Example 1. Calculate the mass of the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $x / 10+y / 8+z / 3=1$, if the density distribution of mass at each point is given by the function $\rho=(1+x / 10+y / 8+z / 3)^{-6}$.
Solution. We have $m=\iint_{W} \int \rho d V$. The triple integral is reduced to a double and a definite one (see point $4^{\circ}$):

$$
m=\iint_{D} d x d y \int_{0}^{z} \frac{d z}{\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{6}}
$$

The upper limit, or the exit point from the region, is the ordinate of the point on the plane $\frac{x}{10}+\frac{y}{8}+\frac{z}{3}=1$ (Fig. 3.16), i.e., $z=3\left(1-\frac{x}{10}-\frac{y}{8}\right)$. The double integral is extended over the triangle, which is the base of the tetrahedron formed by

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-061.jpg?height=424&width=477&top_left_y=180&top_left_x=2005)

Fig. 3.16 the lines $x=0, y=0, \frac{x}{10}+\frac{y}{8}=1$; therefore (some steps are omitted):

$$
\begin{gathered}
m=\left.\iint_{D} d x d y\left[-\frac{3}{5}\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{-5}\right]\right|_{0} ^{3(1-x / 10-y / 8)}= \\
=\frac{3}{5} \int_{0}^{10} d x \int_{0}^{8(1-x / 10)}\left[2^{-5}-\frac{1}{\left(1+\frac{x}{10}+\frac{y}{8}\right)^{5}}\right] d y= \\
=-\left.\frac{3}{5} \int_{0}^{10} d x\left(\frac{y}{32}+\frac{8}{4}\left(1+\frac{x}{10}+\frac{y}{8}\right)^{-4}\right)\right|_{0} ^{8(1-x / 10)}= \\
=\frac{3}{5} \int_{0}^{10}\left(\frac{3}{8}-\frac{x}{40}-2\left(1+\frac{x}{10}\right)^{-4}\right) d x= \\
=-\left.\frac{3}{5}\left(\frac{3}{8} x-\frac{x^{2}}{80}+\frac{20}{3}\left(1+\frac{x}{10}\right)^{-3}\right)\right|_{0} ^{10}=2
\end{gathered}
$$

Answer. $m=2$.