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PROFESSOR: Welcome
back to recitation.

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In this video we're going to
do another solid of revolution

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problem.

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So what I'd like to
do in this problem is

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to find the volume of the
solid generated by rotating

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the region bounded by the
following curves: y equals 0,

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x equal 4, and y equals square
root of x, around the line x

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equals 6.

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And you can choose your
favorite method to do this.

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What I would like
first, when you're

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doing this kind
of problem, is get

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a rough sketch of the region.

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So you have some
picture of what's

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actually going to happen.

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You don't necessarily need
a three-dimensional picture.

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But at least have the
two-dimensional region

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and understand the where the
rotation line is, with respect

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to that region.

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So I will give you
a little bit of time

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to work on that problem.

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And when I come back I'll
show you how I do it.

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OK, welcome back.

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So again, what we're
doing in this video is

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we're going to be looking,
finding the volume

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of a solid of revolution.

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And as I mentioned, the
first thing you want to do

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is get a rough picture
of what the region looks

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like that's being rotated.

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So I'm going to draw a
rough sketch of that.

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Right here.

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So my axes, here's my x-axis
and my y-axis, and I will draw,

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first the curve y equals 0.

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That's y equals 0.

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And I'm next going to draw the
curve y equals square root of x

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to make this a little easier for
myself, to do that one first.

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So y equals square root of
x looks something like this,

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roughly.

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And then, x equal 4, because
it's a very rough sketch,

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I can just draw
this line right here

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and say that line's x equal 4.

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And I want the region bounded
by those three curves.

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So, this region right here.

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And my rotating line, in
this case, is x equals 6.

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So to give myself
some perspective,

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I remember this is x equals
4, here's x equals 0.

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So x equals 6, which
may be right about here,

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I'll draw a dotted line.

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And I'm going to draw a
little arrow around there

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so I know that's the
line I'm rotating about.

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Now we notice there's a
little bit of a gap here.

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But you can actually work on
this problem quite simply.

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If we use the shell
method, we won't have

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to worry about the gap at all.

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We will have to be careful in
how we determine our radius.

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That's actually where you'll
see this sort of gap coming up.

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So when I rotate
this around, I'm

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going to get some
object that's going

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to have a nice curve here.

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And there's going to be a little
cut-out, a cylinder missing

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in the middle when I rotate it.

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So I'm going to use
the shell method.

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I like the shell method
for this problem.

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You could use the washer
method for this problem,

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but I like the shells and
that's what I'm going to use.

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So to do the shell method,
remember we are going to be

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interested in 2*pi*r*h.

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That's really what we need.

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And now what we
want to figure out

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is, do I want to do this in
terms of x or in terms of y?

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All right?

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When I talk about
these shells is

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it better to think about
them as functions of x or y?

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And it will, you'll
see we sort of

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have to think about
them in a certain way.

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So if I draw, let's think about
if I draw a straight line down

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and I rotate it
around the line x

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equals 6, which I should
have labeled here,

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then I get my shell.

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So I need these lines
to go from 0 to 4.

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I need to have those
lines go from 0 to 4.

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And so I see that it's the
x-value I'm varying over.

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So I know this is all going to
be an integral in terms of x.

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So let's keep that in mind.

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So r I should write as a
function of x, and h I should

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write as a function of x.

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So if I examined this
segment again here,

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the height is fairly simple.

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Because what is this curve?

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This is the curve y
equals square root of x.

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So the height is just
square root of x.

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So h is equal to
square root of x.

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And so now I just
need r in terms of x.

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Well let's look at
this segment again.

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It's just a generic position.

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How would I find that radius?

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What is the radius?

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Well I'm going to draw the
three-dimensional picture over

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here.

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That that's this
curve, this segment,

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rotated about the
line x equals 6.

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Right?

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And the radius is not x.

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It's the distance from x
equals 6 to my segment.

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That's an important distinction
we should understand.

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If we come back and
look at this picture,

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this is not the radius.

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Right?

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The radius is the distance from
this value to this value in x.

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And how do I measure that?

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Well that's fairly
straightforward.

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That's the difference in x
values between these two.

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The x value here is 6.

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And the x value here,
this is a generic x.

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So the difference, the
distance, is just 6 minus x.

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Right?

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This is just a simple case
of the distance formula.

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Because I'm not interested
in the changes in y, at all.

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I'm just interested in the
difference between these two

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x values.

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So it's simply 6 minus x.

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That's our radius.

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So now I have the
radius in terms of x.

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And I have the
height in terms of x.

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So I can completely
set up my integral.

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So let me come a
little bit further over

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and set up the integral.

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So I know I'm
integrating-- we even

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said what we were integrating
from and to already.

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I'm starting x at 0.

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And I'm stopping
it at x equal 4.

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And then I want to
integrate 2*pi*r*h.

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So I'll put the
2*pi out in front.

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The radius is 6 minus x.

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And the height is
square root of x.

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And then I'm integrating
it all in dx.

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And at this point,
solving the problem

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is fairly straightforward.

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I can write square root
of x as x to the 1/2

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to make myself feel a
little better about it.

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I can then distribute.

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And then I just use the power
rule to do the integration

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and evaluate to get an actual
number, to get that volume.

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So I'm going to go
back and just make

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sure we remember what
all the pieces are

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and that we feel
comfortable with doing

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this type of problem.

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And I'll let you
finish and evaluate

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that if you would like
to know an actual number.

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So let's come back over.

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We were finding the volume
of a solid generated

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by rotating a certain region
around the line x equals 6.

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The way this might be
different from some

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of the problems
you've seen before,

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is it's not the x or y axis.

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And the region has a gap.

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The region is not connected
to the line of rotation.

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So if we come to
see our picture,

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we can see that clearly.

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The rotating region
is the blue region.

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The line of rotation
is the dotted line.

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So there's a little
bit of a gap there.

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I let you pick your
favorite method.

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I chose shells,
because I didn't have

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to worry about subtracting off
pieces, this little piece here.

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It was inherent in
the radius, the way

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I determined the radius and
the x values that I chose.

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So I used shell method.

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And so how did I, I had to
figure out a couple things.

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I had to figure out
first, what variable

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I was doing this in, x or y.

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And then I had to figure
out the radius and height

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in terms of that variable.

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You figure out what variable
you need by saying, OK

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if I'm doing shells, I'm looking
at these kinds of segments.

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And I'm varying
the x-values, then,

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as I move and look at
those different segments.

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So I know that this is
a dx type of problem.

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I know I'm integrating
something in x.

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So then I have to find radius
and height in terms of x.

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The height's the freebie.

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It's just from 0
up to the function.

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The radius is maybe
a little harder.

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But that's still
just the distance

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between the line of rotation
and the point where you are.

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And so that's 6 minus x.

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And then I just set up
the integral from there.

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So that's really
what the idea is

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in doing these
types of problems.

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And because we get to pick
our simplest way, for me

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it was shells.

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Maybe you picked washers.

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But for me, this problem,
the simplest way is shells.

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So I guess that's
where I'll stop.