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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

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the order of integration.

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And how you can look
at a given region

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and you can integrate over
it by integrating dx dy

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or by 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

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vertices are the origin,
the point (0, 2),

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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

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like you to write down
what a double integral

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over these regions
looks like, but I'd

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like you to do it
two different ways.

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

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dx dy.

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

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in the order dy dx.

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So I'd like you to
express the integrals

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over these regions in
terms of iterated integrals

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in both possible orders.

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

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come back, and we can
work on it together.

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So the first thing
to do whenever

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you're given a
problem like this--

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and in fact, almost anytime you
have 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,

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really the first 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.

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So in part a, you
have a triangle,

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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

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and we can say, what
are the boundaries

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of 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,

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the right boundary is
the line x equals 0,

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and this sort of lower left
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

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is we want to figure out, OK,
if you're integrating this

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

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with respect to y
and then x, what

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does that integral look
like when you set it up

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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

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over this 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,

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and those bounds 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

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

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y, what is the
leftmost point and what

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is 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

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x is going to take is
along this line x equals 0.

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When you fix some value
of y, the rightmost point

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that x can 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 leftmost boundary value?

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So what we want to
do is we want to take

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that equation for that boundary
and we want to solve it

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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

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the line x 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,

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and what's the 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

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is going to become when we
evaluate it in the order dx dy.

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So now let's talk
about evaluating it

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in 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

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going to be for y in terms
of x, and the outer bounds

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are going to be
absolute bounds on x.

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So for y in terms
of x, that means we

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look at this region-- we want
to know, for a fixed value of x,

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what's the bottom
boundary of this region,

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and what's the top boundary?

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

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

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is this line 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

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a point in this
region are the values

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that y is at least
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,

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Those are the
absolute bounds on x.

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So we need to know what
the absolute leftmost

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point and the absolute rightmost
point in this region are.

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And so the absolute leftmost
point is this 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

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this right edge 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,

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and the 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

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the line that's the x-axis.

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

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of the circle in here.

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So this region inside the
circle, below the line y

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equal 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,

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and this is the
point square root

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of 2 comma square root of 2.

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That common point of the line
y equals x and the circle

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x squared plus y
squared equals 4.

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That's what this
boundary curve is:

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x squared plus y
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

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has the equation 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

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can talk about expressing it
as an iterated integral in two

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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, something dx and then

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dy.

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OK, so we need
bounds on x, which

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means for a fixed
value of y, we need

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to know what is the
leftmost boundary

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and what's the rightmost bound.

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So for a fixed
value of y, we want

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to know what the left edge
is and the right edge is.

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And it's easy to see because
we've drawn this picture,

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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

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is our actual circle.

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Yeah?

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So those are the left
and right boundaries,

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so what we 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

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going to be the rightmost
bound-- the largest

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value that x takes--
is when x squared

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plus y squared equals 4.

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So when x is equal to the square
root of 4 minus y squared.

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Now you might say
to me, why do I

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know that it's the
positive square root here

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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

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is the top half of the
circle and it's also

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the 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

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want the positive 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

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is the outermost variable that
we're integrating with respect

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to, so we want the absolute
bounds-- the absolute lowest

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value that y takes
in this region,

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and the absolute largest
value that y takes.

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So the smallest
value that y takes

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in this region-- that's
the lowest point-- that's

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

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And the largest
value that y takes--

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that's when y is as
large as possible

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as it can get in
this region-- is

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up at this point of
intersection there,

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so that's 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-- the integral over
this region R-- dy dx.

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OK.

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So we're going to
look at this region

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and we want to say-- 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-- right?--

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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 intersection--

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

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square root of 2--
when you're over there,

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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.

250
00:11:44,410 --> 00:11:47,860
So this is complicated, and what
does this complication mean?

251
00:11:47,860 --> 00:11:49,840
Well, it means that it's
not easy to write this

252
00:11:49,840 --> 00:11:51,770
as a single iterated integral.

253
00:11:51,770 --> 00:11:53,280
If you want to do
this in this way,

254
00:11:53,280 --> 00:11:55,520
you have to break the
region into two pieces,

255
00:11:55,520 --> 00:11:58,600
and write this double
integral as a sum of two

256
00:11:58,600 --> 00:12:00,090
iterated integrals.

257
00:12:00,090 --> 00:12:00,590
OK?

258
00:12:00,590 --> 00:12:03,930
So one iterated integral will
take care of the left part

259
00:12:03,930 --> 00:12:07,020
and one will take care
of the right part.

260
00:12:07,020 --> 00:12:08,665
So let's do the left part first.

261
00:12:11,550 --> 00:12:18,550
So here we're going to have a
iterated integral integrating

262
00:12:18,550 --> 00:12:20,120
with respect to y first.

263
00:12:20,120 --> 00:12:25,290
So to fixed value of x, we want
to know what the bounds on y

264
00:12:25,290 --> 00:12:25,834
are.

