Find the centroid of the homogeneous tetrahedron bounded
Find the centroid of the homogeneous tetrahedron bounded by the planes $x+2y+z=1$, $x=0$, $y=0$, $z=0$.
Explanation of Method
The four vertices are found as pairwise-triple intersections of the bounding planes; the volume and moments follow from a straightforward iterated triple integral, and the whole computation is cross-checked against the well-known fact that a tetrahedron's centroid is the average of its four vertices.
- Find the four vertices.
- Compute the volume $V$.
- Compute $\bar x,\bar y,\bar z$ via the moment integrals.
- Verify via the vertex-average rule.
Step 1 · Vertices
$O=(0,0,0)$, $A=(1,0,0)$, $B=\left(0,\tfrac12,0\right)$, $C=(0,0,1)$. For $x\in[0,1]$, $y\in\left[0,\dfrac{1-x}2\right]$, $z\in[0,1-x-2y]$.
Step 2 · Volume
$$ V = \int_0^1\int_0^{(1-x)/2}(1-x-2y)\,dy\,dx = \int_0^1\frac{(1-x)^2}4\,dx = \frac1{12}. $$
(Matches $V=\dfrac{|\det M|}6=\dfrac{1\cdot\frac12\cdot1}6=\dfrac1{12}$ for the edge vectors from $O$.)
Step 3 · Coordinates $\bar x,\bar y,\bar z$
$$ M_{yz} = \int_0^1\frac{x(1-x)^2}4\,dx = \frac1{48}, \qquad \bar x = \frac{M_{yz}}V = \frac14. $$
$$ M_{xz} = \int_0^1\frac{(1-x)^3}{24}\,dx = \frac1{96}, \qquad \bar y = \frac{M_{xz}}V = \frac18. $$
$$ M_{xy} = \int_0^1\frac{(1-x)^3}{12}\,dx = \frac1{48}, \qquad \bar z = \frac{M_{xy}}V = \frac14. $$
Step 4 · Verification via vertex average
For any tetrahedron the centroid is the average of its four vertices:
$$ \bar x = \frac{0+1+0+0}4 = \frac14, \qquad \bar y = \frac{0+0+\frac12+0}4 = \frac18, \qquad \bar z = \frac{0+0+0+1}4 = \frac14.\ \checkmark $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically by direct triple integration: $V=1/12$, $\bar x=1/4$, $\bar y=1/8$, $\bar z=1/4$, matching exactly.
Result
The centroid of the tetrahedron is: