We want to find the surface area of revolution
about the Y axis of Y equals X squared divided by nine
over the interval where X is greater or equal to zero
and less or equal to four.
The graph of the function over the given interval
is shown here.
We have rotation about the Y axis.
So if you rotate this curve about the Y axis,
you get this surface.
Our goal is to find the area of this surface
or the surface area.
Looking at our notes, to find the surface area
of revolution of the rotation about the Y axis,
we have a choice of two formulas.
Notice in the first formula, we integrate
with respect to Y.
The second formula, we integrate with respect to X.
These two integrals are equivalent
and let's talk about why.
Remember F of Y is equal to X
so both of these give us the distance
from the curve to the axis of rotation.
For example, if we consider this horizontal distance here,
we know this horizontal distance is equal to X
which also equals F of Y.
And then the square root of one plus square
of F prime of Y Dy is equal to Ds.
And so is the square root of one plus square
of F prime of X Dx.
While this represents the arc length,
whether we integrate along the X axis or Y axis,
of course we get the same arc length.
So you might be asking which formula we should use.
You want to use the formula where the derivative function
would be simpler.
So looking at the given function Y which equals F of X,
in order to get F prime of X which will be 2/9 X
which is pretty simple,
if you wanted to find F prime of Y,
you'd first write this as a function of Y by solving for X.
Notice how we'd have to multiply by nine
and take the square root of both sides of the equation
which means F prime of Y would involve rational exponents.
So because F prime of X is simpler than F prime of Y,
for this example, we'll use this formula here
to find the surface area.
So we know that F of X equals X squared
divided by nine.
And therefore F prime of X is equal to 2/9 X.
And we already know the interval of X is from zero to four.
So we have all the information we need.
The surface area is equal to 2Pi
times the integral from zero to four
of X times the square root of one plus
F prime of X squared.
That would be the square of 2/9 X dx.
Let's go ahead and square 2/9 X so we have 2 Pi
times the integral from zero to four of x.
Let's also write this using a rational exponent.
So we'd have quantity one plus four over 81x squared
raised to the one half power Dx.
Now to evaluate this on the next slide.
Notice how we'll perform, use substitution
where u is equal to 1 plus 4 over 81x squared.
So differential of u is equal to zero plus,
and we're going to multiply by two so that'd be
eight over 81 x dx.
Let's go and solve this for xdx
and multiplying both sides by the reciprocal
of eight over 81 which would be 81 over eight.
So we'll have 81 over 8du equals xdx.
So performing this substitution,
xdx equals 81 over 8du.
And all this would be u to the 1/2.
So we'd have 2Pi.
The antiderivative with respect to U
would be 81 eigths u to the 3/2, divided by 3/2
which is the same as 2/3 u to 3/2.
So with respect to X we'll have 81 over eight
times two thirds times the quantity
one plus four over 81 x squared raised to 3/2.
Let's go ahead and factor these fractions out.
Two times 81 over eight times 2/3 equals 27/2.
So we have 27Pi halves times,
when X is four, we'd have quantity one plus
four over 81 times four squared raised to 3/2
minus when X is zero we just have one to the 3/2.
Simplifying, we have 27 Pi divided by two
times, here we have 16 times four over 81
and then plus one.
Which comes out to 145 over 81
just raised to the 3/2 power.
And then one to the 3/2 is just one.
So this would be the exact surface area
which we'll leave in this form.
This should be square units.
Let's also get a decimal approximation.
So we have 27Pi divided by two.
And we have open parenthesis, another open parenthesis
for fraction of 145 divided by 81.
Close parenthesis.
This is all raised to the 3/2 power
so here's the exponent key.
Three divided by two, right arrow
minus one, close parenthesis and enter.
So the surface area is approximately 59.1685 square units.
So we have both the exact and approximate surface area
of this surface here generated by rotating
this curve over the given interval about the Y axis.
I hope you found this helpful.