We want to find the length of the entire
perimeter of the region inside r equals five sine theta,
graphed here in blue, but outside r equals three,
graphed here in red.
Let's first identify the region, and then we'll find
the length of the entire perimeter.
Again the region that is inside the blue circle
but outside the red circle,
would be this region here.
Our goal is to find the entire length of the perimeter
of this region, which would be this length
and this length.
We'll find this length using the arc length formula,
L equals the integral of the square root of r squared,
plus dr d theta squared, integrated with respect to theta,
from alpha to beta.
For alpha and beta, the limits of integration
would be the angle that passes through
this point of intersection
in the first quadrant,
and this point of intersection in the second quadrant.
So alpha would be this angle here, in the first quadrant,
and beta would be this angle here, in the second quadrant.
But notice how this piece of the perimeter
will be an arc length from r equals five sine theta,
and this length will be an arc length from r equals three.
So our integral will actually have two square roots
to find the length of the entire perimeter.
Let's begin by determining alpha and beta,
which will give us the limits of integration.
Again the angles would have to pass through
these two points of intersection,
which would occur when these two equations
have the same value of r.
Since they're both solved for r,
these angles would satisfy the equation
five sine theta equals three.
If we divide both sides by five,
we have sine theta equals 3/5.
Having a sine function value of 3/5 will not give us
a nice reference angle where we hope we can find
the unit circle, so we'll have to use the calculator.
So to solve this for theta, we'll take inverse sine
of both sides.
So we'd have inverse sine of sine theta,
equals inverse sine of 3/5.
And therefore theta equals inverse sine of 3/5.
Now remember though, the output or range of inverse sine
is from negative pi over two to pi over two radians,
so the calculator's only going to give us angle alpha,
we'll have to determine angle beta on our own.
Let's first verify we're in radian mode,
so we'll press the mode key,
notice how radian is highlighted.
So I'll go back to the home screen,
press second sine for inverse sine,
3/5, close parenthesis and enter.
So angle theta in the first quadrant
is approximately 0.6435 radians.
Which again would be this angle here.
Now here we have an option, we could use angle beta here
by taking pi radians half a revolution and subtracting
0.6435 radians,
but notice how this region and perimeter
are symmetrical across the y axis.
So instead of using this angle for angle beta,
what we could do is integrate from alpha
to pi over two radians, this angle here.
And notice how that would give us half of the perimeter.
If we multiplied that by two we could then find
the entire perimeter.
So let's go ahead and do that.
Let's let pi over two radians be equal to beta,
and then we'll just double that length rather than
use this angle here.
Now as I mentioned earlier, this is actually two arc lengths
based upon the two equations.
So let's find r and dr d theta for both equations.
So if we have r, let's call it r sub one,
equals five sine theta.
Notice that dr sub one d theta
would be equal to five cosine theta.
And if we let r sub two be equal to three,
notice that dr sub 2 d theta would be equal to zero.
Now we have all the information we need
in order to find the entire perimeter.
Let's call it p for perimeter, we'll have p equals two,
times the integral, from 0.6435 to pi over two,
of the square root
of r sub one squared.
That would be five sine theta squared.
Plus dr d theta squared,
which would be five cosine theta squared.
Plus the square root
of r sub two squared,
plus dr sub 2 d theta squared, which would be zero squared.
Now that we have this set up, let's go an evaluate this
on the next slide.
So here we have the square root of this would be
25 sine squared theta.
Plus here we'd have 25 cosine squared theta.
And here we just have the square root of nine,
which would be three.
Now looking at these two terms here,
if we factored out 25 we'd have 25 times the quantity
sine squared theta,
plus cosine squared theta.
We should recognize this would be one,
so this simplifies to the square root of 25,
which equals five.
So we have two times the integral of five plus three
which is equal to eight.
So if we want we could factor out the eight
and write this as 16 times the integral of one d theta.
So the integral of one with respect to theta
would just be theta.
So we have 16 theta.
So when theta is pi over two, of course we just have
pi over two.
And then when theta is approximately 0.6435,
we just have 0.6435.
Now we'll go to the calculator.
16 times pi divided by two.
Minus .6435.
Which is approximately 14.8367,
which would be the length of the entire perimeter
of our region.
So again this is the length
of the perimeter of the yellow region
outlined here in blue.
I hope you found this helpful.