Mass of a cylinder.

We consider an elementary hollow cylinder with radius $r$ , thickness $dr$ and height $dz$ located at an elevation of $z$ .

Then the differential volume of the elementary cylinder is

$dV = 2\pi rdrdz$

We know that $\Large\frac{dM}{dV}$ $= \rho$ where $M$ is the mass, $V$ is the volume and $\rho$ is the density of any given body.

In our question, $\rho = z(x^2 + y^2)^2 = z(r^2)^2 = zr^4$ . So $dM = \rho dV\newline \Rightarrow dM = (zr^4)(2\pi rdrdz)$

$\Rightarrow M = 2\pi\int_{0}^{h}\int_{0}^{R}zr^5drdz$

$\Rightarrow M = 2\pi\int_{0}^{h}z\Big(\int_{0}^{R}r^5dr\Big)dz$

$\Rightarrow M = 2\pi\int_{0}^{h}\Large\frac{zR^6}{6}$ $dz$

$\Rightarrow M = \Large\frac{2\pi R^6}{6}$ $\int_{0}^{h}zdz$

$\Rightarrow M = \Large\frac{2\pi R^6}{6}\cdot \frac{h^2}{2}$

$\Rightarrow M = \Large\frac{\pi R^6h^2}{6}$

Putting $R = \Large\frac{1}{2}$ and $h = 1$ we get

$M = \Large\frac{\pi}{2^6\cdot 6} = \frac{\pi}{3\cdot 2^7}$

So, $a = 3$ and $b = 2 \Rightarrow a + b = \boxed{5}$

We require the triple integral in cylindrical coordinates:

$M = \int_{0}^{2\pi} \int_{0}^{1/2} \int_{0}^{1} z(r^2)^{2} \cdot r dzdrd\theta = \int_{0}^{2\pi} \int_{0}^{1/2} \int_{0}^{1} zr^5 dzdrd\theta = \frac{\pi}{384} = \boxed{\frac{\pi}{3\cdot2^{7}}}.$