Shell method to rotate around y-axis - Mathematics Video Lecture - Engineering Mathematics

We've been doing a lot of
rotating around the x-axis, so
let's start rotating around the
y-axis and see what we can do.
Or at least attempt to.
Let's me draw my axes.
That's y-axis.
That's my x-axis.
Well let's just do it with an
example, but we'll call it f
of x too because it'll
be generalizable.
Let's just draw y
equals x squared.
Let me just draw the positive
because we're going to rotate
it around the y-axis and it's
symmetric anyway, so that's
y equals x squared.
This is y-axis.
This is x-axis.
Actually, no I'm going to keep
it general, then we'll actually
solve it particularly.
So we'll call this f of
x, but clearly this is
y equals x squared.
This is f of x.
And we know how to take the
volume if I were to rotate
this around the x-axis.
But what if I wanted to say-- I
guess we could call it the area
between 0 and-- I'm trying to
determine how general to be.
Well let's just say
between 0 and 1.
I think the boundaries
might make sense to you.
Roughly this area, and
I'm going to rotate it
around the y-axis now.
So what's that final figure
going to look like?
The base of it-- let me see
how well I can draw it.
Nope that's not what
I wanted to do.
The base is going to
look something like a
cylinder like that.
And then the top of it is also
going to be-- no, that's
not what I wanted to do.
Let me draw the side lines.
So it's going to look
something like that.
And then the top of it
looks something like that.
But it's not just going
to be cylinder, right?
If I was doing this entire
block it would be a cylinder.
But the inside of it is going
to be kind of hollowed out.
Let me see how effective
I am at drawing that.
I'll do it in a
different color.
So the inside is going
to be hollowed out.
I don't know if that
makes sense to you that.
It's kind of like on the inside
it'll look like a bowl.
On the outside it'll look
like a cylinder or a can.
Hopefully that makes sense.
You take this and you
rotate this around.
And the curve that specifies
the inside would be y
is equal to x squared.
It would rotate all
the way around.
I think that makes sense.
The drawing is the
hardest part.
So how do we do it?
Well even the shape
might give you an idea.
We can't use that disk method,
what we were doing before when
we were rotating the x-axis,
that was the disk method,
because we were essentially
imagining each of these
particular disks and
then summing them up.
Now we're going to do something
called the shell method.
So what's the shell method?
Instead of taking a bunch of
disks and figuring out their
combined volumes, we're going
to take a bunch of shells.
So what's a shell?
So imagine a rectangle
right here.
Hope you can see
it right there.
Let's say it's at
the point x,1.
What's its height going to be?
Its height going
to be f of x,1.
That's its height.
Now imagine taking that
sliver and rotating
it around the y-axis.
What's it going to look like?
Well, it's going to look like a
shell, it's going to look like
a cylinder, just like the
outside of a cylinder.
It's going to look not too
different then that but I want
to draw it well because
intuition is the most important
thing that, not getting
the problem right.
Let me see if I can
draw this respectably.
And then we're going to have
the bottom of the shell, it'll
look something like that.
Let me finish these lines up.
I think you get the point.
OK.
So it's going to look
like a shell like that.
The outside of the shell
is going to be solid.
And it'll have some width,
but the inside is hollow.
Let me do a different color.
Maybe a darker color to show
that that's the inside.
You know it's like a
ring essentially.
And so what's the
height of this ring?
The height is going
to be f of x,1.
So let me do a brighter color
so you know what I'm saying.
The height of this
ring is f of x,1.
f of x evaluated at that
arbitrary point we picked up.
What is going to be the
surface area of this ring?
You know, this outside.
Well let's think about it.
It'll be the circumference of
this ring times it's height.
So what's the circumference
of this ring?
Let's go back to our
basic geometry.
Circumference is equal to
2 pi times the radius.
So if we know the radius of it,
we know the circumference.
Well what's the radius?
Well the radius is how far
we went from the axis of
rotation to that point.
So that's the radius.
So in our particular
example the radius is x,1.
It's that x point that
we're evaluating it at.
So circumference is going to be
equal to 2 pi times that point
that we're evaluating at.
And so the surface area-- this
magenta thing that I filled
in-- that's going to be equal
to the circumference times this
height, which we already
said is f of x,1.
Let's call it area surface.
Surface area is equal to
circumference times height,
which is equal to 2 pi
x,1 times f of x,1.
We figured out the
surface are of this.
Now how do we figure
out the volume?
Well what's the width of it?
How thick is this ring?
What's this thickness
right here?
It's a very small thickness.
But we took this sliver, and
this sliver as we learned in
previous calculus, the width of
this little rectangle is dx.
And you know when we take the
integral, it's going to get
infinitely smaller and smaller
and we'll have infinitely
more and more of them.
So the width of this is dx.
Let me draw it big, not
so horrible looking.
So if this is a sliver,
it's width is dx.
It's height is f of x,1.
x,1 will be right
in the center.
And then it's distance from
the center is of course x,1.
Hopefully that make sense.
So what's the volume
of this shell?
So the volume of the shell--
this shell, not this one-- the
volume of the shell is going to
be equal to the surface area
of the shell times how
wide that surface is.
And that width is dx, so it's
going to equal this times dx.
So the volume of that
shell is 2 pi x,1 times
f of x,1 times dx.
I think you see where I'm
going with this now.
So what would be the volume
of the entire rotated
figure, this thing here?
Well I'm just going to sum
up each of these shells.
I have one shell there, then
here I'll have a slightly less
high shell, and up here I would
have a much bigger shell,
and I'll add them up.
Here's one shell
that goes around.
Then they'll be another shell
here, and I'll add them all up.
And that's taking the integral.
So the total volume of the
figure when I rotate it around
the y-axis is going to be-- and
my boundary is from 0 to 1-- 2
pi-- this one I just told you a
particular x,1 but we're going
to sum them over
all of the x's.
So it's going to be
2 pi x f of x dx.
This is just a constant,
so you could call it
2 pi times x f of x.
So let's take a
particular example.
Let's do it for x squared.
Let's say the function
is x squared.
So in this case the volume is
going to equal-- let's take the
2 pi out-- 2 pi integral 0 to 1
x times f of x-- f of x in our
case is x squared, which I drew
earlier-- dx equals 2 pi.
This is just x to
the third, right?
x to the third.
So it's going to be 2 pi
times the antiderivative
of x to the third.
Well that's x to
the fourth over 4.
Evaluate it at 1 minus
evaluate it at 0.
Well that equals 2 pi times
1 to the fourth is 1, so
1/4 and then minus 0.
So it's 2 pi times 1/4.
So that's pi over 2.
That's the volume, and we just
rotated it around the y-axis.
I will see you in
the next video.