Cylinder and cube on plank

Consider the given cylinder at first. Its density is given by:

$\rho = 768z(x^2+y^2)^2$

Given that the origin of the cylinder is at the bottom of the cylinder, transforming the above equation into cylindrical coordinates gives:

$x = r\cos{\theta}$ $y = r\sin{\theta}$ $\rho = 768zr^4$

The total mass of the cylinder is:

$M = \int_{V} \rho \ dV =\int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{0.5} 768zr^4 (r \ dr \ d\theta \ dz)$

Evaluating and simplifying gives:

$M = 2 \pi$

The COM coordinates of the cylinder are (using definition of COM):

$x_c = \frac{1}{M}\left(\int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{0.5} (r\cos{\theta})\left(768zr^4 (r \ dr \ d\theta \ dz)\right)\right)=0$ $y_c = \frac{1}{M}\left(\int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{0.5} (r\sin{\theta})\left(768zr^4 (r \ dr \ d\theta \ dz)\right)\right)=0$ $z_c = \frac{1}{M}\left(\int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{0.5} (z)\left(768zr^4 (r \ dr \ d\theta \ dz)\right)\right) \ne 0$

The COM is located on the axis of the cylinder. Moreover, the COM of the plank is at its midpoint (5m from A) and the COM of the cube also coincides with its geometrical centre. The mass of the cube is also $m = 2$ and that of the plank is $m_p=2$ . Performing a torque balance about point A gives(free body diagram not shown):

$\frac{5}{2}Mg + 5m_pg + 7mg = 4V_B$

Solving:

$V_B = \left(\frac{5 \pi}{4} + 6\right)g$