The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:

[latex]\large{A(R)=\displaystyle\int_0^2\displaystyle\int_0^31dx \ dy =\displaystyle\int_0^2\left[x\bigg|_0^3\right]dy = \displaystyle\int_0^23dy=3\displaystyle\int_0^2dy=3y\bigg|_0^2=3(2)=6}.[/latex]

First notice the graph of the surface [latex]z=27-2x^2-y^2[/latex] in Figure 1(a) and above the square region [latex]R_1=[-3,3]\times[-3,3][/latex]. However, we need the volume of the solid bounded by the elliptic paraboloid [latex]2x^2+y^2+z=27[/latex], the planes [latex]x = 3[/latex] and [latex]y = 3[/latex], and the three coordinate planes.

Figure 1. (a) The surface [latex]\small{z=27-2x^{2}-y^{2}}[/latex] above the square region [latex]\small{R_{1}=[-3,3]\times[-3,3]}[/latex]. (b) The solid [latex]\small{S}[/latex] lies under the surface [latex]\small{z=27-2x^{2}-y^{2}}[/latex] above the square region [latex]\small{R_{2}=[0,3]\times[0,3]}[/latex].

Now let’s look at the graph of the surface in Figure 1(b). We determine the volume [latex]V[/latex] by evaluating the double integral over [latex]R_2[/latex]:

[latex]\hspace{5cm}\begin{align} V&=\underset{R}{\displaystyle\iint}z \ dA =\underset{R}{\displaystyle\iint}(27-2x^2-y^2) \ dA \\ &=\displaystyle\int_{y=0}^{y=3}\displaystyle\int_{x=0}^{x=3}(27-2x^2-y^2)dx \ dy &\quad &\text{Convert to iterated integral.} \\ &=\displaystyle\int_{y=0}^{y=3}\left[27x-\frac23x^3-y^2x\right]\Bigg|_{x-0}^{x=3}dy &\quad &\text{Integrate with respect to }x. \\ &=\displaystyle\int_{y=0}^{y=3}(63-3y^2)dy=63y-y^3\bigg|_{y=0}^{y=3}=162. \end{align}[/latex]

Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles [latex](m = 2 \ and\ n = 3)[/latex], as shown in Figure 3. Assume [latex]f(x,y)[/latex] denotes the storm rainfall in inches at a point approximately [latex]x[/latex] miles to the east of the origin and [latex]y[/latex] miles to the north of the origin. Let [latex]R[/latex] represent the entire area of [latex]250 \times 300 = 75,000[/latex] square miles. Then the area of each subrectangle is

[latex]\large{{\Delta{A}} = {\frac{1}{6}} {(75,000)} = {12,500}}[/latex].

Assume [latex]({x^*_{ij}},{y^*_{ij}})[/latex] are approximately the midpoints of each subrectangle [latex]R_{ij}[/latex]. Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:

At [latex](x_{11}, y_{11})[/latex] the rainfall is 0.08.

At [latex](x_{12}, y_{12})[/latex] the rainfall is 0.08.

At [latex](x_{13}, y_{13})[/latex] the rainfall is 0.01.

At [latex](x_{21}, y_{21})[/latex] the rainfall is 1.70.

At [latex](x_{22}, y_{22})[/latex] the rainfall is 1.74.

At [latex](x_{23}, y_{23})[/latex] the rainfall is 3.00.

Figure 3. Storm rainfall with rectangular axes and showing the midpoints of each subrectangle.

According to our definition, the average storm rainfall in the entire area during those two days was

[latex]\begin{align} f_{\text{ave}}&=\frac1{\text{Area }R}\underset{R}{\displaystyle\iint}f(x,y)dxdy=\frac1{75,000}\underset{R}{\displaystyle\iint}f(x,y)dxdy \\ &\cong\frac1{75,000}\sum_{i=1}^3\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\Delta{A} \\ &\cong\frac1{75,00}\left[f(x_{11}^*,y_{11}^*)\Delta{A}+f(x_{12}^*,y_{12}^*)\Delta{A}+f(x_{13}^*,y_{13}^*)\Delta{A}+f(x_{21}^*,y_{21}^*)\Delta{A}+f(x_{22}^*,y_{22}^*)\Delta{A}+f(x_{23}^*,y_{23}^*)\Delta{A}\right] \\ &\cong\frac1{75,000}[0.08+0.08+0.01+1.70+1.74+3.00]\Delta{A} \\ &\cong\frac1{75,00}[0.08+0.08+0.01+1.70+1.74+3.00]12,500 \\ &\cong\frac5{30}[0.08+0.08+0.01+1.70+1.74+3.00] \\ &\cong1.10. \end{align}[/latex]

During September 22–23, 2010 this area had an average storm rainfall of approximately [latex]1.10[/latex] inches.