Answer
The average value is $\frac{2}{b-a}$.
Work Step by Step
In polar coordinates, the annular region $R$ can be written as $R=\{(r,\theta)|a\leq r\leq b,0\leq \theta\leq 2\pi\}$.
Find the area of the annular region:
$A=\iint_R dA=\int_a^b\int_0^{2\pi}rd\theta dr=\int_a^b 2\pi r dr=\pi r^2]_a^b=(b-a)^2\pi$
Evaluate $\iint_R f(x,y) dA$:
$\iint_R f(x,y) dA=\int_a^b\int_0^{2\pi} \frac{1}{\sqrt{r^2}}\cdot rd\theta dr$
$=\int_a^b \int_0^{2\pi} d\theta dr$
$=\int_a^b2\pi dr$
$=2\pi r]_a^b$
$=2(b-a)\pi$
Find the average value of $f(x,y)$ on the annular region:
$\bar{f}=\frac{\iint_R f(x,y) dA}{A}=\frac{2(b-a)\pi }{(b-a)^2\pi}=\frac{2}{b-a}$
