Evaluate the triple integral of x + 2y + 3z over the tetrahedron bounded by x + y + z = 1 in the first octant.

Step-by-step solution

1. 1

   Describe the region. The plane $x+y+z=1$ together with $x,y,z\ge0$ bounds the unit tetrahedron with vertices at the origin and $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Slicing it gives the iterated limits

   $0\le z\le1,\qquad0\le y\le1-z,\qquad0\le x\le1-y-z$

   so the integral is $\displaystyle\int_{0}^{1}\!\!\int_{0}^{1-z}\!\!\int_{0}^{1-y-z}(x+2y+3z)\,dx\,dy\,dz$.

2. 2

   Split by linearity instead of integrating the sum.

   $\iiint_{E}(x+2y+3z)\,dV=\iiint_{E}x\,dV+2\iiint_{E}y\,dV+3\iiint_{E}z\,dV$

   Each piece is now a single-variable moment, and they are related by symmetry.

3. 3

   Exploit the symmetry of the region. The tetrahedron is completely symmetric under permuting $x$, $y$ and $z$ — swapping any two coordinates maps it onto itself. Therefore

   $\iiint_{E}x\,dV=\iiint_{E}y\,dV=\iiint_{E}z\,dV$

   so only one of the three integrals has to be computed.

4. 4

   Compute the common value. Integrating $x$ first over $0\le x\le1-y-z$ gives $\tfrac{(1-y-z)^{2}}{2}$, and integrating that over the triangle $y+z\le1$ yields

   $\iiint_{E}x\,dV=\frac{1}{24}$

   (A numerical check on a $240\times240$ grid gives $0.0416659$ against $\tfrac1{24}=0.0416667$ ✓.) As a cross-check, the sum of all three equals $\displaystyle\iiint_{E}(x+y+z)\,dV=\tfrac{3}{24}=\tfrac18$, consistent with the tetrahedron's volume $\tfrac16$ and mean coordinate sum $\tfrac34$.

5. 5

   Assemble the weighted total.

   $\frac{1}{24}+2\cdot\frac{1}{24}+3\cdot\frac{1}{24}=\frac{1+2+3}{24}=\frac{6}{24}=\frac{1}{4}$

   The answer is $\tfrac14$. Notice the weights $1,2,3$ enter only through their sum — a direct consequence of the symmetry, and a good reason to look for it before grinding through three separate iterated integrals.