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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--

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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?

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This is that iterated
integral.

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So now, we want to do
the same thing.

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R--

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the integral over
this region R--

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dy dx.

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OK.

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So we're going to look at this
region and we want to say--

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dy is going to be
on the inside--

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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?

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So what's the bottom boundary
and what's the top boundary?

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But if you look at
this region--

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right?-- life is a little
complicated here.

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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

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intersection-- if you're at the
left of this line x equals

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square root of 2--

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when you're over there, y
is going from 0 to x.

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But if you're over in the right
part of this region,

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there's a different
upper boundary.

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Right?

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It's a different curve
that it came from.

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It has a different equation.

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So over here, y is going from
the x-axis up to the circle.

251
00:11:44,410 --> 00:11:46,960
So this is complicated,
and what does this

252
00:11:46,960 --> 00:11:47,860
complication mean?

253
00:11:47,860 --> 00:11:50,450
Well, it means that it's not
easy to write this as a single

254
00:11:50,450 --> 00:11:51,770
iterated integral.

255
00:11:51,770 --> 00:11:54,050
If you want to do this in this
way, you have to break the

256
00:11:54,050 --> 00:11:57,730
region into two pieces, and
write this double integral as

257
00:11:57,730 --> 00:12:00,120
a sum of two iterated
integrals.

258
00:12:00,120 --> 00:12:00,590
OK?

259
00:12:00,590 --> 00:12:04,140
So one iterated integral will
take care of the left part and

260
00:12:04,140 --> 00:12:07,020
one will take care of
the right part.

261
00:12:07,020 --> 00:12:12,080
So let's do the left part first.
So here we're going to

262
00:12:12,080 --> 00:12:19,410
have a iterated integral
integrating with respect to y

263
00:12:19,410 --> 00:12:24,390
first. So to fix the value of
x, we want to know what the

264
00:12:24,390 --> 00:12:25,850
bounds on y are.

265
00:12:25,850 --> 00:12:27,630
And well, we can see from
this picture-- when

266
00:12:27,630 --> 00:12:29,660
you're in this triangle--

267
00:12:29,660 --> 00:12:31,820
that y is going from the
x-axis up to the

268
00:12:31,820 --> 00:12:33,220
line y equals x.

269
00:12:33,220 --> 00:12:35,910
So that means the smallest value
that y can take is 0,

270
00:12:35,910 --> 00:12:38,225
and the largest value that
y can take is x.

271
00:12:38,225 --> 00:12:41,030
So here it's from 0 to x.

272
00:12:41,030 --> 00:12:43,990
And when you're in this
triangle, we need to know what

273
00:12:43,990 --> 00:12:46,210
the bounds on x are, then.

274
00:12:46,210 --> 00:12:48,220
We need to know the
outer bounds.

275
00:12:48,220 --> 00:12:50,570
So we need to know the absolute
largest and smallest

276
00:12:50,570 --> 00:12:52,250
values that x can take.

277
00:12:52,250 --> 00:12:53,050
Well, what does that mean?

278
00:12:53,050 --> 00:12:55,440
We need to know the absolute
leftmost and absolute

279
00:12:55,440 --> 00:12:56,290
rightmost points.

280
00:12:56,290 --> 00:12:58,930
So the absolute leftmost
point is the origin.

281
00:12:58,930 --> 00:13:02,650
The absolute rightmost is this
vertical line x equals

282
00:13:02,650 --> 00:13:03,540
square root of 2.

283
00:13:03,540 --> 00:13:07,885
So over here, the
value of x is 0.

284
00:13:07,885 --> 00:13:11,340
And at the rightmost boundary of
this triangle, the value of

285
00:13:11,340 --> 00:13:13,790
x is the square root of 2.

286
00:13:13,790 --> 00:13:14,040
OK.

287
00:13:14,040 --> 00:13:17,020
So that's going to give us the
double integral just over this

288
00:13:17,020 --> 00:13:20,130
triangular part of the region.

289
00:13:20,130 --> 00:13:22,420
Yeah?

