3. Given that $x, y$ are distinct natural numbers, and $x^{2}+19 y$ $=y^{3}+19 x$. Then the integer part of $\sqrt{x^{2}+y^{2}}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3.C.

Assume $x>y$.
$\because x^{3}-y^{3}=19(x-y)$, and $x, y$ are natural numbers, $\therefore x^{2}+x<x^{2}+xy+y^{2}=19<3x^{2}$. From this, we get $x=3$, and then $y=m$. Therefore, $\sqrt{x^{2}+y^{2}}=\sqrt{13}$, and its integer part is 3.