WEBVTT
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Hi.
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Welcome back to recitation.
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In class, one of the things
you've talked about recently
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was computing surface areas
of solids of rotation.
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So I have a nice problem
relating to that here.
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So the circle with center
(R, 0) and radius little r--
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so this is center big
R, 0, and radius little
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r, which is less than big R--
is rotated around the y-axis.
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And the question is,
what's the surface
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area of the resulting solid?
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So we have here the circle.
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Its center is at
the point big R, 0,
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and its radius is
little r, so this
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is the equation of that circle.
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And we're going to rotate
this circle around this axis.
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So we're going to
spin it around.
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And what you're going
to get is a donut,
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or what mathematicians
call a torus.
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So here's a little
schematic of it
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here, with one dotted little
cross section corresponding
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to this circle.
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So the question is, what is
the surface area of this torus?
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So why don't you
pause the video,
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take a few minutes to work
this problem out yourself,
00:01:07.069 --> 00:01:08.860
come back, and we can
work it out together.
00:01:17.270 --> 00:01:21.970
So when we solved surface
area problems in class,
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we took the curve that
we were going to rotate,
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and we cut it into lots of
little pieces with length ds.
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So let me draw that, just
very quickly, up here.
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So we had, you took
whatever your curve was,
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and you cut it into lots
of these little segments.
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And then for each segment,
you rotated it around an axis.
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And so if the segment has
this little length ds,
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a little piece of arc
length-- so in our case,
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we're going to rotate it around
the y-axis, so the length,
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the area that this thing
traces out as it spins around,
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is going to be this
little piece of area, dA,
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which is equal to 2*pi*x*ds.
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Now, this is a
little bit different
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than most of the examples
Professor Jerison
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did in class, because here we're
rotating around the y-axis, not
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around the x-axis.
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So if you rotated
around the x-axis,
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what you would get is 2*pi*y*ds.
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And here we get 2*pi*x*dx.
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The idea of the x and the
y in this formula, this x,
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it's just telling
you what the radius
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is between your little segment
and the axis around which
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you're rotating it.
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So here this is, you know,
2*pi*x is the circumference
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of this circle that it traces
out, and ds is its length,
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because it's giving you a
little ribbon as it goes around.
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So we have this formula,
dA equals 2*pi*x*ds.
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And so in order to get the
surface area, what we do,
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is we have to integrate this
over an appropriate region.
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So in order to do
that, we first need,
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you know, all the variables in
the integrand to be the same,
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so we need to write everything
in terms of x, or everything
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in terms of y, or everything
in terms of some variable
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that we can integrate against.
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So in our particular
case with this torus,
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I think we can take
advantage of a little bit
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of symmetry here, which
is that this, you know,
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torus is top-bottom
symmetric, right?
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As the top half of the
circle goes around,
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it traces out one surface.
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As the bottom half of
the circle goes around,
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it traces out another surface.
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But those two surfaces
have exactly the same area.
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They're just mirror
images of each other,
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because the circle is symmetric.
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So we can just
consider the problem
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of spinning the top half
of the circle around.
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And so for the top
half of the circle,
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we can write down
an equation for y
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in terms of x, and so then
we can integrate-- you know,
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that sets up a nice
integral with respect to x.
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So in order to do this, well
we're going to need two things.
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So we're going to need
to know what ds is.
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And so you had a couple
of different formulas
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for this in class.
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So you had ds-- so one
easy mnemonic that I
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like is to write ds equals
the square root of dx squared
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plus dy squared.
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So for me, this always--
I can remember this,
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because it's just the
Pythagorean theorem, right?
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So you have a little
dx horizontal distance,
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and a little dy
vertical distance,
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and so the ds is just a
hypotenuse of that triangle.
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So that's how I remember this.
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And so then you also have
the equivalent formula
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if you factor out
a dx from here.
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You know, divide through by
dx and multiply outside by dx.
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You can write this as the
square root of 1 plus dy by dx
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squared times, outside
the square root, dx.
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So this is our little piece ds.
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And OK, we have an x
in the formula already.
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So ds-- so we now have
ds with a dx here.
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So the only thing
we have left, is
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we need to figure out
what dy/dx is in order
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to put all this into the
formula, in order to integrate.
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So OK.
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So dy/dx.
