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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 ex3 part 1

Breaking apart a larger surface into its components. Created by Sal Khan.

## Want to join the conversation?

- Why are we taking the integral of z?(12 votes)
- the question is just set up that way. We are taking the integral of z because it is the function in the given equation- if the z were zy^2 we would take the surface integral of zy^2.(13 votes)

- But the surface area of surface 1 is πr²=π, not 0, right?(8 votes)
- It gets multiplied by z, which is 0 for surface 1, so you end up with 0.(10 votes)

- Can we evaluate S2 by using a line integral concept since the surface is like an area of curtain between the x-y plane and z = 1 - x plane ?(4 votes)
- I was wondering how you would go about calculating the surface area of an intersection of two surfaces. Say you want to find the surface area of the surface cut out by x² + y² = 2x and x² + y² = z². You then know that z is the square root of 2x, but how do you find boundaries for x and y?(3 votes)
- What does taking a surface integral do? Does it give you the surface area?(1 vote)
- It gives you the function for the surface area by throwing in a function in the formula.(2 votes)

- what is the meaning of taking the surface integral of z? Surface integral is the area of surface defined by r(s,t) right? whats z doing there?(1 vote)
- at5:06i could have done also the inverse? r_v x r_u?(1 vote)
- Hi, the way to solve S2 is really interesting. At4:45, I just don't understand why z= v, with the domain of 0<=v<= 1-cosu. Shouldn't there be 2 symmetrical triangle-like things wrapping around the chopped cylinder? Could anyone please tell me how these 2 parts are
**both**included in this method?(1 vote)- I will try to make it clear that indeed both of those "triangle-like things" are included in this method. The parameter u is the angle made by r with the positive x-axis, and as u goes from 0 to pi, the x value goes from 1 to -1 on this interval, and z = 1 - x = 1 - cos(u) goes from 0 to 2 on this interval. This corresponds to the right side of the S2 as Sal drew it in the video, or you could call it the first "triangle-like thing." The second one comes as u completes it's turn, going from pi to 2pi. Here x goes from -1 back to 1 where it began, and z goes from 2 back to 0 where it began. Each "triangle-like thing" has one half the unit circle as a base, and 2 as a height. Hope this helps picture it correctly.(1 vote)

