02:39.180 --> 00:02:42.160
Now in the last video, we broke
it up into two functions of

00:02:42.160 --> 00:02:44.600
x's, two y's as a
function of x.

00:02:44.600 --> 00:02:45.920
Now we want to deal with y's.

00:02:45.920 --> 00:02:48.530
Let's break it up into
two functions of y.

00:02:48.530 --> 00:02:54.560
So if we break this path into
two paths, those are kind of

00:02:54.560 --> 00:02:59.720
our extreme points, let's call
this past right here, let's

00:02:59.720 --> 00:03:06.110
call that path right there,
let's call that y, let's call

00:03:06.110 --> 00:03:10.930
that x of-- so here, along this
path, x is equal to-- or I

00:03:10.930 --> 00:03:16.350
could just write path
2, or a call it c2.

00:03:16.350 --> 00:03:18.956
We could say it's,

00:03:18.956 --> 00:03:22.210
x is equal to x2 of y.

00:03:22.210 --> 00:03:23.550
That's that path.

00:03:23.550 --> 00:03:26.250
And then the first path, or
it doesn't have to be the

00:03:26.250 --> 00:03:27.740
first path, depending
where you start.

00:03:27.740 --> 00:03:28.670
You can start anywhere.

00:03:28.670 --> 00:03:30.410
Let's say this one in magenta.

00:03:30.410 --> 00:03:33.050
We'll call that path 1, and we
could say, that that's defined

00:03:33.050 --> 00:03:36.480
as x is equal to x1 of y.

00:03:36.480 --> 00:03:39.395
It's a little confusing when
you have x as a function of y,

00:03:39.395 --> 00:03:41.290
but it's really completely
analogous to what we

00:03:41.290 --> 00:03:42.470
did in the last video.

00:03:42.470 --> 00:03:44.630
We're literally just
swapping x's and y.

00:03:44.630 --> 00:03:47.140
We're now expressing x as
a function of y, instead

00:03:47.140 --> 00:03:48.870
of y as a function of x.

00:03:48.870 --> 00:03:50.560
So we have these two curves.

00:03:50.560 --> 00:03:52.680
You can imagine just flipping
it, and we're doing the exact

00:03:52.680 --> 00:03:54.460
same thing that we did in
the last videos, just

00:03:54.460 --> 00:03:55.820
now in terms of y.

00:03:55.820 --> 00:03:59.610
But if you look at it this way,
this line integral can be

00:03:59.610 --> 00:04:07.590
rewritten as being equal to
the integral, let's

00:04:07.590 --> 00:04:11.650
just do c2 first.

00:04:11.650 --> 00:04:15.380
This is the integral
from b to a.

00:04:15.380 --> 00:04:17.860
We start at b, and we go to a.

00:04:17.860 --> 00:04:21.290
This is, we're coming back
from a high y to a low y.

00:04:21.290 --> 00:04:30.240
The integral from b to a of
Q of-- in that gray color.

00:04:30.240 --> 00:04:34.430
Q of, instead of having an x
there, we know along this curve

00:04:34.430 --> 00:04:37.650
right here, x is equal to, we
want everything in terms of y.

00:04:37.650 --> 00:04:40.210
So here, x is equal to x2 of y.

00:04:40.210 --> 00:04:47.800
So Q of x2 of y, x2 of y comma
y, maybe I'm using too many

00:04:47.800 --> 00:04:50.680
colors here, but I think
you get the idea.

00:04:50.680 --> 00:04:51.045
dy.

00:04:54.320 --> 00:04:56.580
So this is the part of the
line integral, just over

00:04:56.580 --> 00:04:57.800
this left hand curve.

00:04:57.800 --> 00:05:03.940
And then we're going to add to
that the line integral, or

00:05:03.940 --> 00:05:07.980
really just a regular integral
now, from y is equal to a to y

00:05:07.980 --> 00:05:18.325
is equal to b of Q of, instead
of x being equal to x2, now x

00:05:18.325 --> 00:05:21.170
is equal to x1 of y, it's equal
to this curve, this

00:05:21.170 --> 00:05:22.780
other function.

00:05:22.780 --> 00:05:33.060
So x1 of y, x1 of y comma y,
dy, and we can do exactly what

00:05:33.060 --> 00:05:34.900
we did in the previous video.

00:05:34.900 --> 00:05:37.370
Instead of, we don't like the
larger number on the bottom, so

00:05:37.370 --> 00:05:39.760
let's swap these two around.

00:05:39.760 --> 00:05:42.870
So if you swap these two, if
you make this into an a, and

00:05:42.870 --> 00:05:46.800
this into a b, that makes it
the negative of the integral,

00:05:46.800 --> 00:05:50.200
when you swap the two,
change the direction.

00:05:50.200 --> 00:05:52.150
This is exactly what we did in
the last video, so hopefully

00:05:52.150 --> 00:05:53.000
it's nothing too fancy.

00:05:53.000 --> 00:05:55.490
But now that we have the same
boundaries of integration,

00:05:55.490 --> 00:05:58.690
these two definite integrals,
we can just write them as

00:05:58.690 --> 00:06:00.280
one definite integral.

00:06:00.280 --> 00:06:05.470
So this is going to be equal
to the integral from a to b.

00:06:05.470 --> 00:06:08.450
And I'll write this one
first, since it's positive.

00:06:08.450 --> 00:06:10.370
Of, I'll write in this one.

00:06:10.370 --> 00:06:18.500
Q of x1 of y comma
y, minus this one.

00:06:18.500 --> 00:06:18.730
Right?

00:06:18.730 --> 00:06:19.710
We have the minus sign here.

00:06:19.710 --> 00:06:28.590
Minus q of x2 of y and y dy.

00:06:28.590 --> 00:06:31.540
Let me do that in
that neutral color.

00:06:31.540 --> 00:06:35.590
dy, that's multiplied by
all of these things.

00:06:35.590 --> 00:06:38.060
I distributed out the dy,
I think you get the idea.

