Triple Integral with a tetrahedron as the domain (given vertices of the tetrahedron)

Question

Evaluate the triple integral: $$\iiint_Txz \, dV$$ where $T$ is the solid tetrahedron with vertices at $(0, 0, 0)$, $(1, 0, 1)$, $(0, 1, 1)$, and $(0, 0, 1)$. The answer that I calculated was $\frac{1}{12}$, but I am not sure if my bounds were correct in the calculation.

Answer 1

The three vertices you list last have $z=1$; the first has $z=0$.

For the ones with $z=1$, the $(x,y)$ components have vertices at $(0,0)$, $(1,0)$, and $(0,1)$. The line from $(0,0)$ to $(1,0)$ runs along the $x$-axis; that from $(0,0)$ runs along the $y$-axis, and that from $(1,0)$ to $(0,1)$ runs along the line $x+y=1$. Thus when $z=1$, you can have $x$ anywhere between $0$ and $1$, and for that fixed value of $x$, $y$ can be anywhere between $0$ and $1-x$, i.e. it can go from the point where $x+y$ is only $x$ to the point where $x+y=1$.

For a smaller value of $z$ than $1$, the sum $x+y$ goes only from $0$ up to $z$ rather than up to $1$. Any value of $x+y$ that is larger than $z$ is not between the four points you specify initially. You can see that by drawing the picture, but perhaps it will be objected that that is not logically rigorous. So try writing the point $(x,y,z)$ as a weighted average of your four vertices. That it's between them means the weights are not negative. If you call the weights $a,b,c,d$, assigned to $(0, 0, 0)$, $(1, 0, 1)$, $(0, 1, 1)$, and $(0, 0, 1)$ in that order, you have \begin{align} & a+b+c+d=1 \\ & a(0,0,0) + b(1,0,1) + c(0,1,1) + d(0,0,1) = (x,y,z). \end{align} Four equations in four variables, $a,b,c,d$.

So you can write: \begin{align} & 0 \le z \le 1 \\ & 0 \le x \\ & 0 \le y \\ & x+y \le z \end{align}

Therefore you have $$ \iiint_T xz \, dV = \int_0^1 \left( \int_0^z \left( \int_0^{z-x} xz \, dy \right) \, dx \right)\, dz. $$

You can also integrate in any of the other five orders, and you have to figure out how to write the bounds for any of those that you use.