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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,

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come back, and we can
work it out together.

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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

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say-- we want to simplify
this to a usable form.

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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,

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and so when you add it
to x minus big R squared

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over little r squared minus
x minus big R squared,

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this part cancels.

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And so you're left with-- OK.

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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.

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OK.

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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,

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so we need dA, and dA
is 2*pi*x times ds.

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So-- OK.

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So that's easy to write
down now that I've got ds.

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So this is dA equal to 2*pi*r*x
over r squared minus x minus

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big R squared, square
root of that, dx.

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All right.

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This is dA.

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You know, we haven't done
any calculus, actually.

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Oh, I guess we took a
derivative somewhere.

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We haven't done any
integration yet.

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Now, to compute the surface
area, we just integrate this.

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00:10:05,072 --> 00:10:10,350
So we get to, you know, the
calculus step of this problem.

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So integrate this.

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But of course, you know,
I'm expecting a number out

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at the end.

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00:10:15,170 --> 00:10:17,020
I'm taking a definite integral.

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So I need bounds.

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Well, what bounds do I need?

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Well, I'm integrating
with respect to x.

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So I need to integrate over
the relevant values of x.

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What are the
relevant values of x?

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Well, come back to
our picture over here.

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We have this circle.

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Its center is at x
equals R, y equals 0.

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And it has radius little r.

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So, you know, the
relevant values of x

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are just from the leftmost
point of the circle

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to the rightmost
point of the circle.

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00:10:49,770 --> 00:10:51,640
And so this leftmost
point is just-- well,

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the radius is
little r, so this is

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00:10:53,460 --> 00:10:56,280
big R minus-- x equals
big R minus little r,

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and the rightmost point is x
equals big R plus little r.

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00:11:00,520 --> 00:11:07,130
So the bounds-- so I'm going
to go up here-- so I have area

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is what I get when
I integrate dA.

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00:11:09,645 --> 00:11:14,720
And I want to integrate it from
x equals big R minus little

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r to big R plus little
r, and dA is this thing

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I found, just a moment ago.

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So this is-- well, OK.

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So 2*pi little r is a constant.

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I'm just going to factor
that out in front.

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So this is 2*pi little r times
x over the square root of little

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00:11:39,890 --> 00:11:51,130
r squared minus big R minus--
sorry-- x minus big R squared.

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That's all under
the square root.

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ds.

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OK.

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So now we have to figure out
how to integrate this thing.

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Right?

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So this is a little ugly.

231
00:12:04,730 --> 00:12:05,970
It's not horrible, though.

232
00:12:05,970 --> 00:12:06,470
Right?

233
00:12:06,470 --> 00:12:08,550
So down here we have
something that's

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really reminiscent of one of
these trig integral things

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00:12:13,270 --> 00:12:13,790
we've done.

236
00:12:13,790 --> 00:12:14,290
Right?

237
00:12:14,290 --> 00:12:16,400
We've got a square root
of a something squared

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minus a something else squared.

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00:12:18,280 --> 00:12:21,510
So that, reminds, you
know, what does that

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00:12:21,510 --> 00:12:23,740
remind, maybe some
sine substitution.

241
00:12:23,740 --> 00:12:24,950
Something like that.

242
00:12:24,950 --> 00:12:27,300
There's some there's some
trig substitution waiting

243
00:12:27,300 --> 00:12:30,240
to happen here.

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00:12:30,240 --> 00:12:30,960
But, so OK.

245
00:12:30,960 --> 00:12:36,040
We could sort of shoot to do
it all in one substitution.

246
00:12:36,040 --> 00:12:38,040
I like, my life
always feels simpler

247
00:12:38,040 --> 00:12:40,500
when I do one little
substitution at a time.

248
00:12:40,500 --> 00:12:42,470
And so one little
substitution I could do

249
00:12:42,470 --> 00:12:46,700
is to simplify this
x minus r thing.

250
00:12:46,700 --> 00:12:48,600
I could just shift this by r.

251
00:12:48,600 --> 00:12:51,560
So I'm just going to do a
little linear substitution.

252
00:12:51,560 --> 00:12:57,600
I'm going to do u
equals x minus big R,

253
00:12:57,600 --> 00:12:59,970
or I'm going to want to
substitute the other way,

254
00:12:59,970 --> 00:13:06,050
so that's the same thing as
saying x equals u plus big R.

