9.3.22:
The lemniscate with polar equation
r
2
= 4 cos2
θ
is shown in Fig. 9.3.15. We will find the area of
the right-hand loop. First,
r
= 0 when cos 2
θ
= 0, so that
θ
is an odd integral multiple of
π/
4. Hence the
area of the right-hand loop is
A
=
1
2
π/
4
−
π/
4
4 cos2
θ dθ
=
π/
4
0
4 cos2
θ dθ
=
2 sin 2
θ
π/
4
0
= 2
.
9.3.23:
Given: The polar equation
r
2
= 4 sin
θ
.
One way to construct its graph is first to construct the
Cartesian
graph
y
= 4 sin
x
, which is shown next for 0
x
2
π
.
3
6
x
-4
-2
2
4
y
Then construct the Cartesian graph of
y
=
±
√
4 sin
x
, shown next, also for 0
x
2
π
.
Note that there
is no graph for
π < x <
2
π
but
two
graphs for 0
x
π
.
966

3
6
x
-4
-2
2
4
y
Finally, use the last graph to construct the polar graph
r
2
= 4 sin
θ
by sketching both
r
=
√
4 sin
θ
and
r
=
−
√
4 sin
θ
.
As
θ
varies from 0 to
π
,
r
begins at 0, increases to a maximum
r
= 2 when
θ
=
π/
2, then
decreases to 0 as
θ
runs through the values from
π/
2 to
π
.
Meanwhile,
−
r
sweeps out the mirror image of
the previous curve, and thus we obtain the “double oval” shown in the next figure.
-1
1
x
-2
-1
1
2
y
The area of the upper loop is therefore
A
=
1
2
π
0
4 sin
θ dθ
=
−
2 cos
θ
π
0
= 2
−
(
−
2) = 4
.
9.3.24:
The area of one loop of this rose is
A
=
1
2
π/
12
−
π/
12
36 cos
2
6
θ dθ
= 36
π/
12
0
1
2
(1 + cos 12
θ
)
dθ
= 18
θ
+
1
12
sin 12
θ
π/
12
0
=
3
π
2
.
9.3.25:
See the graph below.
967

-1
1
x
-1
1
2
y
Find the area
A
of the region both inside the circle
r
= 2 sin
θ
and outside the circle
r
= 1. The circles
cross where 2 sin
θ
= 1, thus where
θ
=
π/
6 and where
θ
= 5
π/
6. So
A
=
1
2
5
π/
6
π/
6
(4 sin
2
θ
−
1)
dθ
=
1
2
5
π/
6
π/
6
2(1
−
cos 2
θ
)
−
1
dθ
=
1
2
θ
−
sin 2
θ
5
π/
6
π/
6
=
1
2
5
π
6
−
π
6
+
√
3
2
+
√
3
2
=
2
π
+ 3
√
3
6
≈
1
.
913222954981
.
9.3.26:
See the graph below.
-2
-1
1
2
3
4
x
-2
-1
1
2
y
The circles
r
= 4 cos
θ
and
r
= 2 cross where 4 cos
θ
= 2, thus where
θ
=
−
π/
3 and where
θ
=
π/
3. To
find the area
A
of their intersection, we double the area of its top half; thus
A
=
π/
3
0
4
dθ
+
π/
2
π/
3
16 cos
2
θ dθ
=
4
π
3
+ 4
π/
2
π/
3
2(1 + cos 2
θ
)
dθ
=
4
π
3
+ 4 2
θ
+ sin 2
θ
π/
2
π/
3
=
4
π
3
+ 4
π
−
2
π
3
−
√
3
2
=
4
π
3
+ 4
π
−
8
π
3
−
2
√
3 =
8
π
−
6
√
3
3
≈
4
.
91347879
.
9.3.27:
See the figure below.
968

-0.5
0.5
x
-0.5
1
y
The two circles
r
= cos
θ
and
r
=
√
3 sin
θ
cross where
cos
θ
=
√
3 sin
θ
;
tan
θ
=
√
3
3
;
θ
=
π
6
;
they also cross at the pole. The region inside both is divided by the ray
θ
=
π/
6 into a lower region of area
B
and an upper region of area
C
. Thus we find the area of the region inside both circles to be
A
=
B
+
C
.
Now
B
=
1
2
π/
6
0
3 sin
2
θ dθ
=
3
4
π/
6
0
(1
−
cos 2
θ
)
dθ
=
3
4
θ
−
sin
θ
cos
θ
π/
6
0
=
3
4
π
6
−
√
3
4
and
C
=
1
2
π/
2
π/
6
cos
2
θ dθ
=
1
4
π/
2
π/
6
(1 + cos 2
θ
)
dθ
=
1
4
θ
+ sin
θ
cos
θ
π/
2
π/
6
=
1
4
π
3
−
√
3
4
.
Therefore
A
=
B
+
C
=
5
π
−
6
√
3
24
≈
0
.
221485767606.

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- Fall '09
- Cartesian Coordinate System, Slope, Cos, Polar coordinate system, dθ