CHRISTINE BREINER: Welcome
back to recitation.
In this video, we want to
work on using the change
of variables technique.
In particular, we're going to
look at the following problem.
It says, using the change
of variables u is equal to x
squared minus y squared and
v is equal to y divided by x,
supply the limits and integrand
for the following integral,
which is the double integral
over region R of 1 over x
squared, dx*dy.
And R is the infinite region
in the first quadrant that
is both under the
curve y equals 1
over x, and to the
right of the curve
x squared minus y
squared equals 1.
So this is a
challenging problem.
Again, I want to
point out we just
want to find the limits
and the integrand.
You don't actually have
to compute the integral.
But it is going to be
tough, but stick with it.
Pause the video, give it
your best shot-- hopefully
you find the appropriate
limits and integrand--
and then when you
feel comfortable,
bring the video back up, and
I'll show you how I do it.
OK.
Welcome back.
So once again,
what we want to do
is this change of
variables problem
where we've defined u to be
x squared minus y squared,
v to be y divided by x, and
we have this region that
is in the first
quadrant and it's
under the curve y
equals 1 divided by x
and it's to the right of the
curve x squared minus y squared
equals 1.
And we want to
compute the limits
and integrand for that
particular integral.
So what I'm going to
do, to try and make
this as organized
as possible, is
I'm going to try first
to graph the region R,
or to figure out what the
region R is in the xy-plane.
Then I'm going to try and
figure out what the region
R is mapped to in the uv-plane.
So what it looks
like in the uv-plane.
That will give me my limits.
And then I'm going to try
and determine the Jacobian.
And then I will determine
from that and the fact
that I started with 1 divided by
x squared as my function I was
integrating, I will
put those two together
to figure out the integrand.
So there are a bunch of
steps to these problems.
But the first one, again is I'm
going to graph the region R.
So I'm going to give you a
very rough sketch, over here,
of the region R. And I know
it's in the first quadrant
and I know it's infinite.
I was already told that.
OK.
So in the xy-plane, the region
R is below the curve y equals 1
over x.
So let me draw that curve.
Again, this is very rough.
This is a rough sketch.
I'm putting up no
scale on purpose.
I'll put in one value,
maybe, in this whole thing.
OK?
And so this is the curve
y equals one over x.
And then I need
the curve which is
part of the hyperbola that
is x squared minus y squared
equals 1.
So I'll draw in
the part we need,
which looks roughly like this.
Something like that.
Again, this is not meant
to be an exact graph,
but to give you an idea of
what the region looks like.
And the only thing
I'm going to mention
is that this point we know
is x equals 1 and y equals 0.
So the region we're
interested in that
is both to the right of x
squared minus y squared equals
1 and below y equals 1 over
x and in the first quadrant
is exactly this region
I'm shading here.
So we want to understand
what the values of u and v
are along these bounds.
We need to understand
where this region maps
to when I do the
change of variables
in order to understand
what the limits are.
So let me put the graph of
this region in the uv-plane
so that we can really
understand what our bounds are.
And I know already
where it's going.
So I'm going to just make
the first quadrant, because I
know this is going into
the first quadrant.
So it doesn't always
work that something
in the first quadrant maps
into the first quadrant,
but in this case, I already did
the work, so I know it does.
So let me point out a few things
about where this region R maps.
The first thing I
want to point out
is that we actually
know that this curve,
under the change of
variables, maps to u equals 1.
Because if you remember, u
is equal to x squared minus y
squared.
So this whole curve is
going to map to u equals 1.
Now, I don't want the
whole curve for my region.
I only want this
little piece of it.
So I'm going to
have-- in my uv plane,
I'm going to have
some segment at 1.
And actually, I'll
just know that it's
some part of the line u equals
1 is going to show up in there.
But if you notice, I know
where it starts right away.
Because at x equal
1, y equals 0,
if I look at what v is-- if
we come back here and remember
what v is-- at x equal
1, y equals 0-- v is 0.
And so my starting point on this
segment-- if we come back here,
my starting point
on this segment
is actually also at (1, 0).
OK?
So I know there's
some point right
here that maps down to here
where the segment will stop.
I'll find that point
later, algebraically.
Right?
And then now we need to figure
out where these two curves go.
And then we can get
a picture, and then
we'll figure out
what that point is,
and we'll understand
all the limits.
So the first thing I want to
point out is along this curve,
we have y equals 0
and x is non-zero.
And just to help
ourselves, I'm going
to rewrite what the
change of variables
is here, so I don't have to keep
walking over to the other side.
Our change of variables
was u is equal to x
squared minus y squared, and
v was equal to y divided by x.
So this whole curve
has y equals 0.
So what happens to u and what
happens to v along that curve?
Well, u is going
to be x squared,
and v is going to equal 0.
And so the point
of this, really,
is that even though
in u, this curve maybe
is mapping at a different speed
in some form to this curve
here, it's still-- it's just
taking that segment goes--
or that infinitely long ray goes
to an infinitely long ray here
along the u-axis.