00:06:38.060 --> 00:06:39.930
This is identical to what
we did in the last video.

00:06:39.930 --> 00:06:43.810
And this could be rewritten as,
this is equal to the integral

00:06:43.810 --> 00:06:49.120
from a to b of, and inside of
the integral, we're evaluating

00:06:49.120 --> 00:06:57.490
the function of Q of xy from
the boundaries, the upper

00:06:57.490 --> 00:07:02.880
boundary, where, the upper
boundary is going to be from x

00:07:02.880 --> 00:07:08.696
is equal to x1 of y, and
the lower boundary is

00:07:08.696 --> 00:07:11.560
x is equal to x2 of y.

00:07:11.560 --> 00:07:11.970
Right?

00:07:11.970 --> 00:07:14.950
All the x's we substituted with
that, and then we get some

00:07:14.950 --> 00:07:18.490
expression, and then from that,
we subtract this with x

00:07:18.490 --> 00:07:19.650
substituted as x2 of y.

00:07:19.650 --> 00:07:22.110
That's exactly what we did, and
just like I said in last video,

00:07:22.110 --> 00:07:24.160
we're going kind of the reverse
direction that we normally

00:07:24.160 --> 00:07:24.990
go in definite integrals.

00:07:24.990 --> 00:07:26.640
We normally get to this,
and then the next

00:07:26.640 --> 00:07:27.580
step is, we get this.

00:07:27.580 --> 00:07:29.210
But now we're going in the
reverse direction, but it's

00:07:29.210 --> 00:07:30.420
all the same difference.

00:07:30.420 --> 00:07:31.860
And all of that times dy.

00:07:35.540 --> 00:07:38.740
And just like we saw in the
last video, this, let me do it

00:07:38.740 --> 00:07:41.870
in orange, this expression
right here, actually let me

00:07:41.870 --> 00:07:43.590
draw that dy a little
further out so it doesn't

00:07:43.590 --> 00:07:45.460
get all congested.

00:07:45.460 --> 00:07:49.830
Let me do the dy out here.

00:07:49.830 --> 00:07:56.100
This expression, this entire
expression, is the same exact

00:07:56.100 --> 00:08:07.410
thing as the integral from x is
equal to, I can just write it

00:08:07.410 --> 00:08:11.280
here, let me write it
in the same colors.

00:08:11.280 --> 00:08:27.450
x2 of y to x1 of y to x1
of y the partial of Q

00:08:27.450 --> 00:08:31.760
with respect to x dx.

00:08:35.220 --> 00:08:36.020
I want to make it very clear.

00:08:36.020 --> 00:08:38.620
This is, at least in my
mind, the first part,

00:08:38.620 --> 00:08:39.680
a little confusing.

00:08:39.680 --> 00:08:41.620
But if you just saw an integral
like this, this is the

00:08:41.620 --> 00:08:42.680
inside of a double integral.

00:08:42.680 --> 00:08:43.390
And it is.

00:08:43.390 --> 00:08:46.250
The outside is what we
saw there, the integral

00:08:46.250 --> 00:08:50.380
from a to b, dy.

00:08:50.380 --> 00:08:53.280
But if you just saw this in a
double integral, what you would

00:08:53.280 --> 00:08:55.620
do is you would take the
antiderivative of this, the

00:08:55.620 --> 00:08:58.640
antiderivative of this
with respect to x, the

00:08:58.640 --> 00:09:01.430
antiderivative of the partial
of Q with respect to x with

00:09:01.430 --> 00:09:04.510
respect to x, is going
to be just Q of xy.

00:09:04.510 --> 00:09:07.120
And since it's a definite
integral, you would evaluate it

00:09:07.120 --> 00:09:11.850
at x1 of y, and then subtract
from that, this function

00:09:11.850 --> 00:09:14.370
evaluated x2 of y, which
is exactly what we did.

00:09:14.370 --> 00:09:16.220
So hopefully you
appreciate that.

00:09:16.220 --> 00:09:18.520
And then we got our result,
which is very similar

00:09:18.520 --> 00:09:19.410
to the last result.

00:09:19.410 --> 00:09:24.220
What does this double
integral represent?

00:09:24.220 --> 00:09:28.390
It represents, well, anything,
if you have any double integral

00:09:28.390 --> 00:09:31.970
that goes from-- if you
imagine, this is some function,

00:09:31.970 --> 00:09:33.640
let me draw it in
three dimensions.

00:09:33.640 --> 00:09:35.430
This is really almost a
review of what we did

00:09:35.430 --> 00:09:36.360
in the last video.

00:09:36.360 --> 00:09:42.240
If that's the y-axis, that's
our x-axis, that's our z-axis.

00:09:42.240 --> 00:09:45.720
This is some function of x
and y, so some surface you

00:09:45.720 --> 00:09:48.595
can imagine on xy plane.

00:09:48.595 --> 00:09:50.750
It's some surface.

00:09:50.750 --> 00:09:56.160
So we could call that the
partial of q with respect to x.

00:09:56.160 --> 00:09:59.260
And what this double integral
is, this is essentially

00:09:59.260 --> 00:10:02.490
defining a region, and you can
kind of view this dx times

00:10:02.490 --> 00:10:05.550
dy as kind of a small
differential of area.

00:10:05.550 --> 00:10:09.690
So the region under question,
the boundary points, are from y

00:10:09.690 --> 00:10:14.460
going from, y goes from, at the
bottom, it goes from x2 of y,

00:10:14.460 --> 00:10:17.130
which we saw was a curve that
looks something like this.

00:10:17.130 --> 00:10:19.620
That's the lower y, and over
here, if we draw it in two

00:10:19.620 --> 00:10:22.600
dimensions, this was
the lower y-curve.

00:10:22.600 --> 00:10:27.730
The upper y-curve is x1 of y,
so the upper y-curve looks

00:10:27.730 --> 00:10:29.210
something like that.

00:10:29.210 --> 00:10:32.570
The upper y-curve goes
something like that.

