3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$
3.11.

Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\end{aligned}
\]

Thus, when \( a^{2}+10=11 \), i.e., \( a= \pm 1 \), \( y^{2} \) is minimized. At this point, \( v_{\text {min }}=11 \).