We find the mass, \(M\text{,}\) of the tetrahedron by the triple integral
\begin{equation*}
M = \iiint_S \delta(x,y,z) \, dV,
\end{equation*}
where \(S\) is the solid tetrahedron described above. In this example, we choose to integrate with respect to \(z\) first for the innermost integral. The top of the tetrahedron is given by the equation
\begin{equation*}
x + 2 y + 3 z = 6;
\end{equation*}
solving for \(z\) then yields
\begin{equation*}
z = \frac{1}{3}(6 - x - 2y).
\end{equation*}
The bottom of the tetrahedron is the
\(xy\)-plane, so the limits on
\(z\) in the iterated integral will be
\(0 \leq z \leq \frac{1}{3}(6-x-2y)\text{.}\)
To find the bounds on \(x\) and \(y\) we project the tetrahedron onto the \(xy\)-plane; this corresponds to setting \(z = 0\) in the equation \(z = \frac{1}{3}(6 - x - 2y)\text{.}\) The resulting relation between \(x\) and \(y\) is
\begin{equation*}
x + 2 y = 6.
\end{equation*}
The right image in
FigureΒ 3.6.9 shows the projection of the tetrahedron onto the
\(xy\)-plane.
If we choose to integrate with respect to \(y\) for the middle integral in the iterated integral, then the lower limit on \(y\) is the \(x\)-axis and the upper limit is the hypotenuse of the triangle. Note that the hypotenuse joins the points \((6,0)\) and \((0,3)\) and so has equation \(y = 3 - \frac{1}{2}x\text{.}\) Thus, the bounds on \(y\) are \(0 \leq y \leq 3 - \frac{1}{2}x\text{.}\) Finally, the \(x\) values run from 0 to 6, so the iterated integral that gives the mass of the tetrahedron is
\begin{equation}
M = \int_{0}^{6} \int_{0}^{3-(1/2)x} \int_{0}^{(1/3)(6-x-2y)} x+y+z \, dz \, dy \, dx.\tag{3.6.2}
\end{equation}
Evaluating the triple integral gives us
\begin{align*}
M \amp = \int_{0}^{6} \int_{0}^{3-(1/2)x} \int_{0}^{(1/3)(6-x-2y)} x+y+z \, dz \, dy \, dx\\
\amp = \int_{0}^{6} \int_{0}^{3-(1/2)x} \left[xz+yz+\frac{z}{2}\right]\biggm|_{0}^{(1/3)(6-x-2y)} \, dy \, dx\\
\amp = \int_{0}^{6} \int_{0}^{3-(1/2)x} \frac{4}{3}x - \frac{5}{18}x^2 - \frac{}{9}xy + \frac{2}{3}y - \frac{4}{9}y^2 + 2 \, dy \, dx\\
\amp = \int_{0}^{6} \left[\frac{4}{3}xy - \frac{5}{18}x^2y - \frac{7}{18}xy^2 + \frac{1}{3}y^2 - \frac{4}{27}y^3 + 2y \right]\biggm|_{0}^{3-(1/2)x} \, dx\\
\amp = \int_{0}^{6} 5 + \frac{1}{2}x - \frac{7}{12}x^2 + \frac{13}{216}x^3 \, dx\\
\amp = \left[5x + \frac{1}{4}x^2 - \frac{7}{36}x^3 + \frac{13}{864}x^4 \right] \biggm|_{0}^{6}\\
\amp = \frac{33}{2}.
\end{align*}