We want to find the area inside one leaf
or one petal of the rose given
by r equals four sine five theta graphed here in blue.
So we want to find the area of one petal
or the area of the shaded region here.
To do this we'll be using the formula,
area equals 1/2 times integral of r squared
integrated with respect to theta, from alpha to beta.
Where the interval from alpha to beta
would trace out one petal of our rose.
So, the area of one petal is equal to 1/2
times the integral of r squared,
which would be four sine
five theta squared integrated with respect to theta
from alpha to beta.
To determine the limits of integration
we want to find the interval that would trace out one petal.
Notice when theta equals zero
we'd have r equals four times sine zero,
which would be four times zero, or zero.
Which means when theta equals zero
would be at the pole here
which we'll use as the lower limit of integration.
Now to find the upper limit of integration
we want to determine
what's the next smallest positive value for theta
where r would be zero again.
There's a couple ways of doing this.
One way would be to analyze the graph
of r equals four sine five theta.
The other way would be to set r equal to zero
and solve theta.
Let's show both methods.
The horizontal axis would be theta,
and the vertical axis would be r.
Notice how the period would be two pi divided by five,
which let's say would be here.
So if we want to graph one period of our sign function here,
we would divide this into four equal intervals.
Here would be pi over five radians,
half of this would be pi over 10 radians,
and here we'd be at three pi divided by 10 radians.
Notice how the amplitude is positive four,
so r has a maximum value of four,
and a minimum value of negative four.
Now because we have the sine function
at zero radians it starts at the mid-line,
which would be here.
Then here we'd be at a maximum of four,
here we'd be at the mid-line,
here we'd be a at a minimum of negative four,
here would be back at the mid-line.
So the graph of r equals four sine five theta
would look something like this.
Again notice that r is zero at zero radians
and notice how it returns back to zero
at pi over five radians
which means the point would be back at the pole.
We would have traced out one petal of our rose.
So, pi over five radians is the upper limit of integration.
Again, this petal is traced out from theta equals zero
to theta equals pi over five radians.
As we see here.
Now the second method for determining
our limits of integration
would be to set r equal to zero and solve for theta,
let's also show that method.
If we set r equal to zero,
we'd have zero equals
four sine five theta.
If we divide both sides by four
we have zero equals sine five theta.
Let's perform a substitution here
and let x equal five theta.
So the equation would now be zero equals sine x.
We know we want to start at zero radians.
So if we let x equal zero,
notice how the next positive angle
where sine x would be zero,
would be x equals pi radians.
But x is really equal to five theta,
so we'd have five theta equals zero,
five theta equals pi.
If we divide both sides by five
notice how we get theta equals zero
and theta equals pi over five,
which we already found graphically.
This isn't the only possible interval
but this interval does trace out one petal
and therefore we can use this interval
to find the area of one petal.
Now to evaluate this def integral,
notice how if we square four sine five theta
we'd have 16, 16 times 1/2 is eight.
So let's write this as eight times the integral of,
we'd have sine squared
five theta.
Now from here we'll perform a substitution
for sine squared five theta
which is a power reducing formula given here below.
Notice how if we have sine squared x here
on the right we have cosine two x, so the angle is doubled.
So, sine squared five theta is equal to
1/2 times the quantity one minus cosine 10 theta.
Since eight times 1/2 is four,
we can write this as four times integral
of the quantity one minus cosine
10 theta.
Now, when we find the anti-derivative
we have to be a little careful here.
We have to perform u substitution
to integrate cosine 10 theta,
where u would be 10 theta
and therefore differential u equals 10d theta.
So if we divide both sides by 10,
notice how we have 1/10 du
equals d theta.
Which means we'll have an extra factor of 1/10
when we integrate cosine 10 theta.
So we'd have four times
the anti-derivative of one,
with respect to theta would be theta,
minus the anti-derivative of cosine 10 theta.
Again after performing u substitution
we'd have 1/10 sine 10 theta.
So when theta is equal to pi over five
we'd have pi over five
minus 1/10 times sine of,
10 times pi over five would be two pi.
Then when theta is zero
we'd have zero minus 1/10 times sine zero.
Notice how sine two pi is zero,
so this part is zero, this is zero,
and sine zero is also zero.
So the exact value, or the exact area,
would be four times pi over five
or four pi
divided by five,
and this would be square units.
Let's also get our decimal approximation
on the graphing calculator.
Four pi
divided by five
is approximately 2.5133.
Since we do have our graphing calculator
let's go and evaluate the original def integral
to check out work.
Before we do this though, let's check our mode.
Notice how we are in radian mode,
notice how we're also in function mode.
If we want to we can change this to polar mode.
It doesn't really matter except if we change to polar mode
the variable will be theta instead of x.
Let's go and change this to polar mode by pressing enter.
Now we'll go back to the home screen.
Now we'll enter .5 for 1/2 and then math.
We want option nine for function integration,
which is here, enter.
Now I do have a newer operating system.
Notice how this looks like an integral.
If you have an older version it will look different.
But in this case I'm going to enter zero
for the lower limit of integration.
Right arrow, upper limit of integration
is pi divided by five,
right arrow,
in parentheses I'll enter four sine five theta squared.
So open parenthesis,
four sine
five theta,
close parenthesis for the angle.
Close parenthesis and then squared, right arrow.
The variable we're using is theta
so I'll enter theta and enter.
Notice how this does verify that our work is correct.
The area is approximately 2.5133.
I hope you found this helpful.