WEBVTT

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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about computing

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double integrals and about
changing the order of

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integration.

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And how you can look at a
given region and you can

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integrate over it by integrating
dx dy or by

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integrating dy dx.

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So here I have some examples.

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I have two regions.

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So one region is the triangle
whose vertices are the origin,

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the point 0, 2, and the
point minus 1, 2.

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And the other one is a
sector of a circle.

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So the circle has a radius 2 and
is centered at the origin.

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And I want the part of that
circle that's above the x-axis

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and below the line y equals x.

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So what I'd like you to do is
I'd like you to write down

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what a double integral over
these regions looks like, but

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I'd like you to do it
two different ways.

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I'd like you to do it as
an iterated integral in

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the order dx dy.

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And I'd also like you to do it
as an iterated integral in the

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order dy dx.

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So I'd like you to express the
integrals over these regions

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in terms of iterated integrals
in both possible orders.

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So why don't you pause the
video, have a go at that, come

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back, and we can work
on it together.

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So the first thing to do
whenever you're given a

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problem like this--

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and in fact, almost
anytime you have

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to do a double integral--

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is to try and understand
the region in question.

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It's always a good idea
to understand

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the region in question.

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And by understand the region in
question, really the first

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thing that I mean is
draw a picture.

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All right.

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So let's do part a first.
So in part a, you have a

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triangle, it has vertices at the
origin, at the point 0, 2,

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and at the point minus 1, 2.

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So this triangle is our
region in question.

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So now that we've got a picture
of it, we can talk and

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we can say, what are
the boundaries of

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this region, right?

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And we want to know what
its boundaries are.

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So the top boundary is the line
y equals 2, the right

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boundary is the line x equals 0,
and this sort of lower left

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boundary-- the slanted line--

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is the line y equals minus 2x.

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OK.

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So those are the boundary
edges of this triangle.

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And so now what we want to
figure out is we want to

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figure out is, OK, if you're
integrating this with respect

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to x and then y, or if you're
integrating this with respect

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to y and then x, what does that
integral look like when

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you set it up as a
double integral.

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So let's start on one of them.

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It doesn't matter which one.

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So let's try and write the
double integral over this

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region R in the order dx dy.

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OK, so we have inside
bounds dx dy.

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So OK.

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So we need to find the bounds
on x first, and those bounds

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are going to be in terms of y.

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So the bounds on x.

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So that means when we look at
this region, what we want to

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figure out is we want to figure
out for a given value

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y, what is the leftmost
point and what is

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the rightmost point?

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What are the bounds on x?

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So for given value y, the
largest value x is going to

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take is along this
line x equals 0.

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When you fix some value of y,
the rightmost point that x can

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reach in this region is at
this line x equals 0.

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So x is going to go up to 0.

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That's going to be
its upper bound.

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The lower bound is going to be
the left edge of our region.

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For a given value of
y, what is that

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leftmost boundary value?

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So what we want to do is we want
to take that equation for

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that boundary and we want to
solve it for x in terms of y.

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So that's not hard to
do in this case.

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The line y equals minus
2x is also the line x

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equals minus 1/2 y.

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So that's that left boundary:
minus 1/2 y.

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OK?

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So then our outer
bounds are dy.

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So we want to find the
absolute bounds on y.

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What's the smallest value that
y takes, and what's the

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largest value that y takes?

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So that means what's the lowest
point of this region

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and what's the highest?

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And so the lowest point
here is the origin.

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So that's when y takes
the value of 0.

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And the highest point-- the
very top of this region--

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is when y equals 2.

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OK.

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So this is what that double
integral is going to become

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when we evaluate it in
the order dx dy.

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So now let's talk about
evaluating it in

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the opposite order.

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So let's switch our
bounds for dy dx.

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So we want the double integral
over R, dy dx.

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OK, so this is going to be
an iterated integral.

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And this time the inner bounds
are going to be for y in terms

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of x, and the outer bounds
are going to be

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absolute bounds on x.

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So for y in terms of x, that
means we look at this region--

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we want to know for a
fixed value of x--

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what's the bottom boundary of
this region, and what's the

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top boundary?

