Problem 88 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

ARCHITECTURAL DESIGN A building is to have a curved roof above a rectangular base. In relation to a rectangular grid, the base is the rectangular region \(-30 \leq x \leq 30,-20 \leq y \leq 20\), where \(x\) and \(y\) are measured in meters. The height of the roof above each point \((x, y)\) in the base is given by $$ h(x, y)=12-0.003 x^{2}-0.005 y^{2} $$ a. Find the volume of the building. b. Find the average height of the roof.

Step-by-Step Solution

Answer

The volume of the building is approximately 21280.2 cubic meters. The average height of the roof is approximately 8.8676 meters.

Step 1: Set Up the Integral for the Volume

To find the volume of the building, the volume under the surface defined by the roof function must be calculated over the given rectangular base. This requires setting up a double integral over the region \tilde{R}:\[ V = \int\int_{R} h(x, y) \, dx \, dy \]where \({-30 \leq x \leq 30}, {-20 \leq y \leq 20}\)

Step 2: Integrate the Height Function

Integrate the height function \(h(x, y) = 12 - 0.003x^2 - 0.005y^2\) with respect to \(y\) first:\[ \int_{-20}^{20} \left (12 - 0.003x^2 - 0.005y^2 \right ) dy \]This will yield:\[ \int_{-20}^{20} 12 \, dy - \int_{-20}^{20} 0.003x^2 \, dy - \int_{-20}^{20} 0.005y^2 \, dy \]

Step 3: Compute the Integrals

Evaluate the integrals one by one:1) \[ \int_{-20}^{20} 12 \, dy = 12y \bigg|_{-20}^{20} = 12(20 - (-20)) = 12 \cdot 40 = 480 \]2) \[ \int_{-20}^{20} 0.003x^2 \, dy = 0.003x^2 y \bigg|_{-20}^{20} = 0.003x^2 (20 - (-20)) = 0.003x^2 \cdot 40 = 0.12 x^2 \]3) \[ \int_{-20}^{20} 0.005y^2 \, dy = 0.005 \int_{-20}^{20} y^2 \, dy = 0.005 \cdot \frac{2}{3} y^3 \bigg|_{-20}^{20} = 0.005 \cdot \frac{2}{3} (20^3 - (-20)^3) \]Note: \[20^3 = 8000, (-20)^3 = -8000 \]Therefore,\[ 0.005 \cdot \frac{2}{3} \cdot (8000 - (-8000)) = 0.005 \cdot \frac{2}{3} \cdot 16000 = 0.005 \cdot 10666.67 = 53.33 \]

Step 4: Combine the Results

Combine the results of the integrals to get:\[ \int_{-20}^{20} \left (12 - 0.003x^2 - 0.005y^2 \right ) dy = 480 - 0.12x^2 - 53.33 \]Thus, the function to integrate with respect to \(x\) is:\[ \int_{-30}^{30} (480 - 0.12 x^2 - 53.33) dx \]

Step 5: Integrate with Respect to x

Evaluate the integral with respect to \(x\):\[ \int_{-30}^{30} (426.67 - 0.12 x^2) dx \]Separating, we get:\[ \int_{-30}^{30} 426.67 \, dx - \int_{-30}^{30} 0.12 x^2 \, dx \]These integrals are:1) \[ \int_{-30}^{30} 426.67 \, dx = 426.67x \bigg|_{-30}^{30} = 426.67(30 - (-30)) = 426.67 \cdot 60 = 25600.2 \]2) \[ \int_{-30}^{30} 0.12 x^2 \, dx = 0.12 \int_{-30}^{30} x^2 \, dx = 0.12 \cdot \frac{2}{3} x^3 \bigg|_{-30}^{30} = 0.12 \cdot \frac{2}{3} \cdot (30^3 - (-30)^3) \]\[ \int_{-30}^{30} x^2 \, dx = \frac{2}{3} \cdot (30^3 - (-30)^3) = \frac{2}{3} \cdot 54000 = 36000 \]Therefore,\[ 0.12 \cdot 36000 = 4320 \]So, the volume integral evaluates to:\[ V = 25600.2 - 4320 = 21280.2 \, \text{cubic meters} \]

Step 6: Calculate the Average Height

To find the average height of the roof, use the formula for the average value of a function over a region \(R\):\[ \text{Average height} = \frac{1}{\text{Area of R}} \int\int_{R} h(x, y) \ dx \ dy \]The area of the rectangular base \(R\) is given by\tilde{A} = (60 \times 40) = 2400 \, \text{square meters}So,\[ \text{Average height} = \frac{21280.2}{2400} = 8.8676 \, \text{meters} \]

Key Concepts

Double Integrals

A double integral allows us to calculate the volume under a surface in a multivariable calculus context. We use it here to find the volume under the roof of a building defined by a function over a specific region. For rectangular regions, which are defined by constant boundaries, computing double integrals simplifies, as the sides are just lines.

Volume Calculation

To determine the volume, we integrate the height function over the given base region. The building's roof height is given as a function: \( h(x, y) = 12 - 0.003x^2 - 0.005y^2 \). The integration involves setting up and evaluating an integral in two variables, first with respect to one variable and then the other. By integrating between the specified limits, from \( x = -30 \) to \( x = 30 \) and \( y = -20 \) to \( y = 20 \), we find the entire volume under the surface.

Average Value of a Function

Calculating the average value of a function over a region helps find the mean height of the roof. The formula is: \( \text{Average value} = \frac{1}{\text{Area of } R} \int \int_{R} h(x, y) \, dx \, dy \). This gives a single value that represents the average height over the entire base area. For our rectangular base with dimensions 60 by 40 meters, the area is 2400 square meters. We use this area in the formula with our previously calculated volume.

Rectangular Region

A rectangular region simplifies setting up double integrals because its boundaries are constants. In our example, the region is defined by: \( -30 \leq x \leq 30 \) and \( -20 \leq y \leq 20 \). With these boundaries, the integration limits are straightforward, making it easier to execute the calculations for volume.

Integration Steps

Step-by-step integration involves breaking down the process:

• First, set up the double integral over the region.
• Second, integrate the height function with respect to \( y \), treating \( x \) as a constant.
• Third, evaluate the resulting integrals.
• Fourth, you'll combine the results and then integrate with respect to \( x \).
• Lastly, compute the final volume and average height. Each step must be performed meticulously to ensure accuracy. By substituting the values and solving integrals methodically, we arrive at correct and meaningful results.