In this example, we're asked to find
the area of the surface, or surface area,
generated by rotating the curve
given by the parametric equations on the given interval of t
when the curve is rotated about the x-axis.
So the given parametric equation that's on this interval
would produce this blue curve in the first quadrant.
Then if we rotate this about the x-axis here,
it would produce this red surface.
And our goal is to find the area of this surface,
or the surface area.
To do this we're going to be using this formula here.
And notice because we're rotating about the x-axis,
the integrand contains y.
This y represents the radius.
Notice how the radius would be a vertical distance.
When rotating about the y-axis, the integrand contains x.
An example of one,
we did discuss where this formula comes from,
but for a quick review, the important thing to remember is
when finding the surface area of revolution
in parametric form for rotating about the x-axis...
Again the integrand contains the variable y,
where the y represents the radius,
but in this case the radius is a vertical distance.
And when rotating about the y-axis,
the integrand contains x
because the radius is now a horizontal distance.
So now let's go back to our example
and find the surface area.
So the surface area is going to be equal to two pi
times the integral from zero to pi over two.
And the integrand is going to be y times this square root.
Well, in this case y is equal to sine cubed t.
And then we'll have the square root, d t,
where the raticand is dx/dt squared plus dy/dt squared.
And notice to find these derivatives,
we will have to apply the chain rule.
So for dx/dt, since x is equal to cosine cubed t,
the derivative would be three cosine squared t
times the derivative of cosine t.
So we'd have three cosine squared t
times the derivative of cosine t, which is negative sine t.
And all of this is squared.
And then plus dy/dt squared,
again where y is sine cubed t.
So dy/dt would be three times sine squared t
times the derivative of sine t.
So that would be three sine squared t
times cosine t.
And again all of this is squared.
Now let's begin to simplify the integrand.
So we'll start by simplifying underneath the square root.
So here again we're squaring this,
so three squared is nine, so we'll have nine.
And then we're squaring cosine squared.
That's cosine to the fourth t,
and then we're squaring negative sine t,
which would be sine squared t.
And then plus, again three squared is nine.
And then we're squaring sine squared t,
so sine to the fourth t,
and then cosine squared t.
And now for the next step we'll factor out
the greatest common factor from the raticand,
which would be nine cosine squared, sine squared.
So we're going to factor out nine cosine squared t,
sine squared t.
That's going to leave us with
cosine squared plus sine squared.
But cosine squared t plus sine squared t is equal to one.
So this simplifies nicely to just
nine cosine squared, sine squared.
But now nine cosine squared t,
sine squared t is a perfect square,
so this simplifies nicely to just
three cosine t, sine t.
Now let's go ahead and factor out the three,
then multiply the trig functions.
Factor out the three, that would give us six pi.
And then we have four factors of sine
and one factor of cosine,
so sine to the fourth t, cosine t, dt.
And now we should recognize that
the derivative of sine t is equal to cosine t,
so we can to integrate using u-substitution,
where if u is equal to sine t,
which is by u to the fourth,
and differential u is equal to cosine t dt,
which we have here.
So again if u equals sine t, this becomes u to the fourth.
And if differential u equals cosine t dt, this becomes d u.
When writing in terms of u,
we'll leave off the limits of integration,
so let's write this as six pi times the integral of,
this would be u to the fourth, d u.
The integral of u to the fourth would be
u to the fifth divided by five, or 1/5 u to the fifth.
Let's go and factor out the 1/5.
Six times 1/5 would be 6/5,
so we have six pi over five, and then times.
And then u to the fifth is sine to the fifth t.
And now because we're back in terms of t,
the limits of integration were from zero to pi over two.
So now we have six pi over five,
times sine to the fifth pi over two,
minus sine to the fifth, zero.
Well sine pi over two is equal to one,
so this is one to the fifth, or one.
Sine zero is zero.
Of course zero to the fifth is still zero.
So this is just six pi divided by five times one,
or six pi over five.
And this is square units for our surface area.
So if we go back to our graph,
this red surface area is equal to 6/5 pi,
or six pi divided by five square units.
Okay, I hope you found this helpful.