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So here, it's easy to see that
the bottom boundary is this

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line y equals minus 2x, and the
top boundary is this line

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y equals 2.

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So y is going from
minus 2x to 2.

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Yeah?

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So for a fixed value of x, the
values of y that give you a

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point in this region are the
values that y is at least

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minus 2x and at most 2.

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So OK.

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And now we need the
outer bounds.

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So the outer bounds have to be
some real numbers, Those are

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the absolute bounds on x.

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So we need to know what the
absolute leftmost point and

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the absolute rightmost point
in this region are.

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And so the absolute leftmost
point is this

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point minus 1, 2.

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So that has an x-value
of minus 1.

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And the absolute rightmost point
is along this right edge

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at x equals 0.

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OK.

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So here are the two integrals.

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The double integral with respect
to x then y, and the

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double integral with respect
to y and then x.

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OK.

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So that's the answer
to part a.

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Let's go on to part b.

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So for part b, our region is we
take a circle of radius 2,

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and we take the line y equals x,
and we take the line that's

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the x-axis.

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And so we want a circle, and
we want this sector of the

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circle in here.

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So this region inside the
circle, below the line y equal

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x, and above the x-axis.

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So this wedge of this circle.

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Let's see.

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This value is at x equals 2,
this is the origin, and this

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is the point square root of
2 comma square root of 2.

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A common point of the line y
equals x in the circle x

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squared plus y squared
equals 4.

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That's what this boundary curve
is: x squared plus y

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squared equals 4.

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And of course, this boundary
curve is the line y equals x.

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And this boundary line is the
x-axis, which has the equation

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y equals 0.

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So those are our boundary
curves for our region.

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We've got this nice picture,
so now we can talk about

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expressing it as an iterated
integral in

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two different orders.

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So let's again start off with
this with respect to x first,

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and then with respect to y.

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So we want the double integral
over R, dx dy.

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So this should be an
iterated integral,

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something dx and then dy.

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OK, so we need bounds on x,
which means for a fixed value

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of y, we need to know what is
the leftmost boundary and

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what's the rightmost bound.

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So for a fixed value of y, we
want to know what the left

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edge is and the right edge is.

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And it's easy to see
because we've drawn

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this picture, right?

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Drawing the picture makes this
a much easier process.

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The left edge is this line y
equals x and the right edge is

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our actual circle.

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Yeah?

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So those are the left and right
boundaries, so what we

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put here are just the equations
of that left edge

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and the equation of
that right edge.

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But we want their equations in
the form x equals something.

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And that's the something
that we put there.

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So for this left edge, it's
the line x equals y.

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So the left bound is y there.

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In this region, x
is at least y.

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And the upper bound here,
which is going to be the

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rightmost bound-- the largest
value that x takes--

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is when x squared plus
y squared equals 4.

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So when x is equal to
the square root

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of 4 minus y squared.

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Now you might say to me, why
do I know that it's the

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positive square root here and
not the negative square root?

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And if you said that
to yourself,

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that's a great question.

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And the answer is that this part
of the circle is the top

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half of the circle
and it's also the

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right half of the circle.

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So here we have positive
values of x.

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So it's the right half
of the circle.

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We want the positive values of
x, so we want the positive

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square root.

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OK.

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Good.

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And so those are the
bounds on x.

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Now we need the bounds on y.

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So the bounds on y, well,
what are they?

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Well, we want the absolute
bounds on y. y is the

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outermost variable that we're
integrating with respect to,

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so we want the absolute bounds--
the absolute lowest

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value that y takes in this
region, and the absolute

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largest value that y takes.

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So the smallest value that y
takes in this region-- that's

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the lowest point--

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that's along this line, and
that's when y equals 0.

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And the largest value that y
takes-- that's when y is as

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large as possible as it can
get in this region--

00:10:27.060 --> 00:10:29.650
is up at this point of
intersection there, so that's

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when y is equal to the
square root of 2.

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OK, three quarters done.

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Yeah?

00:10:36.940 --> 00:10:39.290
This is that iterated
integral.

00:10:39.290 --> 00:10:42.902
So now, we want to do
the same thing.

