JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been
learning about computing
double integrals
and about changing
the order of integration.
And how you can look
at a given region
and you can integrate over
it by integrating dx dy
or by integrating dy dx.
So here I have some examples.
I have two regions.
So one region is
the triangle whose
vertices are the origin,
the point (0, 2),
and the point minus 1, 2.
And the other one is
a sector of a circle.
So the circle has a radius 2
and is centered at the origin.
And I want the part of that
circle that's above the x-axis
and below the line y equals x.
So what I'd like
you to do is I'd
like you to write down
what a double integral
over these regions
looks like, but I'd
like you to do it
two different ways.
I'd like you to do it as an
iterated integral in the order
dx dy.
And I'd also like you to do
it as an iterated integral
in the order dy dx.
So I'd like you to
express the integrals
over these regions in
terms of iterated integrals
in both possible orders.
So why don't you pause the
video, have a go at that,
come back, and we can
work on it together.
So the first thing
to do whenever
you're given a
problem like this--
and in fact, almost anytime you
have to do a double integral--
is to try and understand
the region in question.
It's always a good
idea to understand
the region in question.
And by understand the
region in question,
really the first thing that
I mean is draw a picture.
All right.
So let's do part a first.
So in part a, you
have a triangle,
it has vertices at the
origin, at the point (0, 2),
and at the point minus 1, 2.
So this triangle is
our region in question.
So now that we've got a
picture of it, we can talk
and we can say, what
are the boundaries
of this region, right?
And we want to know
what its boundaries are.
So the top boundary is
the line y equals 2,
the right boundary is
the line x equals 0,
and this sort of lower left
boundary-- the slanted line--
is the line y equals minus 2x.
OK.
So those are the boundary
edges of this triangle.
And so now what we
want to figure out
is we want to figure out, OK,
if you're integrating this
with respect to x and then y,
or if you're integrating this
with respect to y
and then x, what
does that integral look
like when you set it up
as a double integral.
So let's start on one of them.
It doesn't matter which one.
So let's try and write
the double integral
over this region R
in the order dx dy.
OK, so we have
inside bounds dx dy.
So OK.
So we need to find
the bounds on x first,
and those bounds are
going to be in terms of y.
So the bounds on x.
So that means when we
look at this region, what
we want to figure out is we want
to figure out for a given value
y, what is the
leftmost point and what
is the rightmost point?
What are the bounds on x?
So for given value
y, the largest value
x is going to take is
along this line x equals 0.
When you fix some value
of y, the rightmost point
that x can reach in this region
is at this line x equals 0.
So x is going to go up to 0.
That's going to be
its upper bound.
The lower bound is going to be
the left edge of our region.
For a given value of y, what is
that leftmost boundary value?
So what we want to
do is we want to take
that equation for that boundary
and we want to solve it
for x in terms of y.
So that's not hard
to do in this case.
The line y equals
minus 2x is also
the line x equals minus 1/2 y.
So that's that left
boundary: minus 1/2 y.
OK?
So then our outer bounds are dy.
So we want to find the
absolute bounds on y.
What's the smallest
value that y takes,
and what's the largest
value that y takes?
So that means what's the
lowest point of this region
and what's the highest?
And so the lowest point
here is the origin.
So that's when y
takes the value of 0.
And the highest point-- the
very top of this region--
is when y equals 2.
OK.
So this is what
that double integral
is going to become when we
evaluate it in the order dx dy.
So now let's talk
about evaluating it
in the opposite order.
So let's switch our
bounds for dy dx.
So we want the double
integral over R, dy dx.
OK, so this is going to
be an iterated integral.
And this time the
inner bounds are
going to be for y in terms
of x, and the outer bounds
are going to be
absolute bounds on x.
So for y in terms
of x, that means we
look at this region-- we want
to know, for a fixed value of x,
what's the bottom
boundary of this region,
and what's the top boundary?
So here, it's easy to see
that the bottom boundary is
this line y equals minus
2x, and the top boundary
is this line y equals 2.
So y is going from
minus 2x to 2.
Yeah?
So for a fixed value of x,
the values of y that give you
a point in this
region are the values
that y is at least
minus 2x and at most 2.
So OK.
And now we need
the outer bounds.
So the outer bounds have
to be some real numbers,
Those are the
absolute bounds on x.