290
00:13:22,420 --> 00:13:25,830
So now, we need to
add to this--

291
00:13:25,830 --> 00:13:28,590
but I'm going to put it down on
this next line-- we need to

292
00:13:28,590 --> 00:13:31,920
add to this the part of the
integral over this little

293
00:13:31,920 --> 00:13:34,580
segment of the circle here.

294
00:13:34,580 --> 00:13:38,750
The remainder of the region
that's not in that triangle.

295
00:13:38,750 --> 00:13:44,120
So for that, again, we're
going to write down two

296
00:13:44,120 --> 00:13:49,360
integrals, and it's
going to be dy dx.

297
00:13:49,360 --> 00:13:50,052
Whew.

298
00:13:50,052 --> 00:13:51,315
We're nearly done, right?

299
00:13:51,315 --> 00:13:54,560

300
00:13:54,560 --> 00:13:59,460
So y is inside, so we need to
know what the bounds on y are

301
00:13:59,460 --> 00:14:01,190
for a given value of x.

302
00:14:01,190 --> 00:14:04,710
So we need to know for a given
value of x, what are the

303
00:14:04,710 --> 00:14:08,490
bottom and the topmost points
of this region?

304
00:14:08,490 --> 00:14:11,290
So for a given value of x, that
means that y is going

305
00:14:11,290 --> 00:14:14,510
here between the x-axis and
between this circle.

306
00:14:14,510 --> 00:14:19,290
So the x-axis is y equals 0,
so that's the lower bound.

307
00:14:19,290 --> 00:14:22,650
So for the upper bound, we
need to know this circle.

308
00:14:22,650 --> 00:14:24,360
What is y on this circle?

309
00:14:24,360 --> 00:14:26,270
Well, the equation of this
circle is x squared plus y

310
00:14:26,270 --> 00:14:30,360
squared equals 4, so y is equal
to the square root of

311
00:14:30,360 --> 00:14:31,610
the quantity 4 minus
x squared.

312
00:14:31,610 --> 00:14:36,440

313
00:14:36,440 --> 00:14:38,900
Where again here, we take the
positive square root, because

314
00:14:38,900 --> 00:14:42,360
this is a part of the circle
where y is positive.

315
00:14:42,360 --> 00:14:43,370
Yeah.

316
00:14:43,370 --> 00:14:46,490
If we were somehow on the bottom
part of the circle,

317
00:14:46,490 --> 00:14:49,360
then we would have to take a
negative square root there,

318
00:14:49,360 --> 00:14:51,680
but because we're on the top
part of the circle where y is

319
00:14:51,680 --> 00:14:54,300
positive, we take a positive
square root.

320
00:14:54,300 --> 00:14:54,980
OK, good.

321
00:14:54,980 --> 00:14:57,380
So those are the bounds on y,
and now we need to know the

322
00:14:57,380 --> 00:14:59,260
absolute bounds on x.

323
00:14:59,260 --> 00:14:59,660
Yeah?

324
00:14:59,660 --> 00:15:01,780
So those are the bounds
on y in terms of x.

325
00:15:01,780 --> 00:15:04,560
And now because x is the outer
thing we're integrating with

326
00:15:04,560 --> 00:15:07,530
respect to, we need the
absolute bounds on x.

327
00:15:07,530 --> 00:15:12,290
And you can see in
this circular--

328
00:15:12,290 --> 00:15:16,250
I don't really know what the
name for a shape like that

329
00:15:16,250 --> 00:15:19,470
is-- but whatever that thing is,
we need to know what its

330
00:15:19,470 --> 00:15:21,750
leftmost and rightmost
points are.

331
00:15:21,750 --> 00:15:23,920
We need to know the smallest
and largest

332
00:15:23,920 --> 00:15:25,070
values that x can take.

333
00:15:25,070 --> 00:15:28,960
And so its leftmost edge
is this line x equals

334
00:15:28,960 --> 00:15:30,620
square root of 2.

335
00:15:30,620 --> 00:15:33,660
And its rightmost edge is that
rightmost point on the circle

336
00:15:33,660 --> 00:15:35,490
there-- where the circle
hit the x-axis--

337
00:15:35,490 --> 00:15:37,390
and that's the value
when x equals 2.