00:10:42.902 --> 00:10:44.262
R--

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the integral over
this region R--

00:10:46.900 --> 00:10:49.800
dy dx.

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OK.

00:10:50.920 --> 00:10:53.510
So we're going to look at this
region and we want to say--

00:10:53.510 --> 00:10:55.790
dy is going to be
on the inside--

00:10:55.790 --> 00:10:57.910
so we're going to say, OK, so
we need to know for a fixed

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value of x, what's the smallest
value that y can take

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and what's the largest value
that y can take?

00:11:03.270 --> 00:11:06.680
So what's the bottom boundary
and what's the top boundary?

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But if you look at
this region--

00:11:08.030 --> 00:11:10.290
right?-- life is a little
complicated here.

00:11:10.290 --> 00:11:12.740
Because if you're in the left
half of this region--

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what do I mean by left half--

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I mean if you're to the
left of this point of

00:11:16.950 --> 00:11:19.450
intersection-- if you're at the
left of this line x equals

00:11:19.450 --> 00:11:21.150
square root of 2--

00:11:21.150 --> 00:11:26.660
when you're over there, y
is going from 0 to x.

00:11:26.660 --> 00:11:31.610
But if you're over in the right
part of this region,

00:11:31.610 --> 00:11:33.120
there's a different
upper boundary.

00:11:33.120 --> 00:11:34.250
Right?

00:11:34.250 --> 00:11:36.300
It's a different curve
that it came from.

00:11:36.300 --> 00:11:38.200
It has a different equation.

00:11:38.200 --> 00:11:44.410
So over here, y is going from
the x-axis up to the circle.

00:11:44.410 --> 00:11:46.960
So this is complicated,
and what does this

00:11:46.960 --> 00:11:47.860
complication mean?

00:11:47.860 --> 00:11:50.450
Well, it means that it's not
easy to write this as a single

00:11:50.450 --> 00:11:51.770
iterated integral.

00:11:51.770 --> 00:11:54.050
If you want to do this in this
way, you have to break the

00:11:54.050 --> 00:11:57.730
region into two pieces, and
write this double integral as

00:11:57.730 --> 00:12:00.120
a sum of two iterated
integrals.

00:12:00.120 --> 00:12:00.590
OK?

00:12:00.590 --> 00:12:04.140
So one iterated integral will
take care of the left part and

00:12:04.140 --> 00:12:07.020
one will take care of
the right part.

00:12:07.020 --> 00:12:12.080
So let's do the left part first.
So here we're going to

00:12:12.080 --> 00:12:19.410
have a iterated integral
integrating with respect to y

00:12:19.410 --> 00:12:24.390
first. So to fix the value of
x, we want to know what the

00:12:24.390 --> 00:12:25.850
bounds on y are.

00:12:25.850 --> 00:12:27.630
And well, we can see from
this picture-- when

00:12:27.630 --> 00:12:29.660
you're in this triangle--

00:12:29.660 --> 00:12:31.820
that y is going from the
x-axis up to the

00:12:31.820 --> 00:12:33.220
line y equals x.

00:12:33.220 --> 00:12:35.910
So that means the smallest value
that y can take is 0,

00:12:35.910 --> 00:12:38.225
and the largest value that
y can take is x.

00:12:38.225 --> 00:12:41.030
So here it's from 0 to x.

00:12:41.030 --> 00:12:43.990
And when you're in this
triangle, we need to know what

00:12:43.990 --> 00:12:46.210
the bounds on x are, then.

00:12:46.210 --> 00:12:48.220
We need to know the
outer bounds.

00:12:48.220 --> 00:12:50.570
So we need to know the absolute
largest and smallest

00:12:50.570 --> 00:12:52.250
values that x can take.

00:12:52.250 --> 00:12:53.050
Well, what does that mean?

00:12:53.050 --> 00:12:55.440
We need to know the absolute
leftmost and absolute

00:12:55.440 --> 00:12:56.290
rightmost points.

00:12:56.290 --> 00:12:58.930
So the absolute leftmost
point is the origin.

