Question

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

Answer 1

`"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` 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)

Answer 2

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

Explanation:

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....