Welcome to a lesson on the shell method
for determining volume.
This video will only show
rotation about the x-axis.
There are other videos that show rotation
about other axes.
To develop the idea of the shell method,
let's go ahead and take
a look at an animation.
If we wanted to determine
the volume of the solid
generated by rotating this gray region
about the x-axis, we could determine it
by shells created by very thin rectangles
bounded by these two functions rotated
about the x-axis.
If we rotated this very thin rectangle
about the x-axis, it
would produce this shell
as we see here on the right.
And the idea behind the shell method is
if we start to accumulate
the volume of the shells
as the number of shells increases,
it will approach the volume of the solid
as we see here on the right.
So as the number of shells
approaches infinity,
it will approach the
volume of the actual solid.
So to use the shell method
to determine the volume,
we really have to consider a single shell
and think about how we
would determine the volume
of this shell.
What we'd have to do is
calculate the surface area
and then multiply it by its thickness.
So let's take this idea
back to the presentation.
If we were to consider the region bounded
by this function and the x- and y-axis
and then rotate it about the x-axis,
it would produce a solid
that looks similar to this.
Now, if we can understand how to determine
the volume of a shell
that would be generated
by a rotating rectangle
from the original region
parallel to the axis of rotation,
we'll have a much better understanding
of how the shell method actually works.
So if we were to cut
this shell and unroll it,
it would be a long rectangle
where this distance here
would come from the
circumference of the circle.
And that's equal to two pi r.
And this distance here would come
from the height of that rectangle.
So to determine the volume of this shell,
we'd have to take this surface area
and then multiply it by the
thickness of that shell.
And with this orientation,
the thickness of this
rectangle would be delta y.
So the volume for a single shell
would be equal to two pi r times h times w
for the width or thickness of that shell.
And that's going to be
equal to two pi times r of y
for the radius function
times h of y for the height function
and then times delta y.
And so, the actual volume will equal
the sum of those shells
as the number of shells
approach infinity.
So to formalize this,
we have the volume is equal to two pi
times the definite integral from c to d
of r of y times h of y with respects to y
where r of y is the distance
between the axis of revolution
and the center of the rectangle
and h of y is the height of the rectangle.
So the representative
rectangle is really key
to setting up and solving
these volume problems.
And for the shell method,
the rectangle is going
to be parallel to the axis of rotation.
So if we're rotating about the x-axis,
our rectangle must be
parallel to the x-axis.
So it might look something like this.
And since the thickness of
that rectangle is delta y,
we'll be integrating with respects to y.
So everything else must be in terms of y.
R of y, the distance
from the axis of rotation
to the center of the rectangle,
would be this distance here.
And the height of the rectangle
would be this distance here.
And this horizontal
distance must be expressed
in terms of y.
Let's go and take a look at an example.
We want to use the shell method
to determine the volume formed
by the bounded region
rotated about the x-axis.
So we have y equals x-squared in blue.
We have y equals zero and x equals two.
So this would be our region
rotated about the x-axis.
So this is what the solid would look like
rotated about the x-axis.
And since we'll be using the shell method,
the rectangle that
would generate one shell
would be parallel to the x-axis.
So it might look something like this.
And since the thickness of this rectangle
would be delta y, that's
telling us we integrate
with respects to y as we see here.
So let's go ahead and see
if we can set this up.
Our volume is going to be equal to two pi
times the definite integral from c to d.
Now, the limits of integration
must be in terms of y.
So even though the x
interval is from zero to two,
we want the y interval which will be
from zero to four.
Now we have to determine
the radius or the distance
from the center of the rectangle
to the axis of rotation.
That's going to be this distance here
which would be equal to y.
I'm going to multiply this by
the height of that rectangle.
Now, the height of the rectangle will be
the distance from here to here
determined by the line x equals two
and the blue function.
It's going to take some
thought to determine
how to express this.
It must be expressed in terms of y.
So let's start by considering the distance
from here to here.
We know that's equal to two.
So if we take this total distance two
and then subtract the distance from here,
that will leave us with
the height function.
So the question becomes, how do we express
this distance here?
Well, this distance is x
determined by the function.
However, remember this must be expressed
in terms of y, so we have
to take the function y
equals x-squared and solve it for x.
We can do that by taking the
square root of both sides.
So we'll have the square
root of y is equal to x.
And since we're in the first quadrant,
we'll only use the principal square root.
So x is equal to the square root of y.
So now, if we take the value of two
and subtract the square root of y,
that will equal the
height of that rectangle.
So h of y is going to
be equal to two minus
the square root of y.
And again, we're integrating
with respects to y.
So let's go ahead and simplify this.
So we're going to have two y
minus y times, this would
be y to the one-half.
That would be y to the three halves.
Now let's go ahead and
find the antiderivative.
So we'll have two times
y-squared over two minus.
Well, three halves plus
one, that'd be five halves.
So we'll have minus y to the five halves
divided by five halves
evaluated at four and zero.
Let's simplify this one more time
and then we'll perform the substitution.
So we'll have y-squared
and this will be minus.
Well, dividing by five halves is the same
as multiplying by the reciprocal.
So I'll have two-fifths
y to the five halves.
So now, we'll replace y with four
and then y with zero.
Take all of this and
then subtract the value
when y is zero.
Well, when y is zero, both
of these will be zero.
And to save some time,
I've already determined
this value is 32 pi over five
which again would be
the volume of the solid
generated by rotating this region
about the x-axis using the shell method.
Okay, I hope you found this helpful.
There will be some additional examples
in the next video.