255
00:13:06,050 --> 00:13:11,460
And so du equals dx.

256
00:13:11,460 --> 00:13:14,110
This is a simple
little substitution.

257
00:13:14,110 --> 00:13:15,980
And I'm going to have
to move the bounds,

258
00:13:15,980 --> 00:13:22,680
so when x is big R minus little
r, then u is minus little r,

259
00:13:22,680 --> 00:13:25,600
and this top bound here, u is
going to be equal to little r.

260
00:13:25,600 --> 00:13:26,620
Positive little r.

261
00:13:26,620 --> 00:13:30,810
So let me just make
that substitution.

262
00:13:30,810 --> 00:13:35,340
So area is 2 pi little
r, integral from

263
00:13:35,340 --> 00:13:37,880
minus little r to plus little r.

264
00:13:37,880 --> 00:13:44,390
So x is u plus big
R divided by, now

265
00:13:44,390 --> 00:13:48,930
this thing becomes square root
of little r squared minus u

266
00:13:48,930 --> 00:13:51,510
squared du.

267
00:13:54,697 --> 00:13:55,280
OK, all right.

268
00:13:55,280 --> 00:14:01,990
So now-- this is really kind
of two separate pieces, right,

269
00:14:01,990 --> 00:14:04,190
for purposes of
difficulty of integrating.

270
00:14:04,190 --> 00:14:07,154
There's the u over the
square root of r squared

271
00:14:07,154 --> 00:14:08,570
minus u squared
piece, and there's

272
00:14:08,570 --> 00:14:10,280
the big R over the
square root of r

273
00:14:10,280 --> 00:14:12,640
squared minus u squared piece.

274
00:14:12,640 --> 00:14:15,440
So let's think about
them separately.

275
00:14:15,440 --> 00:14:18,050
So for this first piece,
the u over the square root

276
00:14:18,050 --> 00:14:21,550
of little r squared
minus u squared,

277
00:14:21,550 --> 00:14:23,610
this is something
you can integrate.

278
00:14:23,610 --> 00:14:27,260
This is a-- you don't need a
trig substitution to do that.

279
00:14:27,260 --> 00:14:31,230
But it's actually-- you don't
need to do any work to do that.

280
00:14:31,230 --> 00:14:35,460
Because that function u
divided by square root

281
00:14:35,460 --> 00:14:38,280
of little r squared
minus u squared,

282
00:14:38,280 --> 00:14:40,110
that's an odd function.

283
00:14:40,110 --> 00:14:40,860
Right?

284
00:14:40,860 --> 00:14:44,820
This part, the denominator
is even, u is odd,

285
00:14:44,820 --> 00:14:46,870
and we're integrating
over an interval that's

286
00:14:46,870 --> 00:14:48,260
symmetric around the origin.

287
00:14:48,260 --> 00:14:51,610
So when we integrate this u
divided by this denominator

288
00:14:51,610 --> 00:14:57,700
part between u from minus r to
r, that's just going to give 0.

289
00:14:57,700 --> 00:15:00,410
So I can forget
about that entirely.

290
00:15:00,410 --> 00:15:02,450
So this is nice.

291
00:15:02,450 --> 00:15:04,000
And then I'll have
a constant here,

292
00:15:04,000 --> 00:15:05,740
so I'm going to factor
that out as well.

293
00:15:05,740 --> 00:15:09,780
So I get 2 pi little
r big R, integral

294
00:15:09,780 --> 00:15:17,890
from minus r to r of du over the
square root of little r squared

295
00:15:17,890 --> 00:15:18,780
minus u squared.

296
00:15:21,400 --> 00:15:23,820
OK.

297
00:15:23,820 --> 00:15:24,320
Good.

298
00:15:24,320 --> 00:15:26,700
So far, so good.

299
00:15:26,700 --> 00:15:28,770
So we've got this
nice simplified

300
00:15:28,770 --> 00:15:29,710
form for the integral.

301
00:15:29,710 --> 00:15:34,090
So now this is screaming
out trig substitution to me.

302
00:15:34,090 --> 00:15:34,590
Right?

303
00:15:34,590 --> 00:15:35,839
There's nothing else I can do.

304
00:15:35,839 --> 00:15:37,650
So there's sort of two
things you could do.

