Answer
The average value of the given function is given by:
$$
\begin{aligned}
f_{\mathrm{ave}} &=\frac{1}{A(D)} \iint_{D} \frac{1}{\sqrt{x^{2}+y^{2}}} d A \\
&=\frac{2}{a+b}.
\end{aligned}
$$
Work Step by Step
Find the average value of the function $f(x,y)=\frac{1}{\sqrt {x^{2}+y^{2}}}$ on the annular region $a^{2}\leq x^{2}+y^{2} \leq b^{2},\quad 0 \lt a \lt b . $
The average value of a function $f(x,y)$ of two variables defined on region $D$ is
$$
f_{\mathrm{ave}}=\frac{1}{A(D)} \iint_{R} f(x, y) d A
$$
Here
$$ D=\left\{(r, \theta) | a \leq r \leq b, \quad 0 \leq \theta \leq 2\pi \right\}
$$
and hence,
$$
A(D)=\pi b^{2}- \pi a^{2}= \pi (b^{2}- a^{2}).
$$
So, the average value of the given function is given by:
$$
\begin{aligned}
f_{\mathrm{ave}} &=\frac{1}{A(D)} \iint_{D} \frac{1}{\sqrt{x^{2}+y^{2}}} d A \\
&=\frac{1}{\pi\left(b^{2}-a^{2}\right)} \int_{0}^{2 \pi} \int_{a}^{b} \frac{1}{\sqrt{r^{2}}} r d r d \theta \\
&=\frac{1}{\pi\left(b^{2}-a^{2}\right)} \int_{0}^{2 \pi} d \theta \int_{a}^{b} d r \\
&=\frac{1}{\pi\left(b^{2}-a^{2}\right)}[\theta]_{0}^{2 \pi}[r]_{a}^{b} \\
& =\frac{1}{\pi\left(b^{2}-a^{2}\right)}(2 \pi)(b-a) \\
&=\frac{2(b-a)}{(b+a)(b-a)} \\
&=\frac{2}{a+b}.
\end{aligned}
$$