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So OK, so that means I
need y in terms of x.
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Now let-- OK.
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So if we're focusing only on
the top half of this torus, that
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means y-- well, we can
solve this equation for y,
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but when we take
the square root,
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we're only taking the positive
square root, because we're only
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looking at the top
half of the torus,
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and then we'll just
double at the end
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to account for the
bottom half as well.
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So this is y equals
the square root
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of-- so you solve,
you subtract x minus r
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squared from both sides.
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So this is little r squared
minus x minus big R squared.
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So that's y.
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And now you can differentiate
to get dy by dx.
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All right.
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So now we have to do
our chain rule right.
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So we've got a 1/2 power.
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So this is 1/2
times-- well, it's
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going to be the inside
to the minus 1/2.
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So this is over the
square root of r squared
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minus x minus R quantity
squared, and now I
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need to multiply by the
derivative of the inside, which
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is minus 2 times x minus R. OK.
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And the 2's cancel
out a little bit,
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so we can rewrite this as
minus x minus R divided
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by the square root
of little r squared
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minus x minus R squared.
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OK.
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So that's dy/dx.
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The square root--
that's a little ugly,
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but it's OK, because we're
about to square it out again.
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So all right.
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So we've got dy/dx.
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So now we go back,
now we can compute ds,
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and then with ds, we can
go even further back,
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and we can compute dA.
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And once we've got
dA, to get a, we just
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integrate dA over the
appropriate bounds.
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So-- which we haven't
figured out yet, by the way.
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We will have to talk
about that at the very,
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you know, in a minute.
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OK.
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So we take this,
so we have dy/dx.
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So from dy/dx, we
get ds is equal
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to-- it's the square root of,
well, let's use this formula.
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It's the square root of 1
plus, now, dy/dx squared.
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OK.
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So I put this in
and I square it.
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So I get x minus
R squared on top.
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You know, you square the
minus sign, gives you 1.
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And on the bottom, we
square the square root,
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so we get little r squared
minus x minus R squared, OK, dx.
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Great.
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Good.
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So now we add these two
things together, right?
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I mean, well we
want to-- I should
00:08:20.840 --> 00:08:24.050
say-- we want to simplify
this to a usable form.
00:08:24.050 --> 00:08:25.470
And so to put it
in a usable form,
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well we're going
to manipulate it
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until it looks nice, or as
nice as we could hope for.
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And in this case, right.
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So we can, there's an
obvious simplification,
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or simplifying step,
which is we can
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add this 1 into the fraction.
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OK.
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And so this is little r
squared minus x minus big R
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squared, over little r squared
minus x minus big R squared,
00:08:44.500 --> 00:08:48.050
and so when you add it
to x minus big R squared
00:08:48.050 --> 00:08:52.220
over little r squared minus
x minus big R squared,
00:08:52.220 --> 00:08:53.300
this part cancels.
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And so you're left with-- OK.
00:08:55.900 --> 00:08:59.430
And now I'm just going to pass
the square root through right
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away, so this is little r over
the square root of little r
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squared minus x minus
big R squared dx.
00:09:13.940 --> 00:09:14.440
OK.
00:09:14.440 --> 00:09:16.023
So just a little bit
of algebra there.
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So this is ds.
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Now, OK, so now we're going--
we want to compute surface area,
00:09:21.180 --> 00:09:26.730
so we need dA, and dA
is 2*pi*x times ds.
00:09:26.730 --> 00:09:27.530
So-- OK.
00:09:27.530 --> 00:09:31.080
So that's easy to write
down now that I've got ds.
00:09:31.080 --> 00:09:46.300
So this is dA equal to 2*pi*r*x
over r squared minus x minus
00:09:46.300 --> 00:09:52.730
big R squared, square
root of that, dx.
00:09:52.730 --> 00:09:53.230
All right.
00:09:53.230 --> 00:09:54.850
This is dA.
00:09:54.850 --> 00:09:56.950
You know, we haven't done
any calculus, actually.
00:09:56.950 --> 00:09:58.741
Oh, I guess we took a
derivative somewhere.
00:09:58.741 --> 00:10:01.460
We haven't done any
integration yet.
00:10:01.460 --> 00:10:05.072
Now, to compute the surface
area, we just integrate this.