00:10:32.570 --> 00:10:35.510
So x varies from the lower
y-curve to the upper

00:10:35.510 --> 00:10:36.650
y-curve, right?

00:10:36.650 --> 00:10:38.540
That's what we're
doing right here.

00:10:38.540 --> 00:10:42.180
And then y varies from a to b.

00:10:42.180 --> 00:10:44.340
And so this is essentially
saying, let's take the double

00:10:44.340 --> 00:10:51.020
integral over this region
right here of this function.

00:10:51.020 --> 00:10:53.950
So it's essentially the volume,
if this is the ceiling,

00:10:53.950 --> 00:10:56.300
and this boundary is
essentially the wall.

00:10:56.300 --> 00:10:57.680
It's the volume of that room.

00:10:57.680 --> 00:10:58.750
And I don't know what it
would look like when

00:10:58.750 --> 00:11:00.140
it comes up here.

00:11:00.140 --> 00:11:01.840
But you can kind of imagine
something like that.

00:11:01.840 --> 00:11:02.990
It would be the volume of that.

00:11:02.990 --> 00:11:04.690
So that's what we're taking.

00:11:04.690 --> 00:11:07.490
This is the identical result
we got in the last video.

00:11:07.490 --> 00:11:09.060
And this is a
pretty neat thing.

00:11:09.060 --> 00:11:13.600
So all of a sudden, this
vector, that-- and Q of xy, I

00:11:13.600 --> 00:11:16.060
didn't draw it out like I did
the last time, Q of xy

00:11:16.060 --> 00:11:16.630
only has [? things ?]

00:11:16.630 --> 00:11:19.475
in the j-direction, so it only
has, if I were to draw its

00:11:19.475 --> 00:11:21.920
vector field, the vectors
only go up and down.

00:11:25.960 --> 00:11:28.650
They have no horizontal
component to them.

00:11:28.650 --> 00:11:32.270
But we saw, when you start with
a vector field like this, you

00:11:32.270 --> 00:11:35.810
take the line integral around
this closed loop, and I'll

00:11:35.810 --> 00:11:39.000
rewrite it right here, you take
this line integral around this

00:11:39.000 --> 00:11:44.200
closed loop of q dot dr, which
is equal to the integral around

00:11:44.200 --> 00:11:49.560
the closed loop of Q of xy dy.

00:11:49.560 --> 00:11:53.240
We just figured out that that's
equivalent to the double

00:11:53.240 --> 00:11:58.490
integral over the region.

00:11:58.490 --> 00:12:00.660
This is the region.

00:12:00.660 --> 00:12:00.920
Right?

00:12:00.920 --> 00:12:02.380
That's exactly what
we're doing over here.

00:12:02.380 --> 00:12:03.950
If I just gave you the region,
you'd have to define it, you'd

00:12:03.950 --> 00:12:07.116
say, well, x is going from,
this is going from this

00:12:07.116 --> 00:12:09.780
function to that function, and
y is going from a to b, and you

00:12:09.780 --> 00:12:11.310
might want to review the double
integral videos, if

00:12:11.310 --> 00:12:11.970
that confuses you.

00:12:11.970 --> 00:12:15.270
So we're taking the double
integral over the region of the

00:12:15.270 --> 00:12:20.370
partial of Q with respect to x,
d-- well, you could write dx

00:12:20.370 --> 00:12:23.490
dy, or you can even right
a little da, right?

00:12:23.490 --> 00:12:26.110
The differential of area,
right, that we can imagine

00:12:26.110 --> 00:12:28.670
as a da, which is the
same thing as a dx dy.

00:12:28.670 --> 00:12:31.390
And if we combine that with the
last video, and this is kind of

00:12:31.390 --> 00:12:35.030
the neat bringing it all
together part, the result of

00:12:35.030 --> 00:12:36.760
the last video was this.

00:12:36.760 --> 00:12:39.940
That if I had a function that's
defined completely in terms of

00:12:39.940 --> 00:12:42.270
x, we had this, right here.

00:12:42.270 --> 00:12:43.090
We had that result.

00:12:43.090 --> 00:12:46.280
Actually, let me copy and paste
both of these to a nice clean

00:12:46.280 --> 00:12:51.100
part of my whiteboard,
and then we can do the

00:12:51.100 --> 00:12:53.030
exciting conclusion.

00:12:53.030 --> 00:12:57.170
Let me copy and paste that.

00:12:57.170 --> 00:12:59.300
So that's what we got
in the last video.

00:12:59.300 --> 00:13:03.200
And this video, we
got this result.

00:13:03.200 --> 00:13:07.080
I'll just copy and paste
that part right there.

00:13:07.080 --> 00:13:12.140
You might already predict
where this is going.

00:13:12.140 --> 00:13:13.490
And then let me
paste it over here.

00:13:13.490 --> 00:13:16.240
This is the result
from this video.

00:13:16.240 --> 00:13:20.390
Now let's think about an
arbitrary vector field but is

00:13:20.390 --> 00:13:27.160
defined as, I'll do that in
pink, let's say F is a vector

00:13:27.160 --> 00:13:32.780
field defined over the xy
plane, and F is equal to P

00:13:32.780 --> 00:13:39.920
of xy i plus Q of xy j.

00:13:39.920 --> 00:13:43.520
You can almost imagine F being
the addition of our vector

00:13:43.520 --> 00:13:46.040
fields, P and Q, that we did
in the last two videos.

00:13:46.040 --> 00:13:49.550
Q was this video, and we did
P in the video before that.

00:13:49.550 --> 00:13:53.570
But this is really any
arbitrary vector field.

00:13:53.570 --> 00:13:56.710
And let's say we want to take
the vector field, or sorry, the

00:13:56.710 --> 00:14:00.680
line integral of this vector
field, along some path.

00:14:00.680 --> 00:14:02.730
It could be the same one
we've done, which has been

00:14:02.730 --> 00:14:03.870
a very arbitrary one.

00:14:03.870 --> 00:14:05.440
It's really any arbitrary path.

