Find the centroid of a homogeneous truncated prism
Find the centroid of a homogeneous truncated prism bounded by the coordinate planes, $x=1$, $y=1$, and $x+y+z=4$.
Explanation of Method
Over the square base $D=[0,1]\times[0,1]$, the solid's height is $z_{\text{top}}(x,y)=4-x-y$; the volume and the three moments then follow from stacking $\iint_D(\cdot)\,z_{\text{top}}\,dA$-type double integrals (with an extra factor of $\tfrac12 z_{\text{top}}$ for the moment about the $xy$-plane, from integrating $z$ in the vertical direction).
- Set up the solid over the base $D=[0,1]^2$.
- Compute the volume $V$.
- Compute $\bar x$ and (by symmetry) $\bar y$.
- Compute $\bar z$.
Step 1 · Setup
For $(x,y)\in D=[0,1]\times[0,1]$, $z$ ranges from $0$ to $4-x-y\ (\ge2>0)$.
Step 2 · Volume
$$ V = \iint_D(4-x-y)\,dx\,dy = \int_0^1\left(\frac72-y\right)dy = 3. $$
Step 3 · Coordinates $\bar x,\bar y$
$$ M_{yz} = \iint_Dx(4-x-y)\,dx\,dy = \int_0^1\left(\frac53-\frac y2\right)dy = \frac{17}{12}, \qquad \bar x = \frac{M_{yz}}V = \frac{17}{36}. $$
Since $D$ and $4-x-y$ are both symmetric under $x\leftrightarrow y$: $\bar y=\bar x=\dfrac{17}{36}$.
Step 4 · Coordinate $\bar z$
$$ M_{xy} = \iint_D\frac{(4-x-y)^2}2\,dx\,dy = \frac12\int_0^1\left(\frac{37}3-7y+y^2\right)dy = \frac{55}{12}, \qquad \bar z = \frac{M_{xy}}V = \frac{55}{36}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically by direct double integration: $V=3$, $\bar x=\bar y\approx0.472222$, $\bar z\approx1.527778$, matching $\dfrac{17}{36}$ and $\dfrac{55}{36}$ exactly.
Result
The centroid of the truncated prism is: