Problem 41
Question
Let \(R\) be the region bounded by the following curves. Let \(S\) be the solid generated when \(R\) is revolved about the given axis. If possible, find the volume of \(S\) by both the disk/washer and shell methods. Check that your results agree and state which method is easier to apply. \(y=x, y=x^{1 / 3}\) in the first quadrant; revolved about the \(x\) -axis
Step-by-Step Solution
Verified Answer
Question: Determine the volume of the solid generated when the region enclosed by the graphs of \(y=x\) and \(y=x^{1/3}\) is revolved about the x-axis.
Answer: The volume of the solid generated is \(\frac{3}{5}\pi\).
1Step 1: Find points of intersection
To find the points of intersection between \(y=x\) and \(y=x^{1/3}\), we can set them equal to each other and solve for x.
\(x = x^{1/3}\)
\(x^3 = x\)
\(x^3 - x = 0\)
\(x(x^2 - 1) = 0\)
\(x=0, x=1\)
So the points of intersection are \((0, 0)\) and \((1, 1)\).
2Step 2: Set up disk/washer method integral
To use the disk/washer method, we must slice the volume perpendicular to the axis of rotation, which is the x-axis in this case. We have two functions for \(y\), so we can subtract the smaller one from the larger one to find the height of each disk:
\(y_1 = x\)
\(y_2 = x^{1/3}\)
Height of each disk: \(y_1 - y_2 = x - x^{1/3}\)
Now, we can set up an integral for the disk/washer method. Since we're revolving about the x-axis, we need to integrate with respect to x. The volume of the solid can be found by integrating the area of each disk from \(x=0\) to \(x=1\):
\(V_{disk} = \pi \int_{0}^{1} (x - x^{1/3})^2dx\)
3Step 3: Set up shell method integral
To use the shell method, we must slice the volume parallel to the axis of rotation, which is the x-axis in this case. In order to use the shell method, we need to express x in terms of y for both functions:
\(x_1 = y\)
\(x_2 = y^3\)
The height of each cylindrical shell is the distance between them: \(x_1 - x_2 = y - y^3\)
Now, we can set up an integral for the shell method. Since we're revolving about the x-axis, we need to integrate with respect to y. The volume of the solid can be found by integrating the circumference times the height of each shell from \(y=0\) to \(y=1\):
\(V_{shell} = 2\pi \int_{0}^{1} y(y - y^3)dy\)
4Step 4: Evaluate and compare both integrals
Now we can evaluate both integrals and compare the results.
For the disk/washer method:
\(V_{disk} = \pi \int_{0}^{1} (x - x^{1/3})^2dx = \frac{3}{5}\pi\)
For the shell method:
\(V_{shell} = 2\pi \int_{0}^{1} y(y - y^3)dy = \frac{3}{5}\pi\)
Both methods give the same result: \(\frac{3}{5}\pi\).
5Step 5: Determine the easier method
Both methods produce the correct volume of the solid, but the shell method's integral is slightly easier to evaluate due to the simpler integrand (\(y(y - y^3)\)) compared to the square in the integrand of the disk/washer method (\((x - x^{1/3})^2\)). However, the difference is not significant in this case, and both methods can be applied easily.
Key Concepts
Disk/Washer MethodShell MethodVolume IntegrationSolid of Revolution Calculus
Disk/Washer Method
The disk/washer method is a technique used in calculus to determine the volume of a solid of revolution. When a region in the plane is revolved around an axis, disks or washers are formed, and the volume of each slice is approximated by these shapes. For example, in the exercise provided, to calculate the volume when the region is revolved about the x-axis, we consider a typical disk with a small thickness \(\text{dx}\). The outer radius of the disk is given by the larger function \(\text{y}_1\) and the inner radius by the smaller function \(\text{y}_2\), if present. To find the volume of the entire solid, we integrate these areas along the axis of revolution from the boundaries of the region.
The volume \(\text{V}_{disk}\) is therefore given by the integral \(\text{V}_{disk} = \pi \int_{a}^{b} (\text{y}_1 - \text{y}_2)^2 \text{dx}\), where \(\text{a}\) and \(\text{b}\) are the points of intersection. In our exercise, after setting up the appropriate integral, the volume by the disk/washer method is calculated as \(\text{V}_{disk} = \pi \int_{0}^{1} (x - x^{1/3})^2 \text{dx}\). The nature of the resulting integrand, which is a square of a binomial, can sometimes make this method more challenging.
