Welcome to a lesson that will compare
using the washer and shell methods
to determine the volume of revolution.
Let's start with a quick review.
One of the most important
things to remember
when setting up these types of problems
is to sketch a representative rectangle.
Which would represent a
washer for the washer method
and would represent a
shell for the shell method.
The rectangle is perpendicular
to the axis of rotation
for the washer method
and the rectangle is parallel
to the axis of rotation
for the shell method.
Let's go and take a look at some examples.
In this video, we're not going
to evaluate the integrals,
we're just going to set
them up for practice.
We have the region bounded by
y equals the square root of x plus 2,
y equals zero, x equals
zero, and x equals four.
And that would be this region here.
And we're rotating about the x axis.
And first we'll solve this
using the washer method.
The rectangle that would
represent one washer,
or in this case looks
like it will be a disc,
is going to be perpendicular
to the axis of rotation
so it would look something like this.
And since the width of this
rectangle would be delta x,
that tells us that we
integrate with respects to x,
so we'll be using this integral
to determine the volume.
We have to determine the outer radius
and the inner radius
for the washer method.
Well, the outer radius is the distance
from the red function to the x axis,
as we see here.
This would just be equal
to the function value
which is the square root of x plus two.
And there is no inner radius,
so this would be a disc.
Or we could say a little
r of x is equal to zero.
That's pretty much all we
need to set this one up
using the washer method.
The volume is equal to pi
times your definite integral
of big R squared minus little r squared,
where big R is a square
root of x plus two,
and little r is equal to zero.
Lastly, we are integrating
with respects to x,
and the interval along the x axis
will be from zero to four,
so those are our limits of integration.
Again, we're not going to evaluate this.
What we're going to do is compare
how to solve the same problem
now using the shell method.
So this is the same problem,
so we have the same region
rotated about the x axis.
But now we want to use the shell method.
So to use the shell method,
the representative
rectangle would have to be
parallel to the axis of rotation,
and that's a little more involved here
because if we drew a
rectangle in this region,
it's bounded by two vertical lines.
But if we drew a rectangle
above y equals two,
it's bounded by the
vertical line x equals four
and the function.
What this is telling us
is we're going to have to set it up
and solve two different
integrals to find this volume
using the shell method.
Notice that the thickness
or width of these rectangles
are delta y now,
which means we'll be
integrating with respects to y.
Let's go ahead and divide this
into two different regions
since we'll have two
different definite integrals.
Let's set up the integral for
the lower rectangle first.
The volume will be equal to
two pi times your definite integral of the
radius times the height
of this first rectangle.
Well, the radius would be
the distance from the center
of this rectangle to the x axis.
Or r of y.
This distance would be equal to y.
We need to multiply this by
the height of this rectangle.
Where the distance from here
to here will always be four
in this lower region.
So the height is constant.
Now that we know we're
integrating with respects to y,
the limits of integration
for this region here
will be from zero to two.
That'll be the volume rotating
this lower region about the x axis.
Now to determine the total volume,
we have to add the volume
that would be generated
by rotating this upper region.
So let's see if we can
set that up as well.
So it'd be this volume plus
the limits of integration
would be from two to four
along the y axis.
Now the radius,
the distance from this
rectangle to the x axis
would still be y,
but what's going to challenging here
is to determine what the height
of this rectangle would be.
We know the height would be
the distance from here to here,
so that would be h of y.
But we have to think about
how we're going to express this.
We know the total distance from the y axis
to x equals four would be four units.
So if we took four units
and then subtract it,
the distance from here to here.
That would give us h of y,
or the height of that rectangle.
Well, this distance here is equal to x
determined by the function
y equals the square root of x plus two,
but remember we have to express
this height in terms of y.
So we have to take this equation
y equals the square root of x plus two
and solve it for x.
Let's go ahead and do that down here.
We'll subtract two on both sides.
And then square both sides.
We have x equals the
quantity y minus two squared.
So if this x is equal to the
quantity y minus two squared,
we can use that to determine h of y.
h of y will be equal to the total distance
four minus the quantity
y minus two squared.
Now we're going to stop here,
but you can see if you
had a choice between
which method to use,
the result would be the same,
but the shell method
is a lot more involved.
Let's go and take a look
at the same problem now,
rotated about the y axis.
So again, it's the same region,
but now we're rotating
it about the y axis.
And we'll first start by
using the washer method.
Which means the rectangle
is going to be perpendicular
to the y axis.
And now we're going to have
the same issue with this method
because we're rotating about the y axis.
Meaning, if we drew a
rectangle in the lower region,
it would look like this,
bounded by two vertical lines.
But if we drew a rectangle
above y equals two,
it's bounded by the function
and one vertical line.
That tells us to use this method,
we'll have to use two
different definite integrals.
Next these rectangles
have a height of delta y,
which means we have to
integrate with respects to y.
So this is the proper integral to use.
Again, let's go ahead
and divide this up into
two different regions.
Let's first integrate from zero to two
and then from two to four
using the washer method.
We're going to have pi
times the integral from,
again we'll start from zero to two,
and the outer radius will be four,
so we'll have four squared.
Minus the inner radius,
but again that would be zero.
Now this wouldn't be too bad,
but that's only going to be
the volume of the lower region,
but we still have to add the volume
that we would generate
from the upper region
or from y equals two to four.
Now the outer radius would still be four,
minus the inner radius
would be the distance
from here to here,
and it must be in terms of y.
Well this distance here
would be equal to x
based upon the function
y equals the square root of x plus two.
so we have to take this
equation and solve it for x
so we can express the
radius in terms of y.
We just did this on the previous screen,
but we'll go ahead and do it again.
So again, we have x equals the
quantity y minus two squared,
and this would be the inner
radius for this rectangle here.
So we have four squared
minus y minus two squared
for the second definite integral.
So the sum of these two definite integrals
would give us a total
volume of this region
rotated about the y axis
using the washer method.
Now let's finish by taking a look
at the same problem
using the shell method.
Same region.
With the shell method.
So we're rotating about the y axis,
which means the rectangle would
be parallel to the y axis,
so it's going to look like this,
and the width is delta x.
Which means we integrate
with respects to x.
So here's the proper formula
for the shell method.
The volume is equal to two pi times the
definite integral from along the x axis
from zero to four.
The radius would be equal to x.
And the height would be
equal to the function value.
Which is the square root of x plus two.
So when we rotate it about the y axis,
you can see that the shell method
was actually a lot less work
than the washer method.
So as you can see from these problems,
if you have a choice
on which method to use,
if you're rotating about the x axis,
it's often easier to
use the washer method.
If you're rotating about the y axis,
the shell method is often
a little bit easier.
Hope you found this review helpful.
