Stokes' theorem proof part 3

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:04.590
We've now laid the groundwork
so we can express this surface

00:00:04.590 --> 00:00:06.390
integral, which is
the right-hand side

00:00:06.390 --> 00:00:08.430
of the way we've
written Stokes' theorem.

00:00:08.430 --> 00:00:12.020
We can now express this
as a double integral

00:00:12.020 --> 00:00:15.442
over the domain of the
parameters that we care about.

00:00:15.442 --> 00:00:17.150
And we're going to do
that in this video.

00:00:17.150 --> 00:00:19.270
And then in the next
series of videos,

00:00:19.270 --> 00:00:21.310
we'll do the same thing
for this expression.

00:00:21.310 --> 00:00:23.560
We're actually going to do
that using Green's theorem.

00:00:23.560 --> 00:00:24.730
What we're going to
do is we're going

00:00:24.730 --> 00:00:26.938
to see we're going to get
the same expressions, which

00:00:26.938 --> 00:00:28.910
will show us that
Stokes' theorem is true,

00:00:28.910 --> 00:00:31.487
at least for this
special class of surfaces

00:00:31.487 --> 00:00:32.820
that we are studying right here.

00:00:32.820 --> 00:00:34.510
But they're pretty general.

00:00:34.510 --> 00:00:36.894
Now let's now try to do that.

00:00:36.894 --> 00:00:38.310
So our surface
integral-- I'm just

00:00:38.310 --> 00:00:39.560
going to rewrite it down here.

00:00:39.560 --> 00:00:43.800
It's the surface integral,
so over our surface

00:00:43.800 --> 00:00:44.920
of the curl of f.

00:00:44.920 --> 00:00:46.970
Actually, let me go
a little bit lower.

00:00:46.970 --> 00:00:57.420
So we have our surface integral
of the curl of F dot ds.

00:01:00.260 --> 00:01:02.390
Well, we've already figured
out what our curl of F

00:01:02.390 --> 00:01:04.060
is here two videos ago.

00:01:04.060 --> 00:01:06.250
And we've almost
figured out what ds is.

00:01:06.250 --> 00:01:09.660
ds is the cross product of
these two vectors times dA.

00:01:09.660 --> 00:01:11.290
The cross product
of these two vectors

00:01:11.290 --> 00:01:12.780
is this right over here.

00:01:12.780 --> 00:01:16.330
So we could just
write that ds is

00:01:16.330 --> 00:01:19.570
going to be equal to
this thing times dA.

00:01:19.570 --> 00:01:21.470
This is the cross
product of the partial

00:01:21.470 --> 00:01:24.490
of r with respect to x and the
partial of r with respect to y.

00:01:24.490 --> 00:01:27.780
And then we have to multiply
that times dA right over there.

00:01:27.780 --> 00:01:29.995
So this expression is
just going to be the dot

00:01:29.995 --> 00:01:36.140
product of the curl of F, which
is this stuff up here dotted

00:01:36.140 --> 00:01:37.720
with this stuff down here.

00:01:37.720 --> 00:01:38.530
And essentially, we're
going to take the dot

00:01:38.530 --> 00:01:40.620
product of this
vector and that vector

00:01:40.620 --> 00:01:42.840
and then multiply it times
this-- we can actually

00:01:42.840 --> 00:01:44.840
consider this to
be a scalar value.

00:01:44.840 --> 00:01:45.870
So let's do that.

00:01:45.870 --> 00:01:47.940
So this is going to
be equal to-- and

00:01:47.940 --> 00:01:50.090
when we do this, all
of these we are now

00:01:50.090 --> 00:01:53.490
going to start operating in
the domain of our parameters.

00:01:53.490 --> 00:01:55.850
So it's going to turn
from a surface integral

00:01:55.850 --> 00:01:58.280
into a double integral
over that domain,

00:01:58.280 --> 00:02:00.560
over that region
that we care about.

00:02:00.560 --> 00:02:03.760
This is the domain of our
parameters, this region R.

00:02:03.760 --> 00:02:09.163
And this is how we've
manipulated any of the surface

00:02:09.163 --> 00:02:10.830
integrals that we've
come across so far.

00:02:10.830 --> 00:02:13.430
We've turned them
into double integrals

00:02:13.430 --> 00:02:15.440
over the domain
of the parameters.

00:02:15.440 --> 00:02:19.240
So this is going to turn into a
double integral over the domain

00:02:19.240 --> 00:02:21.720
of our parameters, which is
the region R. It's the region

00:02:21.720 --> 00:02:25.120
R in the xy plane right up here.

00:02:25.120 --> 00:02:27.610
And now we can take the dot
product of the curl of F

00:02:27.610 --> 00:02:31.062
with dotted ds, which is all of
this business right over here.

00:02:31.062 --> 00:02:32.770
And let me see if I
can show both of them

00:02:32.770 --> 00:02:34.330
on the screen at the same time.

00:02:34.330 --> 00:02:35.260
There you go.

00:02:35.260 --> 00:02:38.790
So first let's think
about the x components.

00:02:38.790 --> 00:02:41.260
So you have that
right over there.

00:02:41.260 --> 00:02:43.830
And then you have
this right over here.

00:02:43.830 --> 00:02:44.925
You multiply the two.

00:02:44.925 --> 00:02:47.091
The negative we can swap
the order right over there.

00:02:47.091 --> 00:02:51.225
You have the partial of z
with respect to x times--

00:02:51.225 --> 00:02:53.350
and we're going to swap
this order right over here.

00:02:53.350 --> 00:02:57.350
The partial of Q with respect
to z minus the partial of R

00:02:57.350 --> 00:02:59.150
with respect to y.

00:02:59.150 --> 00:03:05.780
Now let's think about
the j component.

00:03:05.780 --> 00:03:09.750
You have negative z sub y
times all of this up here,

00:03:09.750 --> 00:03:12.330
at least the stuff that's
multiplied times the j

00:03:12.330 --> 00:03:13.430
component.

00:03:13.430 --> 00:03:16.350
This negative can cancel out
with that negative, so you have

00:03:16.350 --> 00:03:20.800
plus z sub y, the partial
of z with respect to y,

00:03:20.800 --> 00:03:23.790
times the partial
of R with respect

00:03:23.790 --> 00:03:27.800
to x minus the partial
of P with respect to z.

00:03:27.800 --> 00:03:31.360
Let me make that clear, that
is an R right over here.

00:03:31.360 --> 00:03:33.265
And then we have
the k component.

00:03:33.265 --> 00:03:35.890
And the k component is actually
the easiest because it's just 1

00:03:35.890 --> 00:03:36.490
here.

00:03:36.490 --> 00:03:38.380
So it's just going
to be 1 times--

00:03:38.380 --> 00:03:41.720
and I'll just do it in
that same color-- 1 times

00:03:41.720 --> 00:03:46.390
the partial of Q with respect
to x minus the partial of P

00:03:46.390 --> 00:03:48.540
with respect to y.

00:03:48.540 --> 00:03:51.170
And then finally, we just
have this dA over here.

00:03:51.170 --> 00:03:53.690
And this dA is
multiplied by everything.

00:03:53.690 --> 00:03:56.840
So we'll put some parentheses,
and we'll write dA.

00:03:56.840 --> 00:03:57.790
So we're done.

00:03:57.790 --> 00:04:00.430
We've expressed our surface
integral as a double integral

00:04:00.430 --> 00:04:02.929
over the domain
of our parameters.

00:04:02.929 --> 00:04:04.970
And what we're going to
do in the next few videos

00:04:04.970 --> 00:04:07.178
is do the same thing with
this using Green's theorem.

00:04:07.178 --> 00:04:10.760
And we're going to see that
we get the exact same value.


Stokes' theorem proof part 2

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:03.090
Let's now parameterize
our surface.

00:00:03.090 --> 00:00:06.550
And then, we can figure out what
ds would actually look like.

00:00:06.550 --> 00:00:10.130
And so I will define my
position vector function

00:00:10.130 --> 00:00:11.422
for our surface as r.

00:00:11.422 --> 00:00:13.630
And I'm going to make it a
function of two parameters

00:00:13.630 --> 00:00:15.750
because we're going to have
to define a surface right over

00:00:15.750 --> 00:00:16.410
here.

00:00:16.410 --> 00:00:20.650
And I can actually use
x and y as my parameters

00:00:20.650 --> 00:00:22.580
because the surface
can be defined

00:00:22.580 --> 00:00:24.600
as a function of x and y.

00:00:24.600 --> 00:00:27.350
So I'll say that
my parameterization

00:00:27.350 --> 00:00:29.700
is going to be a
function of x and y.

00:00:29.700 --> 00:00:34.210
And in my i direction,
it's going to be x times i.

00:00:34.210 --> 00:00:37.340
In my j direction, it's
going to be y times j.

00:00:37.340 --> 00:00:39.250
And then, in the k
direction, well, that's

00:00:39.250 --> 00:00:40.370
just going to be z.

00:00:40.370 --> 00:00:42.180
And z is a function of x and y.

00:00:45.584 --> 00:00:47.750
And whenever you do a
parameterization of a surface,

00:00:47.750 --> 00:00:49.208
you have to think
about, well, what

00:00:49.208 --> 00:00:51.909
are the constraints on the
domain for your parameters?

00:00:51.909 --> 00:00:54.200
And so the constraints of
the domain for my parameters,

00:00:54.200 --> 00:00:58.310
I'm going to say that
every pair of xy-- every xy

00:00:58.310 --> 00:01:00.540
coordinate-- it has
to be a member-- this

00:01:00.540 --> 00:01:02.490
is the symbol for
member-- it has

00:01:02.490 --> 00:01:06.180
to be a member of this little
region right over here.

00:01:06.180 --> 00:01:10.280
We could call this the
domain of the parameters.

00:01:10.280 --> 00:01:11.685
It has to be a member of r.

00:01:11.685 --> 00:01:13.060
Actually, we assume
that up here.

00:01:13.060 --> 00:01:14.518
And actually, I
should have written

00:01:14.518 --> 00:01:17.580
this as the coordinates
of xy-- the xy pairs that

00:01:17.580 --> 00:01:20.810
are a member of r-- that's going
to help define our surface.

00:01:20.810 --> 00:01:23.130
An xy that is not a
member of r, then we're

00:01:23.130 --> 00:01:25.870
not going to consider
that the z of that

00:01:25.870 --> 00:01:27.470
xy as part of the surface.

00:01:27.470 --> 00:01:32.150
Only the z of xy's where the
xy's are part of this region.

00:01:32.150 --> 00:01:35.030
So now that we have
a parameterization

00:01:35.030 --> 00:01:37.290
for our surface,
we're ready to start

00:01:37.290 --> 00:01:41.509
thinking about what ds might be.

00:01:41.509 --> 00:01:43.800
And we need to think about
this a little bit carefully.

00:01:43.800 --> 00:01:45.330
So first, I'll just
write something.

00:01:45.330 --> 00:01:49.210
And then, we can confirm whether
this really will be the case.

00:01:49.210 --> 00:01:52.550
ds-- and we've seen
this before about why

00:01:52.550 --> 00:01:54.100
this is the case--
is going to be

00:01:54.100 --> 00:01:58.630
the cross product of the partial
derivative of this with respect

00:01:58.630 --> 00:02:01.680
to each of the parameters
times the little

00:02:01.680 --> 00:02:05.970
chunk of area in that domain.

00:02:05.970 --> 00:02:10.820
You could view it as the partial
of r of this parameterization

00:02:10.820 --> 00:02:17.540
with respect to x crossed with
the partial of r with respect

00:02:17.540 --> 00:02:18.057
to y.

00:02:18.057 --> 00:02:19.890
And then, that whole
thing-- and we actually

00:02:19.890 --> 00:02:21.320
want this to be a
vector, not just

00:02:21.320 --> 00:02:23.220
the absolute value or the
magnitude of this vector.

00:02:23.220 --> 00:02:24.920
We actually want
this to be a vector.

00:02:24.920 --> 00:02:28.810
That thing times-- we could
put in some order dx dy

00:02:28.810 --> 00:02:30.715
or we could write dy dx.

00:02:30.715 --> 00:02:32.090
And if we want to
be general just

00:02:32.090 --> 00:02:34.640
to say that it's a little
differential of our region

00:02:34.640 --> 00:02:36.340
right over here,
we can just write--

00:02:36.340 --> 00:02:42.650
instead of writing dx dy or
dy dx-- we will write da.

00:02:42.650 --> 00:02:44.810
And the reason why I said
we need to be careful

00:02:44.810 --> 00:02:47.470
is we need to make sure, based
on how we've parameterized

00:02:47.470 --> 00:02:52.030
this position vector function,
based on how we parameterized

00:02:52.030 --> 00:02:53.949
it, whether this cross
product really points

00:02:53.949 --> 00:02:55.490
in the right direction,
the direction

00:02:55.490 --> 00:02:57.450
that we need to be oriented in.

