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

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#

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 #