find the area of the loop of the curve
Answers
Answer:
a
2
(
4
−
π
)
2
Explanation:
Converting to polar using
x
=
r
cos
θ
and
y
=
r
sin
θ
:
r
cos
θ
(
r
2
cos
2
θ
+
r
2
sin
2
θ
)
=
a
(
r
2
cos
2
θ
−
r
2
sin
2
θ
)
r
3
cos
θ
(
cos
2
θ
+
sin
2
θ
)
=
a
r
2
(
cos
2
θ
−
sin
2
θ
)
Using
cos
2
θ
+
sin
2
θ
=
1
:
r
3
cos
θ
=
a
r
2
(
cos
2
θ
−
(
1
−
cos
2
θ
)
)
r
3
cos
θ
−
a
r
2
(
2
cos
2
θ
−
1
)
=
0
r
2
(
r
cos
θ
−
a
(
2
cos
2
θ
−
1
)
)
=
0
Note that
r
2
=
0
is just the point
(
0
,
0
)
, so we're left with:
r
cos
θ
=
a
(
2
cos
2
θ
−
1
)
r
=
a
(
2
cos
θ
−
sec
θ
)
The loop will begin and end when
r
=
0
, which is when
2
cos
θ
=
sec
θ
2
cos
2
θ
=
1
cos
θ
=
±
1
√
2
θ
=
±
π
4
The area of a polar curve
r
from
θ
=
α
to
θ
=
β
is given by
1
2
∫
β
α
r
2
d
θ
, so here the integral for the area is:
1
2
∫
π
4
−
π
4
[
a
(
2
cos
θ
−
sec
θ
)
]
2
d
θ
Since the loop is symmetric across the
x
axis:
=
∫
π
4
0
a
2
(
2
cos
θ
−
sec
θ
)
2
d
θ
=
a
2
∫
π
4
0
(
4
cos
2
θ
−
4
cos
θ
sec
θ
+
sec
2
θ
)
d
θ
Using
cos
2
θ
=
1
2
(
1
+
cos
2
θ
)
:
=
a
2
∫
π
4
0
(
2
(
1
+
cos
2
θ
)
−
4
+
sec
2
θ
)
d
θ
=
a
2
∫
π
4
0
(
2
cos
2
θ
+
sec
2
θ
−
2
)
d
θ
Integrating term by term:
=
a
2
(
sin
2
θ
+
tan
θ
−
2
θ
)
∣
∣
π
4
0
=
a
2
(
sin
(
π
2
)
+
tan
(
π
4
)
−
π
2
)
−
a
2
(
sin
0
+
tan
0
−
0
)
=
a
2
(
1
+
1
−
π
2
)
=
a
2
(
4
−
π
)
2
Answer linka
2
(
4
−
π
)
2
Explanation:
Converting to polar using
x
=
r
cos
θ
and
y
=
r
sin
θ
:
r
cos
θ
(
r
2
cos
2
θ
+
r
2
sin
2
θ
)
=
a
(
r
2
cos
2
θ
−
r
2
sin
2
θ
)
r
3
cos
θ
(
cos
2
θ
+
sin
2
θ
)
=
a
r
2
(
cos
2
θ
−
sin
2
θ
)
Using
cos
2
θ
+
sin
2
θ
=
1
:
r
3
cos
θ
=
a
r
2
(
cos
2
θ
−
(
1
−
cos
2
θ
)
)
r
3
cos
θ
−
a
r
2
(
2
cos
2
θ
−
1
)
=
0
r
2
(
r
cos
θ
−
a
(
2
cos
2
θ
−
1
)
)
=
0
Note that
r
2
=
0
is just the point
(
0
,
0
)
, so we're left with:
r
cos
θ
=
a
(
2
cos
2
θ
−
1
)
r
=
a
(
2
cos
θ
−
sec
θ
)
The loop will begin and end when
r
=
0
, which is when
2
cos
θ
=
sec
θ
2
cos
2
θ
=
1
cos
θ
=
±
1
√
2
θ
=
±
π
4
The area of a polar curve
r
from
θ
=
α
to
θ
=
β
is given by
1
2
∫
β
α
r
2
d
θ
, so here the integral for the area is:
1
2
∫
π
4
−
π
4
[
a
(
2
cos
θ
−
sec
θ
)
]
2
d
θ
Since the loop is symmetric across the
x
axis:
=
∫
π
4
0
a
2
(
2
cos
θ
−
sec
θ
)
2
d
θ
=
a
2
∫
π
4
0
(
4
cos
2
θ
−
4
cos
θ
sec
θ
+
sec
2
θ
)
d
θ
Using
cos
2
θ
=
1
2
(
1
+
cos
2
θ
)
:
=
a
2
∫
π
4
0
(
2
(
1
+
cos
2
θ
)
−
4
+
sec
2
θ
)
d
θ
=
a
2
∫
π
4
0
(
2
cos
2
θ
+
sec
2
θ
−
2
)
d
θ
Integrating term by term:
=
a
2
(
sin
2
θ
+
tan
θ
−
2
θ
)
∣
∣
π
4
0
=
a
2
(
sin
(
π
2
)
+
tan
(
π
4
)
−
π
2
)
−
a
2
(
sin
0
+
tan
0
−
0
)
=
a
2
(
1
+
1
−
π
2
)
=
a
2
(
4
−
π
)
2
Answer link