Question

Volume of a Storage Shed A storage shed has a circular base of diameter 80 feet (see figure). Starting at the center, the interior height is measured every 10 feet and recorded in the table. \begin{tabular}{|l|c|c|c|c|c|} \hline\(x\) & 0 & 10 & 20 & 30 & 40 \\ \hline Height & 50 & 45 & 40 & 20 & 0 \\ \hline \end{tabular} (a) Use Simpson's Rule to approximate the volume of the shed. (b) Note that the roof line consists of two line segments. Find the equations of the line segments and use integration to find the volume of the shed.

Short answer

The volume of the shed, when calculated using Simpson's rule, is approximately 20364.5 cubic feet. Using integration with the equations of the line segments for the roof, the volume is approximately 20944.4 cubic feet.

Step-by-step solution

- Understanding Simpson's Rule

Simpson's Rule is a method of numerical integration used to approximate the integral of a function. It is used when you have a table of values for a function at evenly spaced intervals and want to estimate the area under the curve. The formula for Simpson's rule is: \( V = \frac{\pi}{3}\Delta x(f(x_0) + 4f(x_1) + 2f(x_2) + ... + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)) \) where \( \Delta x \) is the width of each interval (in this case 10 feet), and \( f(x) \) is the function (in this case, the height of the shed) at each point.

- Applying Simpson's Rule

Given the radius values \( x \) and corresponding heights, the formula can be applied to approximate volume: \( V = \frac{\pi}{3} * 10 * (50 + 4*45 + 2*40 + 4*20 + 0) \) which gives approximately 20364.5 cubic feet.

- Modeling the Roof Line

The roof line appears to consist of two line segments. Starting from \( x = 0 \) to \( x = 20 \), the height decreases linearly from 50 ft to 40 ft. Hence, one can model this segment as \( y = -0.5x + 50 \). For \( x \) from 20 to 40, the height decreases from 40 ft to 0. Therefore, the second segment can be modeled as \( y = -2x + 80 \).

- Calculating Volume with Integration

Calculating the volume of the shed in terms of \( x \) requires the integration of the cross-sectional area with respect to \( x \). Thus, the volume is given by \( V = \int_{0}^{40} \pi [ r(x) ]^2 dx \) where \( r(x) \) is the given roof line equation. Splitting the integral according to the two line segments gives \( V = \int_{0}^{20} \pi [ -0.5x + 50 ]^2 dx + \int_{20}^{40} \pi [ -2x + 80 ]^2 dx \). Evaluating these integrals gives approximately 20944.4 cubic feet.

Key concepts

Numerical Integration

Numerical integration is a cornerstone of computational mathematics, allowing us to approximate the area under a curve when finding an exact solution is either impossible or impractical. Techniques, such as Simpson's Rule, trapezoidal rule, and others, come to the rescue when we're faced with irregular shapes or when we rely solely on data points rather than a continuous function.

In the context of the exercise, Simpson's Rule is employed because the storage shed's volume derives from uneven height measurements across its radius. By using the formula
\[V = \frac{\pi}{3}\Delta x(f(x_0) + 4f(x_1) + 2f(x_2) + ... + 4f(x_{n-1}) + f(x_n))\]
we obtain an estimate of the volume by calculating the weighted sum of heights at regular intervals, capturing the curve's shape. Each selected height corresponds to a cross-sectional area of the shed, and the interval width (\( \Delta x \)) signifies the spacing between measurements.

Interestingly, the real power of numerical integration shines when we have limited information. Although its results are approximations, in engineering and the sciences, these 'close-enough' calculations often provide sufficient accuracy for practical applications.

Calculus in Engineering

Calculus may at times seem like an abstract mathematical tool, but it's undoubtedly essential in the field of engineering. Engineers use calculus concepts for a plethora of applications, such as design, optimization, and modeling phenomena in real-world scenarios.

Take our storage shed volume problem as an example of calculus in action. The challenge here involved determining the volume, which necessitates integrating across the shed's radius—essentially adding up an infinite number of infinitesimally small cross-sectional areas.

Modeling with Functions

In our problem, we had to establish the equations of the roof line segments to compute the volume. Such modeling translates real-world shapes into mathematical terms, facilitating the application of calculus to solve practical problems.

After establishing the equations, we integrated those functions over the radius to find the shed's volume, demonstrating calculus' capability to convert geometry and algebra into tangible outcomes in engineering.

Volume Approximation

We encounter various shapes and structures within engineering tasks, and measuring their volume often requires creative approaches. The volume approximation is necessary when dealing with irregular shapes, or when precision takes a back seat to speed and simplicity in the design process.

In our shed example, the volume approximation kicked off with the application of Simpson's Rule, which provided a quick estimate despite the irregular roofline. However, to enhance precision, we moved on to integrate the equations that modeled the roof segments separately, accounting for the distinct slope in each section.

Accuracy vs. Simplicity

The balance between accuracy and simplicity is a recurring theme in volume approximation. While numerical methods offer speed and flexibility, direct integration can incorporate complexity for better accuracy, as demonstrated by comparing the results from Simpson's Rule with those from the integration of the roofline models. This highlights the importance of selecting an appropriate method based on the problem's requirements.