00:02:57.450 --> 00:03:00.150
Because, remember,
if we're traversing

00:03:00.150 --> 00:03:02.220
a boundary like
this, we want to make

00:03:02.220 --> 00:03:04.384
sure that the surface is
oriented the right way.

00:03:04.384 --> 00:03:05.800
And the way we
think about-- if we

00:03:05.800 --> 00:03:08.540
were to twist a cap like
this, the cap would move up.

00:03:08.540 --> 00:03:12.140
Or if you were to walk around
this boundary in this direction

00:03:12.140 --> 00:03:15.090
with the surface to your left,
your head would point up.

00:03:15.090 --> 00:03:18.680
And so we need to make sure
that this vector, which really

00:03:18.680 --> 00:03:20.720
defines the orientation
of the surface,

00:03:20.720 --> 00:03:26.120
is definitely going to be
pointed up or above the surface

00:03:26.120 --> 00:03:28.059
as opposed to going
below the surface.

00:03:28.059 --> 00:03:29.850
And so let's think
about that a little bit.

00:03:29.850 --> 00:03:33.110
The partial derivative
with respect to x-- well,

00:03:33.110 --> 00:03:36.240
as x gets bigger,
it's going to go

00:03:36.240 --> 00:03:40.660
in that direction
along the surface.

00:03:40.660 --> 00:03:43.410
And the partial with respect
to y, as y gets bigger,

00:03:43.410 --> 00:03:47.495
it's going to go in that
direction along the surface.

00:03:47.495 --> 00:03:50.830
If I take r cross y-- and we
could use the right-hand rule

00:03:50.830 --> 00:03:51.370
here.

00:03:51.370 --> 00:03:53.560
We put our index
finger in the direction

00:03:53.560 --> 00:03:55.960
of the first thing we're
taking the cross product of,

00:03:55.960 --> 00:03:59.350
our middle finger in the
direction of the second thing.

00:03:59.350 --> 00:04:00.180
So just like that.

00:04:00.180 --> 00:04:01.160
We bend our middle finger.

00:04:01.160 --> 00:04:02.710
We don't care what the
other fingers are doing.

00:04:02.710 --> 00:04:04.168
So I'll just draw
them right there.

00:04:04.168 --> 00:04:07.270
Then, the thumb will go in the
direction of the cross product.

00:04:07.270 --> 00:04:10.930
So in this case, the thumb
is going to point up.

00:04:10.930 --> 00:04:12.910
And that's exactly what
we wanted to happen.

00:04:12.910 --> 00:04:15.490
So this actually is
the right ordering.

00:04:15.490 --> 00:04:18.404
The partial with respect to y
cross the partial with respect

00:04:18.404 --> 00:04:20.070
to x actually would
not have been right.

00:04:20.070 --> 00:04:22.028
That would have given us
the other orientation.

00:04:22.028 --> 00:04:24.070
We would have done that
if this boundary actually

00:04:24.070 --> 00:04:25.560
went the other way around.

00:04:25.560 --> 00:04:28.010
But this is the right
orientation given the way

00:04:28.010 --> 00:04:30.520
that we are going to
traverse the boundary.

00:04:30.520 --> 00:04:32.080
Now, with that out
of the way, let's

00:04:32.080 --> 00:04:36.240
actually calculate
this cross product.

00:04:36.240 --> 00:04:38.040
So let me just rewrite it.

00:04:38.040 --> 00:04:40.430
So this is going to
be equal to-- well,

00:04:40.430 --> 00:04:42.410
I'll just focus on the
cross product right now.

00:04:45.400 --> 00:04:50.345
The partial of r with respect to
x crossed with the partial of r

00:04:50.345 --> 00:04:53.576
with respect to y is
going to be equal to--

00:04:53.576 --> 00:04:55.200
and we've done this
many times already.

00:04:55.200 --> 00:04:57.910
We'll just do it in
more general terms now.

00:04:57.910 --> 00:04:59.930
It's going to be equal
to the determinant

00:04:59.930 --> 00:05:03.400
of this matrix-- i, j, and k.

00:05:03.400 --> 00:05:04.900
Let me do those in
different colors.

00:05:04.900 --> 00:05:07.430
I think that's a helpful
way of thinking about it.

00:05:07.430 --> 00:05:12.510
So i, j, and k.

00:05:17.590 --> 00:05:20.660
Actually, I'll use
this magenta color.

00:05:20.660 --> 00:05:23.180
i, j, and k.

00:05:23.180 --> 00:05:26.290
Let me write this
line right over here.

00:05:30.107 --> 00:05:32.440
We want to write the different
components of the partial

00:05:32.440 --> 00:05:33.750
with respect to x.

00:05:33.750 --> 00:05:37.450
And so the partial of the i
component with respect to x

00:05:37.450 --> 00:05:40.120
is just going to be 1.

00:05:40.120 --> 00:05:42.310
The partial of the j
component with respect to x

00:05:42.310 --> 00:05:44.040
is going to be 0.

00:05:44.040 --> 00:05:45.870
And the partial of
this with respect to x,

00:05:45.870 --> 00:05:48.036
well, we can just write
that as a partial derivative

00:05:48.036 --> 00:05:51.780
of the function z
with respect to x.

00:05:51.780 --> 00:05:54.670
And so r sub x-- or the
partial with respect

00:05:54.670 --> 00:06:00.210
to x-- is the vector 1, i
plus 0, j plus z sub x, k.

00:06:00.210 --> 00:06:03.140
And we'll do the same thing
for this piece right over here.

00:06:03.140 --> 00:06:06.160
The partial with respect
to y-- it's i component,

00:06:06.160 --> 00:06:08.100
well, this is going to be 0.

00:06:08.100 --> 00:06:12.140
It's j component
is going to be 1.

00:06:12.140 --> 00:06:14.270
The partial of this
with respect to y is 1.

00:06:14.270 --> 00:06:18.120
And the partial of
this with respect to y

00:06:18.120 --> 00:06:20.930
is the partial of z
with respect to y.

00:06:20.930 --> 00:06:25.690
And actually, I forgot to
write k right over here

00:06:25.690 --> 00:06:27.480
in our parameterization.

00:06:27.480 --> 00:06:29.520
So now with all
of this set up, we

00:06:29.520 --> 00:06:32.170
are ready to figure out
what the cross product is.

00:06:32.170 --> 00:06:38.590
The cross product is going to
be equal to-- so our i component

00:06:38.590 --> 00:06:41.810
is going to be 0 times
the partial with respect

00:06:41.810 --> 00:06:47.870
to y minus 1 times the partial
of z with respect to x.

00:06:47.870 --> 00:06:52.420
So we get negative partial
of z with respect to x.

00:06:52.420 --> 00:06:54.570
And then, checkerboard pattern.

00:06:54.570 --> 00:06:59.730
We'll have minus j times--
so ignore that column, that

00:06:59.730 --> 00:07:02.750
row-- 1 times the partial
with respect to y.

00:07:02.750 --> 00:07:04.740
So that's z sub y--
the partial of z

00:07:04.740 --> 00:07:07.360
with respect to y--
minus 0 times this.

00:07:07.360 --> 00:07:09.400
So we're just left with
that right over there.

00:07:09.400 --> 00:07:13.470
And then, finally, we're
going to have plus k.

00:07:13.470 --> 00:07:17.070
And here we have 1
times 1 minus 0 times 0.

00:07:17.070 --> 00:07:18.720
So it's going to be k times 1.

00:07:18.720 --> 00:07:22.240
So we can just
write k over there.

00:07:22.240 --> 00:07:25.770
And so we can write,
the cross product really

00:07:25.770 --> 00:07:35.360
is just equal to negative z sub
x times i minus z sub y times

00:07:35.360 --> 00:07:38.930
j plus k.


Stokes' theorem proof part 1

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:02.760
- [Instructor] In this video,
I will attempt to prove,

00:00:02.760 --> 00:00:04.310
or actually this and
the next several videos,

00:00:04.310 --> 00:00:08.570
attempt to prove a special
case version of Stokes' theorem

00:00:08.570 --> 00:00:10.520
or essentially Stokes'
theorem for a special case.

00:00:10.520 --> 00:00:11.900
And I'm doing this because the proof

00:00:11.900 --> 00:00:12.890
will be a little bit simpler,

00:00:12.890 --> 00:00:15.870
but at the same time
it's pretty convincing.

00:00:15.870 --> 00:00:17.920
And the special case we're
going to assume is that

00:00:17.920 --> 00:00:20.690
the surface we're dealing
with is a function of x and y.

00:00:20.690 --> 00:00:23.400
So if you give me any particular x and y,

00:00:23.400 --> 00:00:26.500
it only determines one
point on that surface.

00:00:26.500 --> 00:00:28.440
So a surface like this would be the case.

00:00:28.440 --> 00:00:31.620
So it's kind of a mapping of
this region in the xy-plane

00:00:31.620 --> 00:00:33.060
into three dimensions.

00:00:33.060 --> 00:00:35.430
So for any xy, we can
figure out the height.

00:00:35.430 --> 00:00:38.170
So essentially z is going
to be a function of x and y,

00:00:38.170 --> 00:00:40.450
and we can get a point on the surface.

00:00:40.450 --> 00:00:42.480
So this proof would not apply to a surface

00:00:42.480 --> 00:00:44.850
that's like a sphere
or something like that,

00:00:44.850 --> 00:00:46.730
where any point on the xy-plane would,

00:00:46.730 --> 00:00:49.110
could actually determine
two points on our surface,

00:00:49.110 --> 00:00:51.540
but this is a pretty good start.

00:00:51.540 --> 00:00:53.680
The other thing that
we're going to assume,

00:00:53.680 --> 00:00:55.500
we're going to assume that z,

00:00:55.500 --> 00:00:57.100
which is essentially
a function of x and y,

00:00:57.100 --> 00:00:59.650
that this function of x and y is,

00:00:59.650 --> 00:01:02.960
has continuous second-order derivatives,

00:01:02.960 --> 00:01:05.040
so continuous,

00:01:05.040 --> 00:01:07.290
continuous second,

00:01:07.290 --> 00:01:08.900
second derivatives.

00:01:08.900 --> 00:01:11.040
And the reason why I'm
gonna make that assumption

00:01:11.040 --> 00:01:13.160
is it's going to help us
in our proof later on.

00:01:13.160 --> 00:01:15.220
It's going to allow us to say

00:01:15.220 --> 00:01:17.440
that the partial of our surface

00:01:17.440 --> 00:01:20.050
or the partial of z with respect to x,

00:01:20.050 --> 00:01:22.720
and then taking the derivative
of that with respect to y,

00:01:22.720 --> 00:01:24.100
is going to be the same

00:01:24.100 --> 00:01:26.400
as the partial of z with respect to y

00:01:26.400 --> 00:01:29.240
and then taking the derivative
of that with respect to x.

00:01:29.240 --> 00:01:31.120
And in order to be able
to make this statement,

00:01:31.120 --> 00:01:33.980
we have to assume that z or our,

00:01:33.980 --> 00:01:36.710
the z right over here, z
is a function of x and y,

00:01:36.710 --> 00:01:39.380
has continuous second-order derivatives.

00:01:39.380 --> 00:01:41.930
And over here, we've just
written our vector field F

00:01:41.930 --> 00:01:44.100
that we're going to deal
with when we're trying to

00:01:44.100 --> 00:01:46.330
play with Stokes'
theorem, and we'll assume

00:01:46.330 --> 00:01:49.610
that it has continuous
first-order derivatives.

00:01:49.610 --> 00:01:50.470
Now with that out of the way,

00:01:50.470 --> 00:01:52.790
let's think about what
Stokes' theorem tells us.

00:01:52.790 --> 00:01:56.070
And then we'll think about,
for this particular case,

00:01:56.070 --> 00:01:58.720
how we can write it out, and
then hopefully we will see

00:01:58.720 --> 00:02:01.660
that the two things are equal.

00:02:01.660 --> 00:02:02.970
So let me write it out.

00:02:02.970 --> 00:02:05.163
So Stokes' theorem tells us that if,

00:02:06.160 --> 00:02:07.453
that F,

00:02:07.453 --> 00:02:09.030
F dot dr

00:02:09.030 --> 00:02:11.070
over some path,

00:02:11.070 --> 00:02:12.480
and the path that we care about

00:02:12.480 --> 00:02:14.330
is essentially this path right over here.

00:02:14.330 --> 00:02:15.630
I'll do it in blue.

00:02:15.630 --> 00:02:18.290
It's this path right over here.

