Example 2. Calculate the static moments of the tetrahedron bounded by the coordinate planes and the plane $x+y+z=1$, if the density of the tetrahedron is $\rho=x y$.
Solution.

$$
\begin{gathered}
M_{x y}=\iiint_{W} \rho d V=\iiint_{W} x y z d x d y d z=\int_{0}^{1} x d x \int_{0}^{1-x} y d y \int_{0}^{1-x-y} z d z= \\
=\int_{0}^{1} x d x \int_{0}^{1-x} y \cdot \frac{(1-x-y)^{2}}{2} d y= \\
=\frac{1}{2} \int_{0}^{1} x d x \int_{0}^{1-x}\left(y(1-x)^{2}-2(1-x) y^{2}+y^{3}\right) d y= \\
=\left.\frac{1}{2} \int_{0}^{1} x d x\left((1-x)^{2} \frac{y^{2}}{2}-2(1-x) \frac{y^{3}}{3}+\frac{y^{4}}{4}\right)\right|_{0} ^{1-x}= \\
M_{y z}=\iint_{W} \rho x d x d y d z=\int_{0}^{1} x^{2} d x \int_{0}^{1-x} y d y \int_{0}^{1-x-y} d z= \\
=\left.\int_{0}^{1} x^{2} d x\left[(1-x) \frac{y^{2}}{2}-\frac{y^{3}}{3}\right]\right|_{0} ^{1-x}=\frac{1}{6} \int_{0}^{1} x^{2}(1-x)^{3} d x=\frac{1}{360} \\
M_{x z}=\int_{W} \int_{0} \rho y d x d y d z=\int_{0}^{1} x d x \int_{0}^{1-x} y^{2} d y \int_{0}^{1-x-y} d z= \\
\left.\int_{0}^{1} x d x\left((1-x) \frac{y^{3}}{3}-\frac{y^{4}}{4}\right)\right|_{0} ^{1-x}=\frac{1}{12} \int_{0}^{1} x(1-x)^{4} d x=\frac{1}{720}
\end{gathered}
$$

Answer. $M_{x y}=\frac{1}{720}, M_{y z}=\frac{1}{360}, M_{x z}=\frac{1}{360}$.