Problem 1 - Thomas Calculus

Question

In Exercises 1 and \(2,\) find a formula for the area \(A(x)\) of the cross- sections of the solid perpendicular to the \(x\) -axis. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) In each case, the cross-sections perpendicular to the \(x\) -axis between these planes run from the semicircle \(y=-\sqrt{1-x^{2}}\) to the semicircle \(y=\sqrt{1-x^{2}}\) a. The cross-sections are circular disks with diameters in the \(x y\) -plane. b. The cross-sections are squares with bases in the \(x y\) -plane. c. The cross-sections are squares with diagonals in the \(x y\) -plane. (The length of a square's diagonal is \(\sqrt{2}\) times the length of its sides.) d. The cross-sections are equilateral triangles with bases in the \(x y\) -plane.

Step-by-Step Solution

Answer

a. \( \pi (1-x^2) \); b. \(4(1-x^2)\); c. \(2(1-x^2)\); d. \(\sqrt{3}(1-x^2)\).

Step 1: Identify the Boundaries

The solid lies between planes perpendicular to the x-axis at points \(x = -1\) and \(x = 1\). The cross-sections span from \(y = -\sqrt{1-x^2}\) to \(y = \sqrt{1-x^2}\), forming semicircles in the xy-plane.

Step 2: Formula for Diameter of Circular Disks

For part (a), the diameter of the circular disk is the vertical distance between the curves: \(D = \sqrt{1-x^2} - (-\sqrt{1-x^2}) = 2\sqrt{1-x^2}\). The radius \(r\) is thus \(\sqrt{1-x^2}\).

Step 3: Area of Circular Disks

The area of a circular disk, given radius \(r\), is \(A(x) = \pi r^2\). Substitute \(r = \sqrt{1-x^2}\) to find \(A(x) = \pi(\sqrt{1-x^2})^2 = \pi(1-x^2)\).

Step 4: Formula for Side Length of Square with Base

For part (b), the side length of the square equals the diameter, i.e., \(s = 2\sqrt{1-x^2}\).

Step 5: Area of Square with Base

The area of the square is \(A(x) = s^2 = (2\sqrt{1-x^2})^2 = 4(1-x^2)\).

Step 6: Formula for Side Length of Square with Diagonal

For part (c), the diagonal \(d\) equals the diameter: \(d = 2\sqrt{1-x^2}\). The side length \(s\) is then \(\frac{d}{\sqrt{2}} = \frac{2\sqrt{1-x^2}}{\sqrt{2}} = \sqrt{2(1-x^2)}\).

Step 7: Area of Square with Diagonal

The area of this square is \(A(x) = s^2 = (\sqrt{2(1-x^2)})^2 = 2(1-x^2)\).

Step 8: Formula for Side Length of Equilateral Triangle

For part (d), the base of the equilateral triangle is \(b = 2\sqrt{1-x^2}\).

Step 9: Height of Equilateral Triangle

The height \(h\) of an equilateral triangle is \(\frac{\sqrt{3}}{2} \times b\), hence \(h = \sqrt{3}\sqrt{1-x^2}\).

Step 10: Area of Equilateral Triangle

The area of the triangle is \(A(x) = \frac{1}{2} b h = \frac{1}{2} (2\sqrt{1-x^2})(\sqrt{3}\sqrt{1-x^2}) = \sqrt{3}(1-x^2)\).

Key Concepts

Cross-Sectional Area

When studying calculus integration, understanding cross-sectional area is essential. This involves calculating the area of a shape that results from slicing a solid perpendicular to a specific axis, usually the x-axis. For the given exercise, the solid lies between two planes at points where the cross-sections are semicircles in the xy-plane. The cross-sectional area gives crucial information about the solid's geometry and is vital when finding volumes.

• In part (a), the cross-sections are circular disks. The diameter of these disks comes from the semicircles' span, equaling twice the radius of the semicircle, which is determined as the vertical distance between these curves.
• In part (b), with square cross-sections having their bases along the diameter, understanding the area as a function of side length is pivotal. The side length is equivalent to the disk diameter.

Cross-sectional themes come up in diverse calculus problems and are central in understanding complex solids.

Solids of Revolution

Solids of revolution are formed when a two-dimensional shape is rotated around an axis, creating a three-dimensional object. In this exercise, the concept indirectly relates to the way we perceive the symmetry and geometry of the solid formed by cross-sections. The semicircles given by the boundaries form a core part of solids of revolution, typically analyzed by problems revolving around cylinders, spheres, or similar shapes. When integrating to find volume, the cross-sectional area from the axis is pivotal as each infinitesimally thin slice's area adds up. This invention of solids using fundamental shapes showcases how calculus can bridge simple geometric shapes to complex three-dimensional forms, allowing for intensive analysis and computation of physical volumes.

Geometric Shapes

Geometric shapes, such as circles, squares, and triangles, are integral when calculating areas within calculus problems. Each shape has its area formula, essential for solving problems involving integration.

• For a circle, the area is calculated as \(A = \pi r^2\), utilizing the radius derived from the semicircle span.
• For squares, the area formula \(A = s^2\) adapts based on side lengths either given across bases or diagonals.
• Triangles, particularly equilateral ones, depend on height derived from unique geometric properties, importantly using the height formula \(h = \frac{\sqrt{3}}{2} \text{base}\) for accurate area calculations.

Understanding these fundamental formulas enhances one's ability to compute the cross-sectional areas pivotal for volume computations in calculus.

Coordinate Geometry

Coordinate geometry combines algebra and geometry to represent and solve problems using the coordinate plane. In this problem, the semicircles are defined by equations in the xy-plane. These represent the geometric framework needed to determine the cross-sections.Key to coordinate geometry is recognizing how curves like semicircles relate to broader calculus applications like integration. Here, equations such as \(y = \pm\sqrt{1-x^2}\) depict the boundaries that aid in constructing and visualizing the solid of interest. By understanding how these curves span and interact within the coordinate plane, one can devise ways to bridge algebraic equations to geometric shapes, serving as a foundation for later use in integration and analysis.