6.2.1. Find the integral $\iiint_{V} x d x d y d z$, if the body $V$ is bounded by the planes: $x=0, y=0, x+4 y-2 z=0$ and $x+y+z-6=0$.
Solution. The body V is bounded by the coordinate planes $x \mathrm{Oz}$ and $y \mathrm{Oz}$, and from below and above by the planes: $z=\frac{1}{2}(x+4 y)$ and $z=6-x-y$. Let's find the line of intersection of these planes:

$$
\begin{aligned}
& z=\frac{1}{2}(x+4 y) \\
& z=6-x-y
\end{aligned}
$$

or

$$
\frac{1}{2}(x+4 y)=6-x-y
$$

or

$$
x+2 y-4=0 .
$$

We have $y=\frac{1}{2}(4-x)$. Let's construct the region $D$ on the plane $x \mathrm{Oy}$ (Fig. 62). We compute:

$$
\begin{gathered}
\iiint_{V} x d x d y d z=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y \int_{\frac{1}{2}(x+4 y)}^{6-x-y} d z=\left.\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y z\right|_{\frac{1}{2}(x+4 y)} ^{6-x-y}= \\
=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} d y(6-x-y)-\frac{1}{2}(x+4 y)= \\
=\int_{0}^{4} x d x \int_{0}^{\frac{1}{2}(4-x)} 6-\frac{3}{2} x-3 y d y=3 \int_{0}^{4} x d x 2 y-\frac{1}{2} x y-\left.\frac{1}{2} y^{2}\right|_{0} ^{\frac{1}{2}(4-x)}= \\
=3 \int_{0}^{4} x d x 2 \cdot \frac{1}{2}(4-x)-\frac{1}{2} x \cdot \frac{1}{2}(4-x)-\frac{1}{2} \frac{1}{2}(4-x)^{2}= \\
=3 \int_{0}^{4} x 2-x+\frac{1}{8} x^{2} d x=3 \int_{0}^{4} 2 x-x^{2}+\frac{1}{8} x^{3} d x= \\
=32 \cdot \frac{x^{2}}{2}-\frac{1}{3} x^{3}+\left.\frac{1}{8} \frac{1}{4} x^{4}\right|_{0} ^{4}=316-\frac{64}{3}+8=8
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-266.jpg?height=498&width=635&top_left_y=1292&top_left_x=409)

Fig. 62