So we need to know what
the absolute leftmost
point and the absolute rightmost
point in this region are.
And so the absolute leftmost
point is this point minus 1, 2.
So that has an
x-value of minus 1.
And the absolute
rightmost point is along
this right edge at x equals 0.
OK.
So here are the two integrals.
The double integral with
respect to x then y,
and the double integral with
respect to y and then x.
OK.
So that's the answer to part a.
Let's go on to part b.
So for part b, our region is
we take a circle of radius 2,
and we take the line y
equals x, and we take
the line that's the x-axis.
And so we want a circle,
and we want this sector
of the circle in here.
So this region inside the
circle, below the line y
equal x, and above the x-axis.
So this wedge of this circle.
Let's see.
This value is at x equals
2, this is the origin,
and this is the
point square root
of 2 comma square root of 2.
That common point of the line
y equals x and the circle
x squared plus y
squared equals 4.
That's what this
boundary curve is:
x squared plus y
squared equals 4.
And of course, this boundary
curve is the line y equals x.
And this boundary line
is the x-axis, which
has the equation y equals 0.
So those are our boundary
curves for our region.
We've got this nice
picture, so now we
can talk about expressing it
as an iterated integral in two
different orders.
So let's again start off with
this with respect to x first,
and then with respect to y.
So we want the double
integral over R, dx dy.
So this should be an iterated
integral, something dx and then
dy.
OK, so we need
bounds on x, which
means for a fixed
value of y, we need
to know what is the
leftmost boundary
and what's the rightmost bound.
So for a fixed
value of y, we want
to know what the left edge
is and the right edge is.
And it's easy to see because
we've drawn this picture,
right?
Drawing the picture makes
this a much easier process.
The left edge is this line y
equals x and the right edge
is our actual circle.
Yeah?
So those are the left
and right boundaries,
so what we put here are just
the equations of that left edge
and the equation
of that right edge.
But we want their equations in
the form x equals something.
And that's the something
that we put there.
So for this left edge,
it's the line x equals y.
So the left bound is y there.
In this region, x is at least y.
And the upper bound
here, which is
going to be the rightmost
bound-- the largest
value that x takes--
is when x squared
plus y squared equals 4.
So when x is equal to the square
root of 4 minus y squared.
Now you might say
to me, why do I
know that it's the
positive square root here
and not the negative
square root?
And if you said
that to yourself,
that's a great question.
And the answer is that
this part of the circle
is the top half of the
circle and it's also
the right half of the circle.
So here we have
positive values of x.
So it's the right
half of the circle.
We want the positive
values of x, so we
want the positive square root.
OK.
Good.
And so those are
the bounds on x.
Now we need the bounds on y.
So the bounds on y,
well, what are they?
Well, we want the
absolute bounds on y. y
is the outermost variable that
we're integrating with respect
to, so we want the absolute
bounds-- the absolute lowest
value that y takes
in this region,
and the absolute largest
value that y takes.
So the smallest
value that y takes
in this region-- that's
the lowest point-- that's
along this line, and
that's when y equals 0.
And the largest
value that y takes--
that's when y is as
large as possible
as it can get in
this region-- is
up at this point of
intersection there,
so that's when y is equal
to the square root of 2.
OK, three quarters done.
Yeah?
This is that iterated integral.
So now, we want to
do the same thing.
R-- the integral over
this region R-- dy dx.
OK.
So we're going to
look at this region
and we want to say-- dy is
going to be on the inside--
so we're going to say, OK, so
we need to know for a fixed
value of x, what's the
smallest value that y can take
and what's the largest
value that y can take?
So what's the bottom boundary
and what's the top boundary?
But if you look at
this region-- right?--
life is a little
complicated here.
Because if you're in the
left half of this region--
what do I mean by left half?
I mean if you're to the left
of this point of intersection--
if you're at the left
of this line x equals
square root of 2--
when you're over there,
y is going from 0 to x.
But if you're over in the
right part of this region,
there's a different
upper boundary.
Right?
It's a different curve
that it came from.
It has a different equation.
So over here, y is going from
the x-axis up to the circle.
So this is complicated, and what
does this complication mean?
Well, it means that it's
not easy to write this
as a single iterated integral.
If you want to do
this in this way,
you have to break the
region into two pieces,
and write this double
integral as a sum of two
iterated integrals.