00:02:18.290 --> 00:02:19.820
This is the boundary.

00:02:19.820 --> 00:02:22.570
This is the boundary of our surface.

00:02:22.570 --> 00:02:25.120
So this is c right over here.

00:02:25.120 --> 00:02:28.180
Stokes' theorem tells us that
this should be the same thing,

00:02:28.180 --> 00:02:32.030
this should be equivalent
to the surface integral

00:02:32.030 --> 00:02:33.890
over our surface,

00:02:33.890 --> 00:02:37.430
over our surface of curl of F,

00:02:37.430 --> 00:02:40.150
curl of F

00:02:40.150 --> 00:02:42.320
dot ds,

00:02:42.320 --> 00:02:46.230
dot, dotted with the surface itself.

00:02:46.230 --> 00:02:48.020
And so in this video, I wanna focus,

00:02:48.020 --> 00:02:49.420
or probably this and the next video,

00:02:49.420 --> 00:02:52.060
I wanna focus on the second half.

00:02:52.060 --> 00:02:53.130
I wanna focus this.

00:02:53.130 --> 00:02:54.760
I wanna see how we can express this

00:02:54.760 --> 00:02:56.250
given the assumptions we've made.

00:02:56.250 --> 00:02:57.980
And then after that, we're gonna see

00:02:57.980 --> 00:03:01.000
how we can express this
given the same assumptions.

00:03:01.000 --> 00:03:03.000
And then hopefully we'll
find that we get them

00:03:03.000 --> 00:03:05.160
to be equal to each other.

00:03:05.160 --> 00:03:06.640
So let's just start figuring out

00:03:06.640 --> 00:03:08.970
what the curl of F is equal to.

00:03:08.970 --> 00:03:10.890
So the curl of F,

00:03:10.890 --> 00:03:13.680
the curl of F

00:03:13.680 --> 00:03:14.730
is equal to,

00:03:14.730 --> 00:03:16.840
you could view it as the del operator

00:03:16.840 --> 00:03:20.970
crossed with our vector
field F, which is equal to,

00:03:20.970 --> 00:03:22.790
can write our components.

00:03:22.790 --> 00:03:27.570
So i, I'll do 'em in
different colors, i, j,

00:03:27.570 --> 00:03:30.230
and k components,

00:03:30.230 --> 00:03:32.260
i, j, and k components,

00:03:32.260 --> 00:03:35.380
and then I need to write my del operator

00:03:36.420 --> 00:03:38.780
or my partial operators I
guess I should call 'em.

00:03:38.780 --> 00:03:41.330
So the partial with respect to x,

00:03:41.330 --> 00:03:43.840
the partial with respect to y,

00:03:43.840 --> 00:03:45.940
partial with respect to z,

00:03:45.940 --> 00:03:47.955
and then I have to write
i, j, and k components

00:03:47.955 --> 00:03:49.750
of my vector field F.

00:03:49.750 --> 00:03:53.490
And I will do that in green,
well, I'll do it in blue.

00:03:53.490 --> 00:03:56.610
And so we have P, which
is a function of x, y, z,

00:03:56.610 --> 00:03:58.890
Q, which is a function of x, y, and z,

00:03:58.890 --> 00:04:02.230
and then R, which is a
function of x, y, and z.

00:04:02.230 --> 00:04:04.900
And so this is going to evaluate as,

00:04:04.900 --> 00:04:07.320
it's going to be i,

00:04:07.320 --> 00:04:08.930
i times,

00:04:08.930 --> 00:04:11.140
so blank out that column, that row.

00:04:11.140 --> 00:04:14.310
It's gonna be the partial
of R with respect to y,

00:04:14.310 --> 00:04:16.560
partial of R with respect to y

00:04:16.560 --> 00:04:19.520
minus the partial of Q with respect to z,

00:04:19.520 --> 00:04:22.200
minus the partial of Q with respect to z.

00:04:22.200 --> 00:04:26.960
And then checkerboard pattern minus j,

00:04:26.960 --> 00:04:28.570
gonna make that hat a little bit better,

00:04:28.570 --> 00:04:33.570
minus j and then times the
partial of R with respect to x,

00:04:33.920 --> 00:04:35.920
partial of R with respect to x

00:04:35.920 --> 00:04:38.760
minus the partial of P with respect to z,

00:04:38.760 --> 00:04:41.480
minus the partial of P with respect to z.

00:04:41.480 --> 00:04:44.210
And then finally, plus k,

00:04:44.210 --> 00:04:48.380
plus k, and that's going to be
times the partial of x with,

00:04:48.380 --> 00:04:51.210
or, sorry, the partial
of Q with respect to x,

00:04:51.210 --> 00:04:53.490
partial of Q with respect to x

00:04:53.490 --> 00:04:56.910
minus the partial of P with respect to y,

00:04:56.910 --> 00:04:59.910
minus the partial of P with respect to y.

00:04:59.910 --> 00:05:02.270
So we figured out the curl of
F, and I'll leave you there.

00:05:02.270 --> 00:05:04.110
I'm trying to make shorter videos now.

00:05:04.110 --> 00:05:05.060
And so in the next video,

00:05:05.060 --> 00:05:07.240
we have to figure out how to express ds,

00:05:07.240 --> 00:05:11.150
and then we'll just evaluate
this, this entire operation.

00:05:11.150 --> 00:05:12.763
We'll take the dot product.


Evaluating line integral directly - part 2

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:02.840
All we have left to do is
evaluate this integral,

00:00:02.840 --> 00:00:04.990
which is really an
exercise in kind

00:00:04.990 --> 00:00:07.580
of evaluating trigonometric
integrals more than anything

00:00:07.580 --> 00:00:07.810
else.

00:00:07.810 --> 00:00:09.476
But it never hurts
to get that practice,

00:00:09.476 --> 00:00:11.220
so let's just do
it step by step.

00:00:11.220 --> 00:00:13.880
So we see sine cubed theta.

00:00:13.880 --> 00:00:17.150
No obvious way to directly
take the antiderivative of sine

00:00:17.150 --> 00:00:17.680
cubed theta.

00:00:17.680 --> 00:00:20.950
But if we had some mixtures
of sines and cosines there,

00:00:20.950 --> 00:00:23.950
then we could start essentially
doing U-substitution,

00:00:23.950 --> 00:00:26.460
which at this point you
probably can do in your head.

00:00:26.460 --> 00:00:28.793
So what we could do is we
could write this is a product.

00:00:28.793 --> 00:00:30.700
So we can write this
as the sine of theta--

00:00:30.700 --> 00:00:32.241
so I'll do this part
right over here.

00:00:32.241 --> 00:00:38.516
This is sine of theta
times sine squared theta.

00:00:38.516 --> 00:00:40.140
And sine squared
theta can be rewritten

00:00:40.140 --> 00:00:42.390
as 1 minus cosine squared theta.

00:00:42.390 --> 00:00:47.830
So this is the same thing
as sine of theta times

00:00:47.830 --> 00:00:51.080
1 minus cosine squared theta.

00:00:51.080 --> 00:00:52.530
And if we multiply
this out, this

00:00:52.530 --> 00:01:00.950
gives us sine theta minus sine
theta cosine squared theta.

00:01:00.950 --> 00:01:02.810
And this is much easier
for us to integrate,

00:01:02.810 --> 00:01:04.976
although it looks like a
more complicated expression

00:01:04.976 --> 00:01:07.360
because it's easy take the
antiderivative of sine theta.

00:01:07.360 --> 00:01:10.000
And now it's easy take
the antiderivative of this

00:01:10.000 --> 00:01:12.274
because we have the derivative
of cosine theta sitting

00:01:12.274 --> 00:01:12.940
right over here.

00:01:12.940 --> 00:01:15.050
So this is going to be
cosine cubed theta over 3.

00:01:15.050 --> 00:01:17.310
So, essentially, we're doing
U-substitution right over here.

00:01:17.310 --> 00:01:18.600
But I'll save that for second.

00:01:18.600 --> 00:01:20.540
Let's rewrite all of
these in a way that's

00:01:20.540 --> 00:01:22.850
easy to take the
antiderivative of it.

00:01:22.850 --> 00:01:26.480
Cosine squared theta, we know
this is a common trig identity.

00:01:26.480 --> 00:01:34.680
That's the same thing as 1/2
of 1 plus cosine of 2 theta.

00:01:34.680 --> 00:01:36.790
And once again, this
is a much, much easier

00:01:36.790 --> 00:01:38.170
to take the antiderivative of.

00:01:38.170 --> 00:01:46.800
So I'll write plus 1/2
plus 1/2 cosine of 2 theta.

00:01:46.800 --> 00:01:49.330
And now, all of this
is actually quite easy

00:01:49.330 --> 00:01:51.560
to take the antiderivative
of, and so I'll just

00:01:51.560 --> 00:01:52.600
rewrite it again.

00:01:52.600 --> 00:02:00.610
So minus 4 cosine theta plus
4 cosine theta sine theta

00:02:00.610 --> 00:02:06.240
minus cosine theta sine
squared theta d theta.

00:02:06.240 --> 00:02:07.900
I just was able to sneak it in.

00:02:07.900 --> 00:02:11.259
And so that's our integral
between 0 and 2 pi.

00:02:11.259 --> 00:02:12.800
So let's just take
the antiderivative

00:02:12.800 --> 00:02:13.790
in every one of these steps.

00:02:13.790 --> 00:02:14.810
It's starting to get
a little bit messy.

00:02:14.810 --> 00:02:16.500
I'll try to write a
little bit neater.

00:02:16.500 --> 00:02:18.850
The antiderivative
of sine of theta

00:02:18.850 --> 00:02:22.125
is negative cosine of theta.

00:02:24.229 --> 00:02:26.020
If you take the derivative
of cosine theta,

00:02:26.020 --> 00:02:27.050
you get negative sine theta.

00:02:27.050 --> 00:02:28.633
Then the negatives
cancel out, and you

00:02:28.633 --> 00:02:29.840
get that right over there.

00:02:29.840 --> 00:02:32.410
Then over here, we
have the derivative

00:02:32.410 --> 00:02:34.600
of cosine theta, which
is negative sine theta.

00:02:34.600 --> 00:02:36.230
So we can essentially
kind of treat

00:02:36.230 --> 00:02:38.580
this-- we could kind of
make the substitution that u

00:02:38.580 --> 00:02:39.280
is cosine theta.

00:02:39.280 --> 00:02:41.280
That's essentially what
we're doing in our head.

00:02:41.280 --> 00:02:43.220
So the antiderivative
of this is going

00:02:43.220 --> 00:02:50.620
to be equal to plus
cosine cubed theta over 3.

00:02:50.620 --> 00:02:54.400
And then the antiderivative
of 1/2 with respect to theta

00:02:54.400 --> 00:02:57.920
is just going to
be plus 1/2 theta.

00:02:57.920 --> 00:03:01.780
The antiderivative of
cosine 2 theta-- well,

00:03:01.780 --> 00:03:04.120
we want the derivative of
this thing sitting someplace.

00:03:04.120 --> 00:03:06.480
The derivative of this
thing over here is 2.

00:03:06.480 --> 00:03:10.090
So if we put a 2 here, we can't
just multiply by 2 arbitrarily.

00:03:10.090 --> 00:03:12.390
We'd have to multiply
and divide by 2.

00:03:12.390 --> 00:03:15.330
So we could put a 2 here,
and then we could also--

00:03:15.330 --> 00:03:17.320
but we'd also have
to divide by 2.

00:03:17.320 --> 00:03:19.104
So then that would become a 4.

00:03:19.104 --> 00:03:20.270
And we haven't changed this.

00:03:20.270 --> 00:03:22.470
Notice, this is now
2/4 cosine of 2 theta,

00:03:22.470 --> 00:03:24.859
the exact same thing as
1/2 half cosine 2 theta.

00:03:24.859 --> 00:03:27.400
But this is useful because now,
the way I've written it here,

00:03:27.400 --> 00:03:29.572
we have the derivative of
2 theta right over here.

00:03:29.572 --> 00:03:31.030
And so we can just
say, well, we'll

00:03:31.030 --> 00:03:32.680
just take the antiderivative
to this whole thing, which

00:03:32.680 --> 00:03:34.030
is going to be sine of 2 theta.

00:03:34.030 --> 00:03:37.730
But we still have the 1 so that
the antiderivative of this part

00:03:37.730 --> 00:03:40.302
right over here is
the sine of 2 theta.

00:03:40.302 --> 00:03:41.760
And then we have
the 1/4 out there.

00:03:41.760 --> 00:03:47.959
So plus 1/4 sine of 2 theta.

00:03:47.959 --> 00:03:49.750
And then the antiderivative
of cosine theta

00:03:49.750 --> 00:03:52.760
is just sine theta,
so minus 4 sine theta.