And again, that's because
along this ray, y equals 0.
And so v is equal
to 0 everywhere on
that ray and u is positive--
it's equal to x squared.
OK?
So I'm going to move
the u out of the way,
because we're going to say
this is part of the region,
or that's one bound
of the region.
And now I have to figure
out where this curve goes.
This curve is slightly
more complicated,
but I can still figure it out.
So I'm going to show you how I
do that sort of algebraically.
That curve-- if you
notice, if you remember--
is y equals 1 divided by x.
So that means that
on that curve--
let me even write it down--
on y equals 1 divided by x, v
is equal to 1 divided
by x divided by x.
So v is equal to 1 divided
by x squared, right?
And then what does
that mean about u?
u, then, is equal to-- well,
x squared is 1 divided by v,
and then y squared, because y
squared on that curve is just 1
divided by x squared,
is v. So let me just
make sure we all followed
that one more time.
We're looking at where
the curve y equals 1
over x goes in the change
of variables, right?
So that's the top curve up here.
y equals 1 over x is the
top curve of our region R.
So we want to know
where that goes.
Well, on y equals 1
over x, v is exactly
equal to 1 over x squared,
because v-- we know--
is y over x.
So if I just substitute in
for y, I get 1 over x squared.
Now, if I look at
this relationship,
this means x squared is equal
to 1 over v. So in terms of u,
x squared becomes 1
over v. And then y
squared-- which is
1 over x squared--
become v. So that curve is
u equals 1 over v minus v.
Now that curve,
roughly, is going
to look something like this.
And it might seem strange.
The thing is, I'm graphing
this in the uv-plane,
and I'm writing what looks
like u as a function of v,
and so it's sort of turned
around from how you usually
see these things written.
But this is the equation
that describes this curve.
And that is sufficient
to understand,
because when we use our--
when we determine our bounds,
we can determine our bounds from
u equals 0 now, to u equals 1
over v minus v. So we
now have the bounds in u.
We're actually doing quite well.
So we have this region.
We now have the bounds
completely in u.
u is going from u equals 0
to u equals 1 over v minus v.
But the problem is now
we don't know the bounds
in v. We don't know what
the bounds are in v,
and so we have to be
a little bit careful.
So actually, no.
I think I was wrong.
It's not 0, is it?
I said that twice now,
and that was incorrect.
u is going from 1, to 1 over
v minus v. So I apologize.
Because the slices of u
are going from whatever
the u-value starts--
which is at the value 1--
and it's coming this way.
So I apologize.
I was moving my arm like
I was doing the v-values,
but I actually wanted
to do the u-values.
So I want to go from where
u starts-- which is at u
equals 1-- to where u stops--
which is when it hits the curve
1 over v minus v equals u.
So hopefully I didn't
confuse anyone by that.
I'm glad I caught it, then.
OK, so now we understand
the bounds in u.
And then to
understand the bounds
in v, all we need to
understand is what
is the v-value at this point.
So once I know the
v-value at this point,
then I'm done with the bounds.
So let's see if
we can find that.
Well, the v-value at that point
is going to be at the point
where these two
curves intersect.
So let's see if we can do a
little algebra to understand
what that will look like.
So let me point out that
where those curves intersect,
I have the equation x squared
minus 1 over x squared
is equal to 1.
And if I want to find
x-values that satisfy this,
I'm also looking for
x-values that satisfy
x to the fourth minus 1
is equal to x squared,
which I can rewrite as x to the
fourth minus x squared minus 1
is equal to 0.
So I can actually use
the quadratic formula
on this in terms of x squared.
So what I get is I get x
squared is equal to 1--
I get plus or minus
root 5-- over 2.
And if you look at it,
the one you're actually
interested in-- you can figure
this out pretty quickly--
is the one that is plus.
OK?
I want the one that
is plus root 5 over 2.
So then that means x is the
square root of this quantity
at that point, right?
Or I could actually
think about it this way.
Let me point out
this. v is equal to 1
over x squared at that
point, because it's
on that curve where we
were talking about y
equals 1 over x.
So v is 1 over x squared.
So 1 over x squared is
just 1 over this quantity.
So it's the reciprocal of this.
It's also negative 1
plus root 5, over 2.
You can check that
if you need to.
But I will write it down
this way as the following:
this is the point 1 comma a.
And if I come over here, I will
denote a will equal negative 1
plus root 5, over 2.
And that's really just
1 divided by x squared.
So let me point that
out again, that a
is equal to 1
divided by x squared
at the point of intersection.
So hopefully you
can see all that.
So that tells us our
bounds completely.
We still have some work to do.
So I'm going to
put in the bounds
and I'm going to
leave an empty space.
Actually, no.
I won't do that,
because this can get
a little messy and confusing.
So I'm just going
to do the Jacobian,
and then we'll figure it
all out and write the answer
right at the end, so
there's no confusion.
But hopefully you see at this
point that we have the bounds.
We know that u goes from
1, to 1 over v minus v.
