We want to find the
surface area of revolution about the X-axis
of the linear function Y equals two X plus one
over the closed interval from one to three.
Looking at the graph on the bottom left,
here is our linear function on the
closed interval from one to three.
If we rotate this about the X-axis,
it would give us the surface.
Our goal is to find the surface area of this surface.
Because the surface is a frustum of a right circular cone,
we could find the surface area
using this geometric formula here,
but let's first find the surface area using integration,
and then we will verify it using this formula.
Because we have a function of X,
the surface area, S, is equal to two pi,
times the integral of R of X,
times the square root of the quantity
one plus F prime of X squared,
integrated with respect to X, from A to B,
where R of X is the distance between the graph of F of X,
and the axis of rotation.
Notice how the square root in the integrand
is the same square root we used to find arc length
when we have a function of X.
If we pick any point on our linear function,
let's just say this point here,
notice how this length here would be R of X,
which is the function value of two X plus one.
We know R of X equals two X plus one,
and F of X is also two X plus one,
so if F of X equals two X plus one,
notice how F prime of X would just be two,
and therefore, the surface area, S,
is equal to two pi times the integral from one to three
of R of X, which we know is two X plus one,
times the square root of one
plus F prime of X squared,
that would be two squared, integrated with respect to X.
Notice how here, we just have the
square root of five, which is a constant.
Let's go ahead and factor that out,
and write this as two square root five pi
times the integral of two X plus one,
integrated with respect to X, from one to three.
Let's go ahead and evaluate this on the next slide.
We would have to square root five pi,
and then we will find the anti-derivative.
The anti-derivative of two X would be
two times X squared divided by two,
plus the anti-derivative of one, which is just X.
Let's go ahead and simplify this.
This simplifies to just X squared.
We have two square root five pi,
and then we have X squared plus X.
Now, we will evaluate this at three, and then one,
and then find the difference.
When X is three, we would have three squared plus three,
and when X is one, we would have one squared plus one.
We have two square root five pi times,
here we have nine plus three, that's 12
minus, this would just be two.
We have two square root five pi times 10,
and therefore, the surface area would be
20 square root five pi,
and this would be square units.
Let's also get a decimal approximation for this.
We would have 20 square root five
times pi.
We would have approximately
140 point four nine six three square units.
Now, as I mentioned earlier,
because the surface is a frustum of a right circular cone,
we could also find the surface area using this formula here,
where S equals two pi times R times L,
where R is the average radius,
or one half times the quantity R sub one plus R sub two,
and L would be the length of the line segment.
Because it's a line segment,
and we can find the length using the distance formula
between two points.
Looking at this sketch of our line
over the closed interval from one to three,
this length here would be R sub one,
and this length here
would be R sub two.
Notice how this point is the point one coma three,
and this point is the point three coma seven.
We will begin by finding R, and then L,
and then use our formula to find the surface area.
R is equal to one half times the quantity,
notice how the length of this radius is three units,
the length of this radius is seven units.
Notice how these lengths are the function values,
or the Y coordinates of the two endpoints.
We have three plus seven.
One half of 10 is five.
Now, let's find L, the length of our segment.
It would be equal to the square root
of X sub two minus X sub one squared,
plus Y sub two minus Y sub one squared.
We would have three minus one squared
plus seven minus three squared.
This gives us the square root of,
this would be two squared, or four,
plus, here we have four squared, or 16,
so we have the square root of 20.
Therefore, the surface area is equal to
two pi times R, which is five,
times L which is square root 20.
The square root of 20 does simplify,
so let's finish this on the next slide.
20 is equal to four times five,
so we would have two times pi times five
times the square root of four times five.
The square root of four is equal to two,
so we have two pi times five
times two square root five,
which does give us the same result
of 20 square root five pi.
This was the same result that we had using integration,
as we see here.
Again, this would be square units,
which is also approximately
140 point four nine six three square units.
Just keep in mind, the only reason we can use
this geometric formula for this example
is because the surface is a frustum
of a right circular cone.
I hope you found this helpful.