CHRISTINE BREINER: Welcome
back to recitation.
In this video, I'd like us to
work on the following problem:
So given the region
R shown below,
which I'll mention
more of in a moment,
compute the integral
double integral
over R of the
quantity 4 x squared
minus y squared to
the fourth dx dy,
and I want you to do it
by changing variables to u
is equal 2x minus y and
v is equal to 2x plus y.
And the region,
defined down here,
is just these
three line segments
and the intersection
of the region bounded
by these three line segments.
So one of them is a portion
of the y-axis, one of the them
is a portion of the line
defined by 2x minus y equals 0,
and one of them is a
portion of the line defined
by 2x plus y equals 2.
So why don't you do
this, make sure you
use this change of variables,
and then I'll come back,
and I'll draw the accompanying
region in the u, v plane,
and then we'll see
how I set it up.
OK, welcome back.
Well, the first
thing I'm going to do
is I'm going to draw the
region in the u, v plane that
is determined by this region in
the x, y plane and the change
of variables.
Now, as was mentioned in the
lecture, all of the changes
are linear, and so I'm going
to be taking lines to lines.
So what I'm going to do is
determine at each endpoint
here where that endpoint goes
under the transformation,
and then I will
connect the dots.
So what do I get?
Well, this first one is (0,
0) for x and y, and so that
corresponds to x and y
are 0 here, so u is 0,
and x and y are 0 here,
so it would be a 0,
so that's the point (0,
0) on the u, v plane.
Now, I want to
mention something here
is that based on how
these lines were defined,
this is telling you that all
along this line 2x minus y
equals 0, that means u is equal
to 0 all along that segment.
And if you'll notice also,
2x plus y is equal to 2
here, so v is equal to 2
all along this segment.
So we should expect whatever
u is, v's going to be 2 here,
and whatever v is here,
u is going to be 0.
OK, so I'm going to have
something with u, 0 initially,
and then v, 2, I
can already expect,
it's going to come up here
and move over that way.
But let's just check.
So this is the point (0,
2), 0 for x and 2 for y.
Let's see what it is
in the u, v variables.
So u in that case is negative
2 and v in that case is 2,
and so I'm going to
go left 2 and up 2.
This might not be drawn to
scale, but this is negative 2,
and this is up 2
here, and so that's
where this point goes
in the transformation.
This point is 1/2 comma 1.
You can actually
check what it is.
But actually, as I
mentioned, you already
know what has to happen, because
this segment is connected right
here, and then we said all
along this line, v is 2,
and all along this line, u is 0,
and so that actually carves out
that rectangle.
So if you were worried and
you weren't able to do it
that way, what you
could actually do
is say, well, where does
the point (1/2, 1) go?
If I plug in 1/2
for x and 1 for y,
I notice that I get u
equals 0 and v equals 2,
and that's this point.
And so now, I'm
looking at this region.
Now, these two things
are not drawn to scale,
because what I want to point
out is they're both triangles.
It's easy to find the area
based on how they're sitting
in two-dimensional space.
So what we see
here is if I wanted
to find the area
of this triangle,
let's make this the base.
The base is 2 and
the height is 1/2,
so 1/2 of base times height
is-- that area is equal to 1/2.
So the area of R is 1/2, and
what's the area of this one?
The area this one is-- well,
I've got base 2 and height 2,
and so it's 1/2 of 2 times 2,
and so that gives me area as 2.
So here the area is
1/2, here the area is 2,
and so notice that
I've multiplied
by 4 to get from here to here.
So I can anticipate
that I should
have-- based on the principle
you saw in lecture, I
should have something
like du dv is going
to be equal to 4 times dx dy.
That is what I expect to get.
Now let's see if when
we do the Jacobian,
we get the same thing.
Let me double check, right?
This is area 2, this is area
1/2, so I have 1/2 times
4 is going to give me 2.
Yeah, that's what I should get.
Oops!
That sticks out a little.
So now let's just
check our Jacobian
and see if that is
indeed what we get.
So I'm going to look at u sub
x, u sub y, v sub x, v sub y,
right?
So u sub x is 2, u
sub y is negative 1,
so I get 2, negative 1.
v sub x is 2, v sub y
is 1, so I get 2, 1.
2 times 1 is 2, minus negative
2 gives me, indeed, 4.
So I do get, in fact, du
dv is equal to 4 dx dy.
So I know that I'm going to
have to-- because I'm changing
variables, though,
from dx dy to du dv,
I'm going to divide
by 4, obviously,
when I do the substitution.
The substitution will
be replacing just dx dy.
So I just mention
that, but notice:
Again, we get what
we expect to get.
We got a 4 based on the
picture and we got a 4
based on the Jacobian.
And so now, we need to
finish up the process.
We need to figure out how to
write this in terms of u and v
and then figure out our
bounds in terms of u and v,
and then we're done.
So let me mention, one
thing you should notice
is that u times v is
equal to exactly-- this
is going to be a
difference of two squares,
and it is precisely 4 x
squared minus y squared.
If you need to multiply it out
to check, you can check it,
but that's indeed what it is,
which is why this particular
substitution is quite nice,
because that means this
function I'm supposed to be
integrating is just u*v raised
to the fourth.
So I'm almost done.
So let me write in the
pieces I know, and then we'll
fill in the last two spots,
or four spots, which will be
all the bounds on the integral.
So I'm going to write
it here, give myself
some space to write the bounds.
So I'm replacing the 4 x squared
minus y squared by a u*v,
so I'm going to get u times
v, and then the function,
that is raised to the fourth
in the initial problem,
so I raise that to the fourth.
Then dx dy, as I
mentioned, will be replaced
by a du dv divided by 4.
And so I can just put this
over 4 and write du dv.
I should be careful
which order I want to do.
It doesn't really matter.
I can do either one.
du dv, OK?
And so now I want to
know what u goes from
and to and then what v
has to go from and to.
So if I come over
here, you'll see
it didn't matter,
because I could have
picked either direction to go.
But if I'm going
to go with u, I'm
coming from whatever this
function is over to here,
right?
And this value here is easy.
That's u equals 0.
So the top bound for u is 0,
and the bottom bound for u,
this is v is equal to minus
u, so the bottom bound for u
is when u is equal
to negative v.
So I'm running from-- because
I put the du on the inside,
my first one is
running from minus v
to-- what did I say-- 0, and
then the v-values from there
go between 0 and 2.
You can see this easily
from the picture.
So they go between 0 and 2.
And I am not going to finish it
off from here, because it would
require not too much work.
It's actually quite
simple, because it's just
a polynomial and u and
a polynomial and v,
but there are a lot
of powers and there
will be big powers of 2.
So suffice it to
say, at this point,
we can evaluate this
integral and it's quite easy
to evaluate.
And the main point I
think we should make
is how did it make it simpler?
I mean, the initial problem,
if we come over here,
this is annoying to
find an anti-derivative,
but not impossible.
But the really
annoying part is I
would have to take this
region and split it
into a bunch of pieces,
or at least two pieces,
to evaluate the integral
in a reasonable way
or my y-values would go from
this line here up to this line
here.
It could be complicated.
So what we've really done is
we've simplified the region.
That's the easiest thing.
I mean, we also
simplified the function,
but we've really simplified the
region we're integrating over.
And so we only have-- one is a
function and one is a constant,
and that's quite nice
to have on the inside.
So I think that's
where I'll stop.