And v goes from 0
up to a, where a
is the value I've written here.
So we know the bounds.
So now we have to figure
out the integrand.
So let's first compute
the Jacobian, OK?
So now we're looking
at del u, v del x, y,
using the notation
we've seen in class.
And so del u, v
del x, y is going
to be the determinant
of the following matrix:
2x, negative 2y.
And then the derivative
respect to v of x
is negative y over x squared.
And the derivative of v with
respect to y is just 1 over x.
So if I take the determinant
of that, I get 2 minus 2 y
squared over x squared.
Which if you notice, in terms
of our change of variables,
is exactly equal to 2 minus 2 v
squared, because v is y over x.
And so I can rewrite
this as 2 times
the quantity 1 minus v squared.
OK?
So, so far so good, hopefully.
Now let's figure out how
to do the final step.
So the final step-- I'm
going to come back over
and just remind us what
the integrand was, OK?
If we come over here,
we're reminded that we were
integrating over the region
R of 1 over x squared, dx*dy.
Right?
That's what we were
interested in initially.
So now, if we come back, I'm
going to write that down just
to have it as a reference.
OK, that's what
we had initially.
Let me make sure.
Yes, that's what
we had initially.
And so now we know dx*dy is
equal to du*dv over 2 times 1
minus v squared.
So that is going to
replace the dx*dy.
And now we have to figure
out what to do with the 1
over x squared.
But, what do we have here?
Now I have to remind myself.
I can't remember all
the steps anymore.
We have u is equal to x
squared minus y squared.
Let me come back.
Now I've forgotten
what I was doing.
Ah, yes.
Now I remember, sorry.
OK.
So the point I should have
remembered that I forgot,
is that 1 minus v
squared is equal to u
divided by x squared.
That's what I had
figured out earlier,
that I just forgot when I
was staring at the board.
And to notice that, what
do we have to remember?
u is x squared minus y squared,
so if I divide everything
by x squared, the
first term is 1
and the second
term is v squared.
So, whew, that's good.
So I was a little
nervous there for second,
but I did in fact
do this earlier.
And I'd forgotten what I did.
So now, the 1 minus v
squared is actually the same
as u divided by x squared.
And notice what that
does to this term here.
That tells us that dx*dy over
x squared is actually equal
to du*dv over-- instead
of the 1 minus v squared,
I put u over x squared
and I get-- notice,
I get an x squared times
2, u divided by x squared.
Right?
I just replace the 1 minus v
squared with what I know it is,
the x squareds divide out,
and so I get du*dv over 2u.
So now the good news is
I have all the pieces,
because I'm about to
run out of board space.
So I have all the
pieces, so I'm just
going to put them together,
and then we're done.
So let me come here
in the final spot,
and say this is
our final answer.
Our final answer is that
we're integrating u from 1,
to 1 over v minus v. And then
we're integrating v from 0
to a-- where a is the value
I determined earlier--
of 1 over 2u, du*dv.
So this is the
final, final answer.
This was a long one.
And I'm sorry I had a little
brain freeze in the middle.
I couldn't remember how
I'd fixed that problem.
So what I did at
that point-- I just
want to point out that when I
was working on this problem,
and I had a 1 minus v
squared, I knew somehow
I had to figure out
how to relate that
and the x squared
in terms of u and v.
And so I actually
saw this expression.
I could have written
it better, maybe, as x
squared times this equals u.
OK.
And maybe that would
have been more obvious,
if that's the case.
But that was really
the step that
allowed me to
replace all of this
by things in terms
of u and v. Which
I know I should have
been able to do,
it's just a matter
of figuring it out.
So let me just go
back to the beginning
and remind you of each of
the steps very briefly,
and then we'll be done.
So we come back over
to the beginning.
We were starting with change
of variables supplied for us.
We already had an integral
in terms of x and y,
and we had an infinite region.
And what we were
asked to do is find
the limits and the integrand.
So the first step
for me is I always
find it very helpful to draw
the region in the xy-plane,
and then draw the new
region in the uv-plane.
Neither one of them
has to be perfect,
but the understanding of the
values of the curves in terms
of equations of u and v are very
important, to understand that.
That gives you the
bounds, the limits.
And then, so we
did all this work.
We found the limits.
There was a little
algebra in the middle.
We found the limits.
And then we found
the Jacobian, which
was going to tell us how
the variables were changing.
We found it in terms of x and y.
We rewrote it in
terms of u and v.
And so when we came back and we
compared what our integrand was
initially, we could
compare dx*dy to du*dv.
But then we also
had to figure out
how to replace the
1 over x squared.
So once we did all that, we had
everything in terms of u and v,
and then we finally had what
the integrand was going to be.
So there were a lot of steps,
but this was ultimately
what the problem was.
And again, I'll just point
out, this is the final solution
right here.
We integrated from 1, to
1 over v minus v, for u.
And we integrated from 0 to a
in v, the function 1 over 2u.
OK.
That is where I will stop.