265
00:12:25,834 --> 00:12:27,500
And well, we can see
from this picture--

266
00:12:27,500 --> 00:12:30,180
when you're in this
triangle-- that y

267
00:12:30,180 --> 00:12:33,220
is going from the x-axis
up to the line y equals x.

268
00:12:33,220 --> 00:12:35,910
So that means the smallest
value that y can take is 0,

269
00:12:35,910 --> 00:12:38,225
and the largest value
that y can take is x.

270
00:12:38,225 --> 00:12:41,030
So here it's from 0 to x.

271
00:12:41,030 --> 00:12:43,050
And when you're
in this triangle,

272
00:12:43,050 --> 00:12:46,210
we need to know what the
bounds on x are, then.

273
00:12:46,210 --> 00:12:48,220
We need to know
the outer bounds.

274
00:12:48,220 --> 00:12:50,925
So we need to know the absolute
largest and smallest values

275
00:12:50,925 --> 00:12:51,967
that x can take.

276
00:12:51,967 --> 00:12:53,050
Well, what does that mean?

277
00:12:53,050 --> 00:12:55,791
We need to know the absolute
leftmost and absolute rightmost

278
00:12:55,791 --> 00:12:56,290
points.

279
00:12:56,290 --> 00:12:58,930
So the absolute leftmost
point is the origin.

280
00:12:58,930 --> 00:13:01,950
The absolute rightmost
is this vertical line

281
00:13:01,950 --> 00:13:03,540
x equals square root of 2.

282
00:13:03,540 --> 00:13:07,885
So over here, the
value of x is 0.

283
00:13:07,885 --> 00:13:11,570
And at the rightmost boundary
of this triangle, the value of x

284
00:13:11,570 --> 00:13:13,540
is the square root of 2.

285
00:13:13,540 --> 00:13:14,040
OK.

286
00:13:14,040 --> 00:13:16,090
So that's going to give
us the double integral

287
00:13:16,090 --> 00:13:20,130
just over this triangular
part of the region.

288
00:13:20,130 --> 00:13:22,420
Yeah?

289
00:13:22,420 --> 00:13:26,160
So now, we need to
add to this-- but I'm

290
00:13:26,160 --> 00:13:28,473
going to put it down on
this next line-- we need

291
00:13:28,473 --> 00:13:30,310
to add to this the
part of the integral

292
00:13:30,310 --> 00:13:34,580
over this little segment
of the circle here.

293
00:13:34,580 --> 00:13:38,750
The remainder of the region
that's not in that triangle.

294
00:13:38,750 --> 00:13:44,740
So for that, again, we're going
to write down two integrals,

295
00:13:44,740 --> 00:13:49,360
and it's going to be dy dx.

296
00:13:49,360 --> 00:13:50,052
Whew.

297
00:13:50,052 --> 00:13:51,315
We're nearly done, right?

298
00:13:54,560 --> 00:13:59,140
So y is inside, so we need
to know what the bounds on y

299
00:13:59,140 --> 00:14:01,041
are for a given value of x.

300
00:14:01,041 --> 00:14:02,540
So we need to know
for a given value

301
00:14:02,540 --> 00:14:06,820
of x, what are the bottom
and the topmost points

302
00:14:06,820 --> 00:14:08,490
of this region?

303
00:14:08,490 --> 00:14:10,450
So for a given value
of x, that means

304
00:14:10,450 --> 00:14:14,510
that y is going here between the
x-axis and between this circle.

305
00:14:14,510 --> 00:14:19,290
So the x-axis is y equals 0,
so that's the lower bound.

306
00:14:19,290 --> 00:14:22,650
So for the upper bound, we
need to know this circle.

307
00:14:22,650 --> 00:14:24,090
What is y on this circle?

308
00:14:24,090 --> 00:14:25,465
Well, the equation
of this circle

309
00:14:25,465 --> 00:14:27,080
is x squared plus
y squared equals 4,

310
00:14:27,080 --> 00:14:31,080
so y is equal to the square
root of the quantity 4 minus x

311
00:14:31,080 --> 00:14:31,580
squared.

312
00:14:36,440 --> 00:14:38,640
Where again, here we take
the positive square root,

313
00:14:38,640 --> 00:14:42,360
because this is a part of the
circle where y is positive.

314
00:14:42,360 --> 00:14:43,370
Yeah.

315
00:14:43,370 --> 00:14:46,490
If we were somehow on the
bottom part of the circle,

316
00:14:46,490 --> 00:14:49,222
then we would have to take a
negative square root there,

317
00:14:49,222 --> 00:14:51,180
but because we're on the
top part of the circle

318
00:14:51,180 --> 00:14:54,300
where y is positive, we
take a positive square root.

319
00:14:54,300 --> 00:14:54,980
OK, good.

320
00:14:54,980 --> 00:14:56,860
So those are the
bounds on y, and now we

321
00:14:56,860 --> 00:14:59,161
need to know the
absolute bounds on x.

322
00:14:59,161 --> 00:14:59,660
Yeah?

323
00:14:59,660 --> 00:15:01,780
So those are the bounds
on y in terms of x.

