# Surface integral ex2 part 1: Parameterizing the surface | Multivariable Calculus | Khan Academy

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
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,
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.