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

Oct 20, 2016

$\text{area} = \frac{7 \pi}{9} + \frac{17 \sqrt{3}}{12}$

#### 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 ${\int}_{\alpha}^{\beta} \frac{1}{2} r {\left(\theta\right)}^{2} d \theta$ since we are basically summing the area of infinitesimal width triangles with vertex at the origin, height $r \left(\theta\right)$ and base length $r \left(\theta\right) d \theta$.

Given two curves:

$r = 4 - 2 \cos \left(3 \theta\right)$

$r = 5$

Points of intersection will satisfy:

$4 - 2 \cos \left(3 \theta\right) = 5$

Hence:

$\cos \left(3 \theta\right) = - \frac{1}{2}$

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

$\theta = \frac{1}{3} {\cos}^{- 1} \left(- \frac{1}{2}\right) = \frac{1}{3} \left(\frac{2 \pi}{3}\right) = \frac{2 \pi}{9}$

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 $\frac{2 \pi}{9}$

$\text{area} = {\int}_{0}^{\frac{2 \pi}{9}} \frac{1}{2} \left({5}^{2} - {\left(4 - 2 \cos \left(3 \theta\right)\right)}^{2}\right) \textcolor{w h i t e}{1} d \theta$

$\textcolor{w h i t e}{\text{area}} = {\int}_{0}^{\frac{2 \pi}{9}} \frac{1}{2} \left(9 + 16 \cos \left(3 \theta\right) - 4 {\cos}^{2} \left(3 \theta\right)\right) \textcolor{w h i t e}{1} d \theta$

$\textcolor{w h i t e}{\text{area}} = {\int}_{0}^{\frac{2 \pi}{9}} \frac{9}{2} + 8 \cos \left(3 \theta\right) - 2 {\cos}^{2} \left(3 \theta\right) \textcolor{w h i t e}{1} d \theta$

$\textcolor{w h i t e}{\text{area}} = {\int}_{0}^{\frac{2 \pi}{9}} \frac{9}{2} + 8 \cos \left(3 \theta\right) - \left(\cos \left(6 \theta\right) + 1\right) \textcolor{w h i t e}{1} d \theta$

$\textcolor{w h i t e}{\text{area}} = {\int}_{0}^{\frac{2 \pi}{9}} \frac{7}{2} + 8 \cos \left(3 \theta\right) - \cos \left(6 \theta\right) \textcolor{w h i t e}{1} d \theta$

$\textcolor{w h i t e}{\text{area}} = {\left[\textcolor{w h i t e}{\frac{1}{1}} \frac{7}{2} \theta + \frac{8}{3} \sin \left(3 \theta\right) - \frac{1}{6} \sin \left(6 \theta\right) \textcolor{w h i t e}{\frac{1}{1}}\right]}_{0}^{\frac{2 \pi}{9}}$

$\textcolor{w h i t e}{\text{area}} = \left(\frac{7}{2} \left(\frac{2 \pi}{9}\right) + \frac{8}{3} \sin \left(\frac{2 \pi}{3}\right) - \frac{1}{6} \sin \left(\frac{4 \pi}{3}\right)\right) - \left(0 + 0 - 0\right)$

$\textcolor{w h i t e}{\text{area}} = \frac{7 \pi}{9} + \frac{8}{3} \left(\frac{\sqrt{3}}{2}\right) - \frac{1}{6} \sin \left(- \frac{\sqrt{3}}{2}\right)$

$\textcolor{w h i t e}{\text{area}} = \frac{7 \pi}{9} + \frac{4 \sqrt{3}}{3} + \frac{\sqrt{3}}{12}$

$\textcolor{w h i t e}{\text{area}} = \frac{7 \pi}{9} + \frac{17 \sqrt{3}}{12}$

$\textcolor{w h i t e}{}$
Rough check

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

$\left(2 , 0\right)$, $\left(5 , 0\right)$ and $\left(5 \cos \left(\frac{2 \pi}{9}\right) , 5 \sin \left(\frac{2 \pi}{9}\right)\right)$

which will be:

$\frac{1}{2} \cdot \text{base" * "height} = \frac{1}{2} \cdot 3 \cdot 5 \sin \left(\frac{2 \pi}{9}\right)$

$\textcolor{w h i t e}{\frac{1}{2} \cdot \text{base" * "height}} \approx \frac{15}{2} \cdot 0.64 = 4.8$

Whereas:

$\frac{7 \pi}{9} + \frac{17 \sqrt{3}}{12} \approx 4.897$

OK (finally)

Oct 20, 2016

$= \left(17 \sqrt{3} - \frac{26}{9} \pi\right) = 20.34$ areal units...

#### Explanation:

The curve ${C}_{1} : r = 4 - 2 \cos 3 \theta$ is periodic, with period

$\frac{2}{3} \pi$. So, it repeats thrice for $\theta \in \left[0 , 2 \pi\right]$. Also, within every

period, $r \in \left[2 , 6\right]$.

In ${Q}_{1}$, the circle ${C}_{2} : r = 5$ intersects ${C}_{1}$

at $\left(3 , \frac{2}{9} \pi\right) \mathmr{and} \left(6 , \frac{\pi}{3}\right)$.

The area in between ${C}_{1} \mathmr{and} {C}_{2}$ 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

$\theta = 0 \mathmr{and} \theta = \frac{2}{9} \pi$

$= \left(25 \pi\right) \left(\frac{\frac{2}{9} \pi}{2 \pi}\right)$

$= \frac{25}{9} \pi$ and

${A}_{2}$ = the area between ${C}_{1}$ below circle in ${Q}_{1}$ and the radius $\theta = \frac{2}{9} \pi$, above

$= \int \int \frac{1}{2} {r}^{2} \mathrm{dr} d \theta$ ,

for r from 0 to $4 - 2 \cos 3 \theta$ and $\theta$ from $0 \to \frac{2}{9} \pi$

$= \frac{1}{6} \int \left[{r}^{3}\right] d \theta$, for the same limits.

$= \frac{1}{6} \int {\left(4 - 2 \cos 3 \theta\right)}^{3} d \theta$, for $\theta$ from $0 \to \frac{2}{9} \pi$

$= \frac{1}{6} \int \left(64 - 96 \cos 3 \theta + 48 {\cos}^{2} 3 \theta - 8 {\cos}^{3} 3 \theta\right) d \theta$, for $\theta$ from $0 \to \frac{2}{9} \pi$

$= \frac{1}{6} \left[88 \theta - 34 \sin 3 \theta + 4 \sin 6 \theta - \frac{2}{9} \sin 9 \theta\right]$, between

$\theta$ from $0 \to \frac{2}{9} \pi$, using cos^2 A = (1 + cos 2A )/2 and

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

$= \frac{88}{27} \pi - \frac{17}{6} \sqrt{3}$

Altogether in the four quadrants, the area is

$6 \left({A}_{1} - {A}_{2}\right)$

$= 6 \left(\frac{25}{9} \pi - \left(\frac{88}{27} \pi - \frac{17}{6} \sqrt{3}\right)\right)$

$= \left(17 \sqrt{3} - \frac{26}{9} \pi\right) = 20.34$ 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

$\theta = \frac{2}{9} \pi \mathmr{and} \theta = \frac{4}{9} \pi$. If this is to be added, I can

make it for Sam....