00:14:05.440 --> 00:14:09.620
So let me draw some
arbitrary path over here.

00:14:09.620 --> 00:14:13.010
Let's say, that is my arbitrary
path, my arbitrary curve.

00:14:13.010 --> 00:14:17.610
Let's say it goes in that
counterclockwise direction,

00:14:17.610 --> 00:14:18.530
just like that.

00:14:18.530 --> 00:14:22.830
And I'm interested in what the
line integral, the closed line

00:14:22.830 --> 00:14:27.570
integral, around that
path of F dot dr is.

00:14:27.570 --> 00:14:29.810
And we've seen it
multiple times.

00:14:29.810 --> 00:14:36.700
dr is equal to dx i
plus dy times j.

00:14:36.700 --> 00:14:40.740
So this line integral can be
rewritten as, this is equal

00:14:40.740 --> 00:14:44.390
to the line integral
around the path c.

00:14:44.390 --> 00:14:48.990
F dot dr, that's going to be
this term times dx, so that's

00:14:48.990 --> 00:15:00.090
p of xy times dx plus this
term, Q of xy times dy.

00:15:00.090 --> 00:15:04.250
And this whole thing,
essentially this is the same

00:15:04.250 --> 00:15:11.470
thing as the line integral of
p of xy dx, plus the line

00:15:11.470 --> 00:15:16.660
integral of Q of xy dy.

00:15:16.660 --> 00:15:17.750
Now what are these things?

00:15:17.750 --> 00:15:20.180
This is what we figured out in
the first video, this is what

00:15:20.180 --> 00:15:22.160
we figured out just
now in this video.

00:15:22.160 --> 00:15:25.190
This thing right here is
the exact same thing

00:15:25.190 --> 00:15:26.510
as that over there.

00:15:26.510 --> 00:15:35.270
So this is going to be equal to
the double integral over this

00:15:35.270 --> 00:15:41.310
region right here, of the minus
partial of P with respect to y.

00:15:45.710 --> 00:15:48.550
Instead of a dy dx, we
could say just over the

00:15:48.550 --> 00:15:50.750
differential of area.

00:15:50.750 --> 00:15:53.870
And then plus this
one, this result.

00:15:53.870 --> 00:15:54.120
Q.

00:15:54.120 --> 00:15:56.930
This thing right here is
exactly what we just proved,

00:15:56.930 --> 00:15:59.480
is exactly what we just
showed in this video.

00:15:59.480 --> 00:16:02.380
So that's plus, I'll leave it
up there, maybe I'll do it in

00:16:02.380 --> 00:16:06.760
the yellow, plus the double
integral over the same region

00:16:06.760 --> 00:16:11.220
of the partial of Q
with respect to x.

00:16:11.220 --> 00:16:15.090
da, where that's just dy
dx, or dx dy, you can

00:16:15.090 --> 00:16:15.950
switch the order, it's

00:16:15.950 --> 00:16:18.010
the differential of area.

00:16:18.010 --> 00:16:20.130
And now, we can add
these two integrals.

00:16:20.130 --> 00:16:21.070
What do we get?

00:16:21.070 --> 00:16:23.220
So this is equal to, and
this is kind of our

00:16:23.220 --> 00:16:25.680
big, grand conclusion.

00:16:25.680 --> 00:16:28.130
Maybe magenta is
called for here.

00:16:28.130 --> 00:16:31.720
The double integral over the
region of, I'll write this one

00:16:31.720 --> 00:16:34.230
first because it's positive,
that one's negative, of the

00:16:34.230 --> 00:16:42.390
partial of Q with respect to x,
minus the partial of p with

00:16:42.390 --> 00:16:48.360
respect to y, d, the
differential of area.

00:16:48.360 --> 00:16:50.960
This is our big takeaway.

00:16:50.960 --> 00:16:52.670
This is our big takeaway.

00:16:52.670 --> 00:16:54.700
Let me write it here.

00:16:54.700 --> 00:16:59.620
The line integral of the closed
loop of F dot dr is equal to

00:16:59.620 --> 00:17:02.420
the double integral
of this expression.

00:17:02.420 --> 00:17:03.720
And it's something,
just remember.

00:17:03.720 --> 00:17:07.545
We're taking the function that
was associated with the

00:17:07.545 --> 00:17:09.540
x-component, or the
i-component, we're taking the

00:17:09.540 --> 00:17:11.880
partial with respect to y.

00:17:11.880 --> 00:17:13.610
And the function that was
associated with the

00:17:13.610 --> 00:17:16.530
y-component, we're taking the
partial with respect to x.

00:17:16.530 --> 00:17:18.580
And that first one, we're
taking the negative of.

00:17:18.580 --> 00:17:19.910
That's a good way
to remember it.

00:17:19.910 --> 00:17:23.140
But this result right here,
this is, maybe I should

00:17:23.140 --> 00:17:30.730
write it in green, this
is Green's Theorem.

00:17:33.850 --> 00:17:38.080
And it's a neat way to relate a
line integral of a vector field

00:17:38.080 --> 00:17:40.770
that has these partial
derivatives, assuming it has

00:17:40.770 --> 00:17:44.460
these partial derivatives, to
the region, to a double

00:17:44.460 --> 00:17:45.330
integral of the region.

00:17:45.330 --> 00:17:48.680
Now, and this is a little bit
of a side note, we've seen in

00:17:48.680 --> 00:17:56.980
several videos before, we've
learned that if F is

00:17:56.980 --> 00:18:02.780
conservative, which means it's
the gradient of some function,

00:18:02.780 --> 00:18:06.280
that it's path-independent,
that the closed integral around

00:18:06.280 --> 00:18:09.550
any path is equal to 0.

00:18:09.550 --> 00:18:10.980
And that's still true.

00:18:10.980 --> 00:18:14.720
So that tells us that if F is
conservative, this thing right

00:18:14.720 --> 00:18:17.560
here must be equal to 0.

