Find the volume of the solid in $\mathbf{R}^{3}$ bounded by $y=x^{2}, x=y^{2}, z=x+y+9$, and $z=0$. \[ \mathrm{V}= \]

Step 1 ：The volume of the solid in $\mathbf{R}^{3}$ bounded by $y=x^{2}, x=y^{2}, z=x+y+9$, and $z=0$ can be found by integrating the function $z=x+y+9$ over the region defined by the other three equations.

Step 2 ：The region in the xy-plane bounded by $y=x^{2}$ and $x=y^{2}$ is a square with vertices at $(0,0), (1,1), (-1,-1), (0,0)$. We can integrate over this region by setting up a double integral with limits of integration from -1 to 1 for both x and y.

Step 3 ：The function $z=x+y+9$ is a plane, so its integral over the region will give the volume of the solid.

Step 4 ：Final Answer: The volume of the solid in $\mathbf{R}^{3}$ bounded by $y=x^{2}, x=y^{2}, z=x+y+9$, and $z=0$ is \(\boxed{36}\).