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### Course: Multivariable calculus > Unit 4

Lesson 11: Surface integrals- Introduction to the surface integral
- Find area elements
- Example of calculating a surface integral part 1
- Example of calculating a surface integral part 2
- Example of calculating a surface integral part 3
- Surface integrals to find surface area
- Surface integral example, part 1
- Surface integral example part 2
- Surface integral example part 3: The home stretch
- Surface integral ex2 part 1
- Surface integral ex2 part 2
- Surface integral ex3 part 1
- Surface integral ex3 part 2
- Surface integral ex3 part 3
- Surface integral ex3 part 4

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# Surface integral ex2 part 1

Parametrizing a surface that can be explictly made a function of x and y. Created by Sal Khan.

## Want to join the conversation?

- Why does y^2 become v not v^2?(10 votes)
- at0:40Sal writes re-writes the surface as z = x + y². At3:53, for his parameterization he just makes the substitution: x=u and y=v. Plug that into the surface function: z = u+ v².

As a position vector valued function that gives you: r(u,v) = (u)i + (v)j + (u+v²)k.(2 votes)

- i didn't get why y was required? all the ds s should give me the area of the surface(2 votes)
- Because you are not calculating the area of the surface in this example. You are calculating the mass of the surface and "y" is the "mass per area", hence "ydS" is the "mass of an infinitesimal surface". Summing all these up, you get the mass of the entire object, which is the surface depicted in the video.(8 votes)

- How do you know how to parametrize a surface? In this video, Sal simply sets x and y directly to their own parameters, whereas in the previous examples, he has to take the sin's and cos's of the parameters to describe x, y, and z. In other words, how do you know when to manipulate the parameters versus simply setting them equal to x, y, or z? Is it just practice/intuition?(5 votes)
- After hours of research, I still confuse about how to choose parameters for parameterization, but I want to share what I though could be correct.

To me, parameterization is a transformation about the inputs of the function from the original one to the new one.

e.g`x,y,z`

as the original inputs,`r,θ,Φ`

as the new inputs.

Or, we could reduce the amount of inputs, e.g:`x,y,z`

->`u,v`

.

The critical difference is about what you are chasing for, in this video, is the surface area, 2 variables is a reasonable choice because the area formula for infinitesimal area only require 2 arguments, not 3.(1 vote)

- Why does he use "u" and "v"? Why not stick with "S" and "t"?(3 votes)
- s and t are traditionally used to represent distance and time, Sal mentioned mass and area for this example so he chose a different but still commonly used pair of letters.(2 votes)

- If we can take x and y as parameters then why have we been parameterizing all surfaces upto now like the torus and the unit sphere?(3 votes)
- Why do we take the integral of y?(2 votes)
- The function being integrated in this problem is y. It could have been x or y² or whatever. It is the function whose surface integral will be taken. The surface in this problem is defined as z=x+y², where 0≤x≤1 and 0≤y≤2.

Review the video from3:00.(3 votes)

- I'm confused. How can a surface have mass? (3:15)(1 vote)
- Think of a sheet of paper. Sure, a sheet of paper does technically have finite thickness to it, but you can define say the mass per unit surface area of the paper.(1 vote)

- At1:30, Sal starts to talk about identifying the surface. Has he done any videos on identification of quadric surfaces (by name) through their equations, rather than working through it visually? The visual aspect is useful but very time-consuming, especially for tests.(1 vote)

## Video transcript

Let's attempt another
surface integral, and I've changed the
notation a little bit. Instead of writing the
surface as a capital sigma, I've written it as a capital S.
Instead of writing d lowercase sigma, I wrote d uppercase
S, which is still a surface integral
of the function y. And the surface we care about
is x plus y squared mins z is equal to 0. X between 0 and 1,
y between 0 and 2. Now, this one might be a
little bit more straightforward than the last one we
did, or at least I hope it's a little bit more
straightforward, because we can explicitly define
z in terms of x and y. And actually we can
even explicitly define x in terms of y and z. But I'll do it the other way. It's a little bit easier
for me to visualize. So if you add z to both sides of
this equation right over here, you get x plus y
squared is equal to z, or z is equal to
x plus y squared. And this is actually
pretty straightforward. this surface is pretty
easy to visualize, or we can give our best
attempt at visualizing it. So if that is our z-axis,
and that is our x-axis, and that is the y-axis,
we care about the region x between 0 and 1. So maybe this is x equals
1, and y between 0 and 2. So let's say this is 1,
this is 2 in the y area. So we essentially care
about the surface over this, over this region
of the xy plane. And then we can think about
what the surface actually looks like. This isn't the surface. This is just kind of the range
of x's and y's that we actually care about. And so let's think
about the surface. When x and y are 0, z is 0. So we're going to
be sitting-- let me do this in a-- let
me do this in green. z is going to be right over there. And now as y increases, or
if when x is equal to 0, if we're just talking
about the zy plane, z is going to be
equal to y squared. So this might be
z is equal to 4. This is z is equal to 2, 1, 3. So z is going to do
something like this. It's going to be a
parabola in the zy plane. It's going to look
something like that. Now, when y is equal to
0, z is just equal to x. So as x goes to 1,
z will also go to 1. So z will go like this. The scales of the axes aren't--
they are not drawn to scale. The z is a little
bit more compressed than the x or y the
way I've drawn them. And then from this
point right over here, you add the y squared. And so you get
something that looks-- so this is this point there. And then this point, when y is
equal to 2 and x is equal to 1, you have z is equal to 5. It's going to look
something like this. And then you're going to have
a straight line like that, at this point, is
right over there. And this surface is the surface
that we are going to take, or the surface over which we're
going to evaluate the surface integral of the function y. And so one way you
could think about it, y could be maybe the mass
density of this surface. And so when you multiply
y times each dS, you're essentially figuring out
the mass of that little chunk, and then you're figuring out
the mass of this entire surface. And so one way you could
imagine as we go more and more in that direction
as y is increasing, this thing is getting
more and more dense. So this part of the
surface is more dense than as y becomes
lower and lower. And then that would
actually give us the mass. With that out of the way,
let's actually evaluate it. And so, as you
know, the first step is to figure out
a parametrization. And it should be
pretty straightforward, because we can write z
explicitly in terms of x and y. And so we can actually use x
and y as the actual parameters, or if we want to just substitute
it with different parameters, we could. But let me-- so let's just
write-- let me do that. So let me just write x is equal
to-- and in the spirit of using different notation, instead
of using s and t, I'll use u and v. X is equal to u,
let's say y is equal to v, and then z is going to be
equal to u plus v squared. And so our surface, written
as a vector position function or position vector
function, our surface, we can write it as
r, which is going to be a function of
u and v. And it's going to be equal to ui plus
vj plus u plus v squared k. And then u is going
to be between 0 and 1, because x is just equal
to u or u is equal to x. So u is going to
be between 0 and 1. And then v is going
to be between 0 and 2. I'm going to leave you there. In the next video,
we'll actually set up the surface
integral now that we have the parametrization done.