Compute $\iiint_R 6z \ dV.$

Question

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?

Answer 1

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 $\frac{A\cdot H}{3}$.

Answer 2

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}