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 is correct.\begin{align}
&\int_0^1 \int_0^{5-5x} \int_0^{-5x-y+5} 6z \,\,\, dz dy dx \\
&=\int_0^1 \int_0^{5-5x}  3(-5x-y+5)^2 \,\,dy dx \\
&=\int_0^1 \int_0^{5-5x}  3(5x+y-5)^2 \,\,dy dx \\
&=\int_0^1   -(5x+0-5)^3 \, dx \\
&=-5^3 \int_0^1(x-1)^3 \, dx \\
&= -5^3 \frac{(x-1)^4\mid_0^1}{4}\\
&=\frac{125}{4}
\end{align}
