Find the volume $V$ of the solid $E$ that is enclosed by the cylinder $y^{2}+z^{2}=4,$ and the planes $x=0,$ $z=0,$ and $x+z=5.$

Figure 50

Solution Figure 50(a) shows that the solid $E$ is $yz$-simple. The front surface is the plane $x_{2}(y, z) =5-z$ and the back surface is the plane $x_{1}(y, z) =0.$ The region $R$ in the $yz$-plane is enclosed by the $y$-axis and the semi-circle $z=\sqrt{4-y^{2}};$ it is $y$-simple. So, $0\leq z\leq \sqrt{4-y^{2}}$ and $-2\leq y\leq 2.$ See Figure 50(b). The volume $V$ of $E$ is given by $$\begin{eqnarray*} V &=&\iiint\limits_{\kern-2ptE}\,{\it dV}=\iint\limits_{\kern-2ptR}\left[ \int_{x_{1}(y,z)}^{x_{2}(y,z)}{\it dx}\right] {\it dz}\,{\it dy}=\int_{-2}^{2}\int_{0}^{ \sqrt{4-y^{2}}}\int_{0}^{5-z}{\it dx}\,{\it dz}\,{\it dy}\\[4pt] &=&\int_{-2}^{2}\int_{0}^{\sqrt{4-y^{2}} }( 5-z) \,{\it dz}\,{\it dy} =\int_{-2}^{2}\left[ 5z-\dfrac{z^{2}}{2}\right] _{0}^{\sqrt{4-y^{2}}}\,{\it dy}\\[4pt] &=&\int_{-2}^{2}\left( 5\sqrt{4-y^{2}}-\dfrac{4-y^{2}}{2}\right){\it dy}\\[4pt] &=&5\int_{-2}^{2}\sqrt{4-y^{2}}\, {\it dy}-\int_{-2}^{2}\left( 2-\dfrac{y^{2}}{2} \right) {\it dy} \notag \\[4pt] &=&5(2\pi) -\left[ 2y-\dfrac{y^{3}}{6}\right] _{-2}^{2}=10\pi -\dfrac{16}{3} \end{eqnarray*}$$