Compute $\iiint_R 6z \ dV.$ 
Integrate the function $$f(x,y,z)=6z$$ over the tetrahedral
  $$R=\{(x,y,z):x\geq0, \ y\geq 0, \ z\geq 0, \ 5x+y+z \leq 5\}.$$

This tetrahedral can obtain hegiths in $z$-axis from 0 to 5 in the first octant. Drawing this out, I get that the bounds are
\begin{array}{lcl}
0 \leq x \leq  1 \\
0 \leq y \leq 5-5x \\
0 \leq z \leq -5x-y+5
\end{array}
So 
$$\iiint_R 6z \ dV=\int_0^1\int_0^{5-5x}\int_0^{-5x-y+5}6z \ dzdydx=\frac{125}{4}.$$
Can anyone confirm this is correct and check for any improvement?

Yes, it seems absolutely correct as set up and also as calculation.
To check the result note that the given integral is equal to
$$6 \cdot z_G \cdot V=6 \cdot \frac54 \cdot \frac{25}{6}=\frac{125}{4}$$
indeed for geometric properties of tetrahedron  the centroid is at $\frac{H}{4}$ and volume is $\frac{A\cdot H}{3}$.