00:12:58.930 --> 00:13:02.650
The absolute rightmost is this
vertical line x equals

00:13:02.650 --> 00:13:03.540
square root of 2.

00:13:03.540 --> 00:13:07.885
So over here, the
value of x is 0.

00:13:07.885 --> 00:13:11.340
And at the rightmost boundary of
this triangle, the value of

00:13:11.340 --> 00:13:13.790
x is the square root of 2.

00:13:13.790 --> 00:13:14.040
OK.

00:13:14.040 --> 00:13:17.020
So that's going to give us the
double integral just over this

00:13:17.020 --> 00:13:20.130
triangular part of the region.

00:13:20.130 --> 00:13:22.420
Yeah?

00:13:22.420 --> 00:13:25.830
So now, we need to
add to this--

00:13:25.830 --> 00:13:28.590
but I'm going to put it down on
this next line-- we need to

00:13:28.590 --> 00:13:31.920
add to this the part of the
integral over this little

00:13:31.920 --> 00:13:34.580
segment of the circle here.

00:13:34.580 --> 00:13:38.750
The remainder of the region
that's not in that triangle.

00:13:38.750 --> 00:13:44.120
So for that, again, we're
going to write down two

00:13:44.120 --> 00:13:49.360
integrals, and it's
going to be dy dx.

00:13:49.360 --> 00:13:50.052
Whew.

00:13:50.052 --> 00:13:51.315
We're nearly done, right?

00:13:54.560 --> 00:13:59.460
So y is inside, so we need to
know what the bounds on y are

00:13:59.460 --> 00:14:01.190
for a given value of x.

00:14:01.190 --> 00:14:04.710
So we need to know for a given
value of x, what are the

00:14:04.710 --> 00:14:08.490
bottom and the topmost points
of this region?

00:14:08.490 --> 00:14:11.290
So for a given value of x, that
means that y is going

00:14:11.290 --> 00:14:14.510
here between the x-axis and
between this circle.

00:14:14.510 --> 00:14:19.290
So the x-axis is y equals 0,
so that's the lower bound.

00:14:19.290 --> 00:14:22.650
So for the upper bound, we
need to know this circle.

00:14:22.650 --> 00:14:24.360
What is y on this circle?

00:14:24.360 --> 00:14:26.270
Well, the equation of this
circle is x squared plus y

00:14:26.270 --> 00:14:30.360
squared equals 4, so y is equal
to the square root of

00:14:30.360 --> 00:14:31.610
the quantity 4 minus
x squared.

00:14:36.440 --> 00:14:38.900
Where again here, we take the
positive square root, because

00:14:38.900 --> 00:14:42.360
this is a part of the circle
where y is positive.

00:14:42.360 --> 00:14:43.370
Yeah.

00:14:43.370 --> 00:14:46.490
If we were somehow on the bottom
part of the circle,

00:14:46.490 --> 00:14:49.360
then we would have to take a
negative square root there,

00:14:49.360 --> 00:14:51.680
but because we're on the top
part of the circle where y is

00:14:51.680 --> 00:14:54.300
positive, we take a positive
square root.

00:14:54.300 --> 00:14:54.980
OK, good.

00:14:54.980 --> 00:14:57.380
So those are the bounds on y,
and now we need to know the

00:14:57.380 --> 00:14:59.260
absolute bounds on x.

00:14:59.260 --> 00:14:59.660
Yeah?

00:14:59.660 --> 00:15:01.780
So those are the bounds
on y in terms of x.

00:15:01.780 --> 00:15:04.560
And now because x is the outer
thing we're integrating with

00:15:04.560 --> 00:15:07.530
respect to, we need the
absolute bounds on x.

00:15:07.530 --> 00:15:12.290
And you can see in
this circular--

00:15:12.290 --> 00:15:16.250
I don't really know what the
name for a shape like that

00:15:16.250 --> 00:15:19.470
is-- but whatever that thing is,
we need to know what its

00:15:19.470 --> 00:15:21.750
leftmost and rightmost
points are.

00:15:21.750 --> 00:15:23.920
We need to know the smallest
and largest

00:15:23.920 --> 00:15:25.070
values that x can take.

