Question

How do you find the area inner loop of r=4−6sinθ?

Answer 1

13
π
−
26
sin
−
1
(
2
3
)
−
12
√
5

Explanation:
First we have to find the values of
θ
that constitute one loop. A loop will begin and end at the pole (origin), when
r
=
0
.
4
−
6
sin
θ
=
0
sin
θ
=
2
3
Then
θ
=
sin
−
1
(
2
3
)
This is the angle in Quadrant
I
. The corresponding angle in Quadrant
II
where
sin
θ
=
2
/
3
is given by
π
−
sin
−
1
(
2
/
3
)
.
So, we want the area from
α
=
sin
−
1
(
2
/
3
)
to
β
=
π
−
sin
−
1
(
2
/
3
)
.
The area of a polar curve
r
from
θ
1
to
θ
2
is given by
1
2
∫
θ
2
θ
1
r
2
d
θ
. The area of the curve is then:
A
=
1
2
∫
β
α
(
4
−
6
sin
θ
)
2
d
θ
Also note that
(
4
−
6
sin
θ
)
2
=
(
2
(
2
−
3
sin
θ
)
)
2
=
4
(
2
−
3
sin
θ
)
2
.
=
2
∫
β
α
(
2
−
3
sin
θ
)
2
d
θ
=
2
∫
β
α
(
4
−
12
sin
θ
+
9
sin
2
θ
)
d
θ
Use the simplification
sin
2
θ
=
1
2
(
1
−
cos
2
θ
)
.
=
2
∫
β
α
(
4
−
12
sin
θ
+
9
2
−
9
2
cos
2
θ
)
d
θ
=
∫
β
α
(
13
−
24
sin
θ
−
9
cos
2
θ
)
d
θ
Integrate term-by-term. Depending on how comfortable you are with integration, you may want to use the substitution
u
=
2
θ
for the last term.
=
[
13
θ
+
24
cos
θ
−
9
2
sin
2
θ
]
β
α
Using
sin
2
θ
=
2
sin
θ
cos
θ
:
=
[
13
θ
+
24
cos
θ
−
9
sin
θ
cos
θ
]
β
α
Recall that
sin
α
=
sin
β
=
2
/
3
. Thus,
cos
α
=
√
1
−
sin
2
α
=
√
5
/
3
. However, since
β
is in Quadrant
II
,
cos
β
=
−
√
5
/
3
.
=
13
β
+
24
cos
β
−
9
sin
β
cos
β
−
13
α
−
24
cos
α
+
9
sin
α
cos
α
=
13
(
β
−
α
)
+
24
(
−
√
5
3
)
−
9
(
2
3
)
(
−
√
5
3
)
−
24
(
5
√
3
)
+
9
(
2
3
)
(
√
5
3
)
=
13
(
π
−
sin
−
1
(
2
3
)
−
sin
−
1
(
2
3
)
)
−
16
√
5
+
4
√
5
=
13
π
−
26
sin
−
1
(
2
3
)
−
12
√
5
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