Find the volume $V$ of the solid $E$ in the first octant that is enclosed by the paraboloid $z=16-4x^{2}-y^{2}$ and the $xy$-plane.

Figure 45

Solution Figure 45(a) shows the solid $E$, which is $xy$-simple. The upper surface is the paraboloid $z=z_{2}(x, y)=16-4x^{2}-y^{2}$ and the lower surface is the plane $z=z_{1}(x, y)=0$. The region $R$ in the $xy$-plane is enclosed by the $x$-axis, the $y$-axis, and part of the ellipse $4x^{2}+y^{2}=16,$ $x\geq 0,$ $y\geq 0.$The region $R$ is both $x$-simple and $y$-simple. We choose to use the iterated integral for an $x$-simple region, so that $0\leq y\leq \sqrt{16-4x^{2}}$ and $0\leq x\leq 2.$ See Figure 45(b). The volume $V$ of $E$ is given by $$\begin{eqnarray*} V&=&\iiint\limits_{\kern-2ptE}\,{\it dV}=\int_{a}^{b}\int_{y_{1}(x)}^{y_{2}(x)} \int_{z_{1}(x, y)}^{z_{2}(x, y)}{\it dz}\, {\it dy}\,{\it dx}= \int_{0}^{2} \int_{0}^{\sqrt{16-4x^{2}} } \int_{0}^{16-4x^{2}-y^{2}}{\it dz}\, {\it dy}\,{\it dx} \\[4pt] &=&\int_{0}^{2}\int_{0}^{\sqrt{16-4x^{2}}}( 16-4x^{2}-y^{2}) \,{\it dy}\,{\it dx}=\int_{0}^{2}\left[ (16-4x^{2})y-\dfrac{y^{3}}{3}\right] _{0}^{\sqrt{ 16-4x^{2}}}\,{\it dx} \\[4pt] &=&\int_{0}^{2}\!\left( (16-4x^2)^{3/2} -\dfrac{( 16-4x^{2}) ^{3/2}}{3}\right)\!{\it dx} = \dfrac{16}{3}\int^2_0 (4-x^2)^{3/2}\,{\it dx} \end{eqnarray*}$$

We use the Table of Integrals, Integral 69 with $a=2$. Then $$\begin{eqnarray*} V=\dfrac{16}{3}\int^2_0 (4-x^2)^{3/2}\,{\it dx} &=&\dfrac{16}{3}\left[ \dfrac{x}{4} ( 4-x^{2}) ^{3/2}+\dfrac{3x}{2 }\sqrt{4-x^{2}}+6\sin ^{-1}\dfrac{x}{2}\right] _{0}^{2}\\[4pt] &=& \dfrac{16}{3} (6) \left(\dfrac{\pi}{2}\right)=16\pi \end{eqnarray*}$$