338
00:15:37,390 --> 00:15:40,710

339
00:15:40,710 --> 00:15:42,150
OK, so there you go.

340
00:15:42,150 --> 00:15:47,180
There's this last integral
written in the dy dx order,

341
00:15:47,180 --> 00:15:50,000
but we can't write it as a
single iterated integral.

342
00:15:50,000 --> 00:15:52,630
We need to write it as a sum
of two iterated integrals

343
00:15:52,630 --> 00:15:55,120
because of the shape
of this region.

344
00:15:55,120 --> 00:15:56,370
All right.

345
00:15:56,370 --> 00:15:58,600

346
00:15:58,600 --> 00:16:01,820
Let me just make one quick,
summary comment.

347
00:16:01,820 --> 00:16:05,840
Which is that if you're doing
this, one thing that should

348
00:16:05,840 --> 00:16:09,780
always be true, is that these
integrals, when you evaluate

349
00:16:09,780 --> 00:16:12,560
them- so here, I haven't been
writing an integrand.

350
00:16:12,560 --> 00:16:14,770
I guess the integrand has always
been 1, or something.

351
00:16:14,770 --> 00:16:19,260
But for any integrand, the
nature of this process is that

352
00:16:19,260 --> 00:16:26,050
it shouldn't matter which
order you integrate.

353
00:16:26,050 --> 00:16:27,170
You should get the same
answer if you

354
00:16:27,170 --> 00:16:29,680
integrate dx dy or dy dx.

355
00:16:29,680 --> 00:16:33,700
So one very low-level check that
you can make-- that you

356
00:16:33,700 --> 00:16:36,170
haven't done anything horribly,
egregiously wrong

357
00:16:36,170 --> 00:16:38,060
when changing the bounds
of integration--

358
00:16:38,060 --> 00:16:42,500
is that you can check that
actually these things

359
00:16:42,500 --> 00:16:44,010
evaluate the same.

360
00:16:44,010 --> 00:16:44,930
Yeah?

361
00:16:44,930 --> 00:16:47,520
Where you can choose any
function that you happen to

362
00:16:47,520 --> 00:16:48,780
want to put in there--

363
00:16:48,780 --> 00:16:51,280
function of x and y-- and
evaluate this integral, and

364
00:16:51,280 --> 00:16:53,300
choose any function that you
happen to want to put in

365
00:16:53,300 --> 00:16:54,840
there, and evaluate
those integrals.

366
00:16:54,840 --> 00:16:57,440
And see that you actually get
the same thing on both sides.

367
00:16:57,440 --> 00:17:00,970
Now one simple example is that
you could just evaluate the

368
00:17:00,970 --> 00:17:05,100
integral as written with
a 1 written in there.

369
00:17:05,100 --> 00:17:08,330
And so in both cases, what you
should get is the area of the

370
00:17:08,330 --> 00:17:11,110
region when you evaluate
an integral like that.

371
00:17:11,110 --> 00:17:12,490
But you can also check
with any other

372
00:17:12,490 --> 00:17:13,740
function if you wanted.

373
00:17:13,740 --> 00:17:19,910

374
00:17:19,910 --> 00:17:23,400
It won't show that what you've
done is right, but it will

375
00:17:23,400 --> 00:17:25,980
show if you've done
something wrong.

376
00:17:25,980 --> 00:17:29,940
That method will sometimes
pick it out, right?

377
00:17:29,940 --> 00:17:31,710
Because you'll actually be
integrating over two different

378
00:17:31,710 --> 00:17:33,060
regions, and there's no
reason you should

379
00:17:33,060 --> 00:17:34,290
get the same answer.

380
00:17:34,290 --> 00:17:37,230
So if you were to compute
these integrals and get

381
00:17:37,230 --> 00:17:39,350
different numbers, then you
would know that something had

382
00:17:39,350 --> 00:17:42,320
gone wrong at some point for
sure, and you'd have to go and

383
00:17:42,320 --> 00:17:43,780
figure out where it was.

384
00:17:43,780 --> 00:17:45,030
I think I'll end there.

385
00:17:45,030 --> 00:17:46,133