00:10:05.072 --> 00:10:10.350
So we get to, you know, the
calculus step of this problem.
00:10:10.350 --> 00:10:12.140
So integrate this.
00:10:12.140 --> 00:10:14.670
But of course, you know,
I'm expecting a number out
00:10:14.670 --> 00:10:15.170
at the end.
00:10:15.170 --> 00:10:17.020
I'm taking a definite integral.
00:10:17.020 --> 00:10:18.650
So I need bounds.
00:10:18.650 --> 00:10:21.190
Well, what bounds do I need?
00:10:21.190 --> 00:10:23.580
Well, I'm integrating
with respect to x.
00:10:23.580 --> 00:10:28.310
So I need to integrate over
the relevant values of x.
00:10:28.310 --> 00:10:29.972
What are the
relevant values of x?
00:10:29.972 --> 00:10:31.680
Well, come back to
our picture over here.
00:10:34.320 --> 00:10:35.310
We have this circle.
00:10:35.310 --> 00:10:39.840
Its center is at x
equals R, y equals 0.
00:10:39.840 --> 00:10:41.280
And it has radius little r.
00:10:41.280 --> 00:10:45.470
So, you know, the
relevant values of x
00:10:45.470 --> 00:10:47.930
are just from the leftmost
point of the circle
00:10:47.930 --> 00:10:49.770
to the rightmost
point of the circle.
00:10:49.770 --> 00:10:51.640
And so this leftmost
point is just-- well,
00:10:51.640 --> 00:10:53.460
the radius is
little r, so this is
00:10:53.460 --> 00:10:56.280
big R minus-- x equals
big R minus little r,
00:10:56.280 --> 00:11:00.520
and the rightmost point is x
equals big R plus little r.
00:11:00.520 --> 00:11:07.130
So the bounds-- so I'm going
to go up here-- so I have area
00:11:07.130 --> 00:11:09.645
is what I get when
I integrate dA.
00:11:09.645 --> 00:11:14.720
And I want to integrate it from
x equals big R minus little
00:11:14.720 --> 00:11:19.320
r to big R plus little
r, and dA is this thing
00:11:19.320 --> 00:11:20.840
I found, just a moment ago.
00:11:20.840 --> 00:11:22.660
So this is-- well, OK.
00:11:22.660 --> 00:11:27.970
So 2*pi little r is a constant.
00:11:27.970 --> 00:11:30.350
I'm just going to factor
that out in front.
00:11:30.350 --> 00:11:39.890
So this is 2*pi little r times
x over the square root of little
00:11:39.890 --> 00:11:51.130
r squared minus big R minus--
sorry-- x minus big R squared.
00:11:51.130 --> 00:11:53.500
That's all under
the square root.
00:11:53.500 --> 00:11:55.791
ds.
00:11:55.791 --> 00:11:56.290
OK.
00:11:56.290 --> 00:11:58.840
So now we have to figure out
how to integrate this thing.
00:11:58.840 --> 00:12:00.510
Right?
00:12:00.510 --> 00:12:04.730
So this is a little ugly.
00:12:04.730 --> 00:12:05.970
It's not horrible, though.
00:12:05.970 --> 00:12:06.470
Right?
00:12:06.470 --> 00:12:08.550
So down here we have
something that's
00:12:08.550 --> 00:12:13.270
really reminiscent of one of
these trig integral things
00:12:13.270 --> 00:12:13.790
we've done.
00:12:13.790 --> 00:12:14.290
Right?
00:12:14.290 --> 00:12:16.400
We've got a square root
of a something squared
00:12:16.400 --> 00:12:18.280
minus a something else squared.
00:12:18.280 --> 00:12:21.510
So that, reminds, you
know, what does that
00:12:21.510 --> 00:12:23.740
remind, maybe some
sine substitution.
00:12:23.740 --> 00:12:24.950
Something like that.
00:12:24.950 --> 00:12:27.300
There's some there's some
trig substitution waiting
00:12:27.300 --> 00:12:30.240
to happen here.
00:12:30.240 --> 00:12:30.960
But, so OK.
00:12:30.960 --> 00:12:36.040
We could sort of shoot to do
it all in one substitution.
00:12:36.040 --> 00:12:38.040
I like, my life
always feels simpler
00:12:38.040 --> 00:12:40.500
when I do one little
substitution at a time.
