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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 10: Surface integral preliminaries# Parametrizing a surface, part 1

Introduction to Parametrizing a Surface with Two Parameters. Created by Sal Khan.

## Want to join the conversation?

- I think it should be "3Pi/2" instead of "3Pi/4" at8:55and9:20.(61 votes)
- Yes. Sal made a mistake there. 3Pi/2 corresponds to 270 deg whereas 3Pi/4 corresponds to 135deg. But if you are able to correctly visualize the torus with respect to its orientation around the three axis, it shouldn't be a problem. :)(26 votes)

- Is there any program or website to help me visualize a torus in 3d space?(4 votes)
- Peter Collingridge made a lovely program on Khan Academy to help with that!

https://www.khanacademy.org/computer-programming/shaded-torus/1071783640

here's the link ^.^

I hope this helps(8 votes)

- Should the axes be labeled differently?

My understanding is that the**bottom right**(where Sal labeled the positive y-axis) should be the**positive x-axis**and the**positive y-axis**should be**behind**it. Because angles in standard position go counter-clockwise from the positive x-axis, I would start the rotations of t from the positive x-axis. I*would like*the origin on his s,t plane to begin on the positive x-axis, rather than the positive y-axis.

Looking at it this way, Sal's shadings of the torus at about18:00make more sense to me.

Please help me to understand what I may have missed.(6 votes)- Notice that the axes you described are the same as the axes Sal drew, just rotated a bit. The real important thing is that your space is what is called a
**right handed coordinate system**. In such a space, x × y = z (that is the cross product).

As for the bit about the angles, all that matters is that the angle goes around counter-clockwise. If you wrap your fingers in the direction of increasing angle, your thumb should point in the direction of the +z axis.(3 votes)

- Is it just me or do the areas he shades in at18:00not match? He is right in that from 0 - pi/2 for S and T only sketches 1/4th of the outer surface for 1/4th of the circumference, or 1/8th of the total torus surface. But on the right diagram he starts shading this parametrization on the y-axis while the left he starts from the x-axis. It is important to maintain the same starting point so parameters and bounds align correctly(3 votes)
- Good catch, based on how he defined t, the diagram on the right is correctly shaded, but the upper left (top view) is not ...

I've put your observation into the Tips&Feedback, I think he is more likely to see those!(3 votes)

- At5:07Sal says S parameter describes the angle between radius and x-z plane. Has he mistaken it for x-y plane, considering the way he uses the parameter afterwards?(2 votes)
- Yes, he meant to say x-y plane. A little box pops up in the lower right corner of the screen with the correction.(2 votes)

- Can someone please explain what the 's' parameter is for in this video? I don't understand.(1 vote)
- I find it helpful to start by thinking of a more familiar circle drawn in 2 dimensions on an x-y coordinate system. This circle can be described with a radius, and the radius rotates through 2pi radians. If we call the radius of the circle 'r', and the angle it rotates through 's', we can parameterize this circle using x = r*cos(s) and y=r*sin(s).

Making the jump to 3 dimensions and describing a torus, you can think of it as a system of two circles. One of the circles describes the ring shape of the torus. The other circle describes the cross-section (if you took a very thin slice of one part of the torus). In the video, the 's' parameter is used to describe the cross section circle, and the 't' parameter is used to describe the overall ring shape of the torus.(3 votes)

- There were basic numerical mistakes Pi 3Pi over 2 (it was stated as 3Pi over 4) why wasn't it corrected(2 votes)
- isn't a doughnut a solid shape? how can we say it's a surface (2 D) ?(1 vote)
- It is a surface in three-dimensional space, just like a sphere is a surface.(3 votes)

- are there videos on arc length parametrization?(2 votes)
- Parametrization is often used in arc length problems. Because you can describe curves or ways better with a parametric function as with a normal functions. But it is a different topic. Look at khan academy for videos.(1 vote)

- how come it is parameters not angles or edges?(1 vote)
- A parameter can be any type of variable. In this case the parameters are angles. Parametrizing means representing the given function in terms of new variables of some convenient type.(2 votes)

