If 3 < y < 7 and 4 < x < 12, what is the largest integer value of x+y2?

Given the inequality for $x$:

$$2 < x < 4$$

Since $x$ is positive, squaring all parts of the inequality will maintain the direction of the inequalities:

$$2^2 < x^2 < 4^2$$ $$4 < x^2 < 16$$

Given the inequality for $y$:

$$3 < y < 7$$

Since $y$ is positive, squaring all parts of the inequality will maintain the direction of the inequalities:

$$3^2 < y^2 < 7^2$$ $$9 < y^2 < 49$$

To find the range for the sum $x^2 + y^2$, we add the corresponding lower bounds and upper bounds of the individual ranges:
Lower bound: $4 + 9 = 13$
Upper bound: $16 + 49 = 65$
So, the range for $x^2 + y^2$ is:

$$13 < x^2 + y^2 < 65$$

The inequality $13 < x^2 + y^2 < 65$ means that $x^2 + y^2$ must be strictly less than 65. The largest integer that is strictly less than 65 is 64.

The largest integer value for $x^2 + y^2$ is $\boxed{\text{64}}$.