00:12:40.500 --> 00:12:42.470
And so one little
substitution I could do
00:12:42.470 --> 00:12:46.700
is to simplify this
x minus r thing.
00:12:46.700 --> 00:12:48.600
I could just shift this by r.
00:12:48.600 --> 00:12:51.560
So I'm just going to do a
little linear substitution.
00:12:51.560 --> 00:12:57.600
I'm going to do u
equals x minus big R,
00:12:57.600 --> 00:12:59.970
or I'm going to want to
substitute the other way,
00:12:59.970 --> 00:13:06.050
so that's the same thing as
saying x equals u plus big R.
00:13:06.050 --> 00:13:11.460
And so du equals dx.
00:13:11.460 --> 00:13:14.110
This is a simple
little substitution.
00:13:14.110 --> 00:13:15.980
And I'm going to have
to move the bounds,
00:13:15.980 --> 00:13:22.680
so when x is big R minus little
r, then u is minus little r,
00:13:22.680 --> 00:13:25.600
and this top bound here, u is
going to be equal to little r.
00:13:25.600 --> 00:13:26.620
Positive little r.
00:13:26.620 --> 00:13:30.810
So let me just make
that substitution.
00:13:30.810 --> 00:13:35.340
So area is 2 pi little
r, integral from
00:13:35.340 --> 00:13:37.880
minus little r to plus little r.
00:13:37.880 --> 00:13:44.390
So x is u plus big
R divided by, now
00:13:44.390 --> 00:13:48.930
this thing becomes square root
of little r squared minus u
00:13:48.930 --> 00:13:51.510
squared du.
00:13:54.697 --> 00:13:55.280
OK, all right.
00:13:55.280 --> 00:14:01.990
So now-- this is really kind
of two separate pieces, right,
00:14:01.990 --> 00:14:04.190
for purposes of
difficulty of integrating.
00:14:04.190 --> 00:14:07.154
There's the u over the
square root of r squared
00:14:07.154 --> 00:14:08.570
minus u squared
piece, and there's
00:14:08.570 --> 00:14:10.280
the big R over the
square root of r
00:14:10.280 --> 00:14:12.640
squared minus u squared piece.
00:14:12.640 --> 00:14:15.440
So let's think about
them separately.
00:14:15.440 --> 00:14:18.050
So for this first piece,
the u over the square root
00:14:18.050 --> 00:14:21.550
of little r squared
minus u squared,
00:14:21.550 --> 00:14:23.610
this is something
you can integrate.
00:14:23.610 --> 00:14:27.260
This is a-- you don't need a
trig substitution to do that.
00:14:27.260 --> 00:14:31.230
But it's actually-- you don't
need to do any work to do that.
00:14:31.230 --> 00:14:35.460
Because that function u
divided by square root
00:14:35.460 --> 00:14:38.280
of little r squared
minus u squared,
00:14:38.280 --> 00:14:40.110
that's an odd function.
00:14:40.110 --> 00:14:40.860
Right?
00:14:40.860 --> 00:14:44.820
This part, the denominator
is even, u is odd,
00:14:44.820 --> 00:14:46.870
and we're integrating
over an interval that's
00:14:46.870 --> 00:14:48.260
symmetric around the origin.
00:14:48.260 --> 00:14:51.610
So when we integrate this u
divided by this denominator
00:14:51.610 --> 00:14:57.700
part between u from minus r to
r, that's just going to give 0.
00:14:57.700 --> 00:15:00.410
So I can forget
about that entirely.
00:15:00.410 --> 00:15:02.450
So this is nice.
00:15:02.450 --> 00:15:04.000
And then I'll have
a constant here,
00:15:04.000 --> 00:15:05.740
so I'm going to factor
that out as well.
00:15:05.740 --> 00:15:09.780
So I get 2 pi little
r big R, integral
00:15:09.780 --> 00:15:17.890
from minus r to r of du over the
square root of little r squared
00:15:17.890 --> 00:15:18.780
minus u squared.
00:15:21.400 --> 00:15:23.820
OK.
00:15:23.820 --> 00:15:24.320
Good.
00:15:24.320 --> 00:15:26.700
So far, so good.
00:15:26.700 --> 00:15:28.770
So we've got this
nice simplified
00:15:28.770 --> 00:15:29.710
form for the integral.