00:18:17.560 --> 00:18:20.070
That's the only way that you're
always going to enforce that

00:18:20.070 --> 00:18:23.350
this whole integral is going to
be equal, is going to be equal

00:18:23.350 --> 00:18:25.970
to 0 over any, any, any region.

00:18:25.970 --> 00:18:27.270
I'm sure you could think
of situations where they

00:18:27.270 --> 00:18:29.880
cancel each other out, but
really over any region.

00:18:29.880 --> 00:18:31.700
That's the only way that
this is going to be true.

00:18:31.700 --> 00:18:33.880
That these two things are
going to be equal to 0.

00:18:33.880 --> 00:18:38.220
So then you could say, partial
of Q with respect to x, minus

00:18:38.220 --> 00:18:42.370
the partial of P with respect
to y, has to be equal to 0, or

00:18:42.370 --> 00:18:44.970
these two things have
to equal each other.

00:18:44.970 --> 00:18:45.300
Or.

00:18:45.300 --> 00:18:47.370
This is kind of a corollary
to Green's Theorem.

00:18:47.370 --> 00:18:49.900
Kind of a low-hanging fruit
you could have figured out.

00:18:49.900 --> 00:18:53.010
The partial of Q with respect
to x is equal to the partial

00:18:53.010 --> 00:18:56.920
of P with respect to y.

00:18:56.920 --> 00:18:59.070
And when you study exact
equations in differential

00:18:59.070 --> 00:19:01.460
equations, you'll
see this a lot more.

00:19:01.460 --> 00:19:04.140
And actually, well, I won't go
into too much, but conservative

00:19:04.140 --> 00:19:07.330
fields, you're actually, the
differential form of what you

00:19:07.330 --> 00:19:10.500
see in the line integrals, if
it's conservative, it would

00:19:10.500 --> 00:19:11.450
be an exact equation.

00:19:11.450 --> 00:19:12.610
But we're not going to
go into that too much.

00:19:12.610 --> 00:19:15.330
But hopefully you might see the
parallels if you've already run

00:19:15.330 --> 00:19:17.790
into exact equations in
differential equations.

00:19:17.790 --> 00:19:20.980
But this is the big takeaway,
and let's maybe do some

00:19:20.980 --> 00:19:25.780
examples using this takeaway
in the next video.


Green's theorem proof part 1

WEBVTT
Kind: captions
Language: en

00:00:00.820 --> 00:00:04.310
Let's say we have a
path in the xy plane.

00:00:04.310 --> 00:00:09.780
That's my y-axis, that is
my x-axis, in my path

00:00:09.780 --> 00:00:11.760
will look like this.

00:00:11.760 --> 00:00:16.280
Let's say it looks like that;
trying to draw a bit of an

00:00:16.280 --> 00:00:19.480
arbitrary path, and let's say
we go in a counter clockwise

00:00:19.480 --> 00:00:23.450
direction like that
along our path.

00:00:23.450 --> 00:00:26.180
And we could call this path--
so we're going in a counter

00:00:26.180 --> 00:00:29.180
clockwise direction --we
could call that path c.

00:00:29.180 --> 00:00:31.730
And let's say we also
have a vector field.

00:00:31.730 --> 00:00:34.335
And our vector field is going
to be a little unusual;

00:00:34.335 --> 00:00:36.190
I'll call it p.

00:00:36.190 --> 00:00:37.270
p of xy.

00:00:37.270 --> 00:00:40.870
It only has an i component, or
all of its vectors are only

00:00:40.870 --> 00:00:43.240
multiples of the i-unit vector.

00:00:43.240 --> 00:00:48.410
So it's capital P of xy
times the unit vector i.

00:00:48.410 --> 00:00:51.920
There is no j component, so if
you have to visualize this

00:00:51.920 --> 00:00:57.880
vector field, all of the
vectors, they're all multiples

00:00:57.880 --> 00:00:59.980
of the i-unit vector.

00:00:59.980 --> 00:01:02.040
Or they could be negative
multiples, so they could

00:01:02.040 --> 00:01:03.960
also go in that direction.

00:01:03.960 --> 00:01:06.780
But they don't go diagonal
or they don't go up.

00:01:06.780 --> 00:01:09.630
They all go left to
right or right to left.

00:01:09.630 --> 00:01:12.660
That's what this vector
field would look like.

00:01:12.660 --> 00:01:15.870
Now what I'm interested in
doing is figuring out the line

00:01:15.870 --> 00:01:20.170
integral over a closed loop--
the closed loop c, or the

00:01:20.170 --> 00:01:27.710
closed path c --of p dot dr,
which is just our standard

00:01:27.710 --> 00:01:29.775
kind of way of solving
for a line integral.

00:01:29.775 --> 00:01:32.140
And we've seen what
dr is in the past.

00:01:32.140 --> 00:01:40.760
dr is equal to dx times i plus
dy times the j-unit vector.

00:01:40.760 --> 00:01:45.250
And you might say, isn't
it dx, dt times dt?

00:01:45.250 --> 00:01:51.530
Let me write that: can't dr
be written as dx, dt times

00:01:51.530 --> 00:01:56.780
dti plus dy, dt times dtj?

00:01:56.780 --> 00:01:59.280
And it could, but if you
imagine these differentials

00:01:59.280 --> 00:02:01.820
could cancel out, and you're
just left with the dx and

00:02:01.820 --> 00:02:03.490
a dy, and we've seen
that multiple times.

00:02:03.490 --> 00:02:06.360
And I'm going to leave it in
this form because hopefully, if

00:02:06.360 --> 00:02:09.390
we're careful, we won't have to
deal with the third

00:02:09.390 --> 00:02:10.540
parameter, t.

00:02:10.540 --> 00:02:13.240
So let's just look at it in
this form right here with

00:02:13.240 --> 00:02:16.120
just the dx's and the dy's.

00:02:16.120 --> 00:02:21.580
So this integral can be
rewritten as the line integral,

00:02:21.580 --> 00:02:25.040
the curve c-- actually let
me do it over down here.