## Video transcript

All the parameterizations
we've done so far have been parameterizing a curve
using one parameter. What we're going to start doing
this video is parameterizing a surface in three dimensions,
using two parameters. And we'll start with an
example of a torus. A torus, or more commonly
known, as a doughnut shape. And we know what a
doughnut looks like. Let me draw it in a suitable,
well, I don't have any suitable doughnut colors,
so I'll just use green. A doughnut looks
something like this. It has a hole in the center,
And maybe the other side of the doughnut looks something
like that, and we could shade it in like that. That is what a
doughnut looks like. So how do we construct that
using two parameters? So what we want to do, is you
can just visualize it, a doughnut, if you were to
draw some axes here. So that's our doughut. Let me draw some axes. So let's say I have the z-axis
that goes straight up and down. So we have drawn it here, the
doughnut's a little at a tilt, so the z-axis, I'll
tilt it a little bit. So our z-axis goes straight
through the center of the doughnut. So that right there, this is
going to be an exercise in drawing more than
anything else. So that is my z-axis, and then
you can imagine the z-axis goes from there, and then this
coming out of here will be my x-axis. That right there is my x-axis,
and then maybe my y-axis comes out like that. And the whole reason why I drew
it this way, is that if you imagine the cross section of
this doughnut, I'll draw a little bit neater, but the
cross section of this doughnut in the x-z axis, is going to
look something like this. If I were to slice it in the
x-z axis, it would look something like that. That would be the slice. It would trace out, and
we're thinking about not a full doughnut, just the
surface of a doughnut. So it would trace out
a circle like this. If you were to cut the doughnut
in the positive z-y axis, it's going to trace out a circle
that looks something like that, right there. And if you go out here, you're
going to get bunch of circles. So if you think about it, it's
a bunch of circles rotated around the z-axis. If you think of it that way,
it'll give us some good intuition for the best way
to parameterize this thing. So let's do it that way. Let's start off with
just the z-y axis. I'll draw it a little bit
neater than I've done here. So that is the z-axis, and
that is y, just like that. And let's say that the center
of these circles, let's say it lies on, you know, it can lie,
when you cross the z-y axis, the center sits on the y-axis. I didn't draw it that
neatly here, but I think you can visualize. So it sits right
there on the y-axis. And let's say that it is a
distance b away from the center of the torus, or from the
z-axis it's a distance of b. It's always going to
be a distance of b. It's always, if you imagine the
top of the doughnut, let me draw the top of the doughnut. If you're looking down on a
doughnut, let me draw a doughnut right here, if you're
looking down on a doughnut, it just looks something like that. The z-axis is just going to
be popping straight out. The x-axis would come down like
this, and then the y-axis would go to the right, like that. So you can imagine, I'm
just flying above this. I'm sitting on the z-axis
looking down at the doughnut. It will look just like this. And if you imagine the cross
section, this circle right here, the top part of the
circle if you're looking down, would look just like that. And this distance b is a
distance from the z-axis to the center of each
of these circles. So this distance, let me draw
it in the same color, from the center to the center of these
circles, that is going to be b. It's just going to keep
going to the center of the circles, b. That's going to be b,
that's going to be b. That's going to be b. From the center of our torus
to the center of our circle that defines the torus,
it's a distance of b. So this distance right here,
that distance right there is b. And from b, we can imagine
we have a radius. A radius of length a. So these circles have
radius of length a. So this distance right here is
a, this distance right here is a, this distance right there
is a, that distance right there is a. If I were look at these
circles, these circles have radius a. And what we're going to do
is have two parameters. One is the angle that this
radius makes with the x-z plane, so you can imagine
the x-axis coming out. Let me do that in
the same color. You can imagine the
x-axis coming out here. So this is the x-z plane. So one parameter is going to
be the angle between our radius and the x-z plane. We're going to call that angle,
or that parameter, we're going to call that s. And so as s goes between 0 and
2 pi, as s goes between 0 in 2 pi, when the 0 is just going to
be at this point right here, and then as it goes to 2 pi,
you're going to trace out a circle that looks
just like that. Now, we only have
one parameter. What we want to do is then
spin this circle around. What I just drew is that
circle right there. What we want to do is spin
the entire circle around. So let's define
another parameter. We'll call this one t, and
I'll take the top view again. This one's getting a
little bit messy. Let me draw another top view. As you can see, this is
all about visualization. So let's say this is my