00:15:29.710 --> 00:15:34.090
So now this is screaming
out trig substitution to me.
00:15:34.090 --> 00:15:34.590
Right?
00:15:34.590 --> 00:15:35.839
There's nothing else I can do.
00:15:35.839 --> 00:15:37.650
So there's sort of two
things you could do.
00:15:37.650 --> 00:15:42.550
One is you could recognize
this as an integral related
00:15:42.550 --> 00:15:45.080
to arcsine, just because
you remember that.
00:15:45.080 --> 00:15:47.850
The other is, you have this r
squared minus u squared thing.
00:15:47.850 --> 00:15:50.890
And so r squared
minus u squared, that
00:15:50.890 --> 00:15:52.860
needs a trig substitution
of some sort,
00:15:52.860 --> 00:15:56.000
and the relevant trig
substitution that you
00:15:56.000 --> 00:15:57.530
would want to do,
is you would want
00:15:57.530 --> 00:16:07.290
to set u equal to r sine t.
00:16:07.290 --> 00:16:07.790
Why?
00:16:07.790 --> 00:16:11.630
Because then down here we'll
have r squared minus r squared
00:16:11.630 --> 00:16:15.380
sine t, or if you factor out
the little r, that's just 1
00:16:15.380 --> 00:16:18.310
minus sine squared t,
which is cosine squared t,
00:16:18.310 --> 00:16:20.810
and then you take a square root,
and you're all good, right?
00:16:20.810 --> 00:16:22.300
Square root of cosine squared t.
00:16:22.300 --> 00:16:23.820
OK.
00:16:23.820 --> 00:16:26.660
So we find this
trig substitution.
00:16:26.660 --> 00:16:30.220
So if we do u equals
r sine t, that's good.
00:16:30.220 --> 00:16:38.400
So du, then, is r
cosine t dt, and I
00:16:38.400 --> 00:16:43.620
need to change the bounds, so
minus u is equal to minus r
00:16:43.620 --> 00:16:46.800
when sine of t is
equal to minus 1.
00:16:46.800 --> 00:16:48.700
So that's at minus pi over 2.
00:16:48.700 --> 00:16:51.810
And u is equal to r when
sine of t is equal to 1.
00:16:51.810 --> 00:16:53.360
So that's a pi over 2.
00:16:53.360 --> 00:17:00.270
So this is equal to 2 pi
little r big R integral
00:17:00.270 --> 00:17:12.060
from minus pi over 2 to
pi over 2 of r cosine t dt
00:17:12.060 --> 00:17:12.980
divided by-- OK.
00:17:12.980 --> 00:17:16.590
So then down here, we
have the square root
00:17:16.590 --> 00:17:22.420
of r squared minus r
squared sine squared t.
00:17:25.030 --> 00:17:29.440
And as we said, so r squared
minus r squared sine squared t,
00:17:29.440 --> 00:17:31.950
this is r squared
cosine squared t.
00:17:31.950 --> 00:17:34.970
And then you take a square root,
and you just get r cosine t.
00:17:34.970 --> 00:17:38.280
And so r cosine t
cancels r cosine t.
00:17:38.280 --> 00:17:42.040
So the integrand here
is actually 1, or 1 dt.
00:17:42.040 --> 00:17:43.670
So that's really
easy to integrate.
00:17:43.670 --> 00:17:45.934
You integrate a constant,
you just get-- well,
00:17:45.934 --> 00:17:47.850
the constant times the
length of the integral.
00:17:47.850 --> 00:17:55.740
So this is equal to 2*pi*r*R
times the constant is 1,
00:17:55.740 --> 00:17:59.250
times the length of the
integral, which is another pi,
00:17:59.250 --> 00:18:00.580
times pi.
00:18:00.580 --> 00:18:01.080
OK.
00:18:01.080 --> 00:18:10.890
So this is equal to 2 pi squared
little r big R. But remember,
00:18:10.890 --> 00:18:13.850
so far we've only
computed-- this is just
00:18:13.850 --> 00:18:16.910
the top half of the torus, is
all we've ever talked about.
00:18:16.910 --> 00:18:20.210
So we want to get the whole
torus, you just double this.