00:02:25.040 --> 00:02:32.600
The line integral over the path
of the curve c of p dot dr. So

00:02:32.600 --> 00:02:35.750
we take the product of each of
the coefficients, let's say the

00:02:35.750 --> 00:02:39.610
coefficient of the i component,
so we get p-- I'll do that in

00:02:39.610 --> 00:02:47.450
green, actually do that purple
color --so we get p of xy times

00:02:47.450 --> 00:02:54.770
dx plus-- well there's no 0
times j times dy; 0 times dy id

00:02:54.770 --> 00:02:58.290
just going to be 0 --so this
our line integral simplified

00:02:58.290 --> 00:02:59.330
to this right here.

00:02:59.330 --> 00:03:02.890
This is equal to this original
integral up here, so we're

00:03:02.890 --> 00:03:06.110
literally just taking the line
integral around this path.

00:03:06.110 --> 00:03:09.890
Now I said that we play our
cards right, we're not going to

00:03:09.890 --> 00:03:12.090
have to deal with the third
variable, t; that we might be

00:03:12.090 --> 00:03:15.230
able just solve this integral
only in terms of x.

00:03:15.230 --> 00:03:16.700
And so let's see if
we can do that.

00:03:16.700 --> 00:03:19.640
So let's look at our minimum
and maximum x points.

00:03:19.640 --> 00:03:22.410
That looks like our
minimum x point.

00:03:22.410 --> 00:03:24.250
Let's call that a.

00:03:24.250 --> 00:03:29.940
Let's call that our maximum
x point; let's call that b.

00:03:29.940 --> 00:03:33.260
What we could do is we can
break up this curve into

00:03:33.260 --> 00:03:36.500
two functions of x.
y is functions of x.

00:03:36.500 --> 00:03:40.750
So this bottom one right here
we could call as y1 of x.

00:03:40.750 --> 00:03:44.570
This is just a standard curve;
you know when we were just

00:03:44.570 --> 00:03:47.940
dealing with standard calculus,
this is just you can

00:03:47.940 --> 00:03:50.570
imagine this is f of x and
it's a function of x.

00:03:50.570 --> 00:03:51.900
And this is y2 of x.

00:03:55.160 --> 00:03:56.090
Just like that.

00:03:56.090 --> 00:04:01.470
So you can imagine two paths;
one path defined by y1 of x--

00:04:01.470 --> 00:04:04.800
let me do that in a different
color; magenta --one path

00:04:04.800 --> 00:04:09.340
defined by y1 of x as we go
from x is equal to a to x is

00:04:09.340 --> 00:04:14.110
equal to b, and then another
path defined by y2 of x as we

00:04:14.110 --> 00:04:19.230
go from x is equal to
b to x is equal to a.

00:04:19.230 --> 00:04:21.330
That is our curve.

00:04:21.330 --> 00:04:24.550
So what we could do is, we
could rewrite this integral--

00:04:24.550 --> 00:04:29.820
which is the same thing as that
integral --as this is equal to

00:04:29.820 --> 00:04:33.790
the integral-- we'll first do
this first path --of x going

00:04:33.790 --> 00:04:40.130
from a to b of p of x.

00:04:40.130 --> 00:04:43.670
And I could to say p of x and
y, but we know along this

00:04:43.670 --> 00:04:47.440
path y is a function of x.

00:04:47.440 --> 00:04:51.290
So we say x and y1 of x.

00:04:51.290 --> 00:04:53.220
Wherever we see a y we
substitute it with

00:04:53.220 --> 00:04:56.910
a y1 of x, dx.

00:04:56.910 --> 00:05:02.030
So that covers that first path;
I'll do it in the same color.

00:05:02.030 --> 00:05:04.430
We could imagine this is c1.

00:05:04.430 --> 00:05:07.180
This is kind of the first half
of our curve-- well it's not

00:05:07.180 --> 00:05:09.820
exactly the half --but that
takes us right from that

00:05:09.820 --> 00:05:10.990
point to that point.

00:05:10.990 --> 00:05:13.430
And then we want to
complete the circle.

00:05:13.430 --> 00:05:16.960
Maybe I'll do that, and
I'll do that in yellow.

00:05:16.960 --> 00:05:19.390
That's going to be equal to--
sorry we're going to have to

00:05:19.390 --> 00:05:25.920
add these two --plus the
integral from x is equal to b

00:05:25.920 --> 00:05:37.290
to x is equal to a of-- do it
in that same color --of p of x.

00:05:37.290 --> 00:05:40.230
And now y is going
to be y2 of x.

00:05:40.230 --> 00:05:42.180
Wherever you see a y, you
can substitute with y2

00:05:42.180 --> 00:05:44.890
of x along this curve.

00:05:44.890 --> 00:05:49.350
y2 of x, dx.

00:05:49.350 --> 00:05:51.650
This is already getting
interesting and you

00:05:51.650 --> 00:05:53.090
might already see where
I'm going with this.

00:05:53.090 --> 00:05:56.130
So this is the curve c2.

00:05:56.130 --> 00:06:00.030
too I think you appreciate if
you take the union of c1 and

00:06:00.030 --> 00:06:02.440
c2, we've got our whole curve.

00:06:02.440 --> 00:06:05.690
So let's see if we can simplify
this integral a little bit.

00:06:05.690 --> 00:06:07.710
Well one thing we want to
do, we might want to make

00:06:07.710 --> 00:06:09.130
their end points the same.

00:06:09.130 --> 00:06:12.150
So if you swap a and b
here, it just turns

00:06:12.150 --> 00:06:13.180
the integral negative.

00:06:13.180 --> 00:06:17.810
So you make this into a b, that
into an a, and then make that

00:06:17.810 --> 00:06:20.290
plus sign into a minus sign.

00:06:20.290 --> 00:06:24.000
And now we can rewrite this
whole thing as being equal to

00:06:24.000 --> 00:06:33.810
the integral from a to b of
this thing, of p of x and y1 of

00:06:33.810 --> 00:06:44.290
x minus this thing, minus p of
x and y2 of x, and then

00:06:44.290 --> 00:06:46.840
all of that times dx.

