Answer
The average temperature along the wire is $8$ $^\circ C$.
Work Step by Step
The circular wire of radius $2$ cm centered at the origin can be parametrized by
${\bf{r}}\left( t \right) = \left( {2\cos t,2\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le 2\pi $
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( { - 2\sin t,2\cos t} \right)\cdot\left( { - 2\sin t,2\cos t} \right)} = 2$
The length of $C$ is
$L = \mathop \smallint \limits_C^{} {\rm{d}}s = \mathop \smallint \limits_0^{2\pi } ||{\bf{r}}'\left( t \right)||{\rm{d}}t = 2\mathop \smallint \limits_0^{2\pi } {\rm{d}}t = 4\pi $
Let $P = \left( {x,y} \right)$ denote a point on the wire. The temperature at $P$ is given by
$T\left( {x,y} \right) = {\left( {x - 2} \right)^2} + {y^2}$
$T\left( t \right) = 4{\left( {\cos t - 1} \right)^2} + 4{\sin ^2}t = 8 - 8\cos t$
Using Eq. (11) we calculate the average value of $T\left( {x,y} \right) = {\left( {x - 2} \right)^2} + {y^2}$ along the segment $C$ as
$Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_C^{} T\left( {x,y,z} \right){\rm{d}}s$
By Eq. (4), the integral becomes
$Av\left( f \right) = \frac{1}{{4\pi }}\mathop \smallint \limits_0^{2\pi } T\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$Av\left( f \right) = \frac{4}{\pi }\mathop \smallint \limits_0^{2\pi } \left( {1 - \cos t} \right){\rm{d}}t$
$Av\left( f \right) = \frac{4}{\pi }\left( {t - \sin t} \right)|_0^{2\pi } = \frac{4}{\pi }\left( {2\pi } \right) = 8$
So, the average temperature along the wire is $8$ $^\circ C$.
