We want to use the Washer Method to determine
the volume of the solid formed when the region
bounded by Y equals two X squared
and Y equals to square root X
is rotated about Y equals three.
Because the axis of rotation is Y equals three
which is parallel to the X axis,
we'll be using this formula here
to determine the volume
When the volume equals pi times
the integral from A to B
of the square of big R of X
minus the square of little R of X
integrated with respect to X,
where big R of X is the outer radius
and little R of X is the inner radius.
So the first step is to graph the bounded region
as well as the axis of rotation
given by Y equals three.
So break down this to save some time.
The green graph is Y equals two X squared,
and the blue graph is Y equals two
times the square root of X.
So notice how the bounded region is this yellow region here,
which are rotating about the line Y equals three,
so rotating this region about this axis here,
which would give us this solid.
So our goal is to find the volume of this solid
shown here on the right.
To help us set up the integral
and determine big R of X and little R of X.
It's always helpful to sketch a representative rectangle,
and if we were to rotate the rectangle
about the axis of rotation,
it would give us one Washer of volume of the solid.
So the representative rectangle is
this red rectangle here,
and again, notice how it's perpendicular
to the axis of rotation.
Notice how the width of the thickness
would be delta X,
which indicates we integrate with respect to X.
So using this rectangle,
notice how the outer radius
would be the distance from Y equals three
to the point on the green function
Y equals two X squared,
this distance here,
and little R of X would be the distance
from Y equals three to the corresponding
point on the blue function,
Y equals two, square root of X,
which would be this distance here.
So let's work on determining big R of X
and little R of X.
Again, big R of X would be this distance here.
You'll notice how the distance from Y equals to three
to Y equals zero,
the X axis would be three units.
So big R of X is equal to three
minus the distance from the X axis
to the point on the function Y equals two X squared,
which means big R of X
is equal to three
minus two X squared.
Again, the distance from Y equals three
to Y equals zero is three units,
and then minus two X squared,
which is minus this distance here,
gives us the distance from Y equals three
to the corresponding point on the function
Y equals two X squared,
and now little R of X
would be equal to,
like in the distance from Y equals three
to Y equals zero would be three units,
and then minus this distance here
would give us little R of X.
This distance here is equal to Y on the function.
Y equals two, square root X,
which means little R of X is equal
to three minus two, square root X.
Remember, we need both big R of X
and little R of X to be functions of X.
So again, little R of X is equal to three,
the distance from here to here,
minus this distance here,
given by two square root X,
which gives us little R of X
equals three minus two, square root X.
Last thing we need would be limits of integration,
and we can see from the graph,
these two functions intersect at the point
zero comma zero, the origin,
and this point here, which is one comma two.
So because we're integrating with respect to X,
we'll integrate from zero to one.
If needed though, sometimes you may have to solve
these two equations as a system of equations
in order to determine the limits of integration,
but here it's pretty easy to see
we integrate from zero to one,
which means the volume V is equal
to pi times the integral from zero to one
of the square of big R of X,
would be the quantity three minus two X squared,
squared, minus little R of X squared,
would be the quantity there minus two,
square root X squared,
and we integrate with respect to X.
So now that we have our integral,
let's evaluate this on the next slide.
So for our next step,
let's go ahead and square these two binomials.
So you first have the quantity three
minus two X squared
times the quantity three minus two X squared,
and we also have the quantity three
minus two X to the power of, let's say, 1/2,
instead of square root,
times the quantity three minus two X to the 1/2.
So multiplying here, we have nine,
and then we have minus six X squared,
minus six X squared.
That's minus 12 X squared.
Then we have plus four X to the fourth.
Here we'll have nine minus six X to the 1/2,
minus six X to the 1/2,
that's minus 12 X to the 1/2,
and then we have plus four,
X to the power of one.
So we have the volume is equal to pi
times the integral from zero to one
of quantity nine minus 12
X to the second,
plus four X to the fourth,
minus the quantity nine minus 12 X
to the 1/2,
plus four X,
integrated with respect to X.
Now let's go ahead and simplify this difference
by combining like terms.
So we have nine minus nine.
That's zero,
and there are no other like terms,
so we have negative 12 X to the second
plus four X to the fourth
plus 12 X to the 1/2
minus four X.
Now we'll find the antiderivative.
So here we'll have negative 12 times
X to the third
divided by three
plus four times X to the fifth divided by five.
So we'll have plus 4/5,
X to the fifth.
Then we have plus 12 times X to the 3/2
divided by 3/2,
which is equivalent to 2/3 times X to the 3/2.
So we have times 2/3 X to the 3/2,
and then minus four times X squared
divided by two.
Let's go ahead and simplify.
So here we have negative four X to the third
plus 4/5 X to the fifth.
Here we'll have plus eight
X to the 3/2,
and then here we'll have minus two X squared.
Now we need to find big F of B
minus big F of A.
So when X is one,
we'd have negative four times one cubed
plus 4/5 times one to the fifth,
plus eight, time one to the 3/2
minus two times one to the second,
and notice when X is zero,
all the terms would be zero.
Simplifying, we're going to have pi times,
we'll have negative four
plus 4/5, plus eight
minus two.
So pi times this quantity is equal
to 14 pi divided by five,
and again, this is volume,
so this would be cubic units.
Let's also get a decimal approximation.
14 pi, divided by five,
is approximately 8.7965,
and again, this would be cubic units.
So going back to our sketch one last time,
if we rotate this bounded region
about Y equals three,
we produce this solid,
and the volume of this solid is 14 pi
divided by five cubic units.
I hope you found this helpful.