lawrence (cdl678) – Homework 6 – ODELL – (56280)
1
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22
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before answering.
001
10.0 points
The shaded region in
lies inside the polar curve
r
= 3 cos
θ
and
outside the polar curve
r
= cos
θ
.
Which of
the following integrals gives the area of this
region?
1.
I
=
1
2
integraldisplay
π/
2
0
2 cos
θ dθ
2.
I
=
1
2
integraldisplay
π/
2
−
π/
2
8 cos
2
θ dθ
correct
3.
I
=
1
2
integraldisplay
π/
2
0
8 cos
2
θ dθ
4.
I
=
integraldisplay
π/
2
−
π/
2
2 cos
θ dθ
5.
I
=
1
2
integraldisplay
π/
2
−
π/
2
2 cos
θ dθ
6.
I
=
integraldisplay
π/
2
−
π/
2
8 cos
2
θ dθ
Explanation:
As the graphs show, the polar curves inter
sect when
3 cos
θ
= cos
θ ,
i.e.
at
θ
=
±
π/
2. Thus the area of the shaded
region is given by
I
=
1
2
integraldisplay
π/
2
−
π/
2
braceleftBig
(3 cos
θ
)
2
−
(cos
θ
)
2
bracerightBig
dθ .
Consequently,
I
=
1
2
integraldisplay
π/
2
−
π/
2
8 cos
2
θ dθ
.
keywords: area, polar cooordinates, definite
integral, circle,
002
10.0 points
Find the area of the region bounded by the
polar curve
r
=
√
4 ln
θ
+ 3
as well as the rays
θ
= 1 and
θ
=
e
.
1.
area =
1
2
(3
e
+ 1)
correct
2.
area =
1
4
(3
e
+ 2)
3.
area = 3
e
+ 2
4.
area = 3
e
+ 1
5.
area =
1
4
(3
e
+ 1)
6.
area =
1
2
(3
e
+ 2)
Explanation:
The area of the region bounded by the
graph of the polar function
r
=
f
(
θ
) as well
as the rays
θ
=
θ
0
, θ
1
is given by the integral
A
=
1
2
integraldisplay
θ
1
θ
0
f
(
θ
)
2
dθ .
When
f
(
θ
) =
√
4 ln
θ
+ 3
,
θ
0
= 1
, θ
1
=
e ,
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lawrence (cdl678) – Homework 6 – ODELL – (56280)
2
therefore, the area of the enclosed region is
thus given by the integral
A
=
1
2
integraldisplay
e
1
(
√
4 ln
θ
+ 3)
2
dθ
=
1
2
integraldisplay
e
1
(4 ln
θ
+ 3)
dθ .
To evaluate this last integral we use Integra
tion by Parts, for then
A
=
1
2
bracketleftBig
4
θ
ln
θ
+ 3
θ
bracketrightBig
e
1
−
1
2
integraldisplay
e
1
4
dθ
=
1
2
bracketleftBig
4
θ
ln
θ
−
1
θ
bracketrightBig
e
1
.
Consequently,
area =
A
=
1
2
(3
e
+ 1)
.
003
10.0 points
Find the area of the region enclosed by the
graph of the polar function
r
= 2
−
sin
θ .
1.
area =
7
2
π
2.
area =
17
2
π
3.
area =
5
2
π
4.
area =
9
2
π
correct
5.
area = 3
π
Explanation:
The area of a region bounded by the graph
of the polar function
r
=
f
(
θ
) and the rays
θ
=
θ
0
, θ
1
is given by the integral
A
=
1
2
integraldisplay
θ
1
θ
0
f
(
θ
)
2
dθ .
On the other hand, the graph of
r
= 2
−
sin
θ
is the cardioid similar to the one shown in
so in this case we can take
θ
0
= 0 and
θ
1
= 2
π
.
Thus the area of the region enclosed by the
graph is given by the integral
A
=
1
2
integraldisplay
2
π
0
(2
−
sin
θ
)
2
dθ .
Now
(2
−
sin
θ
)
2
= 4
−
2 sin
θ
+ sin
2
θ
=
9
2
−
2 sin
θ
−
1
2
cos 2
θ ,
since
sin
2
θ
=
1
2
parenleftBig
1
−
cos 2
θ
parenrightBig
.
But then,
A
=
1
2
integraldisplay
2
π
0
parenleftbigg
9
2
−
2 sin
θ
−
1
2
cos 2
θ
parenrightbigg
dθ
=
1
2
bracketleftBig
9
2
θ
+ 2 cos
θ
−
1
4
sin 2
θ
bracketrightBig
2
π
0
.
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 Spring '10
 odell
 Vector Space, Distance, Vector Motors, Polar coordinate system

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