00:03:55.550 --> 00:03:57.400
Antiderivative of
this right over here,

00:03:57.400 --> 00:03:59.790
we can kind of pick whichever
way we want to do it.

00:03:59.790 --> 00:04:02.210
But we could say, well, the
derivative of sine theta

00:04:02.210 --> 00:04:03.210
is cosine theta.

00:04:03.210 --> 00:04:10.560
So this is going to be the same
thing as 4 sine squared theta

00:04:10.560 --> 00:04:12.799
over 2.

00:04:12.799 --> 00:04:15.090
Or instead of saying over 2,
instead of writing that 4,

00:04:15.090 --> 00:04:21.290
I'll just divide the 4 by
the 2, and I will get a 2.

00:04:21.290 --> 00:04:24.406
So let me erase that and
put a 2 right over here.

00:04:24.406 --> 00:04:25.780
And you can work
it out yourself.

00:04:25.780 --> 00:04:28.029
If you were to take the
derivative of this thing right

00:04:28.029 --> 00:04:30.105
over here, it'd be the
derivative of sine theta.

00:04:30.105 --> 00:04:31.480
If you'd just use
the chain rule,

00:04:31.480 --> 00:04:38.330
which would be cosine theta,
and then times 4 sine of theta.

00:04:38.330 --> 00:04:40.700
So that's exactly what
we have right over here.

00:04:40.700 --> 00:04:42.550
And then we have this last part.

00:04:42.550 --> 00:04:45.482
The derivative of sine
theta is cosine theta.

00:04:45.482 --> 00:04:47.440
And so once again, just
like we've done before,

00:04:47.440 --> 00:04:49.100
the antiderivative
of this whole thing

00:04:49.100 --> 00:04:55.580
is going to be negative
sine cubed of theta over 3.

00:04:55.580 --> 00:04:58.690
And we need to evaluate
this entire expression

00:04:58.690 --> 00:05:02.250
between 0 and 2 pi.

00:05:02.250 --> 00:05:03.820
So let's see how it evaluates.

00:05:03.820 --> 00:05:06.200
So first, let's evaluate
everything at 2 pi.

00:05:06.200 --> 00:05:10.590
So this evaluated at
2 pi is negative 1.

00:05:10.590 --> 00:05:15.080
This evaluated at 2 pi is 1/3.

00:05:15.080 --> 00:05:20.250
This evaluated at 2 pi
is just going to be pi.

00:05:20.250 --> 00:05:23.620
This evaluated at 2 pi is
0, because sine of 4 pi

00:05:23.620 --> 00:05:25.310
is going to be 0.

00:05:25.310 --> 00:05:28.094
This evaluated at 2
pi is going to be 0.

00:05:28.094 --> 00:05:29.760
This evaluated at 2
pi is going to be 0.

00:05:29.760 --> 00:05:31.593
And this evaluated at
2 pi is going to be 0.

00:05:31.593 --> 00:05:33.540
So that's a nice simplification.

00:05:33.540 --> 00:05:35.374
So that's everything
evaluated at 2 pi.

00:05:35.374 --> 00:05:36.790
And from that,
we're going to have

00:05:36.790 --> 00:05:40.300
to subtract everything
evaluated at 0.

00:05:40.300 --> 00:05:43.670
So cosine of 0, well, that's
going to be once again 1.

00:05:43.670 --> 00:05:45.940
And we have a negative
sign, so it's negative 1.

00:05:45.940 --> 00:05:47.967
Then you're going
to have plus 1/3.

00:05:47.967 --> 00:05:49.300
And then you're going to have 0.

00:05:49.300 --> 00:05:51.670
And then all these other
things are going to be 0.

00:05:51.670 --> 00:05:53.930
And so if you
simplify it, you get--

00:05:53.930 --> 00:06:00.400
this is going to be equal to
negative 1 plus 1/3 plus pi.

00:06:00.400 --> 00:06:06.110
And then we have plus
1 plus 1 minus 1/3.

00:06:06.110 --> 00:06:07.585
Well, that cancels with that.

00:06:07.585 --> 00:06:09.090
That cancels with that.

00:06:09.090 --> 00:06:11.150
And we deserve a drum roll now.

00:06:11.150 --> 00:06:12.857
It all simplified
just like when we

00:06:12.857 --> 00:06:14.690
use Stokes' Theorem in
like the four videos.

00:06:14.690 --> 00:06:15.860
Actually, I think it
was a little bit simpler

00:06:15.860 --> 00:06:18.440
to just directly evaluate
the line integral over here.

00:06:18.440 --> 00:06:22.002
We got it simplifying
to just pi.


Stokes example part 4: Curl and final answer

WEBVTT
Kind: captions
Language: en

00:00:00.068 --> 00:00:02.081
- [Instructor] We're
now in the home stretch.

00:00:02.081 --> 00:00:04.049
We just have to evaluate the curl of f

00:00:04.049 --> 00:00:05.116
and then this dot product

00:00:05.116 --> 00:00:07.252
and then evaluate this double integral.

00:00:07.252 --> 00:00:08.728
So let's work on the curl of F.

00:00:08.728 --> 00:00:12.145
So the curl of f is going to be equal to,

00:00:13.127 --> 00:00:15.710
and I just remember it as the determinant,

00:00:15.710 --> 00:00:18.543
so we have our i, j, k components,

00:00:19.667 --> 00:00:21.363
and it's really you could imagine

00:00:21.363 --> 00:00:26.119
it's the del operator crossed
with the actual vector.

00:00:26.119 --> 00:00:28.176
So the del operator, I'll
write this in a different color

00:00:28.176 --> 00:00:31.219
just to ease the monotony,
so this is partial

00:00:31.219 --> 00:00:35.084
with respect to x,
partial with respect to y,

00:00:35.084 --> 00:00:37.251
partial with respect to z,

00:00:38.573 --> 00:00:40.423
and then our vector field, I copied

00:00:40.423 --> 00:00:42.317
and pasted it right over here.

00:00:42.317 --> 00:00:47.317
It is just equal to negative
y squared, is our i component,

00:00:47.736 --> 00:00:51.393
x is our j component, and Z
squared is our k component.

00:00:51.393 --> 00:00:54.741
And so this is going to be equal to,

00:00:54.741 --> 00:00:57.208
this is going to be equal to i,

00:00:57.208 --> 00:01:00.449
is going to be equal to i times

00:01:00.449 --> 00:01:04.032
the partial of Z squared
with respect to y.

00:01:05.210 --> 00:01:07.750
Well, there's the Z
squared is just a constant

00:01:07.750 --> 00:01:09.472
with respect to y so
the partial of Z squared

00:01:09.472 --> 00:01:12.594
with respect to y is
just going to be zero,

00:01:12.594 --> 00:01:13.789
so this is going to be zero.

00:01:13.789 --> 00:01:17.578
Minus the partial of x with respect to z.

00:01:17.578 --> 00:01:19.179
Well, once again this is just a constant

00:01:19.179 --> 00:01:20.706
when you think in terms of z,

00:01:20.706 --> 00:01:22.676
so that's just going to be zero.

00:01:22.676 --> 00:01:24.683
So that's nice simplification,

00:01:24.683 --> 00:01:27.316
and then we're gonna have minus j,

00:01:27.316 --> 00:01:29.086
we need our little checkerboard patterns,

00:01:29.086 --> 00:01:30.437
we put a negative in front of the j,

00:01:30.437 --> 00:01:34.472
minus j and so we'll
have the partial of x,

00:01:34.472 --> 00:01:37.247
the partial of z squared
with respect to x,

00:01:37.247 --> 00:01:39.016
that's zero again, and then minus

00:01:39.016 --> 00:01:43.349
the partial of negative y
squared with respect to z,

00:01:44.439 --> 00:01:47.478
well that's zero again,
and then finally we have

00:01:47.478 --> 00:01:50.561
our k component, k, so plus, plus k,

00:01:53.857 --> 00:01:58.500
and k, we're gonna have the
partial of x with respect to x,

00:01:58.500 --> 00:01:59.777
well that actually gives us a value

00:01:59.777 --> 00:02:02.650
that's just gonna be one minus

00:02:02.650 --> 00:02:07.329
the partial of negative y
squared with respect to y.

00:02:07.329 --> 00:02:10.622
So the partial of negative y
squared with respect to y is

00:02:10.622 --> 00:02:13.221
negative two y and we're subtracting that,

00:02:13.221 --> 00:02:17.103
so it's going to be plus, plus, two y.

00:02:17.103 --> 00:02:19.325
So curl of f simplifies to just,

00:02:19.325 --> 00:02:21.376
all of this is just zero up here,

00:02:21.376 --> 00:02:25.836
is just one plus two y times
k or k times one plus two y.

00:02:25.836 --> 00:02:28.087
And so if we go back
to this right up here,

00:02:28.087 --> 00:02:31.754
if we go back up to
that, we're going to get

00:02:32.936 --> 00:02:36.419
let me re-write the
integral so zero to one

00:02:36.419 --> 00:02:39.285
and that's our r, our
r parameter is gonna go

00:02:39.285 --> 00:02:42.410
from zero to one, theta is
gonna go from zero to two pi.

00:02:42.410 --> 00:02:45.125
And now curl of f has simplified to,

00:02:45.125 --> 00:02:47.511
and I won't skip any steps
although it's tempting,

00:02:47.511 --> 00:02:51.184
it's one plus two y, and actually
instead of writing two y,

00:02:51.184 --> 00:02:53.239
let me write it in
terms of the parameters.

00:02:53.239 --> 00:02:56.406
We saw it up here, y was r sine theta,

00:02:57.315 --> 00:03:00.186
if I remember correctly,
right, y was r sine theta.

00:03:00.186 --> 00:03:01.994
So let me write y that way.

00:03:01.994 --> 00:03:04.077
Two times r sine theta k.

00:03:08.078 --> 00:03:10.010
And we're gonna dot this, we're gonna take

00:03:10.010 --> 00:03:14.177
the dot product of that
with this right over here,

00:03:15.057 --> 00:03:18.640
with r times j plus r
times k, d theto d r.

00:03:23.481 --> 00:03:24.314
And so we take the dot product,

00:03:24.314 --> 00:03:26.350
this thing only has a k component,

00:03:26.350 --> 00:03:29.062
the j component is zero, so
when you take the dot product

00:03:29.062 --> 00:03:31.484
with this j component
you're gonna get zero.

00:03:31.484 --> 00:03:34.436
And neither of them you actually
even have an i component.

00:03:34.436 --> 00:03:37.437
And so the inside is
just going to simplify

00:03:37.437 --> 00:03:41.012
to this piece right over
here is going to simplify

00:03:41.012 --> 00:03:43.362
to, we're just gonna have to
think about the k components,

00:03:43.362 --> 00:03:45.501
cause everything else is
zero, so it's gonna be

00:03:45.501 --> 00:03:49.244
r times this and we're
done! So it's gonna be

00:03:49.244 --> 00:03:53.994
r plus two r squared sine
theta, d theta d r, d theta d r

00:04:02.396 --> 00:04:06.512
and, once again, theta
goes from zero to two pi

00:04:06.512 --> 00:04:09.180
and r goes from zero to one.

00:04:09.180 --> 00:04:11.073
And now this is just a
straight-up double integral.

00:04:11.073 --> 00:04:12.839
We just have to evaluate this thing.

00:04:12.839 --> 00:04:15.388
And so, first we take the antiderivative

00:04:15.388 --> 00:04:18.475
with respect to theta,
so the antiderivative

00:04:18.475 --> 00:04:21.763
with respect to theta is going to give us,

00:04:21.763 --> 00:04:23.079
so this is going to be giving,

00:04:23.079 --> 00:04:24.966
so we're going to focus on theta first,

00:04:24.966 --> 00:04:27.429
so the antiderivative of r
with respect to theta is just

00:04:27.429 --> 00:04:30.600
r theta, you can just do r as a constant,

00:04:30.600 --> 00:04:32.533
and then the antiderivative of this,

00:04:32.533 --> 00:04:36.108
antiderivative of sine of theta
is negative cosine of theta.

00:04:36.108 --> 00:04:41.108
So this is gonna be negative
two r squared cosine of theta.

00:04:41.795 --> 00:04:44.743
And we're gonna evaluate
it from zero to two pi.

00:04:44.743 --> 00:04:46.552
And then we have the outside integral,

00:04:46.552 --> 00:04:51.552
which I will, I'll re-color
in yellow, re-color in yellow,

00:04:52.056 --> 00:04:54.439
so we'll still have to
integrate with respect to r

00:04:54.439 --> 00:04:56.656
and r's gonna go from zero to one.