00:15:25.070 --> 00:15:28.960
And so its leftmost edge
is this line x equals

00:15:28.960 --> 00:15:30.620
square root of 2.

00:15:30.620 --> 00:15:33.660
And its rightmost edge is that
rightmost point on the circle

00:15:33.660 --> 00:15:35.490
there-- where the circle
hit the x-axis--

00:15:35.490 --> 00:15:37.390
and that's the value
when x equals 2.

00:15:40.710 --> 00:15:42.150
OK, so there you go.

00:15:42.150 --> 00:15:47.180
There's this last integral
written in the dy dx order,

00:15:47.180 --> 00:15:50.000
but we can't write it as a
single iterated integral.

00:15:50.000 --> 00:15:52.630
We need to write it as a sum
of two iterated integrals

00:15:52.630 --> 00:15:55.120
because of the shape
of this region.

00:15:55.120 --> 00:15:56.370
All right.

00:15:58.600 --> 00:16:01.820
Let me just make one quick,
summary comment.

00:16:01.820 --> 00:16:05.840
Which is that if you're doing
this, one thing that should

00:16:05.840 --> 00:16:09.780
always be true, is that these
integrals, when you evaluate

00:16:09.780 --> 00:16:12.560
them- so here, I haven't been
writing an integrand.

00:16:12.560 --> 00:16:14.770
I guess the integrand has always
been 1, or something.

00:16:14.770 --> 00:16:19.260
But for any integrand, the
nature of this process is that

00:16:19.260 --> 00:16:26.050
it shouldn't matter which
order you integrate.

00:16:26.050 --> 00:16:27.170
You should get the same
answer if you

00:16:27.170 --> 00:16:29.680
integrate dx dy or dy dx.

00:16:29.680 --> 00:16:33.700
So one very low-level check that
you can make-- that you

00:16:33.700 --> 00:16:36.170
haven't done anything horribly,
egregiously wrong

00:16:36.170 --> 00:16:38.060
when changing the bounds
of integration--

00:16:38.060 --> 00:16:42.500
is that you can check that
actually these things

00:16:42.500 --> 00:16:44.010
evaluate the same.

00:16:44.010 --> 00:16:44.930
Yeah?

00:16:44.930 --> 00:16:47.520
Where you can choose any
function that you happen to

00:16:47.520 --> 00:16:48.780
want to put in there--

00:16:48.780 --> 00:16:51.280
function of x and y-- and
evaluate this integral, and

00:16:51.280 --> 00:16:53.300
choose any function that you
happen to want to put in

00:16:53.300 --> 00:16:54.840
there, and evaluate
those integrals.

00:16:54.840 --> 00:16:57.440
And see that you actually get
the same thing on both sides.

00:16:57.440 --> 00:17:00.970
Now one simple example is that
you could just evaluate the

00:17:00.970 --> 00:17:05.100
integral as written with
a 1 written in there.

00:17:05.100 --> 00:17:08.330
And so in both cases, what you
should get is the area of the

00:17:08.330 --> 00:17:11.110
region when you evaluate
an integral like that.

00:17:11.110 --> 00:17:12.490
But you can also check
with any other

00:17:12.490 --> 00:17:13.740
function if you wanted.

00:17:19.910 --> 00:17:23.400
It won't show that what you've
done is right, but it will

00:17:23.400 --> 00:17:25.980
show if you've done
something wrong.

00:17:25.980 --> 00:17:29.940
That method will sometimes
pick it out, right?

00:17:29.940 --> 00:17:31.710
Because you'll actually be
integrating over two different

00:17:31.710 --> 00:17:33.060
regions, and there's no
reason you should

00:17:33.060 --> 00:17:34.290
get the same answer.

00:17:34.290 --> 00:17:37.230
So if you were to compute
these integrals and get

00:17:37.230 --> 00:17:39.350
different numbers, then you
would know that something had

00:17:39.350 --> 00:17:42.320
gone wrong at some point for
sure, and you'd have to go and

00:17:42.320 --> 00:17:43.780
figure out where it was.

00:17:43.780 --> 00:17:45.030
I think I'll end there.