Welcome to an example
on how to determine a volume of revolution
using the washer method
when we have rotation about Y equals three.
So for this example, we're asked to use a washer method
to determine the volume of the solid
formed when the region bounded by
Y equals 1/2 X squared, Y equals zero, and X equals two,
is rotated about Y equals three.
So because we have rotation about Y equals three,
which is parallel to the X axis,
we'll be using this formula here to determine the volume,
where the volume is equal to pi time integral
from A to B of big R of X squared
minus little R of X squared
integrated with respect to X.
Or Big R of X would be the outer radius,
and little R of X would be the inner radius.
So for the first step, let's go ahead
and graph the bounded region,
and sketch the axis of rotation.
I've write down this to save some time.
Here we have Y equals 1/2 squared.
Here we have Y equals zero, and here we have X equals two.
So this yellow region is a bounded region,
and we'll rotate this region about
Y equals three, which is this axis here.
So if we rotate this region about Y equals three,
it would give us this solid here.
So our goal here is to find the volume of this solid.
To help us set up our integral though,
it's always helpful to sketch a represented rectangle,
which should be a rectangle that's perpendicular
to the axis of rotation,
which would be this rectangle here,
that if we rotated this rectangle,
it would give us one washer of our solid.
Notice how the thickness of this rectangle is delta X,
which is the reason why we integrate with respect to X.
So now let's work on determining big R of X,
the outer radius,
and little R of X, the inner radius.
Because we're rotating about Y equals three,
the outer radius would be this distance here,
so this distance is big R of X.
The inner radius would be the distance from Y equals three
to the graph of Y equals 1/2 X squared,
this distance here.
Notice how the length of the outer radius, big R of X,
has not changed over this bounded region.
The distance from Y equals zero to Y equals three
is always three units,
which means big R of X equals the constant three.
We need to be a little more careful
in determining little R of X.
Again little R of X is the distance
from the line Y equals three
to the graph of Y equals 1/2 X squared,
which should be three units minus this distance here,
which was given by the function value
of Y equals 1/2 X squared,
which means little R of X
is equal to three minus 1/2 X squared.
Again, we take the distance of three units
and subtract the function value
given by Y equals 1/2 X squared.
It leaves us with the inner radius, little R of X.
This is all we need in order to set up the integral.
It will give us the volume of this solid here
when we rotate this bounded region about Y equals three.
We're going to have the volume equals pi times integral of.
We'll notice how the boundary region
goes from X equals zero to X equals two.
So notice integration are from zero to two,
and then we'll have big R of X squared
which would be three squared minus little R of X squared,
which in this case is the quantity three
minus 1/2 X squared, squared.
You integrate this with respect to X.
Now that we have integral,
let's go back to our first slide and evaluate this.
We'll have pi times integral from zero to two.
Of course three squared is nine,
so I have nine minus the quantity.
Let's go ahead and square three minus 1/2 X squared.
So we have three minus one half X squared
times the quantity of three minus 1/2 X squared.
So we have nine,
and then minus 3/2 X squared, minus another 3/2 X squared,
which gives us negative 6/2 X squared
or minus three X squared,
and then we have plus 1/4 X to the fourth.
Remember, we are subtracting the square in this binomial,
so it's important to put this in parenthesis.
We have minus the quantity nine minus three X squared
plus 1/4 X to the fourth.
So, simplifying, we have pi
times the integral from zero to two of...
We have nine minus nine, that's zero.
And then minus negative three X squared
becomes positive three X squared,
and then we have minus 1/4 X to the fourth.
And now we'll find the antiderivatives.
The antiderivative of three X squared respect to X
would be three times X to the third divided by three,
or just X to the third.
And here we're going to have minus 1/4
times antiderivative of X to the fourth.
It should be X to the fifth divided by five.
So let's write this as one over four times five
X to the fifth.
Notice integration are from zero to two.
Let's go and simplify this one more time,
so we have five times.
We have X to the third minus 1/20 X to the fifth.
Now we need to find big F of B minus big F of A,
so we have pi times the quantity.
When X is two, we have two to the third
minus 1/20 times two to the fifth minus...
Notice an X is zero,
both terms would be zero.
So we have pi times two to the third is eight minus
this would be 32/20.
Simplifying, we get 32 pi divided by five.
And again, this is volume,
so this would be cubic units.
So this is the exact volume of this solid here
formed by rotating this yellow region
about the line Y equals three.
Of course if you wanted to,
we could get a decimal approximation using the calculator,
and let's also do that.
We have 32 pi
divided by five,
which would be approximately 20.1062 cubic units.
I hope you found this helpful.