00:04:56.656 --> 00:04:59.357
But inside right over
here, if we evaluate all

00:04:59.357 --> 00:05:01.579
of this business right
over here at two pi,

00:05:01.579 --> 00:05:06.246
we get two pi r, two pi r,
that's that right over there,

00:05:07.657 --> 00:05:10.782
minus... Cosine of two pi is just one.

00:05:10.782 --> 00:05:14.499
So it's minus two r
squared and then from that,

00:05:14.499 --> 00:05:17.990
we're going to subtract from that,

00:05:17.990 --> 00:05:21.367
we're gonna subtract this evaluated zero.

00:05:21.367 --> 00:05:22.965
Well r times zero is just zero,

00:05:22.965 --> 00:05:25.181
and then cosine of zero is one.

00:05:25.181 --> 00:05:26.864
So it's just minus two r squared,

00:05:26.864 --> 00:05:28.509
or negative two r squared,

00:05:28.509 --> 00:05:30.518
negative two r squared.

00:05:30.518 --> 00:05:32.456
And at this negative and this negative,

00:05:32.456 --> 00:05:34.011
you get a positive, and but then you have

00:05:34.011 --> 00:05:36.353
a negative two r squared and
then a plus two r squared

00:05:36.353 --> 00:05:38.075
it's just going to cancel out,

00:05:38.075 --> 00:05:39.638
that and that cancel out,

00:05:39.638 --> 00:05:43.170
and so this whole thing
has simplified quite nicely

00:05:43.170 --> 00:05:48.170
to a simple definite integral,
zero to one of two pi,

00:05:48.266 --> 00:05:51.849
two pi r dr, and the
antiderivative of this

00:05:53.412 --> 00:05:55.598
is just going to be pi r squared,

00:05:55.598 --> 00:05:58.805
so we're just gonna evaluate pi r squared

00:05:58.805 --> 00:06:01.467
from zero to one, when you
evaluate it at one, you get

00:06:01.467 --> 00:06:04.503
pi; when you evaluate it
at zero, you just get zero,

00:06:04.503 --> 00:06:06.925
so you get pi minus
zero, which is equal to,

00:06:06.925 --> 00:06:09.431
and now we deserve a drumroll
'cause we've been doing

00:06:09.431 --> 00:06:13.764
a lot of work over many
videos, this is equal to pi.

00:06:14.604 --> 00:06:16.412
So just to remind ourselves
what we've done over

00:06:16.412 --> 00:06:20.245
the last few videos, we
had this line integral

00:06:21.223 --> 00:06:22.249
that we were trying to figure out,

00:06:22.249 --> 00:06:24.220
and instead of directly
evaluating the line integral,

00:06:24.220 --> 00:06:26.318
which we could do and I
encourage you to do so,

00:06:26.318 --> 00:06:28.331
and if I have time, I might
do it in the next video,

00:06:28.331 --> 00:06:31.248
instead of directly
evaluating that line integral,

00:06:31.248 --> 00:06:32.771
we used Stokes theorem to say,

00:06:32.771 --> 00:06:34.955
oh we could actually instead
say that that's the same thing

00:06:34.955 --> 00:06:38.487
as a surface integral over
a piecewise-smooth boundary

00:06:38.487 --> 00:06:40.871
over piecewise-smooth surface that this

00:06:40.871 --> 00:06:44.852
path is the boundary of,
and so we evaluated this

00:06:44.852 --> 00:06:47.606
surface intergal and
eventually, with a good bit of,

00:06:47.606 --> 00:06:51.268
little bit of calculation,
we got to evaluating it to be

00:06:51.268 --> 00:06:52.351
equal to pi.


Stokes example part 3: Surface to double integral

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:03.770
We're now ready to get into the
meat of evaluating this surface

00:00:03.770 --> 00:00:04.270
integral.

00:00:04.270 --> 00:00:05.728
And we really need
to re-express it

00:00:05.728 --> 00:00:10.152
in terms of a double integral
in the domain of the parameters.

00:00:10.152 --> 00:00:11.610
And the first thing
I'm going to do

00:00:11.610 --> 00:00:16.290
is rewrite this part right
over here using our parameters.

00:00:16.290 --> 00:00:21.300
And we already know that n, our
normal vector times our surface

00:00:21.300 --> 00:00:23.320
differential, can
also be written

00:00:23.320 --> 00:00:25.640
as kind of a vector version
of our surface differential

00:00:25.640 --> 00:00:28.720
that points in the same
direction as our normal vector.

00:00:28.720 --> 00:00:30.591
And this is going to
be the same thing.

00:00:30.591 --> 00:00:32.090
And we need to make
sure that we get

00:00:32.090 --> 00:00:33.960
the order on the
cross product right

00:00:33.960 --> 00:00:35.420
as the partial
derivative-- and I'm

00:00:35.420 --> 00:00:36.836
going to confirm
this in a second.

00:00:39.396 --> 00:00:41.270
The partial derivative
of the parametrization

00:00:41.270 --> 00:00:43.170
with respect to one
of the parameters

00:00:43.170 --> 00:00:45.160
crossed with the
partial derivative

00:00:45.160 --> 00:00:48.142
of the parametrization with
respect to the other parameter.

00:00:48.142 --> 00:00:49.600
And then that whole
thing-- I'm not

00:00:49.600 --> 00:00:51.933
going to take the absolute
value because I need a vector

00:00:51.933 --> 00:00:53.780
right over here--
times differentials

00:00:53.780 --> 00:00:57.510
of the parameters-- d theta
dr. And we can swap these two

00:00:57.510 --> 00:00:59.580
things around depending
on what will make

00:00:59.580 --> 00:01:01.650
our eventual double
integral easier.

00:01:01.650 --> 00:01:03.430
But we can't swap
these two things around

00:01:03.430 --> 00:01:04.846
because this would
actually change

00:01:04.846 --> 00:01:06.110
the direction of the vector.

00:01:06.110 --> 00:01:08.720
So we need to make sure
that this is popping us out

00:01:08.720 --> 00:01:10.500
in the right direction.

00:01:10.500 --> 00:01:12.420
So let's think
about the direction

00:01:12.420 --> 00:01:14.970
that the partial with
respect to r will take us.

00:01:14.970 --> 00:01:17.910
So as r increases, we're
going to be moving radially

00:01:17.910 --> 00:01:21.450
outward from the
center of our surface.

00:01:21.450 --> 00:01:22.990
Let me do this in
a different color.

00:01:22.990 --> 00:01:26.270
As r increases, we'll be
moving radially outwards.

00:01:26.270 --> 00:01:29.550
So this quantity
will be a vector

00:01:29.550 --> 00:01:31.230
that looks something like that.

00:01:31.230 --> 00:01:33.380
And then as theta
increases, we'll

00:01:33.380 --> 00:01:37.745
be going roughly
in that direction.

00:01:37.745 --> 00:01:40.120
And so if we take the cross
product of those two things--

00:01:40.120 --> 00:01:42.040
and we could use
the right-hand rule.

00:01:42.040 --> 00:01:44.420
Take your right hand,
point your index finger

00:01:44.420 --> 00:01:46.190
in the direction of
that yellow vector.

00:01:46.190 --> 00:01:48.440
Let me make it clear-- this
is the orange vector right

00:01:48.440 --> 00:01:48.941
over here.

00:01:48.941 --> 00:01:50.565
Take your index finger
in the direction

00:01:50.565 --> 00:01:51.570
of that yellow vector.

00:01:51.570 --> 00:01:54.260
So that's my index finger,
my shakily drawn yellow index

00:01:54.260 --> 00:01:54.910
finger.

00:01:54.910 --> 00:01:57.580
Put your middle finger in the
direction of the orange vector.

00:01:57.580 --> 00:01:59.080
So my middle, you
bend it and put it

00:01:59.080 --> 00:02:00.500
in the direction of
the orange vector.

00:02:00.500 --> 00:02:02.500
And we don't care what
the other two fingers do.

00:02:02.500 --> 00:02:04.710
And then, your thumb
will be in the direction

00:02:04.710 --> 00:02:05.990
of the cross product.

00:02:05.990 --> 00:02:09.680
So your thumb will
point outward like that.

00:02:09.680 --> 00:02:11.402
That's my best
attempt at drawing it.

00:02:11.402 --> 00:02:13.610
Which is exactly the direction
we needed to point it.

00:02:13.610 --> 00:02:15.276
We need it to point
upward here in order

00:02:15.276 --> 00:02:17.970
to be oriented properly
with the direction

00:02:17.970 --> 00:02:19.690
that we are actually
traversing the path.

00:02:19.690 --> 00:02:21.256
So this is actually
the right order.

00:02:21.256 --> 00:02:23.130
If when we did this we
got the thumb pointing

00:02:23.130 --> 00:02:25.910
into it or below
the plane, then we

00:02:25.910 --> 00:02:28.010
would actually have
to swap these orders.

00:02:28.010 --> 00:02:29.426
So with that out
of the way, let's

00:02:29.426 --> 00:02:31.280
actually evaluate
this cross product.

00:02:34.260 --> 00:02:36.970
So the cross product of the
partial of our parametrization

00:02:36.970 --> 00:02:39.820
with respect to r
crossed with the partial

00:02:39.820 --> 00:02:41.850
of our parametrization
with respect to theta.

00:02:41.850 --> 00:02:45.280
I like to set up the matrix
to take the cross product.

00:02:45.280 --> 00:02:51.850
So we'll put i, j, and k
components, < like that.

00:02:51.850 --> 00:02:56.510
And then, first, I will write
the partial with respect to r.

00:02:56.510 --> 00:02:58.760
So the i-component, if you
take the derivative of this

00:02:58.760 --> 00:03:02.670
with respect to r, is just
going to be cosine theta.

00:03:02.670 --> 00:03:04.350
The derivative of
this with respect to r

00:03:04.350 --> 00:03:07.760
is just sine theta.

00:03:07.760 --> 00:03:09.980
And the derivative of
this with respect to are r

00:03:09.980 --> 00:03:12.400
is just going to be a
negative sine theta.

00:03:17.659 --> 00:03:19.450
And then we're going
to cross it with this,

00:03:19.450 --> 00:03:22.690
so the derivative of this
with respect to theta

00:03:22.690 --> 00:03:25.310
is going to be negative
r sine of theta.

00:03:30.426 --> 00:03:31.800
The derivative of
our j-component

00:03:31.800 --> 00:03:34.305
with respect to theta
will be r cosine theta.

00:03:37.630 --> 00:03:40.065
And the derivative
or our k-component--

00:03:40.065 --> 00:03:42.670
or our z-component
with respect to theta--

00:03:42.670 --> 00:03:46.245
is going to be-- let's see--
negative r cosine theta.

00:03:50.280 --> 00:03:50.880
Is that right?

00:03:50.880 --> 00:03:51.904
Yeah.

00:03:51.904 --> 00:03:53.570
Derivative of sine
theta's cosine theta.

00:03:53.570 --> 00:03:54.070
Yeah.

00:03:54.070 --> 00:03:56.370
So it's negative r cosine theta.

00:03:56.370 --> 00:04:00.310
And now we just evaluate
this determinant over here.

00:04:00.310 --> 00:04:03.710
Our i-component is
going to be-- so

00:04:03.710 --> 00:04:06.290
ignore that row and this column.

00:04:06.290 --> 00:04:11.740
And we get sine times
negative r cosine theta.

00:04:11.740 --> 00:04:13.290
I'll just do this
in a new color.

00:04:13.290 --> 00:04:16.380
So we're going to get negative
r-- that wasn't a new color.

00:04:16.380 --> 00:04:18.320
I'll do it in purple.

00:04:18.320 --> 00:04:27.877
We will get negative r cosine
theta sine theta minus-- well,

00:04:27.877 --> 00:04:29.460
this is going to be
a negative number.

00:04:29.460 --> 00:04:32.043
So when you subtract a negative
number, it's going to be plus.

00:04:32.043 --> 00:04:35.870
So it's going to be plus
r cosine theta sine theta.

00:04:39.130 --> 00:04:41.385
It's always nice when
things cancel out like this.

00:04:41.385 --> 00:04:42.976
This plus this is just 0.

00:04:42.976 --> 00:04:44.600
Negative of it plus
the positive of it.

00:04:44.600 --> 00:04:46.160
So that all canceled out to 0.

00:04:46.160 --> 00:04:47.960
We don't have an i-component.

00:04:47.960 --> 00:04:49.391
Now, let's go to
the j-component.

00:04:49.391 --> 00:04:51.890
And remember, we need to do our
little checkerboard pattern.

00:04:51.890 --> 00:04:54.080
So it's going to be minus j.

00:04:54.080 --> 00:04:57.280
And it's going to be-- ignore
this column, that row-- cosine

00:04:57.280 --> 00:04:58.710
theta times negative r.

