Welcome to a presentation
on surface area of revolution.
Let's start by taking a look at an animation
to develop the idea.
If we have a function and rotate it about the x-axis,
it will produce a solid as we see here.
And our goal is to determine the surface area
of this solid.
Now, you do have to be careful not to confuse this
with the volume of a solid,
and the idea we use to derive the surface area
of this solid will be very similar
to the idea that we use to determine the arc length
of a curve, except now we'll use strips
instead of segments.
So if you take a look at this red band,
the idea is if we were to determine the surface area
of these red bands,
it would approach the surface area of the solid.
So we should be thinking about how we would
determine the surface area of one of these strips.
What we'll do is find the circumference
and then multiply it by the width of the band or strip.
So with that thought in mind,
if we wanted to determine the surface area
of this solid, we could divide it up into multiple strips.
So we might have one here, maybe another here.
Another here, maybe one more here on the end.
Each of these bands would be called a frustum
which we see here above.
And the surface area for our frustum
is equal to two pi r l, where r is the average
of the two radii.
This would be r one, this would be r two.
And l would be the length of this segment here.
So looking at our sketch, we could call this l one,
l two, l three, l four, and so on.
The idea is if we find the surface area
of more and more frustums,
it will approach the surface area of the solid.
Another way to express this would be to say
that the change in the surface area
would be equal to two pi r times delta l,
which is the change in this segment here.
And this would be the change from just one single frustum.
So if we want to sum n frustums,
we could say that the sum from i equals one to n of delta s
would equal the sum from i to one of two pi
times r sub i, times delta l sub i.
Now, let's consider r sub i.
R sub i is the average of the two radii
of the frustum, so we take a look at this blue frustum.
R sub i would be the average radii,
or the average function values from here and here.
But due to the intermediate value theorem,
we can state that there is some function value
in this interval that would be equal to that average radius.
So we're going to go ahead and replace
r sub i with f of x sub i.
And then delta l is the length
of each of these segments here,
which is the same as the segments we use
to determine arc length.
So we're going to go ahead and replace delta l sub i
with delta small s sub i.
Again, there's some function value in this interval
that would be equal to r sub i,
so we'll call it f of x sub i,
and then delta l sub i is the same as delta small s sub i
for the change in arc length.
Now you probably see where we're headed here.
We're going to take the limit as n approaches infinity
of this sum, which will give us our definite integral.
So limit on the left does give us the surface area.
Two pi would be the constants
and then we're left with a definite integral
of f of x times the square root of
one minus f prime of x squared, which, remember,
is equal to our arc length.
Let's go ahead and formalize this idea.
The surface area of revolution formed by revolving the curve
about the x-axis is given as follows.
It's equal to two pi times a def integral from a to b
of f of x ds or two pi
times the definite integral
from a to b of y ds.
Notice this is expressed a little bit differently
than we saw on the previous screen.
And what's happening here is that ds is the arc length
part of the formula, and arc length can be found
two different ways.
One with respect to the x
or with respects to y.
And if we find it in terms of x,
we have to express the radius in terms of x.
And if we find it in terms of y,
we have to express the radius in terms of y.
And if the curve is rotated about the y-axis,
we have a similar option
where ds here and here can be found in terms of x or y,
but again, if it's in terms of x,
the radius would be x,
and if it's in terms of y,
the radius would be expressed in terms of y.
So we do have specific formulas based upon
the axis of rotation.
But we do have a choice when it comes to ds.
Let's go and take a look at an example.
We want to determine the surface area
generated by revolving f of x equals 1/3 x cubed
about the x-axis on the interval from zero to two.
So here's the graph of the function,
and we want to determine the surface area
of this solid rotated about the x-axis.
Well, the surface area is going to be equal to
the def integral from a to b of f of x ds,
as we saw in the previous screen.
And you can see over here in red,
I decided to use the arc length piece
that is written in terms of x.
So before we try to apply this,
notice we do have to find the derivative and then square it.
So f prime of x will be equal to, multiply it by three,
and then subtract one from the exponent,
that will just be x squared.
So f prime of x squared is just going to be equal
to x to the fourth.
So let's go ahead and see if we can apply this now.
Surface area should equal two pi times the def integral
from zero to two of f of x.
Well, f of x is 1/3 x cubed.
Times the square root of one plus x to the fourth.
Let's go ahead and pull this 1/3 out
and also rewrite this in rational exponent form.
So we'd have 2/3 pi, x to the 1/3,
then all this to the 1/2 power.
Now, we're going to have to perform u substitution here
where u is going to be equal to one plus x to the fourth.
So du is going to be equal to 4x cubed dx.
Notice our integrand has x cubed dx in it,
so I'm going to go ahead and divide both sides by four.
So that tells us that 1/4 du is equal to x cubed dx.
Let's go ahead and rewrite this in terms of u.
We'd have 2/3 pi, now x cubed dx is the same as 1/4 du,
so let's pull that 1/4 out and we have the du here,
and then, remember, this is u, so we have u to the 1/2.
So out here, we're going to have 1/6 pi,
and if we integrate this, we'll have
u to the 3/2 divided by 3/2.
So let's go ahead and simplify this
and rewrite it in terms of x.
So here, we're going to have 1/6 times the reciprocal of 3/2.
That'd be 2/3.
1/6 times 2/3, that's going to be 1/9.
So I'll have 1/9 pi.
Then we have u to the 3/2,
but u is one plus x to the fourth.
Need to evaluate this at the limits of integration
which were two and zero.
Let's go and take this over to the next screen to continue.
So first, we'll replace x with two,
so I have one plus two to the fourth to the 3/2 power
minus one plus zero
to the fourth to the 3/2 power.
So we'll have 1/9 pi, looks like we're going to have
two to the fourth, that's 16 plus one.
That's 17 to the 3/2 power minus,
here we'll have one to the 3/2 power,
which would just be one.
And this should be the surface area of that solid.
Let's go back and take a look.
Again, here was the function
rotated about the x-axis, and so that value
is going to be the surface area of this solid.
Okay, that's going to do it for this video.
Thank you for watching.