We want to use the washer method
to determine the volume of the solid formed
when the region bounded by y equals the square root of x
and y equals x to the fourth is rotated
about the y-axis over the closed interval
from zero to one for x.
Because we have rotation about the y-axis
we'll be using this formula here to determine the volume.
The volume equals pi times the integral from c to d
of the square root of big R of y
minus the the square of little r of y
integrated with respect to y.
Where big R of y is the outer radius
and little r of y is the inner radius.
Because we're integrating with respect to y
notice how both the inner radius and outer radius
must be expressed as functions of y.
Let's first look at the bounded region
over this closed interval
and then determine the solid that would be formed
when rotating about the y-axis.
The green graph is y equals x to the fourth
and the blue graph is y equals the square root of x.
Notice how the two graphs intersect
at the point zero comma zero, the origin,
as well as the point one comma one.
So the yellow region is the bounded region
and we're going to rotate this region about the y-axis here,
which would form this solid here on the right.
Our goal is to find the volume of this solid
using the washer method.
Notice how when x is on the closed interval from zero to one
y would also be the closed interval from zero to one.
That's important because remember,
we are going to integrate with respect to y.
Now to help us setup our integral
it's always helpful to sketch a representative rectangle
which is this red rectangle here.
When using the disk or washer method
the representative rectangle will always be perpendicular
to the axis of rotation.
Notice how this rectangle is perpendicular to the y-axis.
We call this a representative rectangle
because if we rotate this about y-axis
it would give us one washer of volume
of the solid shown on the right.
Now let's determine big R of y and little r of y.
Again big R of y would be the outer radius.
So using our rectangle that would be the horizontal distance
from the y-axis to the corresponding point
on the green graph, y equals x to the fourth.
So this horizontal distance here
would be big R of y.
Then little r of y would be the inner radius
which would be the distance from the y-axis
to the corresponding point on the blue function
given by y equals the square root of x.
Therefore this distance here
would be little r of y.
Because these are horizontal distances, or x distances,
we'll have to solve both of these equations for x
in order to find big R of y and little r of y.
Let's first find Big R of y,
which again in controlled by the function
y equals x to the fourth.
If y equals x to the fourth we can solve this for x
by taking the fourth root of both sides of the equation
or take the 1/4 power of both sides.
So x is equal to y raised to the power of 1/4.
So this horizontal distance controlled
by y equals x to the fourth is
big R of y equals y to the 1/4.
Now little r of y is controlled
by y equals the square root of x.
Here if we square both sides of the equation
we can say x equals y squared.
Which means this horizontal distance
controlled by y equals the square root of x
would be the function little r of y equals y squared.
Now we have all the information we need
in order to find this volume.
We'd have the volume equals pi times the integral
from zero to one, remember along the y-axis,
because we're integrating with respect to y,
of the square of big R of y would be
y to the 1/4 squared
minus the square of little r of y
which would be y squared squared
dy.
Now that we have our integral
let's evaluate this on the previous slide.
We have pi times the integral from zero to one of,
y to the 1/4 squared would be y to the 2/4 or y to the 1/2,
minus, y to the second square would be y to the fourth.
Now let's find the anti-derivative.
The integral of y to the 1/2 would be y to the 3/2
divided by 3/2 or 2/3, y to the 3/2,
then minus the integral of y to the fourth
would be y to the fifth divided by five or minus 1/5,
y to the fifth.
Now I need to find big F of b minus big F of a.
So when y equals one we have 2/3
times one to the 3/2
minus 1/5 times one to the fifth.
Then when y is zero both terms would be zero.
We have pi times, this would just be 2/3,
minus 1/5.
The common denominator here would be 15.
Multiply 2/3 by 5/5
and multiply 1/5 by 3/3.
So here we have 10/15 minus 3/15,
so we have 7/15 times pi,
or seven pi divided by 15.
This is volume so it'd be cubic units.
Let's also get a decimal approximation.
Seven pi
divided by 15
is approximately 1.4661
cubic units.
So here we have the exact volume
and the approximate volume
of the solid shown here
formed by rotating this yellow region about the y-axis.
I hope you found this helpful.