Using cylindrical polar coordinates.

$x=r\cos\theta,y=r\sin\theta\\ dx\space dy\space dz=r\space dr\space d\theta\space dz$

$\text{ equation of paraboloid } \\ az=r^2\cos^2\theta+r^2\sin^2\theta \\ \therefore az=r^2$

$\text{ equation of cylinder }\\ r^2\cos^2\theta +r^2\sin^2\theta =a^2\\ r^2=a^2\\ \therefore r=a$

$\therefore $ The equation of paraboloid and cylinder in polar form are $r^2=az \space \& \space r=a$

1. Limit of Z

Z varies from 0 to $(r^2/a)$

$r=0$ to $r=a $

3) limit of $\theta$

To complete whole cylinder

$\theta$ must vary from 0 to $2π$

By $\theta$ will be 0 to $(π/2)×4$ times

Volume $ =4\int\limits_{\theta=0}^{\frac \pi2}\int\limits^a_{r=0}\int\limits_{z=0}^{\frac {r^2}a}r\space dr\space d\theta\space dz$

Integrating w.r.t. z

$V=4\int\limits_0^{\frac \pi2}\int\limits^a_{r=0}r[z]_0^{\frac {r^2}a}d\theta\space dr\\ =4\int\limits_0^{\frac \pi2}\int\limits^a_{r=0}\Bigg[\dfrac {r^3}a\Bigg]dr\space d\theta$

Integrating w.r.t. y

$\therefore v=4\int\limits_0^{\frac \pi2}\Bigg[\dfrac {r^4}{4a}\Bigg]_0^a d\theta\\ v=4\int\limits_0^{\frac \pi2}\dfrac {a^3}4d\theta\\ V=a^3[\theta]_0^{\frac \pi2}\\ =\dfrac \pi2a^3$