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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

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that requires trig
integration in order to solve.

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

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that you get when you take one
hump of the curve y equals sine

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a*x and you revolve
it around the x-axis.

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So you take the curve between
two consecutive roots,

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

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

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vaguely football-shaped thing.

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And so then the
question is, what's

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the volume of that solid?

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

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to 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

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are going to apply
our usual methods

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for computing a volume
of a solid of rotation.

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So in order to do that, remember
that one of the things you need

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is you need to know the region
over which you're integrating,

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and you need to choose
a method of integration.

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So in this case,
looking at this region,

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here, I'm rotating
it around the x-axis.

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It looks to me-- so
we have two choices.

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We could do shells with
horizontal rectangles,

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or we could do disks
with vertical rectangles.

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Looks to me like
vertical rectangles

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are going to be the 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 disks here, like this.

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Vertical rectangles
that are going

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to spin 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,

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and going until the
end of this region.

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

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so we need sine
a*x to be 0, again.

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

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at pi, so we need a*x 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 remember, you know,

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how to do a problem like this.

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So we have each of these disks.

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Well, its height
here is the height

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of the function, which
is y, in this case.

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So the area of a
little disk-- 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-- well, the area
of the disk is pi y squared,

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which is pi sine squared of a*x.

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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

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integrate this over
the appropriate range.

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

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to pi over a of pi times
sine squared of a*x dx.

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So this is just like
the sorts of integrals

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you 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 anti-derivatives,

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but of course, that's
an easy translation

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to make by via the
fundamental theorem.

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So, OK, so we have
this a*x 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 little

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u-substitution,
get rid of the a*x,

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and then I don't have to think
about it very much anymore.

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So I can take u equals
a*x, so du equals a*dx,

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or dx 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

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to pi-- 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

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u 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 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

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where we have an odd power
for either sine or cosine.

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We have sine is appearing
to the even power of 2,

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and cosine, well,
it doesn't appear.

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It appears to the even
power 0, if you like.

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You could say it that way.

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So when we have
a situation where

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sine and cosine both appear in
even powers, what we need to do

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is we need to use 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

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is we want to replace sine
squared u with something

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like a cosine of 2u somehow.

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So in order to do that,
we need to remember

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the 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

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we can rewrite as the half angle
identity sine squared t equals

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1 minus 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

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with 1 minus cosine
of 2u divided by 2.

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

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we've still got the pi over a
in the front-- integral from 0

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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

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should give us a minus sine.

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Right?

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Derivative of sine is cosine,
derivative of minus sine

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is minus cosine.

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

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we're 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,

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you could always check by
differentiating this expression

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and making sure that it matches
that expression, the integrand,

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here.

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

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equals 0 and u equals pi.

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So luckily we changed
our bounds of integration

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

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OK, so this is pi
over a times-- OK,

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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-- OK, now
when we put in 0, here,

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well, we get 0 minus sine
of 0, so that's just 0-- so

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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

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of solids of revolution.

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

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is 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

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going to have to use 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,

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so we got-- then
we-- that reduces

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the integral to something
that's easy to compute,

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where you have one
of sine or cosine

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always 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.