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

##### 2 Answers

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

#### Explanation:

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 = 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

#"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#

**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)

#### Explanation:

The curve

period,

In

at

The area in between

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

difference

for r from 0 to

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

Altogether in the four quadrants, the area is

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

make it for Sam....