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# Surface integral ex2 part 1

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.