OK?
So one iterated integral will
take care of the left part
and one will take care
of the right part.
So let's do the left part first.
So here we're going to have a
iterated integral integrating
with respect to y first.
So to fixed value of x, we want
to know what the bounds on y
are.
And well, we can see
from this picture--
when you're in this
triangle-- that y
is going from the x-axis
up to the line y equals x.
So that means the smallest
value that y can take is 0,
and the largest value
that y can take is x.
So here it's from 0 to x.
And when you're
in this triangle,
we need to know what the
bounds on x are, then.
We need to know
the outer bounds.
So we need to know the absolute
largest and smallest values
that x can take.
Well, what does that mean?
We need to know the absolute
leftmost and absolute rightmost
points.
So the absolute leftmost
point is the origin.
The absolute rightmost
is this vertical line
x equals square root of 2.
So over here, the
value of x is 0.
And at the rightmost boundary
of this triangle, the value of x
is the square root of 2.
OK.
So that's going to give
us the double integral
just over this triangular
part of the region.
Yeah?
So now, we need to
add to this-- but I'm
going to put it down on
this next line-- we need
to add to this the
part of the integral
over this little segment
of the circle here.
The remainder of the region
that's not in that triangle.
So for that, again, we're going
to write down two integrals,
and it's going to be dy dx.
Whew.
We're nearly done, right?
So y is inside, so we need
to know what the bounds on y
are for a given value of x.
So we need to know
for a given value
of x, what are the bottom
and the topmost points
of this region?
So for a given value
of x, that means
that y is going here between the
x-axis and between this circle.
So the x-axis is y equals 0,
so that's the lower bound.
So for the upper bound, we
need to know this circle.
What is y on this circle?
Well, the equation
of this circle
is x squared plus
y squared equals 4,
so y is equal to the square
root of the quantity 4 minus x
squared.
Where again, here we take
the positive square root,
because this is a part of the
circle where y is positive.
Yeah.
If we were somehow on the
bottom part of the circle,
then we would have to take a
negative square root there,
but because we're on the
top part of the circle
where y is positive, we
take a positive square root.
OK, good.
So those are the
bounds on y, and now we
need to know the
absolute bounds on x.
Yeah?
So those are the bounds
on y in terms of x.
And now because x
is the outer thing
we're integrating
with respect to,
we need the absolute
bounds on x.
And you can see
in this circular--
I don't really know what the
name for a shape like that is--
but whatever that thing is, we
need to know what its leftmost
and rightmost points are.
We need to know the smallest and
largest values that x can take.
And so its leftmost edge is this
line x equals square root of 2.
And its rightmost edge is that
rightmost point on the circle
there-- where the
circle hit the x-axis--
and that's the value
when x equals 2.
OK, so there you go.
There's this last integral
written in the dy dx order,
but we can't write it as a
single iterated integral.
We need to write it as a sum of
two iterated integrals because
of the shape of this region.
All right.
Let me just make one
quick, summary comment.
Which is that if you're
doing this, one thing that
should always be true, is
that these integrals, when
you evaluate them--
so here, I haven't
been writing an integrand.
I guess the integrand has
always been 1, or something.
But for any integrand,
the nature of this process
is that it shouldn't matter
which order you integrate.
You should get the same answer
if you integrate dx dy or dy
dx.
So one very low-level
check that you
can make-- that you haven't done
anything horribly, egregiously
wrong when changing the
bounds of integration--
is that you can check that
actually these things evaluate
the same.
Yeah?
Where you can
choose any function
that you happen to
want to put in there--
function of x and y-- and
evaluate this integral,
and choose any function that you
happen to want to put in there,
and evaluate those integrals.
And see that you actually get
the same thing on both sides.
Now one simple example
is that you could just
evaluate the
integral as written,
with a 1 written in there.
And so in both cases,
what you should get
is the area of the
region when you
evaluate an integral like that.
But you can also check with any
other function if you wanted.
It won't show that what
you've done is right,
but it will show if you've
done something wrong.
That method will sometimes
pick it out, right?
Because you'll
actually be integrating
over two different regions,
and there's no reason
you should get the same answer.
So if you were to
compute these integrals
and get different
numbers, then you
would know that something
had gone wrong at some point
for sure, and you'd have to go
and figure out where it was.
I think I'll end there.