00:06:46.840 --> 00:06:48.310
I'll write it in a third color.

00:06:51.480 --> 00:06:53.740
Now, I'm going to do something
a little bit arbitrary, but I

00:06:53.740 --> 00:06:56.190
think you'll appreciate why I
did this by the end of this

00:06:56.190 --> 00:06:58.000
video, and it's just a
very simple operation.

00:06:58.000 --> 00:06:59.580
What I'm going to do is I'm
going to swap these two.

00:06:59.580 --> 00:07:02.490
So I'm essentially going to
multiply this whole thing by

00:07:02.490 --> 00:07:05.020
negative 1, or essentially
multiply and divide

00:07:05.020 --> 00:07:05.910
by negative 1.

00:07:05.910 --> 00:07:08.310
So I can multiply this by
negative 1 and then multiply

00:07:08.310 --> 00:07:10.705
the outside by negative 1, and
I will not have changed the

00:07:10.705 --> 00:07:13.290
integral; I'm multiplying
by negative 1 twice.

00:07:13.290 --> 00:07:16.810
So if I swap these two things,
if I multiply the inside times

00:07:16.810 --> 00:07:20.250
negative 1, so this is going to
be equal to-- do the outside

00:07:20.250 --> 00:07:22.770
of the integral, a to b.

00:07:22.770 --> 00:07:26.460
If I multiply the inside--
I'll do a dx out here --if I

00:07:26.460 --> 00:07:28.790
multiply the inside of the
integral by negative 1,

00:07:28.790 --> 00:07:30.510
these two guys switch.

00:07:30.510 --> 00:07:36.230
So it becomes p
of x of y2 of x.

00:07:36.230 --> 00:07:46.310
And then you're going to have
minus p of x and y1 of x.

00:07:46.310 --> 00:07:47.920
My handwriting's getting
a little messy.

00:07:47.920 --> 00:07:51.420
But I can't just multiply
just the inside by minus 1.

00:07:51.420 --> 00:07:53.250
I don't want to change the
integral, so I multiplied the

00:07:53.250 --> 00:07:55.480
inside by minus 1, let me
multiply the outside

00:07:55.480 --> 00:07:56.140
by minus 1.

00:07:56.140 --> 00:07:59.160
And since I multiplied by
minus 1 twice, these two

00:07:59.160 --> 00:08:00.120
things are equivalent.

00:08:00.120 --> 00:08:04.840
Or you could say this is
the negative of that.

00:08:04.840 --> 00:08:06.400
Either way, I think you
appreciate that I haven't

00:08:06.400 --> 00:08:08.430
changed the integral
at all, numerically.

00:08:08.430 --> 00:08:11.790
I multiplied the inside and
the outside by minus 1.

00:08:11.790 --> 00:08:15.130
And now the next step I'm going
to do, it might look a little

00:08:15.130 --> 00:08:17.780
bit foreign to you, but I
think you'll appreciate it.

00:08:20.660 --> 00:08:23.460
It might be obvious to you
if you've recently done

00:08:23.460 --> 00:08:24.320
some double integrals.

00:08:24.320 --> 00:08:30.270
So this thing can be rewritten
as minus the integral from a to

00:08:30.270 --> 00:08:38.660
b of-- and let me do a new
color --of the function p of x,

00:08:38.660 --> 00:08:49.460
y evaluated at y2 of x minus--
and let me make it very clear;

00:08:49.460 --> 00:08:54.970
this is y is equal to y2 of x
--minus this function evaluated

00:08:54.970 --> 00:08:59.520
at y is equal to y1 of x.

00:08:59.520 --> 00:09:03.820
And of course all
of that times dx.

00:09:03.820 --> 00:09:08.380
This statement and what we
saw right here-- this

00:09:08.380 --> 00:09:14.090
statement right here --are
completely identical.

00:09:14.090 --> 00:09:20.000
And then if we assume that a
partial derivative of capital P

00:09:20.000 --> 00:09:24.590
with respect to y exists,
hopefully you'll realize-- and

00:09:24.590 --> 00:09:26.870
I'll focus on this a little
bit because I don't want to

00:09:26.870 --> 00:09:27.700
confuse you on this step.

00:09:27.700 --> 00:09:28.880
Let me write the outside
of this integral.

00:09:28.880 --> 00:09:32.180
So this is going to be equal
to-- and this is kind of a neat

00:09:32.180 --> 00:09:35.150
outcome, and we're starting
to build up to a very neat

00:09:35.150 --> 00:09:38.490
outcome, which will probably
have to take the next video to

00:09:38.490 --> 00:09:41.210
do --so we do the outside dx.

00:09:43.860 --> 00:09:48.830
If we assume that capital P has
a partial derivative, this

00:09:48.830 --> 00:09:51.410
right here is the
exact same thing.

00:09:51.410 --> 00:09:54.550
This right here is the exact
same thing as the partial

00:09:54.550 --> 00:10:01.860
derivative of P with respect to
y, dy, the antiderivative of

00:10:01.860 --> 00:10:10.010
that from y1 of x to y2 of x.

00:10:10.010 --> 00:10:13.040
I want to make you feel
comfortable that these two

00:10:13.040 --> 00:10:14.740
things are equivalent.

00:10:14.740 --> 00:10:16.800
And to realize they're
equivalent, you'll probably

00:10:16.800 --> 00:10:19.310
just have to start here
and then go to that.

00:10:19.310 --> 00:10:21.540
We're used to seeing this;
we're used to seeing a double

00:10:21.540 --> 00:10:25.100
integral like this, and then
the very first step we say, OK

00:10:25.100 --> 00:10:29.500
to solve this double integral
we start on the inside integral

00:10:29.500 --> 00:10:33.210
right there, and we say, OK
let's take the antiderivative

00:10:33.210 --> 00:10:34.870
of this with respect to y.

