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Chapter 6: Problem 41

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 \(x=5\).

Short Answer

Expert verified
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 by step solution

01

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
02

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
03

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
04

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
05

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
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x.\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \frac{2 \pi d}{\lambda}} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}.\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

A simple model (with different parameters for different people) for the flow of air in and out of the lungs is $$V^{\prime}(t)=-\frac{\pi}{2} \sin \frac{\pi t}{2}$$ where \(V(t)\) (measured in liters) is the volume of air in the lungs at time \(t \geq 0, t\) is measured in seconds, and \(t=0\) corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs in exchanged with each breath. The amount that is exchanged is called the tidal volume. a. Find and graph the volume function \(V\) assuming that $$ V(0)=6 \mathrm{L} $$ b. What is the breathing rate in breaths/min? c. What is the tidal volume and what is the total capacity of the lungs?

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _____. $$v(t)=2 \sin t, \text { for } 0 \leq t \leq \pi$$

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{x}, y=0, x=0,\) and \(x=2\) revolved about the \(x\) -axis

Sketch a solid of revolution whose volume by the disk method is given by the following integrals. Indicate the function that generates the solid. Solutions are not unique. a. \(\int_{0}^{\pi} \pi \sin ^{2} x d x\) b. \(\int_{0}^{2} \pi\left(x^{2}+2 x+1\right) d x\)
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