305
00:15:37,650 --> 00:15:42,550
One is you could recognize
this as an integral related

306
00:15:42,550 --> 00:15:45,080
to arcsine, just because
you remember that.

307
00:15:45,080 --> 00:15:47,850
The other is, you have this r
squared minus u squared thing.

308
00:15:47,850 --> 00:15:50,890
And so r squared
minus u squared, that

309
00:15:50,890 --> 00:15:52,860
needs a trig substitution
of some sort,

310
00:15:52,860 --> 00:15:56,000
and the relevant trig
substitution that you

311
00:15:56,000 --> 00:15:57,530
would want to do,
is you would want

312
00:15:57,530 --> 00:16:07,290
to set u equal to r sine t.

313
00:16:07,290 --> 00:16:07,790
Why?

314
00:16:07,790 --> 00:16:11,630
Because then down here we'll
have r squared minus r squared

315
00:16:11,630 --> 00:16:15,380
sine t, or if you factor out
the little r, that's just 1

316
00:16:15,380 --> 00:16:18,310
minus sine squared t,
which is cosine squared t,

317
00:16:18,310 --> 00:16:20,810
and then you take a square root,
and you're all good, right?

318
00:16:20,810 --> 00:16:22,300
Square root of cosine squared t.

319
00:16:22,300 --> 00:16:23,820
OK.

320
00:16:23,820 --> 00:16:26,660
So we find this
trig substitution.

321
00:16:26,660 --> 00:16:30,220
So if we do u equals
r sine t, that's good.

322
00:16:30,220 --> 00:16:38,400
So du, then, is r
cosine t dt, and I

323
00:16:38,400 --> 00:16:43,620
need to change the bounds, so
minus u is equal to minus r

324
00:16:43,620 --> 00:16:46,800
when sine of t is
equal to minus 1.

325
00:16:46,800 --> 00:16:48,700
So that's at minus pi over 2.

326
00:16:48,700 --> 00:16:51,810
And u is equal to r when
sine of t is equal to 1.

327
00:16:51,810 --> 00:16:53,360
So that's a pi over 2.

328
00:16:53,360 --> 00:17:00,270
So this is equal to 2 pi
little r big R integral

329
00:17:00,270 --> 00:17:12,060
from minus pi over 2 to
pi over 2 of r cosine t dt

330
00:17:12,060 --> 00:17:12,980
divided by-- OK.

331
00:17:12,980 --> 00:17:16,590
So then down here, we
have the square root

332
00:17:16,590 --> 00:17:22,420
of r squared minus r
squared sine squared t.

333
00:17:25,030 --> 00:17:29,440
And as we said, so r squared
minus r squared sine squared t,

334
00:17:29,440 --> 00:17:31,950
this is r squared
cosine squared t.

335
00:17:31,950 --> 00:17:34,970
And then you take a square root,
and you just get r cosine t.

336
00:17:34,970 --> 00:17:38,280
And so r cosine t
cancels r cosine t.

337
00:17:38,280 --> 00:17:42,040
So the integrand here
is actually 1, or 1 dt.

338
00:17:42,040 --> 00:17:43,670
So that's really
easy to integrate.

339
00:17:43,670 --> 00:17:45,934
You integrate a constant,
you just get-- well,

340
00:17:45,934 --> 00:17:47,850
the constant times the
length of the integral.

341
00:17:47,850 --> 00:17:55,740
So this is equal to 2*pi*r*R
times the constant is 1,

342
00:17:55,740 --> 00:17:59,250
times the length of the
integral, which is another pi,

343
00:17:59,250 --> 00:18:00,580
times pi.

344
00:18:00,580 --> 00:18:01,080
OK.

345
00:18:01,080 --> 00:18:10,890
So this is equal to 2 pi squared
little r big R. But remember,

346
00:18:10,890 --> 00:18:13,850
so far we've only
computed-- this is just

347
00:18:13,850 --> 00:18:16,910
the top half of the torus, is
all we've ever talked about.

348
00:18:16,910 --> 00:18:20,210
So we want to get the whole
torus, you just double this.

349
00:18:20,210 --> 00:18:23,900
So you double this, and you
get the area of the whole torus

350
00:18:23,900 --> 00:18:30,330
is 4 pi squared little r big
R, which is a nice formula.

351
00:18:30,330 --> 00:18:34,750
So quickly, just to
summarize what we've done.

