Welcome to a video on
the Washer Method for determining volume.
This video will only show
rotation about the x-axis.
They'll be other videos that show rotation
about other axis.
Now the Washer Method is very
similar to the disc method.
Remember for the disc method,
we combined the idea of
a volume of a cylinder,
as we see here, with
the idea of integration.
The volume of this solid
cylinder, or solid disc,
would be equal to pi r squared h,
where this would be the radius.
For the Washer Method
we'll have a hollow disc.
So to determine the volume
of a hollow cylinder,
we'd have to take the volume
of the larger cylinder,
minus the volume of the smaller cylinder.
As we see here, this will be the volume
of the outer cylinder, and
this will be the volume
of the inner cylinder.
Where this distance here would be big r,
and this distance here will be little r.
And the height would
be this distance here.
So we'll combine this idea
here with integration,
determine the volume of a solid formed
by rotating the area about the x-axis.
So to get a better idea of that,
here is an area bounded by two functions.
If we rotate this about the x-axis,
it would produce this solid here.
And you can see the
inside would be hollow.
So we'll find this volume by accumulating
the volume of very thin
individual washers,
that look something like this,
that will give us the actual volume.
And we'll introduce the
idea of accumulating volume
or area that should remind you
of the idea of integration.
And within this integration
formula to determine
the volume of a solid
rotated about the x-axis,
you can see it's very
similar to the volume formula
on the previous screen,
where we have pi times
the integral of big r squared
minus little r squared dx.
Where big r is the outer radius,
little r is the inner radius,
and dx is going to be
the width or the height
where dx will be the width
of each individual washer.
So let's see how we're
going to set this up.
In order to set up this integral,
we'll consider one washer
that will be formed
by rotating a rectangle
bounded by the two functions
about the x-axis.
So if you take a look
at this one disc here,
it could be formed by this rectangle here.
And this rectangle is going to be key
to setting up this definite integral
to determine the volume.
The first thing we should
notice here is that the width
of this rectangle would be delta x.
And because the width is delta x,
we'll integrate with respects to x.
Next, big r of x is going
to be the outer radius
with the distance from
the axis of rotation,
to the outer function.
Which would be this distance here.
And little r of x will
be the inner radius,
or the distance from the axis of rotation
to the inner function, as we see here.
So let's go ahead and look at a problem
and see if we can set it up.
We want to determine
the volume of the solid
generated by the bounded
region of the given equation
rotated about the x axis.
So y equals negative x squared plus four,
that's the red function.
Y equals two x plus one,
that's the blue function.
Then we have x equals zero,
which would be the y axis.
We're taking this region here,
and rotating it about the x axis.
To get a better feel for
what's actually happening here,
let's take a look at some images.
We rotate this area
bounded by those functions
about the x axis, it would produce a solid
that looks like this.
Now we'll use the Washer
Method to determine the volume.
Which means we'll start
[mumbles] the volume
of very thin washers on this interval,
to determine the actual volume.
So the key to setting up the
integral is going to be to
take a look at one rectangle,
bounded by the two
functions that would produce
a washer that looks like this in blue.
So let's go ahead and do that.
First thing we'll do is draw
representative rectangle
that would form one disc.
That could be from here to here.
And one of the keys here,
is that the width of this rectangle,
would be delta x, which
means we'll integrate
with respects to x.
So our volume is going to be equal to pi
times the integral from a to b.
Well a would be zero and b would
be equal to one in this case.
So we'll integrate from zero to one.
Big r of x is going to
be the outer radius,
or the distance from here to here.
This distance here would be determined by
the parabola, and this is equal to y,
and y is equal to negative
x squared plus four.
So the outer radius will
be negative x squared
plus four.
And we are going to square this.
Minus the inner radius.
And this distance here
is the inner radius.
And this is determined by the blue line.
Well this distance is equal to y,
and y is equal to two x plus one.
So I have two x plus one squared.
And this should give us
the volume of that solid.
So let's go ahead and first
square these binomials.
So we'd have x to the fourth
minus eight x squared,
plus 16 minus, this square
would be four x squared,
plus four x plus one.
Integrated with respects to x.
Now combining like terms,
looks like we have x to the fourth,
minus 12 x squared, minus four x,
and then 16 minus one, that'd be plus 15.
This is from zero to one.
Now we'll go ahead and
find the anti derivative.
So we have x to the fifth over five,
minus 12 times x to the third over three,
minus four times x squared
over two, plus 15 x.
And we'll evaluate this
at the upper or lower
limits of integration.
Let's go and simplify this
and then we'll evaluate it.
So we're going to have pi,
should be 1/5 x to the fifth,
minus four x cubed, minus
two x squared, plus 15 x.
So now we'll go ahead and
substitute the one in for x.
So we'll have 1/5 minus
four, minus two plus 15.
When x equals zero, this
whole thing is zero.
So we have minus zero.
This comes out to 46 pi over five.
So that would be the volume
of the solid that we see here.
And again, the way we did that is,
we considered one disc
formed by a representative
rectangle and then as we
integrate we accumulate
the volumes of an infinite
number of washers,
which would equal the
volume of this solid.
That'll do it for this first video.
Next we'll take a look at
rotating about the y axis.
I hope you found this helpful.
