The number of integer solutions $(x, y)$ of the inequality $x^2 + y^2 \leq 10$ is

Count integer pairs $(x, y)$ with $x^2 + y^2 \leq 10$ by fixing $y$ and finding the allowed $x$ values.

$y = 0$: $x^2 \leq 10$, so $x \in \{-3, -2, -1, 0, 1, 2, 3\}$ gives $7$ values

$y = \pm 1$: $x^2 \leq 9$, so $x \in \{-3, \ldots, 3\}$ gives $7$ values each, total $14$

$y = \pm 2$: $x^2 \leq 6$, so $x \in \{-2, -1, 0, 1, 2\}$ gives $5$ values each, total $10$

$y = \pm 3$: $x^2 \leq 1$, so $x \in \{-1, 0, 1\}$ gives $3$ values each, total $6$

For $|y| \geq 4$, $y^2 \geq 16 > 10$, so no solutions.

Total $= 7 + 14 + 10 + 6 = 37$

Answer $= 37$