00:04:58.710 --> 00:05:04.540
Cosine theta is negative
r cosine squared theta.

00:05:04.540 --> 00:05:06.240
I just multiplied those two.

00:05:06.240 --> 00:05:13.559
And then, from that I'm going
to subtract this times that.

00:05:13.559 --> 00:05:15.600
And so this times that,
the negatives cancel out.

00:05:15.600 --> 00:05:18.970
We get r sine squared theta.

00:05:18.970 --> 00:05:20.740
So this is minus--
let me make sure.

00:05:20.740 --> 00:05:22.948
Yeah, I'm going to subtract
the product of these two.

00:05:22.948 --> 00:05:24.370
The product is positive.

00:05:24.370 --> 00:05:27.276
It's going to be r
sine squared theta.

00:05:27.276 --> 00:05:28.400
This is always a hard part.

00:05:28.400 --> 00:05:30.520
You can make a lot of
careless mistakes here.

00:05:30.520 --> 00:05:32.520
And this looks like we
might be able to simplify

00:05:32.520 --> 00:05:34.020
this in a second, but I'll wait.

00:05:34.020 --> 00:05:35.690
Actually, I'll just distribute
this negative sign just

00:05:35.690 --> 00:05:36.190
for fun.

00:05:36.190 --> 00:05:37.540
So if we distribute
the negative sign,

00:05:37.540 --> 00:05:38.730
these all become positive.

00:05:38.730 --> 00:05:40.680
Helps simplify
things a little bit.

00:05:40.680 --> 00:05:42.990
And now let's worry
about the k-component.

00:05:42.990 --> 00:05:45.430
The k-component I will be
doing in purple-- well,

00:05:45.430 --> 00:05:46.520
I'll do it in blue.

00:05:46.520 --> 00:05:47.340
The k-component.

00:05:47.340 --> 00:05:48.840
Ignore this row,
ignore this column.

00:05:48.840 --> 00:05:54.120
So plus k times cosine
theta times r cosine theta

00:05:54.120 --> 00:05:57.960
is r cosine squared of theta.

00:05:57.960 --> 00:06:02.120
And then from that, I'm going
to subtract negative r sine

00:06:02.120 --> 00:06:03.830
theta times sine theta.

00:06:03.830 --> 00:06:06.102
So that's going to be
negative r sine squared theta,

00:06:06.102 --> 00:06:07.060
but I'm subtracting it.

00:06:07.060 --> 00:06:11.260
So it's going to be plus
r sine squared theta.

00:06:11.260 --> 00:06:13.650
And this looks like we're
going to simplify it as well.

00:06:13.650 --> 00:06:15.130
And so this piece
right over here,

00:06:15.130 --> 00:06:17.200
we can factor out--
let me just rewrite it.

00:06:17.200 --> 00:06:23.500
This can be rewritten as r
times cosine squared theta

00:06:23.500 --> 00:06:26.600
plus sine squared theta.

00:06:26.600 --> 00:06:28.130
Basic trig identity.

00:06:28.130 --> 00:06:29.500
That just evaluates to 1.

00:06:29.500 --> 00:06:31.470
So this is just r times j.

00:06:31.470 --> 00:06:35.000
And this over here simplifies
for the exact same reason.

00:06:35.000 --> 00:06:43.410
This is r times cosine squared
theta plus sine squared theta.

00:06:43.410 --> 00:06:45.370
This also is just 1.

00:06:45.370 --> 00:06:49.400
So this just simplifies
to r times k.

00:06:49.400 --> 00:06:51.850
This whole cross product--
all of this business right

00:06:51.850 --> 00:06:54.610
over here-- simplified
to, quite luckily,

00:06:54.610 --> 00:06:57.560
is equal to r times
our j unit vector

00:06:57.560 --> 00:07:03.480
plus r times our k unit vector.

00:07:03.480 --> 00:07:05.810
And so now we can write
our surface integral.

00:07:05.810 --> 00:07:07.350
Our original surface
integral we can

00:07:07.350 --> 00:07:09.342
write as the double integral.

00:07:09.342 --> 00:07:11.800
Or we might want to change the
order in which we integrate,

00:07:11.800 --> 00:07:14.133
but we'll give ourselves that
option a little bit later.

00:07:14.133 --> 00:07:15.250
Double integral.

00:07:15.250 --> 00:07:18.960
Now, it's going to be over our
parameter domain or the r theta

00:07:18.960 --> 00:07:20.010
domain.

00:07:20.010 --> 00:07:23.600
So it's the double integral of--
we still have the curl of f.

00:07:23.600 --> 00:07:25.810
And we're going to have
to evaluate the curl of f.

00:07:25.810 --> 00:07:30.150
So I'll just write "curl
of our vector field

00:07:30.150 --> 00:07:40.816
f" dotted with this
business, rj plus rk.

00:07:40.816 --> 00:07:42.315
And then we have
our two parameters.

00:07:42.315 --> 00:07:43.910
And we might want
to switch the order.

00:07:43.910 --> 00:07:47.352
So maybe we could
write d theta dr.

00:07:47.352 --> 00:07:48.810
And then if we do
it in this order,

00:07:48.810 --> 00:07:52.742
theta goes from 0 to 2 pi,
and r is going from 0 to 1.

00:07:52.742 --> 00:07:54.450
But if we swap these
two, then obviously,

00:07:54.450 --> 00:07:56.410
we're going to have to
swap these two as well.

00:07:56.410 --> 00:07:57.284
I'll leave you there.

00:07:57.284 --> 00:07:59.622
In the next video, we will
evaluate the curl of f.

00:07:59.622 --> 00:08:01.330
And maybe in that
video, if we have time,

00:08:01.330 --> 00:08:04.340
we will get to the finish line.


Stokes example part 2: Parameterizing the surface

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:02.410
Now that we've set up
our surface integral,

00:00:02.410 --> 00:00:05.760
we can attempt to
parametrise the surface.

00:00:05.760 --> 00:00:07.970
And one way to think about
is we want our x and y

00:00:07.970 --> 00:00:10.930
values to take on all of the
values inside of the unit

00:00:10.930 --> 00:00:13.220
circle, what I'm shading
in right over here.

00:00:13.220 --> 00:00:15.920
And that our z values can be
a function of the y values.

00:00:15.920 --> 00:00:17.870
We can express this
equation right here,

00:00:17.870 --> 00:00:20.180
z is equal to 2 minus y.

00:00:20.180 --> 00:00:23.220
And then we could figure
out how high above to go

00:00:23.220 --> 00:00:24.180
to get our z value.

00:00:24.180 --> 00:00:27.080
And by doing that, we'll
be able to essentially get

00:00:27.080 --> 00:00:30.594
to every point that
sits on our surface.

00:00:30.594 --> 00:00:33.010
And so first let's think about
how we can get every x in y

00:00:33.010 --> 00:00:36.030
value inside of the unit circle.

00:00:36.030 --> 00:00:38.090
So let's just focus
on the xy plane.

00:00:38.090 --> 00:00:39.840
We're kind of rotated
around a little bit,

00:00:39.840 --> 00:00:41.800
so it looks a little
bit more traditional.

00:00:41.800 --> 00:00:48.620
So this is my x-axis and then
my y-axis would look something

00:00:48.620 --> 00:00:49.387
like that.

00:00:49.387 --> 00:00:50.970
Let me draw it a
little bit different.

00:00:50.970 --> 00:00:54.132
This is my y-axis.

00:00:54.132 --> 00:00:55.590
And then if I were
to draw the unit

00:00:55.590 --> 00:00:58.110
circle, some kind of the base
of this thing, or at least

00:00:58.110 --> 00:01:00.670
where it intersects
the xy plane-- actually

00:01:00.670 --> 00:01:02.220
this thing would
keep going down,

00:01:02.220 --> 00:01:03.790
if I wanted to
draw the x squared

00:01:03.790 --> 00:01:04.790
plus y squared equals 1.

00:01:04.790 --> 00:01:08.010
But if I draw where it
intersects the xy plane,

00:01:08.010 --> 00:01:10.050
we get the unit circle.

00:01:10.050 --> 00:01:12.670
So let me just draw it.

00:01:12.670 --> 00:01:14.734
That's my best attempt
at drawing a unit circle.

00:01:14.734 --> 00:01:16.150
We get the unit
circle and we need

00:01:16.150 --> 00:01:18.090
to think of using
parameters so that we

00:01:18.090 --> 00:01:20.450
can get every x and
y-coordinate that's inside

00:01:20.450 --> 00:01:21.502
of the unit circle.

00:01:21.502 --> 00:01:23.710
And to think about that,
I'll introduce one parameter

00:01:23.710 --> 00:01:26.550
that's essentially the
angle with the x-axis.

00:01:26.550 --> 00:01:28.970
And I'll call that
parameter theta.

00:01:28.970 --> 00:01:31.870
So theta is the angle
with the x-axis.

00:01:31.870 --> 00:01:34.150
And so theta will
essentially sweep things

00:01:34.150 --> 00:01:35.150
all the way around.

00:01:35.150 --> 00:01:38.880
So theta can go
between 0 and 2 pi.

00:01:38.880 --> 00:01:41.850
So theta will take on
values between 0 and 2 pi.

00:01:41.850 --> 00:01:44.960
And if we just fix the
radius at some point,

00:01:44.960 --> 00:01:46.850
say radius 1, that
would only give us

00:01:46.850 --> 00:01:48.440
all of the points
on the unit circle.

00:01:48.440 --> 00:01:50.350
But we want all the
points inside of it too.

00:01:50.350 --> 00:01:52.730
So we need to vary
the radius as well.

00:01:52.730 --> 00:01:54.650
So let's introduce
another parameter,

00:01:54.650 --> 00:01:57.920
let's call it r,
that is the radius.

00:01:57.920 --> 00:02:01.280
So for any given r, if
we keep changing theta,

00:02:01.280 --> 00:02:04.299
we would essentially sweep
out a circle of that radius.

00:02:04.299 --> 00:02:06.090
And if you change radius
a little bit more,

00:02:06.090 --> 00:02:07.423
you'll sweep out another circle.

00:02:07.423 --> 00:02:10.144
And if you vary radius
between 0 and 1,

00:02:10.144 --> 00:02:11.560
you'll get all of
the circles that

00:02:11.560 --> 00:02:13.840
will fill out this entire area.

00:02:13.840 --> 00:02:17.090
So the radius is going
to go between 0 and 1.

00:02:17.090 --> 00:02:21.590
Another way of thinking about
it is for any given theta,

00:02:21.590 --> 00:02:23.530
if you keep varying
the radius, you'll

00:02:23.530 --> 00:02:25.730
sweep out all of the
points on this line.

00:02:25.730 --> 00:02:27.460
And then as you
change theta, it'll

00:02:27.460 --> 00:02:29.090
sweep out the entire circle.

00:02:29.090 --> 00:02:30.570
So either way you
think about it.

00:02:30.570 --> 00:02:35.430
So with that, let's actually
define x and y in those terms.

00:02:35.430 --> 00:02:41.870
So we could say that x is equal
to-- so the x value whatever

00:02:41.870 --> 00:02:46.709
r is, the x value is going
to be r cosine theta.

00:02:46.709 --> 00:02:48.250
It's going to be
that component, it's

00:02:48.250 --> 00:02:51.230
going to be r cosine theta.

00:02:51.230 --> 00:02:53.900
And then the y component-- this
is just basic trigonometry-- is

00:02:53.900 --> 00:02:56.870
going to be r sine theta.

00:03:00.190 --> 00:03:01.790
And then the z
component, we already

00:03:01.790 --> 00:03:04.200
said z can be expressed
as a function of y.

00:03:04.200 --> 00:03:07.740
Right over here we can
rewrite this as z is equal to

00:03:07.740 --> 00:03:08.990
2 minus y.

00:03:08.990 --> 00:03:12.920
That'll tell us how high to
go so we end up on that plane.

00:03:12.920 --> 00:03:16.480
So if z is equal to 2 minus
y and if y is r sine theta,

00:03:16.480 --> 00:03:25.366
we can rewrite z as being
equal to 2 minus r sine theta.

00:03:25.366 --> 00:03:26.240
So there, we're done.

00:03:26.240 --> 00:03:27.614
That's our
parametrization, if we

00:03:27.614 --> 00:03:32.550
wanted to write this
as a position vector

00:03:32.550 --> 00:03:35.500
with two parameters--
I'll call it lowercase s,

00:03:35.500 --> 00:03:36.840
it's already used r.

00:03:36.840 --> 00:03:39.100
Lowercase s, this
is our surface,

00:03:39.100 --> 00:03:44.240
and it's going to be
parametrized with r and theta.