## Video transcript

Let's try another
surface integral. And the surface that
we're going to care about, s, is this shape, the outside
of this shape right over here. And you can see we can
kind of decompose it into three separate surfaces. The first surface is its
base, which is really the filled-in unit
circle down here. The second surface,
which we have in blue, is you can kind of view
it as the side of it. You can view it as kind of
like the side of a cylinder, but the cylinder has been
cut by a plane up here. And the plane that cuts
it is the plane z is equal to 1 minus x. And obviously, the plane itself
goes well beyond this shape. But where that plane
cuts the cylinder kind of defines this shape. So the blue surface is above
the boundary of the unit circle and below the plane. And then the third
surface is the subset of the plane, of that
purple plane, of z equals 1 minus x, that
overlaps, that kind of forms the top of this cylinder. And so we can rewrite
this surface integral as the sum of three
surface integrals. It's the surface
integral of z over s1 plus the surface
integral of z over s2 plus the surface
integral of z over s3. And we can just tackle each
of these independently. So let's start with surface 1. And you might immediately want
to start parameterizing things and all the rest, but there's
actually a very fast way to handle this surface integral,
especially because we're taking the surface
integral of z. What value does z take on
throughout this surface, throughout this filled-in
unit circle right over here? Well, that surface
is on the xy plane. When we're on the xy
plane, z is equal to 0. So for this entire surface,
z is going to be equal to 0. You're essentially
integrating 0. 0 times dS is just
going to be 0. So this whole thing is
going to evaluate to 0. So that's about as simple as
evaluating a surface integral can get. But it's always important
to least keep a look out for things like that. And it'll keep you from kind
of going down this wild goose chase-- that wouldn't
be a wild goose chase. You would eventually
get the answer anyway. But you don't have
to waste so much time in parameterizing
things and all the rest. Now let's tackle the
other two surfaces, and we'll focus on surface 2. So the x and y-values,
the valid x and y-values, are the x and y-values along
the unit circle right over here. So we can really
parameterize it the way that we would parameterize
a traditional unit circle. So we could set x is
equal to-- let's set it-- and our radius is 1,
so x is equal to cosine of-- I'll use the parameter u. x is equal to cosine of u. And then let's say y
is equal to sine of u. And then u is the angle
between the positive x-axis and wherever we are essentially
on that unit circle. So that right over there, that
angle right over there is u. And so this would
give us for u-- so as long as u is
between 0 and 2 pi, that we're essentially going
around this unit circle. So those are all the
possible x and y-values that we can take on. And then the z-value is what
takes us up above that boundary and gets it someplace
along the surface. But this is interesting,
because the z-value, it can take on obviously a
bunch of different values, but it always has to be below
this plane right over here. So for this z-value, I'm going
to introduce a new parameter. Let's call it z v. That's
the second parameter. And v is definitely going
to be greater than 0. z and v, they're the same thing. They're always going
to be positive, so they're definitely going to
be greater than or equal to 0. But it's not less than or
equal to some constant thing. This has kind of a
variable roof on it. And so it's always
going to be less than or equal to this
plane right over here. And so we could say is less
than or equal to 1 minus x. We could have said-- so we
know that z is less than or equal to 1 minus x, but if we
use the new parameters, v is z, and 1 minus x is the same
thing as 1 minus cosine u. And so now we have
our parameterization. We are ready to actually
evaluate the surface integral. And to do that, first
let's do the cross product. We want to figure
out what dS is, and we have to take the
magnitude of the cross product of the partial of
our parameterization with respect to u, crossed
with the partial with respect to v. Actually, let me just
write all this stuff down instead of trying
to do shortcuts. So dS, and this is all--
let me do it in blue still because we're talking
about surface 2. dS is going to be equal to the
magnitude of the cross product of the partial of our
parameterization with respect to u, crossed with our partial
of our parameterization with respect to v, du dv. So let's write the partial
of our parameterization with respect to u. And you might say, wait,
where's our parameterization? Well, it's right over here. I just didn't write it in
the traditional ijk form, but I can. I could write r is
equal to-- maybe I call it r2 because we're
talking about surface 2. I shouldn't use
that orange color because I used
that for surface 1. I'll use that in a
still bluish color. So r for surface 2
is equal to cosine of ui plus the
sine of uj plus vk. And this is the range that the
u's and the v's can take on. If I want to take the partial
of r with respect to u, I would get-- let's
see, the derivative of cosine u with respect
to u is negative sine ui. The derivative of sine of u with
respect to u is plus cosine uj. And the derivative of v
with respect to u is just 0. So that's our partial
with respect to u. Our partial-- let' me do
this in a different color just for the sake of it. Our partial with respect
to v is equal to-- well, this is going to be 0,
this is going to be 0, and we're just
going to have one k. So it's just going
to be equal to k. And so now we're ready
to at least evaluate this cross product. So we'll evaluate
the cross product, and then we'll take
the magnitude of that, so just this part, so
just the cross product, the inside of that, before
we take the magnitude. So copy and paste. So just that is going to be
equal to the determinant, the 3 by 3 determinant. Let me write my
components, i, j, k. And then for r sub
u, we have this. Our i component is
negative sine of u. Our j component is cosine of u. And it has no k component,
so we'll put a 0 right there. And then r sub v, no i
component, no j component, and it has 1 for
its k coefficient. And so we can just
evaluate this. And so first, we think
about the i component. Well, we ignore that
column, that row. It's going to be
cosine of u minus 0. So it's going to
be cosine of ui. And then we'll have minus
j times-- ignore column, ignore row-- negative
sine of u times 1 minus 0. So it's negative j times
negative sine of u. So it's just going to
be plus sine of uj. And then for the k, we have
this times 0 minus this times 0. So we're just going to have 0. So this is the cross product. And if we want to take
the magnitude of this, if we just take the
magnitude of this-- so now we're ready to take
the magnitude of this. It's going to be equal
to-- the magnitude of this is going to be equal to
the square root of cosine of u squared plus
sine of u squared. There is no k component. And from the unit circle,
this is the most basic trig identity. This is just going
to be equal to 1. And so we have this part right
over here just simplifies to 1, which is nice. And dS simplifies to du dv
for at least this surface. And so now we are
ready to attempt to evaluate the integral. But I'm almost
running out of time, so I will do that
in the next video.