As a follow-up to Gerry's answer and Day Late Don's comment there, let's consider the triangle formed by the centers of the circles and subtract the sectors. Call the circles 1, 2, and 3, with radii $r_1$, $r_2$, and $r_3$, respectively, and let $\theta_1$, $\theta_2$, and $\theta_3$, be the interior angles of the triangle formed by the centers of the circles, at the centers of circles 1, 2, and 3, respectively.
Applying the Law of Cosines (and simplifying some) to find $\theta_1$, $\theta_2$, and $\theta_3$ gives: $\theta_1=\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$ $\theta_2=\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$ $\theta_3=\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$ so the sector areas are $\frac{1}{2}\theta_1r_1^2=\frac{1}{2}r_1^2\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$ $\frac{1}{2}\theta_2r_2^2=\frac{1}{2}r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$ $\frac{1}{2}\theta_3r_3^2=\frac{1}{2}r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$
Applying Hero's formula to find the area of the triangle (and simplifying) gives $K_\triangle=\sqrt{r_1r_2r_3(r_1+r_2+r_3)}.$
So the area of the target region is $\sqrt{r_1r_2r_3(r_1+r_2+r_3)}-\frac{1}{2}\left(\begin{align} r_1^2&\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right) \\&+r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right) \\&+r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right) \end{align}\right).$