Question

How do you find the area of one petal of `r=cos5theta`?

Answer 1

Ans = `pi/20`

Explanation:

graph{y=cos(5x) [-2.075, 2.076, -1.008, 1.069]}

We first consider the graph y = cos(5x), and considering the roots, this is where each individual petal start and end in r = cos(5`theta`)

Roots; cos(5`theta`) = 0
5`theta` = { `-pi/2`, `pi/2` }
`theta` = { `-pi/10, pi/10` }

Hence to find the area of one petal, we can consider the petal that lies between `theta` = `-pi/10` and `theta` = `pi/10` .

Now we can use
`A =1/2intr^2d theta`

r = cos(5`theta`)
letting z = cis(`theta`), we can use De Moivres theorem
z + `1/z` = 2cos(`theta`)
`(z^5 + 1/z^5)^2` = `4cos^2(5theta)`
`z^10 + 1/z^10 + 2` = `4cos^2(5theta)`
Hence `cos^2 (5theta)` = `1/4 (cos(10theta) + 2 )`

(or via considering `Cos(2theta) = 2cos^2(theta) - 1`)

Hence our integral becomes;

`1/8int_(-pi/10)^(pi/10)cos(10theta)+2d theta`

Hence via evaluating this simple integral;

Area = `(1/8)((2pi)/5)` = `pi/20`

Answer 2

`pi/20`

Explanation:

The polar curve is:

We calculate area in polar coordinates using :

`A = 1/2 \ int_alpha^beta \ r^2 \ d theta`

In order to calculate the area bounded by a single petal we would need to calculate the correct bounding angles, or we can calculate the entire area as we sweep through `pi` radians and divide by `5`, which is the method used.

Thus, the enclosed area is:

`A = 1/2 \ int_(0)^(pi) \ (cos5theta)^2 \ d theta`

Note that the entire area is swept out over the partial region `theta in [0,pi]` as we have a odd number of petals, whereas as we would use `theta in [0,2pi]` if we had an even number of petals. This confusing result is explained in this Scratchpad explanation

Now, Using the identity:

`cos 2A -= 2cos^2A -1`

This becomes:

`A = 1/2 \ int_(0)^(pi) \ (1+cos10theta)/2 \ d theta`
`\ \ \ = 1/4 \ int_(0)^(pi) \ (1+cos10theta) \ d theta`
`\ \ \ = 1/4 \ [theta + (sin10theta)/10]_0^(pi)`
`\ \ \ = 1/4 \ { (pi+(sin10pi)/10) - (0+sin0) }`
`\ \ \ = 1/4 \ (pi+0 - 0)`
`\ \ \ = pi/4`

This is the area of all five petals, so the area of a single petal is

`A_1 = A/5 = pi/20`