Finding a Triple Integral Defined over an \(xy\)-Simple Solid

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*}