Solution:

The solid is bounded by the parabolas $y^{2}=x, x^{2}=y$ in the x y plan

$\ln x-y-z$ plane $x+y+z=2$ is top base.

The volume between this curves is given by,

$\quad v=\iint z d x \ d y=\iint(2-x-y) d x \ d y$

From the diagram we can conclude that the intersection point of both

Parabolas are $(0,0),(1,1) .$

$\begin{aligned} \therefore \mathrm{V} &=\int_{0}^{1} \int_{x^{2}}^{\sqrt{x}}(2-x-y) d x \ d y \\ &=\int_{0}^{1}\left[2 y-x y-\frac{y^{2}}{2}\right]_{x^{2}}^{\sqrt{x}} \mathrm{dx} \end{aligned}$

$=\int_{0}^{1}\left[\left(2 \sqrt{x}-x \sqrt{x}-\frac{x}{2}\right)-\left(2 x^{2}-x^{3}-\frac{x^{4}}{2}\right)\right] \mathrm{dx}$

$\begin{aligned}=&\left[\frac{4 x^{3 / 2}}{3}-\frac{2 x^{\frac{5}{2}}}{5}-\frac{x^{2}}{4}-\frac{2 x^{3}}{3}+\frac{x^{4}}{4}+\frac{x^{5}}{10}\right]_{0}^{1} ] \\ \therefore V &=\frac{11}{30} \end{aligned}$