00:10:34.870 --> 00:10:37.420
So if you take the
antiderivative of the partial

00:10:37.420 --> 00:10:41.830
of p with respect to y, you're
going to end up with p.

00:10:41.830 --> 00:10:44.330
And since this is a definite
integral, the boundaries are

00:10:44.330 --> 00:10:47.420
going to be in terms of x,
you're going to evaluate that

00:10:47.420 --> 00:10:51.020
from y is equal to y2 of x, and
you're going to subtract from

00:10:51.020 --> 00:10:53.210
that y is equal to y1 of x.

00:10:53.210 --> 00:10:55.200
Normally we start with
something like this, and we

00:10:55.200 --> 00:10:56.610
go to something like this.

00:10:56.610 --> 00:10:59.620
This is kind of unusual that we
started, we kind of solved, we

00:10:59.620 --> 00:11:02.030
started with the solution of
the definite integral, and then

00:11:02.030 --> 00:11:05.150
we slowly built back to
the definite integral.

00:11:05.150 --> 00:11:08.400
So hopefully you realize that
this is true, that this is

00:11:08.400 --> 00:11:10.110
just we're kind of going
in a reverse direction

00:11:10.110 --> 00:11:11.020
than we normally do.

00:11:11.020 --> 00:11:13.750
And if you do realize that,
then we've just established

00:11:13.750 --> 00:11:14.850
a pretty neat outcome.

00:11:14.850 --> 00:11:18.880
Because what is
this right here?

00:11:18.880 --> 00:11:21.570
Let me go back, let me see
if I can fit everything.

00:11:21.570 --> 00:11:25.745
I have some function-- and I'm
assuming that the partial of

00:11:25.745 --> 00:11:29.790
P with respect to y exists
--but I have some function

00:11:29.790 --> 00:11:31.510
defined over the xy plane.

00:11:31.510 --> 00:11:33.490
You know, you could
imagine we're dealing in

00:11:33.490 --> 00:11:35.390
three dimensions now.

00:11:35.390 --> 00:11:37.980
We'll draw a little bit neater.

00:11:37.980 --> 00:11:46.110
So that's y, that's x, that's
z, so this, you could imagine,

00:11:46.110 --> 00:11:48.560
is some surface; it just
happens to be the partial

00:11:48.560 --> 00:11:50.350
of P with respect to x.

00:11:50.350 --> 00:11:55.790
So it's some surface on
the xy plane like that.

00:11:55.790 --> 00:11:57.170
And what are we doing?

00:11:57.170 --> 00:12:01.310
We're taking the double
integral under that surface,

00:12:01.310 --> 00:12:02.250
around this region.

00:12:02.250 --> 00:12:04.580
The region's boundaries in
terms of y are defined

00:12:04.580 --> 00:12:06.180
by y2 and y1 of x.

00:12:08.970 --> 00:12:11.160
So you literally
have that curve.

00:12:11.160 --> 00:12:16.500
That's y2 on top,
y1 on the bottom.

00:12:16.500 --> 00:12:18.590
And so we're essentially
taking the volume above.

00:12:22.780 --> 00:12:27.190
So if you imagine with the base
is-- the whole floor of this is

00:12:27.190 --> 00:12:32.060
going to be the area inside of
this curve, and then the height

00:12:32.060 --> 00:12:36.070
is going to be the function
partial of P with respect to y.

00:12:40.490 --> 00:12:42.700
It's going to be a little hard
for me to draw, but this is

00:12:42.700 --> 00:12:44.670
essentially some type of a
volume, if you want to

00:12:44.670 --> 00:12:45.610
visualize it that way.

00:12:45.610 --> 00:12:52.780
But the really neat outcome
here is if you call this region

00:12:52.780 --> 00:12:56.600
r, we've just simplified
this line integral.

00:12:56.600 --> 00:12:57.580
And this was a special one.

00:12:57.580 --> 00:13:01.270
It only had an x-component, the
vector field, but we've just

00:13:01.270 --> 00:13:06.310
simplified this line integral
to being equivalent to-- maybe

00:13:06.310 --> 00:13:09.080
I should write this line
integral because that's what's

00:13:09.080 --> 00:13:10.830
the really neat outcome.

00:13:10.830 --> 00:13:13.130
We've just established that
this thing right here, which

00:13:13.130 --> 00:13:16.970
is the same as our original
one, so let me write that.

00:13:16.970 --> 00:13:22.570
The closed line integral around
the curve c of p of xy, dx,

00:13:22.570 --> 00:13:25.300
we've just established that
that's the same thing as the

00:13:25.300 --> 00:13:31.980
double integral over the region
r-- this is the region

00:13:31.980 --> 00:13:40.333
r --of the partial of
P with respect to y.

00:13:47.990 --> 00:13:52.120
And we could write dy, dx, or
we could write da, whatever you

00:13:52.120 --> 00:13:53.810
want to write, but this is
the double integral

00:13:53.810 --> 00:13:55.310
over that region.

00:13:55.310 --> 00:13:59.560
The neat thing here is using a
vector field that only had an

00:13:59.560 --> 00:14:02.850
x-component, we were able to
connect its line integral to

00:14:02.850 --> 00:14:04.730
the double integral over
region-- oh, and I forgot

00:14:04.730 --> 00:14:06.000
something very important.

00:14:06.000 --> 00:14:07.420
We had a negative
sign out here.

00:14:07.420 --> 00:14:10.700
So this was a minus
sign out here.

00:14:10.700 --> 00:14:12.390
Or we could even put the minus
in here, but I think you

00:14:12.390 --> 00:14:13.940
get the general idea.

00:14:13.940 --> 00:14:16.450
In the next video, I'm going to
do the same exact thing with

00:14:16.450 --> 00:14:20.710
the vector field that only has
vectors in the y-direction.

00:14:20.710 --> 00:14:22.240
And then we'll connect
the two and we'll end up

00:14:22.240 --> 00:14:23.560
with Green's theorem.