What is the area enclosed between the two polar curves: #r = 4 - 2cos 3theta# and #r = 5# ?

There are a couple of ingredients to this:

(1) Determine the #theta# value at which the two curves intersect.

(2) Note that the area of a polar curve is given by #int_alpha^beta 1/2 r(theta)^2 d theta# since we are basically summing the area of infinitesimal width triangles with vertex at the origin, height #r(theta)# and base length #r(theta) d theta#.

Given two curves:

#r = 4 - 2cos(3 theta)#

#r = 5#

Points of intersection will satisfy:

#4 - 2cos(3 theta) = 5#

Hence:

#cos (3 theta) = -1/2#

The smallest positive value of #theta# for which this holds is:

#theta = 1/3 cos^(-1) (-1/2) = 1/3 ((2pi)/3) = (2pi)/9#

So the shaded area will be the difference of two integrals, or equivalently the integral of the difference in values for #r# between the two curves in the range #0# to #(2pi)/9#

#"area" = int_0^((2pi)/9) 1/2(5^2 - (4 - 2cos(3 theta))^2)color(white)(1) d theta#

#color(white)("area") = int_0^((2pi)/9) 1/2(9+16cos(3 theta) - 4cos^2(3 theta))color(white)(1) d theta#

#color(white)("area") = int_0^((2pi)/9) 9/2+8cos(3 theta) - 2cos^2(3 theta)color(white)(1) d theta#

#color(white)("area") = int_0^((2pi)/9) 9/2+8cos(3 theta) - (cos(6theta) + 1)color(white)(1) d theta#

#color(white)("area") = int_0^((2pi)/9) 7/2+8cos(3 theta) - cos(6theta)color(white)(1) d theta#

#color(white)("area") = [color(white)(1/1)7/2 theta +8/3sin(3 theta) - 1/6sin(6theta)color(white)(1/1)]_0^((2pi)/9)#

#color(white)("area") = (7/2 ((2pi)/9) +8/3sin((2pi)/3) - 1/6sin((4pi)/3)) - (0+0-0)#

#color(white)("area") = (7pi)/9 +8/3(sqrt(3)/2) - 1/6sin(-sqrt(3)/2)#

#color(white)("area") = (7pi)/9 +(4sqrt(3))/3 + sqrt(3)/12#

#color(white)("area") = (7pi)/9 +(17sqrt(3))/12#

#color(white)()#
Rough check

The area is similar to that of a triangle with vertices:

#(2, 0)#, #(5, 0)# and #(5cos((2pi)/9), 5sin((2pi)/9))#

which will be:

#1/2 * "base" * "height" = 1/2*3*5sin((2pi)/9)#

#color(white)(1/2 * "base" * "height") ~~ 15/2 * 0.64 = 4.8#

Whereas:

#(7pi)/9 +(17sqrt(3))/12 ~~ 4.897#

OK (finally)

The curve #C_1: r = 4 - 2 cos 3theta# is periodic, with period

#2/3pi#. So, it repeats thrice for #theta in [0, 2pi]#. Also, within every

period, #r in [2, 6]#.

In #Q_1#, the circle #C_2: r = 5# intersects #C_1#

at #(3, 2/9pi) and (6, pi/3)#.

The area in between #C_1 and C_2# in #Q_1#( thanks to George .

for vivid clear illustration in his answer) can be regarded as the

difference #A_1 - A_2#, where

#A_1 # = area of sector of the circle between the radii

#theta = 0 and theta= 2/9pi#

#= (25pi)((2/9pi )/(2pi))#

#=25/9pi# and

#A_2# = the area between #C_1# below circle in #Q_1# and the radius #theta = 2/9pi#, above#

#= int int 1/2r^2 dr d theta# ,

for r from 0 to #4 - 2 cos 3theta# and #theta# from #0 to 2/9pi#

#=1/6int [r^3] d theta#, for the same limits.

#=1/6 int (4-2 cos 3theta)^3d theta#, for #theta# from #0 to 2/9pi#

#= 1/6int(64-96 cos 3theta+48 cos^2 3theta-8cos^3 3theta)d theta#, for #theta# from #0 to 2/9pi#

#=1/6[88theta-34 sin 3theta+4 sin 6theta -2/9 sin 9theta]#, between

#theta# from #0 to 2/9pi#, using #cos^2 A = (1 + cos 2A )/2 and

cos^3A=1/4(cos 9A+3 cos 3A)#

#=88/27pi-17/6sqrt3#

Altogether in the four quadrants, the area is

#6( A_1-A_2)#

#= 6(25/9pi-(88/27pi-17/6sqrt3))#

#=(17sqrt3-26/9pi)=20.34# areal units.

I think this shaded area is referred to as in between area. Here,

the circle is the exterior curve.

There is a set of three other equal in-between areas, for which

the circle is the interior. curve.

In #Q_1#, this is between the radial lines

#theta = 2/9pi and theta = 4/9pi#. If this is to be added, I can

make it for Sam....