Problem 75 - Thomas Calculus

Question

Archimedes 'formula for the volume of a parabolic solid The region enclosed by the parabola \(y=\left(4 h / b^{2}\right) x^{2}\) and the line \(y=h\) is revolved about the \(y\) -axis to generate the solid shown here. Show that the volume of the solid is 3\(/ 2\) the volume of the corresponding cone. (GRAPH NOT COPY)

Step-by-Step Solution

Answer

The volume of the parabolic solid is \(\frac{3}{2}\) of the corresponding cone's volume.

Step 1: Identify the Parabola and Line

The parabola is given by the equation \(y = \frac{4h}{b^2}x^2\). The line is \(y = h\). The region of interest is the area between this parabola and the line along the y-axis.

Step 2: Set up the Limits for Integration

We need to find the volume of the solid generated by revolving the region around the y-axis. First, determine where the parabola intersects the line. Setting \(y = \frac{4h}{b^2}x^2 = h\), find the x-values: \(x = ±\frac{b}{2}\). Thus, the volume should be integrated with respect to \(y\) from \(0\) to \(h\).

Step 3: Write the Expression for x

Solve for \(x\) in terms of \(y\): \(x^2 = \frac{b^2 y}{4h}\), so \(x = \sqrt{ \frac{b^2 y}{4h} }\). This gives the radius \(r = \sqrt{ \frac{b^2 y}{4h} }\) for the disc method.

Step 4: Setup the Volume Integral

Using the disc method formula \(V = \pi \int r^2 \, dy\), substitute \(r = \sqrt{ \frac{b^2 y}{4h} }\) to get \(V = \pi \int_{0}^{h} \frac{b^2 y}{4h} \, dy\).

Step 5: Integrate to Find the Volume

Integrate: \[V = \pi \int_{0}^{h} \frac{b^2 y}{4h} \, dy = \frac{\pi b^2}{4h} \int_{0}^{h} y \, dy = \frac{\pi b^2}{4h} \left[ \frac{y^2}{2} \right]_{0}^{h} = \frac{\pi b^2}{4h} \times \frac{h^2}{2} \].

Step 6: Simplify the Volume Expression

Simplify the expression: \[V = \frac{\pi b^2 h}{8} \]. This is the volume of the parabolic solid.

Step 7: Calculate the Volume of the Corresponding Cone

The corresponding cone has a radius \(r = \frac{b}{2} \) and height \(h\). The formula for the cone volume is \(V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \pi \left(\frac{b}{2}\right)^2 h = \frac{\pi b^2 h}{12}\).

Step 8: Compare the Volumes

The volume of the solid is \(\frac{\pi b^2 h}{8}\) and the cone is \(\frac{\pi b^2 h}{12}\). Calculate the ratio: \[\frac{\text{Volume of solid}}{\text{Volume of cone}} = \frac{\frac{\pi b^2 h}{8}}{\frac{\pi b^2 h}{12}} = \frac{12}{8} = \frac{3}{2}\].

Key Concepts

Archimedes formula

The concept of Archimedes' formula for volumes is foundational in calculus. Archimedes, the ancient Greek mathematician, was fascinated by calculating volumes of solids, especially those with curved surfaces. He discovered that the volume of a solid of revolution can often be related to simpler geometric shapes. In this exercise, we explore how he compared the volume of a parabolic solid to a cone.

This principle is applied by carefully setting up an integral that allows us to determine the volume of a solid. Here, the solid of revolution is generated by the rotation of a parabolic section about the y-axis. By comparing it to the volume of a cone with the same height and base, Archimedes found a simple ratio between their volumes. It turns out, the parabolic solid has \(\frac{3}{2}\) times the volume of the cone. This insight into geometric relationships laid groundwork for integrals used in calculus today.

Volume of solids

Calculating the volume of solids involves understanding how space is occupied by a three-dimensional object. For regular shapes like spheres or cubes, formulas are straightforward. But for irregular or complex shapes, calculus provides a method through integration. In this example, we are finding the volume of a solid formed by rotating a parabola around an axis.

Volume calculations often involve determining the cross-sectional area at various points along the axis of revolution. By integrating these infinitesimally small areas across the desired length or height, we can find the total volume. The volume of the solid in the exercise is calculated by revolving a bounded plane region around the y-axis, resulting in a complex shape whose volume is not immediately obvious without using integration.

Disc integration method

The disc integration method is a powerful technique in calculus to find the volume of solids of revolution. This method involves slicing the solid into infinitesimally thin discs, whose combined volume gives the total volume of the object. Each disc is perpendicular to the axis of revolution.

In our parabola example, the radius of each disc is determined by the distance from the axis to the curve, which changes along the height of the solid. The formula for the volume using the disc method is \ V = \pi \int r^2 \, dy \, where \(r\) represents the radius of the disc. In this problem, integrating from \(0\) to \(h\) accounts for all the discs making up the solid's volume. This method is especially useful for objects with rotational symmetry, as it interpolates the continuous volume from discrete slices.

Parabola

A parabola is a symmetrical, U-shaped curve that is characterized by its quadratic equation of the form \ y = ax^2 + bx + c \. In this exercise, the parabola is defined with a specific equation \ y = \frac{4h}{b^2}x^2 \, meaning it opens upwards and is scaled by the \(\frac{4h}{b^2}\) factor.

The parabola is bounded by a horizontal line \(y = h\), creating an enclosed region that is then revolved around the y-axis in this exercise. This creates a three-dimensional shape. Understanding a parabola's properties, like its vertex and intersection points, is key in using integrals to find the volumes of solids formed by such revolutions. As part of understanding calculus, the parabola offers a diverse range of applications, including optics, physics, and even architecture.