324
00:15:01,780 --> 00:15:03,870
And now because x
is the outer thing

325
00:15:03,870 --> 00:15:05,535
we're integrating
with respect to,

326
00:15:05,535 --> 00:15:07,530
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 --> 00:15:16,480
I don't really know what the
name for a shape like that is--

329
00:15:16,480 --> 00:15:20,020
but whatever that thing is, we
need to know what its leftmost

330
00:15:20,020 --> 00:15:21,750
and rightmost points are.

331
00:15:21,750 --> 00:15:25,070
We need to know the smallest and
largest values that x can take.

332
00:15:25,070 --> 00:15:30,620
And so its leftmost edge is this
line x equals square root of 2.

333
00:15:30,620 --> 00:15:33,660
And its rightmost edge is that
rightmost point on the circle

334
00:15:33,660 --> 00:15:35,490
there-- where the
circle hit the x-axis--

335
00:15:35,490 --> 00:15:37,390
and that's the value
when x equals 2.

336
00:15:40,710 --> 00:15:42,150
OK, so there you go.

337
00:15:42,150 --> 00:15:47,180
There's this last integral
written in the dy dx order,

338
00:15:47,180 --> 00:15:50,000
but we can't write it as a
single iterated integral.

339
00:15:50,000 --> 00:15:52,890
We need to write it as a sum of
two iterated integrals because

340
00:15:52,890 --> 00:15:54,930
of the shape of this region.

341
00:15:54,930 --> 00:15:55,430
All right.

342
00:15:58,600 --> 00:16:01,820
Let me just make one
quick, summary comment.

343
00:16:01,820 --> 00:16:05,570
Which is that if you're
doing this, one thing that

344
00:16:05,570 --> 00:16:09,167
should always be true, is
that these integrals, when

345
00:16:09,167 --> 00:16:10,750
you evaluate them--
so here, I haven't

346
00:16:10,750 --> 00:16:12,520
been writing an integrand.

347
00:16:12,520 --> 00:16:14,770
I guess the integrand has
always been 1, or something.

348
00:16:14,770 --> 00:16:19,000
But for any integrand,
the nature of this process

349
00:16:19,000 --> 00:16:26,050
is that it shouldn't matter
which order you integrate.

350
00:16:26,050 --> 00:16:28,810
You should get the same answer
if you integrate dx dy or dy

351
00:16:28,810 --> 00:16:29,680
dx.

352
00:16:29,680 --> 00:16:32,930
So one very low-level
check that you

353
00:16:32,930 --> 00:16:35,660
can make-- that you haven't done
anything horribly, egregiously

354
00:16:35,660 --> 00:16:38,060
wrong when changing the
bounds of integration--

355
00:16:38,060 --> 00:16:43,090
is that you can check that
actually these things evaluate

356
00:16:43,090 --> 00:16:44,010
the same.

357
00:16:44,010 --> 00:16:44,930
Yeah?

358
00:16:44,930 --> 00:16:46,910
Where you can
choose any function

359
00:16:46,910 --> 00:16:48,780
that you happen to
want to put in there--

360
00:16:48,780 --> 00:16:50,884
function of x and y-- and
evaluate this integral,

361
00:16:50,884 --> 00:16:53,550
and choose any function that you
happen to want to put in there,

362
00:16:53,550 --> 00:16:54,840
and evaluate those integrals.

363
00:16:54,840 --> 00:16:57,440
And see that you actually get
the same thing on both sides.

364
00:16:57,440 --> 00:17:00,360
Now one simple example
is that you could just

365
00:17:00,360 --> 00:17:02,180
evaluate the
integral as written,

366
00:17:02,180 --> 00:17:05,100
with a 1 written in there.

367
00:17:05,100 --> 00:17:07,400
And so in both cases,
what you should get

368
00:17:07,400 --> 00:17:09,320
is the area of the
region when you

369
00:17:09,320 --> 00:17:10,839
evaluate an integral like that.

370
00:17:10,839 --> 00:17:13,380
But you can also check with any
other function if you wanted.

371
00:17:19,910 --> 00:17:22,410
It won't show that what
you've done is right,

372
00:17:22,410 --> 00:17:25,980
but it will show if you've
done something wrong.

373
00:17:25,980 --> 00:17:29,517
That method will sometimes
pick it out, right?

374
00:17:29,517 --> 00:17:31,100
Because you'll
actually be integrating

375
00:17:31,100 --> 00:17:32,800
over two different regions,
and there's no reason

376
00:17:32,800 --> 00:17:34,290
you should get the same answer.

377
00:17:34,290 --> 00:17:36,680
So if you were to
compute these integrals

378
00:17:36,680 --> 00:17:38,140
and get different
numbers, then you

379
00:17:38,140 --> 00:17:40,440
would know that something
had gone wrong at some point

380
00:17:40,440 --> 00:17:43,780
for sure, and you'd have to go
and figure out where it was.

381
00:17:43,780 --> 00:17:45,633
I think I'll end there.