Problem 41 - Calculus Early Transcendentals

Question

For the following regions \(R\), determine which is greater- the volume of the solid generated when \(R\) is revolved about the x-axis or about the y-axis. \(R\) is bounded by \(y=2 x,\) the \(x\) -axis, and \(x=5\).

Step-by-Step Solution

Answer

Answer: The volume generated when the region is revolved about the y-axis is greater than the volume generated when the region is revolved about the x-axis.

Step 1: Determine the limits of integration for the x-axis and y-axis revolutions

Since the region is bounded by y=2x, the x-axis, and x=5, we need to find the intersection points of the boundaries. We know that the x-axis intersects with the function at x=0 and x=5, and the y-axis intersects at y=0 and y=10 (when x=5). So the limits of integration for the x-axis revolution are x=0 to x=5 and for the y-axis revolution is y=0 to y=10. Step 2: Finding the volume generated when the region is rotated about the x-axis

Step 2: Calculate the volume when the region is revolved about the x-axis

Considering the disk/washer method, the outer radius will be \(y = 2x\) and the inner radius will be 0 (as there's no hole). The volume of each disk is \(\pi \cdot (2x)^2\) and to find the total volume, we will integrate this with respect to x: \(V_x = \pi \int_0^5 (2x)^2 dx\) Step 3: Evaluate the integral for the x-axis revolution

Step 3: Find the volume of the solid when rotated about the x-axis

Now we need to evaluate the integral to get the volume: \(V_x = \pi \int_0^5 (4x^2) dx\) \(V_x = \pi \frac{4x^3}{3}\Big|_0^5\) \(V_x = \pi (\frac{4(5)^3}{3} - \frac{4(0)^3}{3})\) \(V_x = \frac{500\pi}{3}\) Step 4: Finding the volume generated when the region is rotated about the y-axis

Step 4: Calculate the volume when the region is revolved about the y-axis

For the y-axis rotation, we need to rewrite the equation of the function for x in terms of y. The outer radius will be \(x = \frac{y}{2}\) and inner radius is 0. The volume of each disc is \(\pi (\frac{y}{2})^2\) and the total volume is found by integrating this with respect to y: \(V_y = \pi \int_0^{10} (\frac{y}{2})^2 dy\) Step 5: Evaluate the integral for the y-axis revolution

Step 5: Find the volume of the solid when rotated about the y-axis

Now we need to evaluate this integral to get the volume: \(V_y = \pi \int_0^{10} \frac{y^2}{4} dy\) \(V_y = \pi \frac{y^3}{12}\Big|_0^{10}\) \(V_y = \pi (\frac{(10)^3}{12} - \frac{(0)^3}{12})\) \(V_y = \frac{1000\pi}{3}\) Step 6: Compare the volumes

Step 6: Determine which volume is greater

Finally, we need to compare the volumes we obtained from rotating around both axes: \(V_x = \frac{500\pi}{3}\) and \(V_y = \frac{1000\pi}{3}\) Since \(\frac{1000\pi}{3} > \frac{500\pi}{3}\), the volume of the solid generated when the region is revolved about the y-axis is greater than the volume generated when the region is revolved about the x-axis.

Key Concepts

Disk/Washer Method

The disk/washer method is a powerful tool used to calculate the volume of a solid of revolution. Imagine taking a thin slice through the solid, creating a disk. The method involves integrating the volume of these small disks over the relevant interval.
The disk is characterized by:

• Outer Radius: This is the distance from the axis of rotation to the outer edge of the region being revolved. It defines the outer boundary of the solid.
• Inner Radius (if applicable): Represents any hollow part within the disk. If no hole exists, the inner radius is simply zero.

To calculate the volume using this method, you determine the area of a typical disk, which is \(\pi \cdot (\text{outer radius})^2\), then integrate this area across the bounds of the region. This approach applies well when revolving regions around either the x-axis or y-axis.

Limits of Integration

Integration limits define the range over which you calculate the volume by the disk/washer method. These limits are crucial since they define the actual section of the curve being revolved.
For example, when revolving around the x-axis:

• Determine where the curve intersects the x-axis. These intersections typically give the lower and upper bounds for your integral.
• Use similar logic for the y-axis, finding intersections or expressing one variable in terms of another to properly set these limits.

In essence, the limits of integration are derived from the boundary conditions of the region you are analyzing. Correct application of these limits ensures accurate computation of the volume.

Rotating Around Axes

Rotating a region around an axis is a fundamental concept in calculating the volume of revolution. This involves sweeping a 2D region through space around a line (such as the x-axis or y-axis), creating a 3D solid.
Key points to consider are:

• Axis of Rotation: Decide whether the region will be revolved around the x-axis or y-axis. Different results will produce varying volumes based on the same region.
• Reorient the function, if necessary, so that integration is straightforward. For example, if rotating around the y-axis, it may be easier to rewrite \(y=2x\) as \(x=\frac{y}{2}\).

Understanding this process is critical because the orientation of your rotation directly impacts the geometry and therefore the final volume of the solid produced.