Welcome back to another video
on the shell method for determining volume.
In this video, we'll take a look at
rotating about axes other than the x and y axis.
Let's start with a quick review.
One of the most important things
to remember about the shell method is
the representative rectangle is
going to be parallel to the axis of rotation
so for a horizontal axis of rotation, as we see here
the thickness of the rectangle is delta y
which tells us we'll integrate with respects to y.
Next, the distance from the center of the rectangle
to the axis of rotation will be the radius, or r of y
and the height of the rectangle will be h of y.
Again, both must be expressed in terms of y
and then lastly, because we're integrating
with respects to y
the limits of integration from c to d
will be along the y axis
and for a vertical axis of rotation
the rectangle will also be vertical
or parallel to the axis of rotation
and since the width of the rectangle is delta x
we integrate with respects to x
so this would be the radius in terms of x
and this would be the height in terms of x.
Let's go ahead and take a look at two examples.
In this problem, we will rotate
this region about the vertical line x equal five
so since we're rotating about a vertical axis
our rectangle will also be vertical
or parallel to this line
so it would look something like this
and since the width of this rectangle would be delta x
we will integrate with respects to x
so see if we can set this up using the shell method.
The volume is equal to two pi
times the definite integral of the radius
times the height.
This is going to take a little bit of thought.
Well, this distance here would be r of x.
We have to think about how we want to express this.
It's determined by the distance
from the red function to the green vertical line.
Well, we do know the distance
from the y axis to x equal five is equal to five units
so if we take five and subtract this distance here
we would have our expression for the radius function
and this distance here would be equal to x
so r of x is going to be equal to five minus x.
Now we have to determine the height of this rectangle
and that would be equal to this distance here
and this is equal to y on the given function
and y is equal to the square root of x
so h of x
is just going to be the square root of x
and since we're integrating with respects to x
the limits of integration will be from zero to four.
Let's go ahead and multiply this out.
Remember, this is the same as x to the one half
so we have five x to the one half
minus, this is x to the first times x to the one half.
That'd be x to the three halves.
Now, we'll go ahead and find the anti-derivative
so adding one to one half, that'll be x to the three halves
divided by three halves
minus, three halves plus one would be five halves
so x to the five halves
divided by five halves
so let's clean this up
and then we'll perform the substitution
so this would be five times two thirds.
That'd be 10 thirds.
Again, dividing by five halves
is the same as multiplying by two fifths.
Now we'll replace x with four.
We have to take all of this
and subtract when x equals zero
but luckily, when x is zero
both of these terms would be zero.
To save some time, I've already worked this out.
It comes out to 416 pi
divided by fifteen
which would be the volume of that solid
and would look like this
and this illustrates what we did using the shell method.
We accumulated the volume of shells
as the number of shells approached infinity
which gave us the actual volume of the solid.
Let's go ahead and take a look at another one.
So this would be the bounded region
and we're going to rotate about the bounded line
y equals negative one
therefore, our rectangle is going to be parallel
to this horizontal axis
so it might look something like this
and since now the thickness is equal to delta y
we have to integrate with respects to y
so the volume will equal two pi
times the definite integral of
r of y times h of y.
Well, r of y would be the distance
from the center of this rectangle
to the axis of rotation.
Well, the distance from the function
to the x axis would be y
but we're one more unit away
so this would be y and then plus one
so r of y is equal to y plus one
and the height of the rectangle would be h of y.
That would be the distance from here to here.
Well, this horizontal distance is equal to x
based upon the function y equals the square root of x
so we have to take y equals the square root of x
and solve it for x
so if we were to take this
and square both sides
we would have y squared equals x
and that will be the function for the height.
H of y is equal to y squared
and lastly, we have to determine our limits of integration.
Because we're integrating with respects to y
our limits of integration will be along the y axis
so the limits of integration will be
from zero to two.
Let's go ahead and finish this.
Now we'll distribute the y squared
so we'll have y cubed plus y squared.
Now we'll find the anti-derivative.
We'll have y to the fourth over four
plus y cubed over three.
Now we'll replace y with two
so two to the fourth
plus two to the third over three
and then when y is zero, both of those terms would be zero
and to save some time, I've already determined this value.
It's going to be 20 thirds
so we have two pi times 20 thirds
which is equal to 40 pi over three
which would be the volume of the solid
formed by rotating this region
about the line y equals negative one
and that'll do it for this video.
Thank you for watching.