The area enclosed between the curves $\mathrm{y}^2=4 x$ and $\mathrm{y}=|x|$ is

The curves $y^2=4x$ and $y=|x|$ intersect at $(0,0)$ and $(4,4)$, and their enclosed region lies in the first quadrant.

For $0 \leq x \leq 4$, the parabola gives $y = 2\sqrt{x}$ (upper curve) and $y = x$ (lower curve).

The area is $A = \int_{0}^{4} (2\sqrt{x} - x) \, dx = \left[ \frac{4}{3}x^{3/2} - \frac{1}{2}x^2 \right]_{0}^{4} = \left( \frac{4}{3}(8) - \frac{1}{2}(16) \right) = \frac{32}{3} - 8 = \frac{8}{3}$.

Final answer: $\boxed{\frac{8}{3}}$