x-axis, that is my y-axis. And we said we started
here on the z-y plane. We are b away from the z-axis,
so that distance is b. In this diagram, the z-axis
is just popping out at us. It's popping out of the page. We're looking straight down. It's just like the same
view as right there. And what I just drew, when s is
equal to 0 radians, we're going to be out here, exactly
one radius further along the y-axis. And then we're going to rotate. As we rotate around, we're
going to rotate and then come all the way over here. That's when we're right over
there, and then come back down. So if you looked on the
top of the circle, it's going to look like that. Now, to make the doughnut,
we're going to have to rotate this whole setup
around the z-axis. Remember, the z-axis
is popping out. It's looking straight up at us. It's coming out of
your video screen. Now to rotate it, we're
going to rotate this circle around the z-axis. And to do that, we'll define a
parameter that tells us how much we have rotated it. So this is when we've
rotated 0 radians. At some point, we're going to
be over here, and we would have rotated it, this is b as
well, and our circle is going to be looking like this. This is maybe this point on
our doughnut, right there. At that point, we would
have rotated it, let's say p radians. So this parameter of how much
have we rotated around the z-axis, how much have we gone
around that way, we're going to call that t. And t is also going to
vary between 0 and 2 pi. And I want to make
this very clear. Let's actually draw the domain
that we're mapping from to our surface, so that we
understand this fully. So let me draw some, and then
we'll talk about how we can actually parameterize
that into a position vector-valued function. So right here, let's
call that the t-axis. That's, remember, how much
we're rotated around the z-axis right there. And let's call this
down here our s-axis. And I think this will help
things out a good bit. So when s is equal to 0, and
we vary just t, so they're both going to vary
between 0 and 2 pi. So this right here is 0,
this right here is 2 pi. Let me do some
things in between. This is pi, this would be pi
over 2 obviously, pi over 2, this would be 3 pi over 4. You do the same thing
on the p-axis. It's going to go up to 2 pi. Let's do that. So we're going to
go up to 2 pi. I really want you to visualize
this, because then the parameterization, I think, will
be fairly straightforward. So that's 2 pi, this is pi,
this is pi over 2, and then this is 3 pi over 4. So let's think about what it
looks like if you just hold s constant at 0, and we just
vary t between zero and 2 pi. And let me do that in
magenta, right here. So we're holding s constant,
and we're just varying the parameter 2 pi. So this, if you think about it,
should just form a curve in three dimensions,
not a surface. Because we're only varying
one parameter right here. So let's think about
what this is. Remember, s is, let
me draw my axes. So that is my x-axis, that
is my y-axis, and then this is my, I'm getting
messier and messier. That is my z-axis, right--
actually, let me draw it a little bit bigger than that. I think it will help all
of our visualizations. All right. So this is my x-axis, that is
my y-axis, and then my z-axis goes up like that. z-axis. Now remember, when s is equal
to 0, that means we haven't rotated around this
circle at all. That means we're out here. We're going to be b away,
and then a away again. Right? We haven't rotated
around this at all. We're setting s as
equal to zero. So essentially, we're going to
be b away, so this is going to be a distance of b away, and
then we're going to be another a away. The b is the center of the
circle, and then we're going to be another a away. We're going to be
right over there. So this is a plus b away. And then we're going to vary t. Remember, t was how much we've
gone around the z-axis. These were top views over here. So this line right here, in our
s-t domain, we can say, when we map it, or parameterize it,
it'll correspond to the curve that's essentially the outer
edge of are doughnut. If this is the top view of the
doughnut, it will be the outer edge of the doughnut,
just like that. So let me draw the outer edge. And to do that a little bit
better, let me draw the axes in both the positive and
the negative domain. It might make my graph a little
bit easier to visualize things. Positive and negative domain,
this is negative z right there. So this line in our t-s plane,
I guess we could say, this magenta line, we hold s at 0
radians and we increase t, this is t is zero, this is t is
equal to 2 pi, that's t is equal to pi, this is t is equal
to 3 pi over 2, all the way back to t is equal to 2 pi. This line corresponds to that
line, as we rotate, as we increase t and hold
s constant at 0. Now let's do another point. Let's say when s is at pi,
right, remember, when s is at pi, we've gone exactly,
pi is 180 degrees. When s is at pi, we've
gone exactly 180 degrees around the circle, around
each of these circles. So we're right over there. And now let's hold it constant
at pi, and then rotate it around to form our doughnut. So we're going to form the
inside of our doughnut. So when s is at pie, and we're
going to take t from 0, so when s is pi and t is 0, we're going
