Find the volume $V$ of the solid in the first octant enclosed by the plane $x+y+z=1$.

Solution Figure 21(a) shows the solid in the first octant enclosed by the plane $x+y+z=1$. The volume $V$ of the solid lies under the plane $z=f(x,y) =1-x-y$ and above the region $R$ in the $xy$-plane bounded by the lines $x=0,$ $y=0,$ and $x+y=1.$ See Figure 21(b). The region $R$ is $x$-simple and $y$-simple. If we treat $R$ as $x$-simple, then $y$ varies from $y=0$ to $y=1-x$, $0\leq x \leq 1$. The volume $V$ of the solid is $$\begin{eqnarray*} V=\displaystyle\iint\limits_{\kern-3ptR}f(x,y) \,{\it dA}& =&\int_{0}^{1}\left[ \int_{0}^{1-x}(1-x-y)\,{\it dy}\right]\! {\it dx}=\int_{0}^{1}\left[ ( 1-x) y- \dfrac{y^{2}}{2}\right] _{0}^{1-x} {\it dx}\\[5pt] &=&\int_{0}^{1}\left[ (1-x)^{2}-\dfrac{(1-x)^{2}}{2}\right] \! {\it dx} \\[5pt] & =&\dfrac{1}{2}\int_{0}^{1}(1-x)^{2}\,{\it dx}=\dfrac{1}{2}\left[ x-x^{2}+\dfrac{x^{3} }{3}\right] _{0}^{1}=\dfrac{1}{6}\hbox{ cubic unit} \end{eqnarray*}$$