00:03:44.240 --> 00:03:51.920
We can write it as r cosine
theta i plus r sine of theta j

00:03:51.920 --> 00:04:01.560
plus r plus 2 minus
r sine of theta k.


Stokes example part 1

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:02.800
Let's now attempt to apply Stokes' theorem

00:00:00.000 --> 00:00:04.466
And so over here we have this little diagram, and we have this path

00:00:00.000 --> 00:00:06.133
that we're calling C, and it's the intersection of the plain Y+Z=2, so that's the plain that

00:00:00.000 --> 00:00:04.333
kind of slants down like that, its the intersection of that plain

00:00:00.000 --> 00:00:04.133
and the cylinder, you know I shouldn't even call it a cylinder

00:00:00.000 --> 00:00:06.066
because if you just have x^2 plus y^2 is equal to one, it would essentially be like a pole,

00:00:00.000 --> 00:00:04.866
an infinite pole that keeps going up forever and keeps going down forever

00:00:00.000 --> 00:00:02.666
so it would never have a top or a bottom

00:00:00.000 --> 00:00:05.333
but we slice that pole, with y plus z is equal to 2, to get where they intersect

00:00:00.000 --> 00:00:04.600
we get our path C. We also have this vector field defined in this way


Conditions for stokes theorem

WEBVTT
Kind: captions
Language: en

00:00:01.014 --> 00:00:04.350
Now that we've explored Stokes' Theorem a little bit, I want to talk about

00:00:04.350 --> 00:00:08.448
the situations in wich we can use it. You'll see that this is pretty general

00:00:08.448 --> 00:00:12.538
theorem. But we do have to thing about what type of surfaces

00:00:12.538 --> 00:00:16.602
and what type of boundary are those surfaces we are actually dealing with

00:00:16.602 --> 00:00:19.834
and the case of Stokes', we need

00:00:19.834 --> 00:00:23.017
surfaces that are piecewise...

00:00:23.017 --> 00:00:25.593
piecewise-smooth

00:00:26.608 --> 00:00:31.372
piecewise-smooth surfaces so this surface right over here

00:00:31.372 --> 00:00:34.644
it is actually smooth not just even piecewise-smooth.

00:00:34.644 --> 00:00:39.856
Sounds like a very fancy term but all the smooth part means

00:00:39.856 --> 00:00:42.332
that you have just continuous derivatives

00:00:42.332 --> 00:00:45.334
and since we are talking about surfaces we're going to have continuous

00:00:45.334 --> 00:00:48.549
partial derivatives regardless of which direction you pick.

00:00:48.549 --> 00:00:51.510
So this is continuous derivatives,

00:00:51.618 --> 00:00:56.114
and another way to think about that conceptually is if you pick a direction

00:00:56.114 --> 00:00:59.347
on the surface if you say that we go in that direction, the slope in that

00:00:59.362 --> 00:01:04.853
direction changes gradually, doesn't jump around. If you pick this direction right over here,

00:01:05.115 --> 00:01:08.957
the slope is changing gradually. So we have a continuous

00:01:08.957 --> 00:01:12.029
derivative. And you're like "what does the 'piecewise' means?"

00:01:12.029 --> 00:01:14.779
Well, the piecewise actually allow us to use Stokes' Theorem with more surfaces.

00:01:14.872 --> 00:01:19.966
Because if you have a surface that looks like... Let's say a surface that looks like

00:01:19.966 --> 00:01:24.219
this. Let's say looks like a cup. So this is the opening of the top of the cup

00:01:24.219 --> 00:01:27.787
let's say that has no opening on top so we can see the backside of the cup

00:01:27.818 --> 00:01:31.319
and this is the side of the cup and this right over here is the bottom

00:01:31.319 --> 00:01:34.665
of the cup and if it was transparent we could actually see through it. So

00:01:34.665 --> 00:01:38.892
surfaces like this is not entirely smooth because it has

00:01:38.892 --> 00:01:41.644
edges. There are points right over here. So this edge

00:01:41.644 --> 00:01:46.114
right over here... If we pick this... let's say if we pick this direction to go and if we go

00:01:46.114 --> 00:01:50.907
this direction along the bottom, then right we get to the edge, all of the sudden the slope changes dramatically

00:01:50.907 --> 00:01:54.808
jumps. So the slope is not continuous at that edge. The slope jumps

00:01:54.808 --> 00:01:58.995
and we start going straight up. And so this entire

00:01:58.995 --> 00:02:02.760
surface is not smooth. But the piecewise actually give us

00:02:02.760 --> 00:02:06.296
an out. This tell us that it's okay as long as we can break

00:02:06.296 --> 00:02:09.655
the surfaces up into pieces that are smooth. And so

00:02:09.655 --> 00:02:13.780
this cup we can break it up but we were doing this wen tackling surfaces integrals

00:02:13.780 --> 00:02:16.411
we can break it up into the bottom part, which is a smooth

00:02:16.411 --> 00:02:22.997
surface, it has continuous derivative, and the sides

00:02:22.997 --> 00:02:24.788
which kind of wraps around is also

00:02:24.788 --> 00:02:29.292
is also a smooth surface

00:02:29.292 --> 00:02:32.610
so most things you'll encounter in a traditional calculus course

00:02:32.610 --> 00:02:36.788
actually do, especially surfaces, do fit piecewise-smooth. And the thing is

00:02:36.788 --> 00:02:41.079
though actually very hard to visualize. I imagine this all outer pointy

00:02:41.079 --> 00:02:44.877
fractely looking things where it's hard to break it up into pieces that are

00:02:44.877 --> 00:02:49.127
actually smooth. That's for surface part but we also

00:02:49.127 --> 00:02:53.138
have to care about the boundary, in order to apply Stokes' Theorem. And that is that

00:02:53.138 --> 00:02:57.296
right over there. The boundary needs to be a simple,

00:02:57.296 --> 00:03:02.892
which means that doesn't cross itself, a simple closed

00:03:02.892 --> 00:03:04.843
piecewise-smooth boundary. So once

00:03:04.843 --> 00:03:08.461
again: simple and closed that just means so this is not a

00:03:08.461 --> 00:03:12.824
simple boundary. If it is really crossing itself or intersecting itself, although you can break it up

00:03:12.824 --> 00:03:16.629
into to tow simple boundaries. But something like this is a simple

00:03:16.629 --> 00:03:20.596
boundary. So that is a simple boundary right over there. It also have to be closed

00:03:20.596 --> 00:03:25.996
wich really means that just loops in on itself. You just have something like that. It actually has to

00:03:25.996 --> 00:03:31.111
close and actually has to loops in on itself. In order to use Stokes' Theorem and once again

00:03:31.111 --> 00:03:34.421
it has to be piecewise-smooth but now we are talking about a path or a

00:03:34.421 --> 00:03:39.039
line or curve like this and a piecewise-smooth just means that you can break it up

00:03:39.039 --> 00:03:42.477
into sections were derivatives are continuous. The way I've drawm

00:03:42.477 --> 00:03:44.878
this one, this one and this one, the slope is changing gradually. So over there

00:03:44.878 --> 00:03:50.183
the slope is like that. It is changing gradually as we go around

00:03:50.183 --> 00:03:53.798
this path. Something that is not smooth, a path that is not

00:03:53.798 --> 00:03:56.176
smooth might look something like this.

00:03:56.176 --> 00:04:01.566
Might look something like that. And the places that this aren't smooth are

00:04:01.566 --> 00:04:06.171
at the edges: not smooth there, not smooth there and not smooth there.

00:04:06.171 --> 00:04:09.634
But we have to be simple-closed and this is simple and closed. And it's not

00:04:09.634 --> 00:04:12.496
smooth but it is piecewise-smooth. We can break it up into

00:04:12.496 --> 00:04:17.445
this section of the path. Which is that line right over there is smooth,

00:04:17.445 --> 00:04:21.532
that line over there is smooth, that line is smooth, and that line is

00:04:21.532 --> 00:04:24.506
smooth. And we've done that when evaluating in line integrals. We broke it up into

00:04:24.506 --> 00:04:28.174
smooth segments that we can then use to actually compute line

00:04:28.174 --> 00:04:33.328
integral. So if you find... if you have a boundary where the...

00:04:33.328 --> 00:04:36.861
if you have a surface that is piecewise-smooth and its

00:04:36.861 --> 00:04:40.732
boundary is a simple-closed piecewise-smooth

00:04:40.732 --> 99:59:59.999
boundary, you're good to go.


Orientation and stokes

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:02.960
I've rewritten Stokes'
theorem right over here.

00:00:02.960 --> 00:00:05.070
What I want to focus
on in this video

00:00:05.070 --> 00:00:08.680
is the question of
orientation because there

00:00:08.680 --> 00:00:12.360
are two different orientations
for our boundary curve.

00:00:12.360 --> 00:00:15.080
We could go in that
direction like that,

00:00:15.080 --> 00:00:17.140
or we could go in the
opposite direction.

00:00:17.140 --> 00:00:19.050
We could be going like that.

00:00:19.050 --> 00:00:21.110
And there are also two
different orientations

00:00:21.110 --> 00:00:22.510
for this normal vector.

00:00:22.510 --> 00:00:24.840
The normal vector might
pop out like that,

00:00:24.840 --> 00:00:28.580
or it could actually go
into the surface like that.

00:00:28.580 --> 00:00:32.466
So we have to make sure that
our orientations are consistent,

00:00:32.466 --> 00:00:33.840
and what I want
to do is give you

00:00:33.840 --> 00:00:35.570
two different ways
of thinking about it.

00:00:35.570 --> 00:00:37.480
And you might think
of others, but these

00:00:37.480 --> 00:00:38.850
are the ones that work for me.

00:00:38.850 --> 00:00:41.080
In order for Stokes'
theorem to hold,

00:00:41.080 --> 00:00:43.400
we have to make sure
that we're not actually

00:00:43.400 --> 00:00:47.410
picking the negative of one
or the other orientations.

00:00:47.410 --> 00:00:49.510
And so the easiest way
for me to remember--

00:00:49.510 --> 00:00:53.420
it is if our normal
vector is-- let's say,

00:00:53.420 --> 00:00:54.890
it goes in that direction.

00:00:54.890 --> 00:00:58.540
And if you have some
hypothetical person traversing

00:00:58.540 --> 00:01:02.450
the boundary of our surface,
and their direction is pointed.

00:01:02.450 --> 00:01:04.620
And their head is pointed
in the same direction

00:01:04.620 --> 00:01:07.520
as the normal vector-- so
this is the normal vector.

00:01:07.520 --> 00:01:09.980
So their head is pointed
in the exact same direction

00:01:09.980 --> 00:01:12.990
as the normal vector-- or you
could say maybe their body

00:01:12.990 --> 00:01:15.060
or, really, their head--

00:01:15.060 --> 00:01:16.120
So that's them.

00:01:16.120 --> 00:01:18.670
Then the direction that you
would have to actually traverse

00:01:18.670 --> 00:01:21.360
the boundary is the direction
that would allow this person

00:01:21.360 --> 00:01:24.460
to keep the surface
to their left.

00:01:24.460 --> 00:01:27.770
So over here, he would have
to go in this direction

00:01:27.770 --> 00:01:31.180
in order to keep the
surface to his left.

00:01:31.180 --> 00:01:35.950
So he would have to
go just like that.

00:01:35.950 --> 00:01:38.110
If we oriented the
surface differently-- so

00:01:38.110 --> 00:01:39.650
let me redraw the
surface right over

00:01:39.650 --> 00:01:41.440
here and draw similar surface.

00:01:41.440 --> 00:01:43.940
So if we had a surface-- so
this surface looks very similar.

00:01:46.640 --> 00:01:48.830
This is a very similar
looking surface

00:01:48.830 --> 00:01:51.103
that I'm drawing
right over here just

00:01:51.103 --> 00:01:53.189
to give a idea of
some of the contours.

00:01:53.189 --> 00:01:55.730
But if we said that the normal
vector for this surface, if we

00:01:55.730 --> 00:01:58.190
orient it in the opposite
way-- so if we said

00:01:58.190 --> 00:02:03.260
that the normal vector here
was actually pointing downward

00:02:03.260 --> 00:02:05.380
like that, then
we would have to,

00:02:05.380 --> 00:02:07.370
in order for Stokes'
theorem to hold,

00:02:07.370 --> 00:02:09.110
we would have to
traverse the boundary

00:02:09.110 --> 00:02:12.280
in a different direction
because, once again, if I draw

00:02:12.280 --> 00:02:13.870
my little character
right over here,

00:02:13.870 --> 00:02:16.850
his head is pointed in the
direction of the normal vector.

00:02:16.850 --> 00:02:18.730
He is now upside down.

00:02:18.730 --> 00:02:20.430
So let me draw him.

