836. Find the center of gravity of a homogeneous truncated prism bounded by the coordinate planes and the planes $x+y+z=4, x=1, y=1$.
Solution. By constructing the given planes (Fig. 170), we observe that the truncated prism bounded by them is symmetric with respect to the plane $x=y$. Consequently,

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-284.jpg?height=473&width=318&top_left_y=231&top_left_x=870)

Fig. 170 this $x_{c}=y_{c}$.

Next, we compute the integrals contained in formulas (5),

$$
I_{1}=\iint_{D} x z d x d y=\iint_{O B C} x(4-x-y) d x d y=
$$

$$
\begin{gathered}
=-\int_{0}^{1} x d x \int_{1}^{1}(4-x-y) d(4-x-y)=-\left.\int_{n}^{1} \frac{(4-x-y)^{2}}{2}\right|_{y=0} ^{y=1} x d x= \\
=\frac{1}{2} \int_{1}^{n}\left(2 x^{2}-7 x\right) d x=\left.\frac{1}{2}\left(\frac{2}{3} x^{3}-\frac{7}{2} x^{2}\right)\right|_{i} ^{n}=\frac{17}{12} \\
I_{2}=\int_{1} \int_{1} z^{2} d x d y=\iint_{n i C C}(4-x-y)^{2} d x d y=
\end{gathered}
$$

$$
\begin{aligned}
= & -\int_{n}^{1} d x \int_{1}^{1}(4-x-y)^{2} d(4-x-y)=-\left.\int_{n}^{1} \frac{(4-x-y)^{3}}{3}\right|_{1=1} ^{1=1} d x= \\
& =\frac{1}{3} \int_{i}^{1}\left|(3-x)^{3}-(4-x)^{3}\right| d x=\left.\frac{1}{3}\left[\frac{(x-3)^{1}}{4}-\frac{(x-4)^{1}}{4}\right]\right|_{1} ^{1}=\frac{55}{6} ; \\
I_{3}= & \int_{D} \int_{0}^{1} z d x d y==\int_{0}^{1} d x \int_{1}^{1}(4-x-y) d y=-\left.\frac{1}{2} \int_{0}^{1}(4-x-y)^{2}\right|_{y==1} ^{u=1} d x= \\
& =-\frac{1}{2} \int_{0}^{1}\left|(3-x)^{2}-(4-x)^{2}\right| d x=\left.\frac{1}{2}\left[\frac{(x-3)^{3}}{3}-\frac{(x-4)^{3}}{3}\right]\right|_{1} ^{0}=3 .
\end{aligned}
$$

Substituting the found values of the integrals into formulas (5),
we get:

$$
x_{c}=y_{c}=\frac{l_{1}}{I_{3}}=\frac{17}{36} ; \quad z_{c}=\frac{l_{2}}{2 I_{3}}=\frac{55}{36}
$$