352
00:18:34,750 --> 00:18:37,220
Standard setup.

353
00:18:37,220 --> 00:18:41,390
Here we did it, we're rotating
around the y-axis instead

354
00:18:41,390 --> 00:18:43,100
of the x-axis.

355
00:18:43,100 --> 00:18:46,940
So this formula for
dA looks a little bit

356
00:18:46,940 --> 00:18:50,340
different than what you
saw in class, mostly.

357
00:18:50,340 --> 00:18:55,180
But the thing to remember
is just that what goes here

358
00:18:55,180 --> 00:18:58,820
is the radius that this,
that your little segment is

359
00:18:58,820 --> 00:19:00,057
traveling.

360
00:19:00,057 --> 00:19:02,140
And this is the circumference
that it's traveling,

361
00:19:02,140 --> 00:19:04,899
and so this is the area
of that little ribbon

362
00:19:04,899 --> 00:19:05,690
that it traces out.

363
00:19:05,690 --> 00:19:06,930
So dA-- OK.

364
00:19:06,930 --> 00:19:10,490
And then we just did, you
know, the sort of usual thing.

365
00:19:10,490 --> 00:19:12,300
For your formula,
you need-- you know,

366
00:19:12,300 --> 00:19:14,440
you remember the formula for ds.

367
00:19:14,440 --> 00:19:18,230
Then you needed to find this
derivative and plug it in.

368
00:19:18,230 --> 00:19:21,460
And so after you've done all
that preparatory work then

369
00:19:21,460 --> 00:19:24,700
you have your integrand set up.

370
00:19:24,700 --> 00:19:26,330
And once you set
up your integrand,

371
00:19:26,330 --> 00:19:31,070
you do whatever integration
tools you can to hit it with.

372
00:19:31,070 --> 00:19:35,450
So in our case, that was
simplifying substitution,

373
00:19:35,450 --> 00:19:39,660
and then a nice observation that
we used here to simplify this,

374
00:19:39,660 --> 00:19:41,330
that this part was odd.

375
00:19:41,330 --> 00:19:43,690
I mean, you could've-- you
didn't need that observation.

376
00:19:43,690 --> 00:19:46,050
You could have done the
problem perfectly well

377
00:19:46,050 --> 00:19:51,810
with another substitution there
to kill off that first part.

378
00:19:51,810 --> 00:19:54,970
And then, OK, and then
a trig substitution,

379
00:19:54,970 --> 00:19:56,525
remembering that
this is an arcsine,

380
00:19:56,525 --> 00:19:59,960
or related to an arcsine,
to finish it off.

381
00:20:03,170 --> 00:20:07,520
And is there anything
else I want to say?

382
00:20:07,520 --> 00:20:08,784
I think that's-- oh, yes.

383
00:20:08,784 --> 00:20:10,450
There's one other
thing I wanted to say,

384
00:20:10,450 --> 00:20:13,340
which is that we could have
done this slightly differently.

385
00:20:13,340 --> 00:20:16,960
Which is, here we solved for
y explicitly in terms of x.

386
00:20:16,960 --> 00:20:19,520
And it would have been
possible to carry this

387
00:20:19,520 --> 00:20:22,690
through using implicit
differentiation instead.

388
00:20:22,690 --> 00:20:24,600
It would have
actually simplified--

389
00:20:24,600 --> 00:20:27,400
I would have had to write
this square root of little r

390
00:20:27,400 --> 00:20:31,340
squared minus x minus big R
quantity squared fewer times

391
00:20:31,340 --> 00:20:33,690
if I'd done implicit
differentiation, starting

392
00:20:33,690 --> 00:20:37,137
from this implicit
equation relating x and y.

393
00:20:37,137 --> 00:20:38,720
You have to be a
little careful there,

394
00:20:38,720 --> 00:20:42,100
whether you're doing the whole
circle all at once, or just

395
00:20:42,100 --> 00:20:43,630
the top half or the bottom half.

396
00:20:43,630 --> 00:20:45,870
But you could do it with
implicit differentiation

397
00:20:45,870 --> 00:20:48,150
instead, and maybe save
yourself a little bit

398
00:20:48,150 --> 00:20:52,640
of messy-looking arithmetic.

399
00:20:52,640 --> 00:20:54,622
So I think I'll end there.