$$y^2 = x ------ (1)$$

$ x^2 = y ------ (2)$

Both are parabola

Put (2) in (1)

$x^4 = x \\ x (x^3 - 1) = 0 \\ x = 0,x = 1 \\ but\space y^2 = x \\ when \space x = 0 ,y = 0 \\ when \space x = 1 ,y = 1$

Now consider he vertical strip

1) Limt of y

a) Upper limit is equation of parabola $y^2 = x\\ \therefore y=\sqrt x$

b) Lower limit is equation of parabola $y = x^2$

2) Limit of z

$Z = 0$ to $z = 1 – x – y$ given

3) Limit of x

$x = 0$ to $x = 1$

$\therefore volume =\int\limits_0^1\int\limits_{x^2}^{\sqrt x}\int\limits_0^{1-y-x} dz\space dy\space dx$

Integrating w.r.t z

$volume =\int\limits_0^1\int\limits_{x^2}^{\sqrt x}[z]_0^{1-y-x} dy\space dx \\ \int\limits_0^1 \int\limits_{x^2}^{\sqrt x}(1-x-y) dy\space dx $

Integrating w.r.t y

$= \int\limits_0^1\Bigg[y-\dfrac {y^2}2-xy\Bigg]_{x^2}^{\sqrt x}dx\\ = \int\limits_0^1\Bigg[\sqrt{x} -\dfrac x2 -x\sqrt x\Bigg]-\Bigg[x^2-\dfrac {x^4}2-x^3\Bigg] dx\\ =\int\limits_0^1\Bigg(x^{1/2}-\dfrac x2-x^{3/2}-x^2+\dfrac {x^4}2+ x^3\Bigg) dx$

Integrating w.r.t x

$Volume =\Bigg[\dfrac {x^{3/2}}{\frac 32}-\dfrac {x^2}{2\times 2}- \dfrac {x^{5/2}}{\frac 52}-\dfrac {x^3}3+ \dfrac {x^5}{2\times 5} +\dfrac {x^4}4\Bigg]_0^1\\ =\dfrac 23-\dfrac 14-\dfrac 25-\dfrac 13+\dfrac 1{10}+\dfrac 14\\ =\dfrac 1{30}$