Find the volume $V$ under the paraboloid $z=f(x,y)=4-x^{2}-y^{2}$ and over the rectangular region $R$ defined by $-1\leq x\leq 1$ and $0\leq y\leq 1$.

Solution We begin by graphing $z=f(x,y)$ and the rectangular region $R$, as shown in Figure 7.

Since $z=f(x,y)$ is continuous and $z\geq 0$ over the rectangular region $-1\leq x\leq 1,$ $0\leq y\leq 1$ , the volume $V$ under the surface $z=f(x,y)$ is given by $$V=\displaystyle\iint\limits_{\kern-3ptR}f(x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}(4-x^{2}-y^{2})\,{\it dA}$$

Using Fubini’s Theorem, we have $$\begin{eqnarray*} V& =&\int_{-1}^{1}\left[ \int_{0}^{1}(4-x^{2}-y^{2})\,{\it dy}\right]\! {\it dx}=\int_{-1}^{1}\left[ 4y-x^{2}y-\dfrac{y^{3}}{3}\right]_{0}^{1}\,{\it dx}\\[4pt] &=&\int_{-1}^{1}\left( 4-x^{2}-\dfrac{1}{3}\right)\!{\it dx}=\int_{-1}^{1}\left( \dfrac{11}{3}-x^{2}\right)\!{\it dx}=\left[ \dfrac{11}{3}x-\dfrac{x^{3}}{3}\right] _{-1}^{1}\\[4pt] &=&\left( \dfrac{11}{3}-\dfrac{1}{3}\right) -\left( -\dfrac{11}{3}+\dfrac{1}{3}\right) =\dfrac{20}{3}\hbox{ cubic units} \end{eqnarray*}$$

$z=f(x,y)=4-x^{2}-y^{2},$ $-1\leq x\leq 1$, $0\leq y\leq 1$