Centroid of polyhedral solid

Let $y_0$ be the $y$ coordinate of the centroid, then $y_0$ is given by the triple integral

$y_0 = \displaystyle \left( \dfrac{1}{V} \right) \int_{x=-2}^{-1} \int_{z = 0}^{-x} \int_{y = 3 x }^{2 z} y\hspace{4pt} dy \hspace{4pt} dz \hspace{4pt} dx$

where $V$ is the volume and is given by $V = \displaystyle \int_{x=-2}^{-1} \int_{z = 0}^{-x} \int_{y = 3 x }^{2 z} dy \hspace{4pt} dz \hspace{4pt} dx = \dfrac{28}{3}$

Integrating with respect to $y$ first,

$y_0 = \displaystyle \left( \dfrac{3}{28} \right) \displaystyle \int_{x=-2}^{-1} \int_{z = 0}^{-x} (2 z^2 - \dfrac{9}{2} x^2 ) \hspace{4pt} dz \hspace{4pt} dx$

Now integrating with respect to $z$ ,

$y_0 = \displaystyle \left( \dfrac{3}{28} \right) \displaystyle \int_{x=-2}^{-1} (- \dfrac{2}{3} x^3 + \dfrac{9}{2} x^3 ) \hspace{4pt} dx = \left( \dfrac{3}{28} \right) \displaystyle \int_{x=-2}^{-1} \dfrac{23}{6} x^3 \hspace{4pt} dx$

Finally, integrating with respect to $x$

$y_0 = \displaystyle \left( \dfrac{3}{28} \right) \left( \dfrac{23}{6} \right)\displaystyle \left( \dfrac{1}{4} \right) ( (-1)^4 - (-2)^4 ) = \left( \dfrac{23}{56} \right ) \left( - \dfrac{15}{4} \right) = - \dfrac{ 345 }{224}$

Therefore, the answer is $345 + 224 = \boxed{569}$

Moment of the solid about the xz-plane, where $\rho$ is the uniform density:

$M_{\text{xz}}=\int _{-2}^{-1}\int _{-x}^0\int _{2 z}^{3 x}\rho ydydzdx$

Mass of the solid:

$m=\int _{-2}^{-1}\int _{-x}^0\int _{2 z}^{3 x}\rho dydzdx$

The y-coordinate of the centre of mass of the solid:

$y=\frac{M_{\text{xz}}}{m}$

Plugging in the numbers: $y=-\frac{345}{224}$ , and the answer is 345+224= $\boxed{569}$