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 easiest to apply. \(y=(x-2)^{3}-2, x=0,\) and \(y=25 ;\) revolved about the \(y\) -axis

Short answer

Based on the analysis and step-by-step solution above, the volume of the solid generated when the region R, bounded by the curves \(y = (x-2)^3 - 2\), \(x = 0\), and \(y = 25\), is revolved about the y-axis is approximately 25,607.51 cubic units. Both the disk/washer method and the shell method yield the same result, but the shell method was easier to apply in this case due to more straightforward height and radius functions and a less complex integral.

Step-by-step solution

Find the points of intersection of the given curves

To find the points of intersection, we need to solve the equations simultaneously. First, let's find when \(y = (x-2)^3 - 2\) intersects the curve \(y = 25\): \((x-2)^3 - 2 = 25\) Solve for x: \((x-2)^3 = 27 \Rightarrow x-2 = 3\) \(x = 5\) Now let's find when the curve \(y = (x-2)^3 - 2\) intersects the line \(x = 0\): \(y = (0-2)^3 - 2 = -10\) Thus, the points of intersection are \((0, -10)\) and \((5, 25)\).

Set up and evaluate the integral for the disk/washer method

For the disk method, we need to vertically slice the region and consider revolving the small segment around the y-axis. We need an expression for the radius in terms of y. The equation of the curve, in terms of x, is given by \(y = (x-2)^3 - 2\). Solving for x in terms of y: \(x - 2 = \sqrt[3]{y+2} \Rightarrow x = \sqrt[3]{y+2} + 2\) Now, we express the volume of the solid in terms of the integral for the disk method: \(V = \pi \int_{-10}^{25} R^2 dy\) where \(R\) is the distance from the y-axis to the curve, which is given by \(R = \sqrt[3]{y+2} + 2\): \(V = \pi \int_{-10}^{25} (\sqrt[3]{y+2} + 2)^2 dy\) Evaluate the integral: \(V = \pi \int_{-10}^{25} (y^{2/3} + 4y^{1/3} + 4y^{2/3} + 8y^{1/3} + 12 + 4) dy\) \(V = \pi \left[\frac{3}{5}y^{5/3} + \frac{12}{4}y^{4/3} + 8y + 6y^2\right]_{-10}^{25}\) \(V = \pi (5236.9 - (-2889.4))\) \(V \approx 25607.51\)

Set up and evaluate the integral for the shell method

For the shell method, we take horizontal slices and revolve them around the y-axis. The height of the shell is the length of the horizontal segment, which can be expressed as a function in terms of x: \(h(x) = y = (x-2)^3 - 2\) We also need to find the radius \(r(x)\), which is the distance from the x-axis to the function: \(r(x) = x\) Now we set up the integral for the shell method: \(V = 2\pi \int_{0}^{5} r(x)h(x) dx\) \(V = 2\pi \int_{0}^{5} x((x-2)^3 - 2) dx\) Evaluate the integral: \(V = 2\pi \left[\frac{1}{5}x^5 - \frac{5}{2}x^4 + 7x^3 - 6x^2\right]_{0}^{5}\) \(V \approx 25607.51\)

Compare the results and state which method is easiest to apply

Both the disk/washer method and the shell method yield the same volume: \(V \approx 25607.51\). Therefore, the results agree. In this case, the shell method is easier to apply, as finding the height and radius functions was more straightforward than finding the radius function for the disk method. Additionally, the integral for the disk method involved more terms, making the computation more complex.

Key concepts

Disk Method: A Way to Slice Through Volume Calculations

When dealing with the volume of solids of revolution, the disk method is a powerful tool that helps us find the volume by slicing the solid into thin disks or washers if there is a hole in the center. This method is often used when revolving a region around a horizontal or vertical axis. To apply the disk method:

• Identify the axis of rotation. In our case, this is the y-axis.
• Set up the integral by expressing the radius of each disk as a function of y, or in some problems, x, depending on the axis of rotation.
• For the above problem, we've expressed the function in terms of y: \( R = \sqrt[3]{y+2} + 2 \).

We then integrate over the range of y-values from the lowest to the highest point within the bounded region. The volume is calculated as:
\[ V = \pi \int_{-10}^{25} (\sqrt[3]{y+2} + 2)^2 \, dy \] By solving this integral, you find the volume of the solid as it is formed by these infinitesimally thin disks.

Shell Method: A Cylinder in Mind

The shell method is another robust approach to finding the volume of solids of revolution, which involves revolving the region around an axis to form cylindrical "shells."

This method is particularly useful when dealing with regions rotated around a vertical or horizontal line that is not a boundary. Here's how it's done:

• Identify the axis of rotation and decide whether slices of the original shapes should be vertical or horizontal. In this problem, we use vertical slices along the x-axis.
• Characterize the height \( h(x) \) of each shell as the original function, where \( h(x) = (x-2)^3 - 2 \).
• The radial distance from the axis of rotation is simply \( x \), or the x-coordinate in this case.

The volume is calculated by integrating over the x-range of interest, described as:
\[ V = 2\pi \int_{0}^{5} x((x-2)^3 - 2) \, dx \]The thin cylindrical shells combine as they revolve to form the volume of the solid. In this task, the shell method simplified the height and radius calculations, making it a quicker approach for this specific problem.

Volume of Solids of Revolution: Twisting Shapes into Reality

The volume of solids of revolution represents the three-dimensional space occupied by a shape formed by revolving a two-dimensional area around an axis. This concept brings geometry to life by revealing how shapes transform into solids:

To find these volumes, two primary methods—disk/washer and shell—are employed, each with their own application depending on the axis of rotation and the orientation of the functions involved.

It's essential to start by sketching the region and identifying its boundaries. Understanding the areas you will revolve makes choosing your method easier.

• Disk/Washer Method: Best for revolving around axes that are boundaries of the region.
• Shell Method: Efficient for axes that are perpendicular to the axis between boundaries or non-boundary axes.

By correctly setting up the required integral using one of these methods, you can accurately calculate how much space that solid will occupy when formed.

Integration Techniques: The Heart of Calculating Volumes

Integration lies at the heart of finding volumes of solids of revolution. This section focuses on understanding some key integration techniques necessary to master these problems. Integrals help quantify the infinite sums of these sliced or shelled pieces of volume:

• Setup: Correctly setting limits of integration is crucial. It ensures that you cover the complete solid, from start-point to end-point.
• Decomposition: Sometimes breaking down complex integrals into simpler parts can streamline the process.
• Use of Formulas: Use available integral formulas and identities to simplify integrals. In our problem, substitution such as \( y+2 \) or polynomial expansion assists in managing the exponentials.

Successfully evaluating integrals unravels the full volume calculation. Practicing the core techniques helps make computations easier, ensuring varied problems yield consistent and accurate results.