Problem 123 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) The cross-sections perpendicular to the \(x\) -axis are a. circles whose diameters stretch from the curve \(y=\) \(-1 / \sqrt{1+x^{2}}\) to the curve \(y=1 / \sqrt{1+x^{2}}\). b. vertical squares whose base edges run from the curve \(y=\) \(-1 / \sqrt{1+x^{2}}\) to the curve \(y=1 / \sqrt{1+x^{2}}\).

Step-by-Step Solution

Answer

a) Volume is \(\frac{\pi^2}{2}\); b) Volume is \(2\pi\).

Step 1: Understand the Problem

We have a solid that lies between the planes at \(x = -1\) and \(x = 1\). Its cross-sections, perpendicular to the \(x\)-axis, are circles and squares defined by the curves \(y = -1/\sqrt{1+x^2}\) and \(y = 1/\sqrt{1+x^2}\). We need to find the volume of each solid.

Step 2: Express the Diameter for Circles

For the circular cross-sections (part a), the diameter stretches between the two curves. Since the diameter is the difference between these two curves \(d = \left( \frac{1}{\sqrt{1+x^2}} \right) - \left(-\frac{1}{\sqrt{1+x^2}}\right) = \frac{2}{\sqrt{1+x^2}}\).

Step 3: Calculate the Area of the Circular Cross-Section

The radius \(r\) of each circle is half of the diameter, so \(r = \frac{1}{\sqrt{1+x^2}}\). Thus, the area of the circle is \(A = \pi r^2 = \pi \left(\frac{1}{\sqrt{1+x^2}}\right)^2 = \frac{\pi}{1+x^2}\).

Step 4: Integrate to Find Volume for Circle Cross-Sections

The volume \(V\) of the solid is the integral of the area of the cross-sections from \(-1\) to \(1\). Thus, \(V = \int_{-1}^{1} \frac{\pi}{1+x^2} \, dx\). This integral evaluates to \(\pi [\tan^{-1}(x)]_{-1}^{1} = \pi \left(\tan^{-1}(1) - \tan^{-1}(-1)\right) = \pi \left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \pi \times \frac{\pi}{2} = \frac{\pi^2}{2}\).

Step 5: Express the Side for Squares

For the square cross-sections (part b), the side length is the same as the diameter in part a, \(s = \frac{2}{\sqrt{1+x^2}}\).

Step 6: Calculate the Area of the Square Cross-Section

The area of a square cross-section is \(A = s^2 = \left(\frac{2}{\sqrt{1+x^2}}\right)^2 = \frac{4}{1+x^2}\).

Step 7: Integrate to Find Volume for Square Cross-Sections

The volume \(V\) of the solid with square cross-sections is \(V = \int_{-1}^{1} \frac{4}{1+x^2} \, dx\), which evaluates to \(4 [\tan^{-1}(x)]_{-1}^{1} = 4 \left(\tan^{-1}(1) - \tan^{-1}(-1)\right) = 4 \times \frac{\pi}{2} = 2\pi\).

Key Concepts

Solid geometry

Solid geometry is the branch of mathematics that deals with three-dimensional figures, like cubes, spheres, and pyramids, among others. In this exercise, we are dealing with solids that extend in three dimensions along the x, y, and z axes. Understanding these figures requires us to see how they are structured in space, which involves looking at their cross-sections and how they span between bounded planes.

For instance, in the given problem, we have a solid bounded by planes at different x-values, specifically between x = -1 and x = 1. The shape of the solid comes from its cross-sections, which are perpendicular to the x-axis. These cross-sections can have different shapes, such as circles or squares. The profile or outline of these cross-sections change as x varies, which determines the overall form of the solid.

Solid geometry allows us to calculate various features of these 3D shapes, such as volume and surface area. By understanding how these geometric properties interact and change, we can solve problems that involve integrating over a given volume, just like this exercise challenges us to do.

Integral calculus

Integral calculus is a core area of calculus focused on the accumulation of quantities, such as areas under a curve, volumes, and other quantities that can be understood as the integral of a continuous function. In the context of this problem, integral calculus is used to find the volume of the solid generated by the described cross-sections. The process involves the following steps:

• Expressing the area of individual cross-sections in terms of a function of x.
• Setting up the integral for the volume as the integral of the area function over the defined range for x, which is from -1 to 1 in this case.
• Evaluating the integral to get the volume of the solid.

For the circular cross-section part of the exercise, we find the area of each cross-section as a function of x, and then integrate this area over the interval to find the total volume. The result of this integral gives us the volume enclosed by these cross-sections between x = -1 and x = 1.

Integral calculus thus helps bridge single-dimensional functions into multi-dimensional concepts such as volume. It's important to comprehend how to set up the integral correctly, apply the right limits, and evaluate the integral accurately to solve these types of geometry problems.

Cross-sections

A cross-section of a solid is a two-dimensional shape obtained by slicing through the solid with a plane. If you imagine slicing through a cake, the face you see is the cross-section of that particular slice. In the context of this problem, cross-sections are used to understand and compute the volume of the entire solid.

In this problem, we explore two different types of cross-sections:

• Circular cross-sections where the diameter is defined by the distance between two curves, namely \( y = -1/\sqrt{1+x^2} \) and \( y = 1/\sqrt{1+x^2} \). This implies that the diameter stretches vertically between these curves.
• Square cross-sections where each side of the square matches the diameter of the circular section, thus having the same span vertically between the two curves.

Understanding these cross-sections is crucial for deriving the formulae to calculate the area at different points along the solid. This further allows us to use integral calculus to compute the total volume of the solid. Each cross-section type translates directly into different integrals, representing the aggregated volume that these infinitely thin slices of area contribute to the entire solid.