to be, this was the center of our circle, we're going
to be a below that. We're going to be
right over there. And then as we vary, as we
increase t, so as we move up along, holding s at pi, and we
increase t, we're going to trace out the inside of our
doughnut, that will look something like that. That was my best
shot at drawing it. And then we can do
that multiple times. When s is pi over 2, I want to
do multiple different colors, when s is pi over 2, we've
rotated up here exactly 90 degrees, right? Pi over 2 is 90 degrees
at this point. And then if we vary t, we're
essentially tracing out the top of the doughnut, right? So let me make sure I draw it . So the cross section, the top
of the doughnut, we're going to start off right over here. So when s is pi over 2, and you
vary it right, and then you very t, I'm having trouble
drawing straight lines. And then you vary t, it's
going to look like this. That's the top of that
circle right there. The top of this circle is
going to be right there. The top of this circle is
going to be right over there. Top of the circle is going
to be right over there. So then I just
connect the dots. It's going to look
something like that. That is the top
of our doughnut. If I was doing this top view,
it would be the top of the doughnut, just like that. And if I wanted to do the
bottom of the doughnut, just to make the picture clear, if I
were to make the bottom of the doughnut, the bottom of the
doughnut would be-- see, if I take s as 3 pi over 4 and I
vary t, that's the bottoms of our doughnuts. So let me draw the circle, it's
right there, the circle is right there, you wouldn't be
able to see the whole thing if this wasn't transparent. So you'd be tracing out the
bottom of the doughnut, just like that. I know this graph is becoming
a little confusing, but hopefully you get the idea. When s is 2 pi again, you're
going to be back to the outside of the doughnut again. That's also going
to be in purple. So that's what happens when we
hold the s constant at certain values and vary the t. Now let's do the opposite. What happens if we hold t
at 0, and we very the s? So t is 0, that means we
haven't rotated at all yet. So we're in the z-y plane. So t is 0, and s will start at
0, and it'll go to pi over 2, that's that point over there. Then it'll go to pi. This point is the same
thing as that point. Then it will go to 3 pi over
4, then it'll come back all the way to 2 pi. So this line corresponds to
this circle, right there. We could keep doing these if we
pick when t is pi -- let me use a different color, that's
not different enough. When t is pi over
2, just like that. We would have rotated around
the z-axis 90 degrees, so now we're over here. And now when we vary s, s will
start off over here, and it'll go all the way
around like that. So this line corresponds
to that circle. We could keep doing
it like this. When t is equal to pi, that
means we've got all the way around the circle like that,
and now when we vary s from 0 to pi over 2, we're going to
start all the way over here, and then we're going to vary,
all the way, we're going to go down and hit all those contours
that we talked about before, and I'll do one more, just
to kind of make this, the scaffold, clear. This dark purple,
hopefully you can see it. When t is 3 pi over 4,
we've rotated all the way. So we're in the x-z plane. And then when you vary s, s
will start off over here, and as you increase s, you're going
to go around the circle, around the circle, just like that. And of course, when you get all
the way back full circle, t over pi over 2, that's
the same thing. You're back over here again. So this is going to be, we can
even shade it the same color. And hopefully you're
getting a sense now of the parameterization. I haven't done any math yet. I haven't actually showed you
how to mathematically represent it as a vector value function,
but hopefully you're getting a sense of what it means to
parameterize by two parameters. And just to get an idea of what
these areas on our s-t plane correspond to onto this
surface, in I guess you could say, in R3, this little square
right here, let's see what's bounded by. This little square, I want to
make sure I picked a square that I can draw neatly. So this square right here, that
is between, when you look at t, it's between 0 and pi over 2. And s is between
0 and pi over 2. So this right here is
this part of our torus. If you're looking at it from
the top, it would look like that, right there. You can imagine, we've
transformed this square. I haven't even shown you how
to do it mathematically yet. But we've transformed
this square to this part of the doughnut. Now, I think we've done about
as much as I can do on the visualization side. I'll stop this video here. In the next video, we're going
to actually talk about, how do we actually parameterize
using these two parameters? Remember, s takes around each
of these circles, and then t takes us around the z-axis. And if you take all of the
combinations of s and t, you're going to have every point
along the surface of this torus or this doughnut. How do you actually go from an
s and a t that goes from 0 to 2 pi, in both cases, and turn
it into a three-dimensional position vector-valued function
that would define this surface? We're going to do that
in the next video.