who knows things about polar curves? :c

alrighty let's look at this

((idk how to put the theta thing in so x = theta from now on))

limacon r = 4 + 2cos x

we know that the area of a polar curve is one half the integral from a to b of the equation squared

so something like

                    _b

A = (1/2)    S   (equation)²dx

                    ^a

limacons ALWAYS go from 0 to 2pi

ALWAYS

NO EXCEPTIONS

0 to 2pi

in this case, the equation would be

                       _2pi

A = (1/2)    S   (4 + 2cosx)²dx

                    ⁰

I would first multiply everything out in the integral (use the FOIL method)

so the inner part becomes 16 + 8cos(x) + 8cos(x) + 4cos²x

which simplifies to 16 + 16cos(x) + 4cos²x

factor out a 4 (it'll make things easier later) and it becomes 4 (4 + 4cos(x) + cos²x)

so the overall equation becomes

                   _2pi
A = (1/2)   S   4(4 + 4cos(x) + cos²x)dx
                   ⁰

you can pull the 4 out in front of the integral and multiply it with the (1/2)

now it looks like:

           _2pi
A = 2   S   (4 + 4cos(x) + cos²x)dx
            ⁰

and now you can integrate (we're gonna focus on JUST the integral)

since it's all addition, we can integrate each part individually

so basically: (I'm not including the bounds for times sake but it's still 0 to 2pi for all of them)

S 4dx      +       S 4cos(x)dx      +      S cos²xdx

S 4dx becomes 4x

S 4cos(x) becomes 4sin(x)

and S cos²x becomes (1/4)sin(2x) + (1/2)x

here's why:

using the identity cos(2x) = 2cos2x - 1 , you can find cos²x

solve for cos²x and you get that it equals (1/2) (cos(2x) +1)

this is much easier to integrate

it now looks like this:

S (1/2) (cos(2x) + 1)dx

once again, bring the (1/2) in front of the integral and it's now

(1/2)  S (cos(2x) + 1)dx

it's addition again, so split up the pieces...

(1/2) [ S cos(2x)dx     +     S 1dx ]

integrate...

(1/2) [ (1/2)sin(2x) + x ]

distribute the (1/2) in front of the brackets and you get:

(1/4)sin(2x) + (1/2)x

BUT WAIT

remember this is only a piece of the original integral.

take this part and plug it back in to the original integral
don't forget to plug in the other parts as well

BEFORE:

              _2pi
A = 2     S     (4 + 4cos(x) + cos2x)dx
              ⁰  

AFTER:

A = 2 [ 4x + 4sin(x) + (1/4)sin(2x) + (1/2)x ] |^2pi

                                                                               |₀

solve for A(2pi) and subtract A(0) from it

so

A = 2 {[ 4(2pi) + 4sin(2pi) + (1/4)sin(4pi) + (1/2)(2pi) ]  -  [ 4(0) + 4sin(0) + (1/4)sin(0) + (1/2)(0) ]}

A = 2 {[ 8pi + 0 + 0 + pi ] - [ 0 + 0 + 0 + 0]}

A = 2 (8pi + pi)

A = 2 (9pi)

A = 18pi

I reallllllly hope this helps

sorry for all the baby steps (I wanted to make sure I didn't make a mistake while typing XD)

sorry it took so long D: