Find the triple integral $\iiint\limits_{\kern-2ptE}4x\,{\it dV}$ over the tetrahedron formed by the coordinate planes and the plane $2x+3y+4z=12$.

Solution The solid $E$ is shown in Figure 44(a). A typical plane section of $E$ using a plane perpendicular to the $x$-axis reveals that the upper surface is $z=z_{2}(x, y)=\dfrac{1}{4}(12-2x-3y)$ and the lower surface is $z=z_{1}(x, y)=0$. So, $E$ is $xy$-simple. The region $R$ in the $xy$-plane is the triangle enclosed by the $x$-axis, the $y$-axis, and the line $y=\dfrac{1}{3}(12-2x)=4-\dfrac{2x}{3}$, as shown in Figure 44(b). The triangle $R$ is both $x$-simple and $y$-simple. If we treat it as an $x$-simple region, so that $R$ is given by $y=0$, $y=4-\dfrac{2x}{3}$ and $\leq x\leq 6$, we have $$\begin{eqnarray*} \iiint\limits_{\kern-2ptE}4x\,{\it dV}&=&\int_{a}^{b}\int_{y_{1}(x)}^{y_{2}(x)} \int_{z_{1}(x, y)}^{z_{2}(x, y)}4x\,{\it dz}\,{\it dy}\,{\it dx}\\ &=&\int_{0}^{6}\int_{0}^{4-2x/3}\int_{0}^{(1/4)(12-2x-3y)}4x\,{\it dz}\,{\it dy}\,{\it dx} \notag \\ &=&\int_{0}^{6}\int_{0}^{4-2x/3} 4x \big[z\big] _{0}^{( 1/4) (12-2x-3y)}\,{\it dy}\,{\it dx}=\int_{0}^{6}\int_{0}^{4-2x/3}~4x\!\left( \dfrac{12-2x-3y}{4 }\right) {\it dy}\,{\it dx} \notag\\ &=&\int_{0}^{6}\int_{0}^{4-2x/3}(12x-2x^{2}-3xy)\,{\it dy}\,{\it dx}=\int_{0}^{6}\left[ ( 12x-2x^{2}) \,y-3x\dfrac{y^{2}}{2}\,\right] _{0}^{4-2x/3}{\it dx} \notag\\ & =&\int_{0}^{6}\left( 24x-8x^{2}+\dfrac{2}{3}x^{3}\right) {\it dx}=\left[ 12x^{2}- \dfrac{8}{3}x^{3}+\dfrac{1}{6}x^{4}\right] _{0}^{6}=72 \end{eqnarray*}$$