The question says to evaluate the given triple integral in cylindrical coordinates over the region $W$. Here's the integrand and the region: 
I'm confused about this question because I keep getting $0$. Here's the triple integral in polar coordinates that represents the region given in question $9$:
Note that I use $x=r\cos\theta$, $y=r\sin\theta$, and $z=z$ to simplify the integrand $f(x,y,z)=x^2+y^2+z^2$.
$$\int_{-1}^1\int_{\pi/4}^{3\pi/4}\int_{0}^4 r(r^2+z^2) \ dr\ d\theta \ dz$$
After integrating the inner intergral we end up with positive powers of $r$. As a result, the function evaluated at $1$ minus the function evaluated at $-1$ ends up being $0$, and the integration stops there. The order of integration turns out not to matter: I always get this result.
Yet the text shows a non-zero number as the answer. What am I missing?
The answer is non-zero. I suspect you're calculating the integrals incorrectly, since your setup looks great. We have \begin{align*} \int_{-1}^1\int_{\pi/4}^{3\pi/4}\int_{0}^4 r\big(r^2+z^2\big) \ dr\ d\theta \ dz &=\frac{\pi}{2}\int_{-1}^1\int_0^4\big(r^3+rz^2\big)\,dr\,dz\\ &=\frac{\pi}{2}\int_{-1}^1\left(\frac{r^4}{4}+\frac{r^2z^2}{2}\right)\bigg|_{0\to 0}^4\,dz\\ &=\frac{\pi}{2}\int_{-1}^1\left(64+8z^2\right)dz\\ &=\pi\int_{0}^1\left(64+8z^2\right)dz\quad\text{(even function on symmetric interval)}\\ &=\pi\left(64z+\frac{8z^3}{3}\right)\bigg|_{0}^{1}\\ &=\frac{200\pi}{3}. \end{align*}