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*}

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

Example 5 by finding \(\int_{0}^{1}\left[ \int_{-1}^{1}(4-x^{2}-y^{2})\,{\it dx}\right] \! {\it dy}.\)

Problem 61.