The volume \(\text{V}_{disk}\) is therefore given by the integral \(\text{V}_{disk} = \pi \int_{a}^{b} (\text{y}_1 - \text{y}_2)^2 \text{dx}\), where \(\text{a}\) and \(\text{b}\) are the points of intersection. In our exercise, after setting up the appropriate integral, the volume by the disk/washer method is calculated as \(\text{V}_{disk} = \pi \int_{0}^{1} (x - x^{1/3})^2 \text{dx}\). The nature of the resulting integrand, which is a square of a binomial, can sometimes make this method more challenging.
Shell Method
Contrastingly, the shell method computes the volume by adding up the volumes of cylindrical shells. This approach is particularly useful when the solid is revolved around the y-axis, or when the functions are more naturally integrated with respect to \(\text{y}\).
For our example, where the region is still revolved about the x-axis, we first express \(\text{x}\) as a function of \(\text{y}\) to create shells parallel to the y-axis. The height of each shell is determined by the difference in the x values of the two functions \(\text{x}_1 - \text{x}_2\), and the radius is simply \(\text{y}\). This gives us the formula for the volume of a single cylindrical shell: \(\text{V}_{shell} = 2 \pi \int_{c}^{d} \text{y}(\text{x}_1 - \text{x}_2) \text{dy}\).
When we apply this method to the exercise, we find that \(\text{V}_{shell} = 2 \pi \int_{0}^{1} \text{y}(\text{y} - \text{y}^3) \text{dy}\), which is generally easier to compute due to the simpler expression for the shell's volume. The choice between the disk/washer and shell method often comes down to which makes the integration simpler.
For our example, where the region is still revolved about the x-axis, we first express \(\text{x}\) as a function of \(\text{y}\) to create shells parallel to the y-axis. The height of each shell is determined by the difference in the x values of the two functions \(\text{x}_1 - \text{x}_2\), and the radius is simply \(\text{y}\). This gives us the formula for the volume of a single cylindrical shell: \(\text{V}_{shell} = 2 \pi \int_{c}^{d} \text{y}(\text{x}_1 - \text{x}_2) \text{dy}\).
When we apply this method to the exercise, we find that \(\text{V}_{shell} = 2 \pi \int_{0}^{1} \text{y}(\text{y} - \text{y}^3) \text{dy}\), which is generally easier to compute due to the simpler expression for the shell's volume. The choice between the disk/washer and shell method often comes down to which makes the integration simpler.
Volume Integration
Volume integration is the cornerstone of finding volumes for solids of revolution. This process involves setting up an integral that represents the sum of infinitesimally small volume elements, which can be disks, washers, or shells, depending on the method used.
The integral bounds are determined by the limits of the region being revolved, and the integrand reflects the area of a cross-sectional slice of the solid. Whether to integrate with respect to \(\text{x}\) or \(\text{y}\) is chosen based on the axis of rotation and the form of the functions involved. When the integral is evaluated, the result gives the volume of the entire solid.
The integral bounds are determined by the limits of the region being revolved, and the integrand reflects the area of a cross-sectional slice of the solid. Whether to integrate with respect to \(\text{x}\) or \(\text{y}\) is chosen based on the axis of rotation and the form of the functions involved. When the integral is evaluated, the result gives the volume of the entire solid.
Solid of Revolution Calculus
Solid of revolution calculus is the field of study concerning the calculation of volumes of shapes generated by rotating a two-dimensional region about an axis. This concept is vital in understanding real-world objects that have circular symmetry, like bowls, vases, or wheels.
Through the use of integration techniques, we can find exact volumes of complex shapes without the need for physical measurement. The methods to calculate these volumes—disk/washer and shell—require an understanding of how a three-dimensional object can be represented as an accumulation of two-dimensional slices. This concept not only serves as a practical application of integral calculus but also deepens the understanding of how calculus can be used to solve tangible engineering and physics problems.
Through the use of integration techniques, we can find exact volumes of complex shapes without the need for physical measurement. The methods to calculate these volumes—disk/washer and shell—require an understanding of how a three-dimensional object can be represented as an accumulation of two-dimensional slices. This concept not only serves as a practical application of integral calculus but also deepens the understanding of how calculus can be used to solve tangible engineering and physics problems.
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