00:18:20.210 --> 00:18:23.900
So you double this, and you
get the area of the whole torus
00:18:23.900 --> 00:18:30.330
is 4 pi squared little r big
R, which is a nice formula.
00:18:30.330 --> 00:18:34.750
So quickly, just to
summarize what we've done.
00:18:34.750 --> 00:18:37.220
Standard setup.
00:18:37.220 --> 00:18:41.390
Here we did it, we're rotating
around the y-axis instead
00:18:41.390 --> 00:18:43.100
of the x-axis.
00:18:43.100 --> 00:18:46.940
So this formula for
dA looks a little bit
00:18:46.940 --> 00:18:50.340
different than what you
saw in class, mostly.
00:18:50.340 --> 00:18:55.180
But the thing to remember
is just that what goes here
00:18:55.180 --> 00:18:58.820
is the radius that this,
that your little segment is
00:18:58.820 --> 00:19:00.057
traveling.
00:19:00.057 --> 00:19:02.140
And this is the circumference
that it's traveling,
00:19:02.140 --> 00:19:04.899
and so this is the area
of that little ribbon
00:19:04.899 --> 00:19:05.690
that it traces out.
00:19:05.690 --> 00:19:06.930
So dA-- OK.
00:19:06.930 --> 00:19:10.490
And then we just did, you
know, the sort of usual thing.
00:19:10.490 --> 00:19:12.300
For your formula,
you need-- you know,
00:19:12.300 --> 00:19:14.440
you remember the formula for ds.
00:19:14.440 --> 00:19:18.230
Then you needed to find this
derivative and plug it in.
00:19:18.230 --> 00:19:21.460
And so after you've done all
that preparatory work then
00:19:21.460 --> 00:19:24.700
you have your integrand set up.
00:19:24.700 --> 00:19:26.330
And once you set
up your integrand,
00:19:26.330 --> 00:19:31.070
you do whatever integration
tools you can to hit it with.
00:19:31.070 --> 00:19:35.450
So in our case, that was
simplifying substitution,
00:19:35.450 --> 00:19:39.660
and then a nice observation that
we used here to simplify this,
00:19:39.660 --> 00:19:41.330
that this part was odd.
00:19:41.330 --> 00:19:43.690
I mean, you could've-- you
didn't need that observation.
00:19:43.690 --> 00:19:46.050
You could have done the
problem perfectly well
00:19:46.050 --> 00:19:51.810
with another substitution there
to kill off that first part.
00:19:51.810 --> 00:19:54.970
And then, OK, and then
a trig substitution,
00:19:54.970 --> 00:19:56.525
remembering that
this is an arcsine,
00:19:56.525 --> 00:19:59.960
or related to an arcsine,
to finish it off.
00:20:03.170 --> 00:20:07.520
And is there anything
else I want to say?
00:20:07.520 --> 00:20:08.784
I think that's-- oh, yes.
00:20:08.784 --> 00:20:10.450
There's one other
thing I wanted to say,
00:20:10.450 --> 00:20:13.340
which is that we could have
done this slightly differently.
00:20:13.340 --> 00:20:16.960
Which is, here we solved for
y explicitly in terms of x.
00:20:16.960 --> 00:20:19.520
And it would have been
possible to carry this
00:20:19.520 --> 00:20:22.690
through using implicit
differentiation instead.
00:20:22.690 --> 00:20:24.600
It would have
actually simplified--
00:20:24.600 --> 00:20:27.400
I would have had to write
this square root of little r
00:20:27.400 --> 00:20:31.340
squared minus x minus big R
quantity squared fewer times
00:20:31.340 --> 00:20:33.690
if I'd done implicit
differentiation, starting
00:20:33.690 --> 00:20:37.137
from this implicit
equation relating x and y.
00:20:37.137 --> 00:20:38.720
You have to be a
little careful there,
00:20:38.720 --> 00:20:42.100
whether you're doing the whole
circle all at once, or just
00:20:42.100 --> 00:20:43.630
the top half or the bottom half.
00:20:43.630 --> 00:20:45.870
But you could do it with
implicit differentiation
00:20:45.870 --> 00:20:48.150
instead, and maybe save
yourself a little bit
00:20:48.150 --> 00:20:52.640
of messy-looking arithmetic.
00:20:52.640 --> 00:20:54.622
So I think I'll end there.