There are two reasonable parameterizations as I see it. The first is as all have recommended, simply translate the origin of polar coordinates to the center of the circle and then =12+12cos
x
=
1
2
+
1
2
cos
t
and =12sin
y
=
1
2
sin
t
, so
=()2+()2‾‾‾‾‾‾‾‾‾‾‾‾‾√=(−12sin)2+(12cos)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=12
d
s
=
(
d
x
)
2
+
(
d
y
)
2
=
(
−
1
2
sin
t
)
2
+
(
1
2
cos
t
)
2
d
t
=
1
2
d
t
So the integral is
=∫20[12+12cos+12sin]⋅12=14(2)(1+0+0)=2
I
=
∫
0
2
π
[
1
2
+
1
2
cos
t
+
1
2
sin
t
]
⋅
1
2
d
t
=
1
4
(
2
π
)
(
1
+
0
+
0
)
=
π
2
Where we have used the average value of 1
1
of 1
1
and the average values of cos
cos
t
and sin
sin
t
of 0
0
over the interval of length 2
2
π
.
The other possibility is using the 'polar coordinate' version of the circle =cos
r
=
cos
t
, =cos=cos2
x
=
r
cos
t
=
cos
2
t
, and =sin=sincos
y
=
r
sin
t
=
sin
t
cos
t
, so
=()2+()2‾‾‾‾‾‾‾‾‾‾‾‾‾√=(−2sincos)2+(cos2−sin2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=(cos2+sin2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=
d
s
=
(
d
x
)
2
+
(
d
y
)
2
=
(
−
2
sin
t
cos
t
)
2
+
(
cos
2
t
−
sin
2
t
)
2
=
(
cos
2
t
+
sin
2
t
)
2
=
d
t
Then
=∫2−2(cos2+sincos)=(12+0)=2
I
=
∫
−
π
2
π
2
(
cos
2
t
+
sin
t
cos
t
)
d
t
=
π
(
1
2
+
0
)
=
π
2
Where we have used the average value of cos2
cos
2
t
of 12
1
2
and of sincos=12sin2
sin
t
cos
t
=
1
2
sin
2
t
of 0
0
over the interval of length
π
.
You knew the answer up front because it should be the average value of
x
plus the average value of
y
over the curve times the arc length of the curve, so
=(12+0)(2)(12)=2
I
=
(
1
2
+
0
)
(
2
π
)
(
1
2
)
=
π
2
Where we have used the mensuration formula =2
s
=
2
π
r
for the perimeter of a circle of radius
r
.
