What Is The Volume Of The Solid Created By Rotating The Region Bounded By Y = (1/4)x^2, X = 0, And Y = 1 Around The X-axis?

In the realm of calculus, determining the volume of a solid formed by revolving a region around an axis is a fundamental concept with applications in various fields. This article delves into the process of calculating such volumes, focusing on a specific example: the region bounded by the equations y = (1/4)x^2, x = 0, and y = 1, revolved around the x-axis. We will explore the method of disks/washers, a powerful technique for tackling these types of problems. By understanding the underlying principles and applying the appropriate formulas, you'll gain the ability to compute the volumes of complex solids with precision.

Understanding the Problem

Before diving into the calculations, let's visualize the problem. We have a parabola defined by y = (1/4)x^2, a vertical line x = 0 (the y-axis), and a horizontal line y = 1. These three boundaries enclose a region in the first quadrant. Imagine taking this region and rotating it 360 degrees around the x-axis. This rotation generates a three-dimensional solid, and our goal is to determine its volume.

To accurately calculate the volume, it's crucial to understand the geometry of the resulting solid. As we revolve the region, each vertical slice perpendicular to the x-axis sweeps out a disk or a washer (a disk with a hole in the center). The method we choose depends on whether the region is directly adjacent to the axis of revolution. In this case, there's a gap between the curve and the x-axis, so we'll be dealing with washers. Each washer has an outer radius determined by the line y = 1 and an inner radius determined by the curve y = (1/4)x^2. The thickness of each washer is an infinitesimal change in x, denoted as dx.

The Disk/Washer Method: A Detailed Explanation

The disk/washer method is a powerful technique rooted in integral calculus for determining the volume of a solid of revolution. It leverages the concept of slicing the solid into infinitesimally thin disks or washers, calculating the volume of each slice, and then summing these volumes using integration. This method is particularly effective when the solid has a consistent cross-sectional shape perpendicular to the axis of revolution.

The Fundamental Principle

The core idea behind the disk/washer method lies in approximating the solid as a stack of thin disks or washers. Each disk or washer has a volume that can be calculated using basic geometric formulas. For a disk, the volume is given by πr^2h, where r is the radius and h is the thickness. For a washer, which has a hole in the center, the volume is given by π(R^2 - r^2)h, where R is the outer radius, r is the inner radius, and h is the thickness. The key to the method is expressing these radii and thickness in terms of the variable of integration, which is typically x or y depending on the orientation of the axis of revolution and the shape of the region.

Setting up the Integral

To calculate the total volume, we need to sum the volumes of all the infinitesimally thin disks or washers. This is where integration comes in. We set up a definite integral where the integrand represents the volume of a single disk or washer, and the limits of integration define the interval over which the solid extends along the axis of revolution. The integral effectively adds up the volumes of all the slices, giving us the total volume of the solid.

Choosing the Right Method: Disks vs. Washers

The choice between using disks or washers depends on the geometry of the region being revolved. If the region is directly adjacent to the axis of revolution, meaning there is no gap between the region and the axis, then we use the disk method. In this case, each slice perpendicular to the axis forms a solid disk. However, if there is a gap between the region and the axis of revolution, we use the washer method. In this scenario, each slice forms a washer, which is essentially a disk with a hole in the center.

Applying the Method to Our Problem

In our specific problem, we are revolving the region bounded by y = (1/4)x^2, x = 0, and y = 1 around the x-axis. As we discussed earlier, there is a gap between the curve and the x-axis, so we will use the washer method. The outer radius of each washer is determined by the line y = 1, and the inner radius is determined by the curve y = (1/4)x^2. The thickness of each washer is dx, as we are integrating with respect to x. This understanding is crucial for setting up the correct integral and accurately computing the volume.

Setting up the Integral for Our Problem

Now that we understand the disk/washer method, let's apply it to our specific problem. We're revolving the region bounded by y = (1/4)x^2, x = 0, and y = 1 about the x-axis. As established, we'll use the washer method. Here's how we set up the integral:

1. Identify the Outer and Inner Radii:

   • Outer radius, R(x): This is the distance from the x-axis to the line y = 1, so R(x) = 1.
   • Inner radius, r(x): This is the distance from the x-axis to the curve y = (1/4)x^2, so r(x) = (1/4)x^2.

2. Determine the Limits of Integration:

To find the limits of integration, we need to determine the x-values where the region begins and ends along the x-axis. The region starts at x = 0 (the y-axis). To find the upper limit, we need to find the x-value where the parabola y = (1/4)x^2 intersects the line y = 1. Setting these equations equal to each other:

(1/4)x^2 = 1
x^2 = 4
x = ±2

Since we're in the first quadrant, we take the positive solution, so the upper limit of integration is x = 2.

3. Write the Integral:

The formula for the volume of a solid of revolution using the washer method is:

V = ∫[a, b] π(R(x)^2 - r(x)^2) dx

Where:

• V is the volume
• a and b are the lower and upper limits of integration
• R(x) is the outer radius
• r(x) is the inner radius

Plugging in our values, we get:

V = ∫[0, 2] π(1^2 - ((1/4)x²⁾2) dx

V = ∫[0, 2] π(1 - (1/16)x^4) dx

This integral represents the volume of the solid. The next step is to evaluate this integral to find the numerical value of the volume. The correct integral accurately represents the problem statement, which is key to solving it.

Evaluating the Integral

Having set up the integral, our next step is to evaluate it to determine the volume of the solid. The integral we derived is:

V = ∫[0, 2] π(1 - (1/16)x^4) dx

Let's break down the evaluation step by step:

1. Apply the Constant Multiple Rule:

   We can pull the constant π out of the integral: V = π ∫[0, 2] (1 - (1/16)x^4) dx

2. Integrate Term by Term:

   We can integrate each term inside the parentheses separately: V = π [∫[0, 2] 1 dx - ∫[0, 2] (1/16)x^4 dx]

3. Apply the Power Rule for Integration:

The power rule states that ∫x^n dx = (x^(n+1))/(n+1) + C. Applying this rule:

V = π [[x]_0^2 - (1/16) * (x^5/5)_02]

4. Evaluate the Definite Integrals:

   Substitute the upper and lower limits of integration and subtract: V = π [(2 - 0) - (1/16) * (2^5/5 - 0^5/5)]

5. Simplify:

   V = π [2 - (1/16) * (32/5)] V = π [2 - 2/5] V = π [10/5 - 2/5] V = π [8/5] V = (8/5)π

Therefore, the volume of the solid formed by revolving the region bounded by y = (1/4)x^2, x = 0, and y = 1 about the x-axis is (8/5)π cubic units. The accurate evaluation of the integral is crucial for arriving at the correct answer.

Conclusion

In this article, we have meticulously walked through the process of computing the volume of a solid formed by revolving a region around the x-axis. We began by visualizing the problem and understanding the geometry of the solid. We then delved into the disk/washer method, a fundamental technique in calculus for calculating volumes of revolution. We carefully set up the integral, identifying the outer and inner radii and the limits of integration. Finally, we evaluated the integral to arrive at the solution: the volume of the solid is (8/5)π cubic units.

This example highlights the power of calculus in solving geometric problems. By understanding the underlying principles and applying the appropriate techniques, we can accurately determine the volumes of complex solids. The application of calculus provides a powerful toolset for solving real-world problems involving shapes and volumes.

Keywords: calculating the volume, compute the volume, disk/washer method, accurate evaluation, application of calculus, correct integral.