Find the area between a large loop and the enclosed small loop of the curve r = 6 + 12 cos(3)
Answers
We can see that it consists of three loops within three larger loops. We will calculate the area inside the two right-hand loops, and subtract the answers. We need the points of intersection with the pole:
6
+
12
cos
3
θ
=
0
cos
3
θ
=
−
1
2
3
θ
=
2
π
3
,
4
π
3
,
8
π
3
,
10
π
3
,
14
π
3
,
16
π
3
θ
=
2
π
9
,
4
π
9
,
8
π
9
,
10
π
9
,
14
π
9
,
16
π
9
6+12cos3θ=0cos3θ=−123θ=2π3,4π3,8π3,10π3,14π3,16π3θ=2π9,4π9,8π9,10π9,14π9,16π9
The values
4
π
9
≤
θ
≤
8
π
9
4π9≤θ≤8π9 cover a big loop, and the values
2
π
9
≤
θ
≤
4
π
9
2π9≤θ≤4π9 cover a small loop. The area of a big loop is
A
=
1
2
∫
8
π
/
9
4
π
/
9
(
6
+
12
cos
3
θ
)
2
d
θ
=
1
2
∫
8
π
/
9
4
π
/
9
36
+
144
cos
3
θ
+
144
cos
2
3
θ
d
θ
=
18
∫
8
π
/
9
4
π
/
9
1
+
9
cos
3
θ
+
9
2
(
1
+
cos
6
θ
)
d
θ
=
18
(
θ
+
3
sin
3
θ
+
9
θ
2
+
sin
6
θ
6
∣
∣
∣
8
π
/
9
4
π
/
9
=
44
π
A=12∫4π/98π/9(6+12cos3θ)2dθ=12∫4π/98π/936+144cos3θ+144cos23θdθ=18∫4π/98π/91+9cos3θ+92(1+cos6θ)dθ=18(θ+3sin3θ+9θ2+sin6θ6|4π/98π/9=44π
The area of a small loop is
A
=
1
2
∫
4
π
/
9
2
π
/
9
(
6
+
12
cos
3
θ
)
2
d
θ
=
1
2
∫
4
π
/
9
2
π
/
9
36
+
144
cos
3
θ
+
144
cos
2
3
θ
d
θ
=
18
∫
4
π
/
9
2
π
/
9
1
+
9
cos
3
θ
+
9
2
(
1
+
cos
6
θ
)
d
θ
=
18
(
θ
+
3
sin
3
θ
+
9
θ
2
+
sin
6
θ
6
∣
∣
∣
4
π
/
9
2
π
/
9
=
22
π
A=12∫2π/94π/9(6+12cos3θ)2dθ=12∫2π/94π/936+144cos3θ+144cos23θdθ=18∫2π/94π/91+9cos3θ+92(1+cos6θ)dθ=18(θ+3sin3θ+9θ2+sin6θ6|2π/94π/9=22π
The difference between these areas is
22
π
.
22π.
