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JOEL LEWIS: Hi.

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Welcome back to recitation.

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You've been doing some work
on trig integration.

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I have a nice example here of
a problem that requires trig

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integration in order to solve.

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So what I'd like you to do is
to compute the volume of the

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solid that you get when you take
one hump of the curve, y

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equals sine ax, and you revolve
it around the x-axis.

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So you take the curve between
two consecutive roots, and

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then you, you know, revolve
that around the x-axis and

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that gives you some, I
don't know, vaguely

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football-shaped thing.

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And so then the question
is, what's the

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volume of that solid?

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So why don't you pause the
video, take a little while to

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work that out, come back, and
we can work it out together.

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Welcome back.

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In order to solve this problem,
we just are going to

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apply our usual methods for
computing a volume of a solid

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of rotation.

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So in order to do that, remember
that one of the

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things you need is you need to
know the region over which

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you're integrating, and you
need to choose a method of

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integration.

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So in this case, looking at
this region, here, I'm

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rotating it around the x-axis.

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i looks to me, so we
have two choices.

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We could do shells with
horizontal rectangles, or we

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could do disks with vertical
rectangles.

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Looks to me like vertical
rectangles are going to be the

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way to go for this function.

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Nice, simple, have their
base on the x-axis.

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You know, this is a nice
setup for disks.

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So we're going to take vertical

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disks here, like this.

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Vertical rectangles that
are going to spin

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into vertical disks.

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And so we're going to integrate
all these disks,

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we're going to add them up
starting at x equals 0, and

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going until the end
of this region.

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So we need to figure out what
the end of that region is, so

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we need sine ax to
be 0, again.

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Well, the first 0, the first
time sine is 0 after 0 is at

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pi, so we need ax to be pi.

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So this value is at pi over a.

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OK.

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So that's the setup
for the problem.

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And now we just need to remembe,
you know, how to do a

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problem like this.

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So we have each of
these disks.

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Well, it's height here is the
height of the function, which

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is y, in this case.

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So the area of a little disk--

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sorry.

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The area of a disk, the
little-- oh, dear.

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The element of volume, the
little bit of volume that we

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get is equal to--

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well, the area of the disk is pi
y squared, which is pi sine

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squared of ax.

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And then the thickness of
the disk is a little dx.

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So this is our little element
dV of volume.

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And now to get the whole volume,
we just integrate this

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over the appropriate range.

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So this means V is equal to the
integral from 0 to, as we

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said, to pi over a of pi times
sine squared of ax dx.

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So this is just like the
sorts of integrals you

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were doing in class.

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It's a definite integral.

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I guess most of the ones
you did were just

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anti-derivatives, but of
course, that's an easy

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translation to make by via
the fundamental theorem.

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So, OK, so we have
this ax here.

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You know, it's up to you.

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I think my life will be simpler
if I just make a

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little u-substitution, get rid
of the ax, and then I don't

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have to think about it
very much anymore.

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So I can take u equals ax,
so du equals a dx, or dx

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equals 1 over a du.

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So I, OK, so I make this
quick substitution.

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When x is 0, u is also 0.

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When x is pi over
a, u is just pi.

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So this becomes the integral
from 0 to pi.

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I can pull this pi out front.

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And I can pull, so I'm going to
get pi times sine squared u

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times dx is 1 over a du.

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So I'm going to pull the 1 over
a out front, as well.

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So it's pi over a times the
integral from 0 to pi of sine

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squared u du.

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OK.

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So now we've just simplified
it to the situation where

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we've just got a trig
integral, no other

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complications at all.

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How do we do this one?

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Well, OK, so this is not one
of those nice ones where we

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have an odd power for either
sine or cosine.

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We have sine is appearing to the
even power, 2, and cosine,

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well, it doesn't appear.

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It appears to the
even power, 0.

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If you like, you could
say it that way.

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So when we have a situation
where sine and cosine both

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appear in even powers, what we
need to do is we need to use

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one of our trig identities.

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We need to use a half
angle identity.

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So the identity in particular
that we want to use is we want

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to replace sine squared u
with something like a

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cosine of 2u somehow.

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So in order to do that, we
need to remember the

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appropriate identity.

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So one of the double angle
identities is cosine of 2t

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equals 1 minus 2 sine squared t,
which we can rewrite as the

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half angle identity, sine
squared t equals 1 minus

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cosine 2t over 2.

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So this is true for
any value, t.

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In particular, it's true
when t is equal to u.

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Back over here.

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So we can rewrite this by
replacing sine squared u with

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1 minus cosine of
2u divided by 2.

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So this integral becomes, so
our integral is equal to--

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well, we've still got the
pi over a in the front--

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integral from 0 to pi of 1 minus
cosine of 2u over 2 du.

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OK, and so now we
integrate it.

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So 1 over 2, that's easy.

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That's just so, OK, so the pi
over a is still out in front.

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1 over 2 integrates, just
gives us u over 2.

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How about cosine of 2u?

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Well, so minus cosine of
2u, so that should

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give us a minus sine.

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Right?

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Derivative of sine is cosine,
derivative of minus sine is

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minus cosine.

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So its minus sine of 2u, and
then because it's 2u, we're

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going to have to divide
by 2 again.

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So over 4.

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All right.

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If you don't believe me, of
course, you could always check

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by differentiating this
expression and making sure

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that it matches that
expression,

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the integrand, here.

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And OK, and so we need to take
that between u equals 0

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and u equals pi.

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So luckily we changed our bounds
of integration and we

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don't have to go all the
way back to x again.

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OK, so this is pi
over a times--

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OK, so u over 2 minus sine, 2u
over 4 when u is pi over 2

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minus sine of 2 pi.

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That's just 0, right?

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Yes, that's just 0.

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Good.

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So this term is just 0.

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pi over 2 minus--

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OK, now when we put in 0, here,
well, we get 0 minus

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sine of 0, so that's just
0-- so just pi over 2.

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OK.

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So the answer, then, is
pi squared over 2a.

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So that's the volume we
were looking for.

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So just to quickly recap, we
have our solid of revolution

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here that we get by spinning
this region around the x-axis.

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We use our typical methods for
computing volumes of solids of

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revolution.

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We've go a, when we do that, the
integral that we set up is

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a trig integral with a
sine squared in it.

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So both sine and cosine appear
to an even power

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in this trig integral.

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When you're in that situation,
you're going to have to use

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your half angle formulas,
like so.

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Once you do that,
you'll simplify.

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Sometimes life is hard, you'll
have to do it more than once.

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In this case, we only had to do
that once, so we got, then

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we, that reduces the integral
to something that's easy to

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compute, where you have one
of sine or cosine always

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appearing to an odd power.

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In this case, very simple.

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You just had a cosine.

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And then you integrate it, and
this was our final answer.

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I'll end there.

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