Problem 36 - Calculus and its applications

Question

The volume of a stadium with a domed roof can be approximated if the graph of \(y=-0.00025 x^{2}+130,\) for \(0 \leq x \leq 400,\) where \(x\) and \(y\) are in feet and the \(x\) -axis represents ground level, is rotated around the \(y\) -axis. Find the volume.

Step-by-Step Solution

Answer

The volume is \(19520000\pi\) cubic feet.

Step 1: Identify the Problem Type

The exercise involves finding the volume of a solid of revolution, where the given curve is revolved around the y-axis. This is a calculus problem typically solved using the method of cylindrical shells or disks/washers.

Step 2: Set Up the Integral

Since the curve is revolved around the y-axis, we use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is \( V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \). For our problem, \( f(x) = -0.00025x^2 + 130 \), and the limits of integration are from 0 to 400.

Step 3: Write the Volume Integral

Substitute the function and limits into the cylindrical shell formula: \( V = 2\pi \int_{0}^{400} x(-0.00025x^2 + 130) \, dx \). This will simplify to \( V = 2\pi \int_{0}^{400} (-0.00025x^3 + 130x) \, dx \).

Step 4: Calculate the Integral

First, integrate each term separately: \( \int_{0}^{400} (-0.00025x^3) \, dx = [-0.00025 \cdot \frac{x^4}{4}]_{0}^{400} = -0.000025[x^4]_{0}^{400} \) and \( \int_{0}^{400} 130x \, dx = [130 \cdot \frac{x^2}{2}]_{0}^{400} = 65[x^2]_{0}^{400} \).

Step 5: Evaluate the Integrals

Evaluate the integrals: \([-0.000025[x^4]]_{0}^{400} = -0.000025[(400)^4 - (0)^4] = -640000\) and \([65[x^2]]_{0}^{400} = 65[(400)^2 - (0)^2] = 10400000\).

Step 6: Compute the Final Volume

Substitute the evaluated integrals into the cylindrical shell formula: \( V = 2\pi (-640000 + 10400000) = 2\pi (9760000) \). Therefore, the volume is \( V = 19520000 \pi \) cubic feet.

Key Concepts

Cylindrical Shell Method

The Cylindrical Shell Method is a powerful technique used in calculus to find the volume of solids of revolution. This method is particularly useful when the region is rotated around an axis that is not easily dealt with using the Disk or Washer methods. To visualize the concept, imagine wrapping a section of a function in such a way that when rotated around an axis, it forms a shell.

• The basic idea is to take infinitesimally thin vertical strips of the region between the function and axis of rotation.
• These strips, when rotated, form cylindrical shells.
• The volume of a single shell can be calculated using the formula: \[ V_{shell} = 2\pi imes ext{radius} imes ext{height} imes ext{thickness} \]
• The integral then sums up the volumes of all these shells to get the total volume of the solid.

For the given exercise, the function describes a curve that, when rotated around the y-axis, forms the domed shape of the stadium. By using the shell method, we focus on the 'shells' formed by rotating vertical fragments of the curve around the y-axis.
This approach handles the symmetry around the y-axis very effectively, providing an accurate measure of the solid's volume.

Calculus Integration Techniques

Integration is the core operation used in the Cylindrical Shell Method. Understanding how to integrate complex functions is key to solving these sort of calculus problems. Here's a breakdown of the crucial steps:

• Begin by recognizing the general form of the integral for cylindrical shells: \[ V = 2\pi \int_{a}^{b} x imes f(x) \, dx \]
• The function \( f(x) \) represents the height of the shell, while \( x \) is the average radius from the axis of rotation to the shell.
• Substitute the limits of integration to cover the portion of the x-axis for the problem, which is from 0 to 400 in this example.
• Calculate the integral by breaking it into simpler parts for each term in the integrand. This often involves basic power rule for integration.

A successful integration process requires attention to detail, especially when applying constants and dealing with negative coefficients as part of the function. In our solution, integration converts geometric intuition into exact figures by fitting an integral setup to observed limits and evaluating under them.
This ensures that every small part of the shape contributes correctly to the final volume.

Applications of Calculus

Calculus plays an incredible role in solving real-world geometry problems by providing tools like the Cylindrical Shell Method. This exercise is a classic illustration of how calculus can be applied beyond textbooks:

• Designers and architects frequently use calculus to predict the quantities needed for large structures.
• It's crucial in optimizing material usage and understanding spatial constraints.
• For a structure like a stadium with a domed roof, determining the volume is practical for systems such as air conditioning and estimating crowd capacities.

Moreover, the elegance of calculus lies in its versatility. It doesn't matter whether a shape is simple or complex, calculus can handle it. While other methods might seem more direct, they often fall short when dealing with nuanced, rotationally-symmetric solids. By using the mathematical prowess of calculus, we not only find exact solutions but often discover more about the fundamental properties of shapes and designs we engage with daily.
It highlights the depth and utility of calculus techniques in various fields beyond mathematics, bridging the gap between theoretical concepts and tangible real-world structures.