00:02:20.430 --> 00:02:23.667
So this is him running
right over here.

00:02:23.667 --> 00:02:24.750
I could draw a better job.

00:02:24.750 --> 00:02:27.100
This is him running
right over here.

00:02:27.100 --> 00:02:29.500
Now, in order to keep--
and from his point of view,

00:02:29.500 --> 00:02:31.870
this would kind of look
like a some type of a pool

00:02:31.870 --> 00:02:33.890
or a ditch of some kind.

00:02:33.890 --> 00:02:34.980
It would actually go down.

00:02:34.980 --> 00:02:36.500
Here, it looks
like a hill to him.

00:02:36.500 --> 00:02:38.300
But since he's
upside down, in order

00:02:38.300 --> 00:02:41.900
for him to keep the
boundary to his left,

00:02:41.900 --> 00:02:47.050
he would have to now go
in the other direction.

00:02:47.050 --> 00:02:50.810
So depending on the orientation
of your normal vector, which

00:02:50.810 --> 00:02:54.500
is really the orientation
of your actual surface,

00:02:54.500 --> 00:02:57.157
will dictate how you need
to traverse the path.

00:02:57.157 --> 00:02:59.240
Now, another way to think
about it-- and this idea

00:02:59.240 --> 00:03:01.310
was introduced by one of
the viewers on YouTube,

00:03:01.310 --> 00:03:03.160
but it's a valid way
of thinking about it--

00:03:03.160 --> 00:03:07.390
is to imagine that the
surface is a bottle cap.

00:03:07.390 --> 00:03:11.130
And so let me draw some
type of a bottle over here.

00:03:11.130 --> 00:03:11.870
So I'll draw.

00:03:11.870 --> 00:03:16.300
Let me draw a bottle.

00:03:16.300 --> 00:03:19.230
You could imagine some type
of a glass soda bottle.

00:03:19.230 --> 00:03:23.970
So what we really care about
is the cap of the bottle--

00:03:23.970 --> 00:03:27.950
so make it feel like it's glass.

00:03:27.950 --> 00:03:33.670
So there, that's our bottle,
And let me draw its cap.

00:03:33.670 --> 00:03:35.340
Let me draw the
cap of the bottle

00:03:35.340 --> 00:03:37.290
because that's
what we care about.

00:03:37.290 --> 00:03:39.630
We can kind of imagine
that being the surface.

00:03:39.630 --> 00:03:42.066
So this is the
cap of our bottle,

00:03:42.066 --> 00:03:43.690
and you just need to
think about, well,

00:03:43.690 --> 00:03:46.210
which way would I have
to twist the cap in order

00:03:46.210 --> 00:03:49.535
to make the cap move up, in
order to take the cap off.

00:03:49.535 --> 00:03:51.910
And you could think of the
normal vector as the direction

00:03:51.910 --> 00:03:57.430
that the cap would move, and
the twisting is the direction

00:03:57.430 --> 00:03:59.300
that you would have
to traverse the path.

00:03:59.300 --> 00:04:01.756
So you would have to
twist the bottle that way,

00:04:01.756 --> 00:04:03.380
or you could think
about the other way.

00:04:03.380 --> 00:04:05.046
If you twisted the
bottle the other way,

00:04:05.046 --> 00:04:06.320
then the cap would move down.

00:04:06.320 --> 00:04:08.770
So the normal vector
is the direction

00:04:08.770 --> 00:04:11.250
that the cap would
move, and the direction

00:04:11.250 --> 00:04:13.800
that you would traverse
the boundary is

00:04:13.800 --> 00:04:15.238
how you would actually twist it.

00:04:15.238 --> 00:04:17.279
So either of these are
ways of thinking about it,

00:04:17.279 --> 00:04:19.320
but they're important to
keep in mind, especially

00:04:19.320 --> 00:04:21.480
once the shapes start
getting a little bit more

00:04:21.480 --> 00:04:25.600
convoluted and oriented
in strange ways.


Dividing positive and negative numbers

WEBVTT
Kind: captions
Language: en

00:00:00.938 --> 00:00:04.559
Now that we know a little bit about multiplying positive and negative numbers,

00:00:04.559 --> 00:00:06.649
Let's think about how how we can divide them.

00:00:06.649 --> 00:00:08.562
Now what you'll see is that it's actually

00:00:08.562 --> 00:00:10.632
a very similar methodology.

00:00:10.632 --> 00:00:12.164
That if both are positive,

00:00:12.164 --> 00:00:15.455
you'll get a positive answer. If one is negative,

00:00:15.455 --> 00:00:18.890
or the other, but not both, you'll get a negative answer.

00:00:18.890 --> 00:00:22.542
And if both are negative, they'll cancel out and you'll get a positive answer.

00:00:22.542 --> 00:00:26.359
But let's apply and I encourage you to pause this video and try these out yourself

00:00:26.359 --> 00:00:29.473
and then see if you get the same answer that I'm going to get.

00:00:29.473 --> 00:00:32.840
So eight (8) divided by negative two (-2).

00:00:32.840 --> 00:00:35.957
So if I just said eight (8) divided by two (2), that would be

00:00:35.957 --> 00:00:40.326
a positive four (4), but since exactly one of these two numbers

00:00:40.326 --> 00:00:45.425
are negative, this one right over here, the answer is going to be negative.

00:00:45.425 --> 00:00:49.556
So eight (8) divided by negative two (-2) is negative four (-4).

00:00:49.556 --> 00:00:53.425
Now negative sixteen (-16) divided by positive four (4)--

00:00:53.425 --> 00:00:54.423
now be very careful here.

00:00:54.423 --> 00:00:59.727
If I just said positive sixteen (16) divided by positive four (4), that would just be four (4).

00:00:59.727 --> 00:01:02.675
But because one of these two numbers is negative,

00:01:02.675 --> 00:01:05.121
and exactly one of these two numbers is negative,

00:01:05.121 --> 00:01:09.104
then I'm going to get a negative answer.

00:01:09.104 --> 00:01:12.174
Now I have negative thirty (-30) divided by negative five (-5).

00:01:12.174 --> 00:01:17.285
If I just said thirty (30) divided by five (5), I'd get a positive six (6).

00:01:17.285 --> 00:01:20.573
And because I have a negative divided by a negative,

00:01:20.573 --> 00:01:25.631
the negatives cancel out, so my answer will still be positive six (6)!

00:01:25.631 --> 00:01:27.332
And I could even write a positive (+) out there, I don't have to,

00:01:27.332 --> 00:01:30.448
but this is a positive six (6).

00:01:30.448 --> 00:01:34.065
A negative divided by a negative, just like a negative times a negative,

00:01:34.065 --> 00:01:37.063
you're gonna get a positive answer.

00:01:37.063 --> 00:01:39.035
Eighteen (18) divided by two (2)!

00:01:39.035 --> 00:01:40.612
And this is a little bit of a trick question.

00:01:40.612 --> 00:01:43.934
This is what you knew how to do before we even talked about negative numbers:

00:01:43.934 --> 00:01:47.067
This is a positive divided by a positive.

00:01:47.067 --> 00:01:49.035
Which is going to be a positive.

00:01:49.035 --> 00:01:52.918
So that is going to be equal to positive nine (9).

00:01:52.918 --> 00:01:55.769
Now we start doing some interesting things,

00:01:55.769 --> 00:01:57.967
here's kind of a compound problem.

00:01:57.967 --> 00:02:01.464
We have some multiplication and some division going on.

00:02:01.464 --> 00:02:03.698
And so first right over here, the way this is written,

00:02:03.698 --> 00:02:06.030
we're gonna wanna multiply the numerator out,

00:02:06.030 --> 00:02:07.794
and if you're not familiar with this little dot symbol,

00:02:07.794 --> 00:02:10.164
it's just another way of writing multiplication.

00:02:10.164 --> 00:02:12.381
I could've written this little "x" thing over here

00:02:12.381 --> 00:02:15.628
but what you're gonna see in Algebra is that the dot become much more common.

00:02:15.628 --> 00:02:17.358
Because the X becomes used for other--

00:02:17.358 --> 00:02:23.383
People don't want to confuse it with the letter X

00:02:23.383 --> 00:02:25.584
which gets used a lot in Algebra.

00:02:25.584 --> 00:02:27.927
That's why they used the dot very often.

00:02:27.927 --> 00:02:31.093
So this just says negative seven (-7) times three (3)

00:02:31.093 --> 00:02:33.447
in the numerator, and we're gonna take that product

00:02:33.447 --> 00:02:35.597
and divide it by negative one (-1).

00:02:35.597 --> 00:02:37.640
So the numerator, negative seven (-7) times three (3),

00:02:37.640 --> 00:02:41.437
positive seven (7) times three (3) would be twenty-one (21),

00:02:41.437 --> 00:02:45.303
but since exactly one of these two are negative,

00:02:45.303 --> 00:02:47.078
this is going to be negative twenty-one (-21),

00:02:47.078 --> 00:02:49.732
that's gonna be negative twenty-one (-21) over negative one (-1).

00:02:49.732 --> 00:02:51.894
And so negative twenty-one (-21) divided by negative one (-1),

00:02:51.894 --> 00:02:54.944
negative divided by a negative is going to be a positive.

00:02:54.944 --> 00:02:57.920
So this is going to be a positive twenty-one (21).

00:02:57.920 --> 00:02:59.510
Let me write all these things down.

00:02:59.510 --> 00:03:03.266
So if I were to take a positive divided by a negative,

00:03:03.266 --> 00:03:07.425
that's going to be a negative.

00:03:07.425 --> 00:03:11.407
If I had a negative divided by a positive,

00:03:11.407 --> 00:03:15.121
that's also going to be a negative.

00:03:15.121 --> 00:03:18.020
If I have a negative divided by a negative,

00:03:18.020 --> 00:03:20.454
that's going to give me a positive,

00:03:20.454 --> 00:03:24.954
and if obviously a positive divided by a positive,

00:03:24.954 --> 00:03:27.770
that's also going to give me a positive.

00:03:27.770 --> 00:03:29.359
Now let's do this last one over here.

00:03:29.359 --> 00:03:30.589
This is actually all multiplication,

00:03:30.589 --> 00:03:31.989
but it's interesting, because we're multiplying

00:03:31.989 --> 00:03:34.539
three (3) things, which we haven't done yet.

00:03:34.539 --> 00:03:37.116
And we could just go from left to right over here,

00:03:37.116 --> 00:03:39.405
and we could first think about negative two (-2) times

00:03:39.405 --> 00:03:41.089
negative seven (-7).

00:03:41.089 --> 00:03:42.627
Negative two (-2) times negative seven (-7).

00:03:42.627 --> 00:03:44.043
They are both negatives, and

00:03:44.043 --> 00:03:45.094
negatives cancel out, so this would give us,

00:03:45.094 --> 00:03:46.844
this part right over here,

00:03:46.844 --> 00:03:49.539
will give us positive fourteen (14).

00:03:49.539 --> 00:03:51.408
And so we're going to multiply positive fourteen (14)

00:03:51.408 --> 00:03:55.207
times this negative one (-1), times -1.

00:03:55.207 --> 00:03:57.908
Now we have a positive times a negative.

00:03:57.908 --> 00:03:59.590
Exactly one of them is negative,

00:03:59.590 --> 00:04:01.470
so this is going to be negative answer,

00:04:01.470 --> 00:04:05.020
it's gonna give me negative fourteen (-14).

00:04:05.020 --> 00:04:06.087
Now let me give you a couple of more,

00:04:06.087 --> 00:04:08.655
I guess we could call these trick problems.

00:04:08.655 --> 00:04:12.070
What would happen if I had zero (0) divided by

00:04:12.070 --> 00:04:14.252
negative five (-5).

00:04:14.252 --> 00:04:16.503
Well this is zero negative fifths

00:04:16.503 --> 00:04:19.219
So zero divided by anything that's non-zero

00:04:19.219 --> 00:04:22.355
is just going to equal to zero.

00:04:22.355 --> 00:04:25.035
But what if it were the other way around?

00:04:25.035 --> 00:04:27.371
What happens if we said negative five divided by zero?

00:04:27.371 --> 00:04:29.837
Well, we don't know what happens when you divide things by zero.

00:04:29.837 --> 00:04:31.522
We haven't defined that.

00:04:31.522 --> 00:04:35.254
There's arguments for multiple ways to conceptualize this,

00:04:35.254 --> 00:04:38.301
so we traditionally do say that this is undefined.

00:04:38.301 --> 00:04:43.213
We haven't defined what happens when something is divided by zero.

00:04:43.213 --> 00:04:46.347
And similarly, even when we had zero divided by zero,

00:04:46.347 --> 00:04:50.431
this is still, this is still, undefined.