$$y^2 = 4ax \to(1) $$ $$x^2 = 4ay ,i.e., \dfrac {x^2}{4a}=y --- (2)$$

Both are parabola

Put (2) in (1)

$\Bigg(\dfrac {x^2}{4a}\Bigg)^2=4ax\\ \therefore \dfrac {x^4}{16a^2}=4ax\\ \therefore x^4=64a^3x\\ \therefore x(x^3-64a^3)=0\\ \therefore x=0 \space or \space x^3-64a^3=0\\ \therefore x=0 \space or \space x=4a \\ But\space \dfrac {x^2}{4a}=y\\ when \space x=0,y=0\\ X=0 \space or \space y=\dfrac {(4a)^2}{4a}=4a$

The two parabola intersect at $(4a, 4a)$ and $(0,0)$

1) Limits of y

The upper limit y will be equation of parabola $y^2 = 4ax$

$\therefore y=\sqrt{4ax}$

Ans lower limit y is equation of parabola $x^2 = 4ay$

$\therefore y=\dfrac {x^2}{4a}$

2) Limit of x as the vertical strip will slide from

$0$ to $4a ,x = 0$ to $x = 4a$ 3) Limits of z given $z = 0$ to $z =3$

$$\therefore Volume =\int\limits_0^{4a}\int\limits_{\frac {x^2}{4a}}^{\sqrt{4ax}}\int\limits_0^{33} dx\space dy\space dz $$ $ \text{ integrating w.r.t. Y}\\ \therefore Volume = 3\int\limits_0^{4a}[y]_{\frac {x^2}{4a}}^{\sqrt{4ax}}dx\\ =3\int\limits_0^{4a} \sqrt{4a-x}-\dfrac {x^2}{4a}\\ =3\int\limits_0^{4a}2\sqrt {a}x^{1/2}-\dfrac {x^2}{4a}dx \\ \text{ integrating w.r.t. Y}\\ =3\Bigg[2\sqrt a\dfrac {x^{3/2}}{\frac 32}-\dfrac 1{4a}\dfrac {(4a)^3}3\Bigg]-[0]\\ =\Big[4\times 8a^{1/2+3/2}-16a^2\Big]\\ =32a^2-16a^2\\ =16a^2$