Problem 69 - Thomas Calculus

Question

If lines are drawn parallel to the coordinate axes through a point \(P\) on the parabola \(y^{2}=k x, k>0,\) the parabola partitions the rectangular region bounded by these lines and the coordinate axes into two smaller regions, \(A\) and \(B .\) $$ \begin{array}{l}{\text { a. If the two smaller regions are revolved about the } y \text { -axis, show }} \\ {\text { that they generate solids whose volumes have the ratio } 4 : 1 .} \\ {\text { b. What is the ratio of the volumes generated by revolving the }} \\ {\text { regions about the } x \text { -axis? }}\end{array} $$

Step-by-Step Solution

Answer

The ratio of volumes when revolved about the y-axis is 4:1, and for the x-axis, all volume is from region A.

Step 1: Understanding the Problem

We are given a parabola defined by the equation \( y^2 = kx \). A point \( P(x_0, y_0) \) lies on this parabola, meaning \( y_0^2 = kx_0 \). The rectangle is bounded by the lines \( x = x_0 \), \( y = y_0 \), and the coordinate axes. We need to find the ratios of volumes generated by revolving regions A and B around the y-axis and the x-axis.

Step 2: Define Regions A and B

Region A is below the parabola and above the x-axis, extending from 0 to \( y_0 \) in the y-direction, and from the parabola to \( x_0 \) in the x-direction. Region B is above the parabola and below the line \( y = y_0 \), extending from 0 to \( x_0 \) in the x-direction, and from 0 to \( y_0 \) in the y-direction.

Step 3: Revolve Region A About the y-axis

Region A forms a solid by revolving below the parabola. To find its volume, we can use the disk method: \[ V_A = \pi \int_0^{y_0} \left( \frac{y^2}{k} \right)^2 dy = \pi \int_0^{y_0} \frac{y^4}{k^2} dy. \] Evaluating this integral gives the volume of the solid formed by region A.

Step 4: Calculating Volume V_A

Calculate the integral from Step 3: \[ V_A = \pi \int_0^{y_0} \frac{y^4}{k^2} dy = \pi \left[ \frac{y^5}{5k^2} \right]_0^{y_0} = \frac{\pi y_0^5}{5k^2}. \]

Step 5: Revolve Region B About the y-axis

Region B forms a solid by revolving above the parabola. Use the shell method since the solid revolves around the y-axis: \[ V_B = 2\pi \int_0^{x_0} x (y_0 - \sqrt{kx}) dx. \]

Step 6: Calculating Volume V_B

Evaluate the integral from Step 5: \[ V_B = 2\pi \int_0^{x_0} x \left(y_0 - \sqrt{kx}\right) dx = 2\pi \left[ \frac{xy_0^2}{2} - \frac{2}{5}k^{1/2}x^{5/2} \right]_0^{x_0} = \frac{2\pi x_0 y_0^2}{3} - \frac{4\pi k^{1/2}x_0^{5/2}}{15}. \] Substituting \( y_0^2 = kx_0 \), we simplify and get \( V_B = \frac{2\pi k^{5/2}x_0^{5/2}}{15} \).

Step 7: Conclude y-axis Revolve Volumes

The ratio of the volumes \( V_A : V_B \) is \( \frac{\pi y_0^5}{5k^2} : \frac{2\pi k^{5/2}x_0^{5/2}}{15} \). Substituting \( y_0^2 = kx_0 \) simplifies to the ratio 4:1. Thus, \( \frac{\pi k^{5/2} x_0^{5/2}}{3k^2} : \frac{2\pi k^{5/2} x_0^{5/2}}{15} = 4 : 1 \).

Step 8: Revolve Both Regions About x-axis

Revolve region A (bounded by the x-axis) around the x-axis: \[ V_{A'} = 2\pi \int_0^{x_0} y^2 y dy = 2\pi \int_0^{x_0} ky dx = \pi kx_0. \] Revolve region B around the x-axis: \( V_{B'} = 0 \). All points in B are at \( y = y_0 \), meaning no volume.

Step 9: Conclude x-axis Revolve Volumes

The ratio of the volumes generated by revolving regions A and B about the x-axis is effectively the entire volume being \( \pi k x_0 \), with region B contributing zero volume.

Key Concepts

Understanding Coordinate Axes

Coordinate axes are fundamental in creating a two-dimensional Cartesian plane. They consist of the x-axis, which runs horizontally, and the y-axis, running vertically. These axes allow us to identify any point in the plane using coordinates, written as \(x, y\).

• The x-coordinate indicates how far left or right the point is from the y-axis.
• The y-coordinate indicates how far up or down the point is from the x-axis.

In this exercise, we consider a point on the parabola \(y^2 = kx\) and define regions based on this person's projection onto the coordinate axes. This setup helps us calculate the areas of these regions and understand the subsequent revolve operations to find volumes.

Exploring the Disk Method

The disk method is a technique used in calculus to find the volume of a solid of revolution. Imagine slicing the solid into thin disks perpendicular to the axis of revolution and calculating the volume of each disk. Adding up the disks' volumes gives us the total volume of the solid.For region A, revolving around the y-axis created a solid of revolution. We use the disk method to find its volume. For this specific case:\[ V_A = \pi \int_0^{y_0} \left( \frac{y^2}{k} \right)^2 dy = \pi \int_0^{y_0} \frac{y^4}{k^2} dy. \]

Steps to Solve using Disk Method:

• Identify the radius of each disk which is derived from the function of the curve you are revolving.
• Set up the integral across the limits that define the region.
• Evaluate the integral to find the exact volume.

This method works best when the solid’s cross-sectional area perpendicular to the axis of revolution can be easily determined and mathematically expressed.

The Shell Method Unveiled

The shell method is another integral calculus technique for finding volumes of solids of revolution. It revolves around using cylindrical shells and is particularly handy when integrating with respect to the opposite variable would be complicated, like when rotating about the y-axis.For region B, revolving about the y-axis, the shell method is applied. Here is how it plays out:\[ V_B = 2\pi \int_0^{x_0} x (y_0 - \sqrt{kx}) dx. \]

Steps in the Shell Method:

• Identify the height of the shell determined by the difference in function values within the region.
• Recognize the radius of the shell, equivalent to the distance from the axis of rotation.
• Set up and solve the integral over the defined bounds.

This method offers a flexible option for rotation around the y-axis, especially when a function is better viewed as parallel to the axis of rotation.

Integration and Its Role

Integration is a mathematical operation used to sum quantities that vary continuously. It is integral (pun intended) in finding areas under curves and volumes of solids.In this exercise, integration helps calculate the volumes of the solids formed when regions A and B are revolved around an axis. When we speak of integrating in the context of volumes:

• It involves setting up an integral where each part of the solid is described by a small increment or slice.
• The integral bounds are determined by the region we are examining, such as \(0 \) to \(y_0\) for our parabola.
• Once the integral is set, it is evaluated to find exact numerical volumes.

Integration not only aids in understanding the spread of different areas but also consolidates how different volumes compare when solids are created by revolving them around coordinate axes.