Problem 2 - 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 \(x=0\) and \(x=4 .\) The cross-sections perpendicular to the \(x\) -axis between these planes run from the parabola \(y=-\sqrt{x}\) to the parabola \(y=\sqrt{x}\) . 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. d. The cross-sections are equilateral triangles with bases in the \(x y\) -plane.

Step-by-Step Solution

Answer

a. \(\pi x\), b. \(4x\), c. \(2x\), d. \(\sqrt{3}x\)

Step 1: Determine the length of the diameter

Since the cross-sections run from the parabola \(y=-\sqrt{x}\) to \(y=\sqrt{x}\), the diameter of these cross-sections in the \(xy\)-plane is given by the vertical distance between these curves. Thus, the length of the diameter is \(\sqrt{x} - (-\sqrt{x}) = 2\sqrt{x}\).

Step 2a: Area formula for circular disks

For circular disks, the diameter is \(2\sqrt{x}\). The radius \(r\) is half of the diameter, so \(r = \sqrt{x}\). The area \(A(x)\) of the disk is \(\pi r^2 = \pi (\sqrt{x})^2 = \pi x\).

Step 2b: Area formula for squares with bases

For squares with bases in the \(xy\)-plane, the side length of a square is equal to the diameter \(2\sqrt{x}\). Therefore, the area \(A(x)\) of a square is \((2\sqrt{x})^2 = 4x\).

Step 2c: Area formula for squares with diagonals

For squares with diagonals in the \(xy\)-plane, the diagonal length is \(2\sqrt{x}\). If \(s\) is the side of the square, then the formula for the diagonal \(d\) is \(d = s\sqrt{2}\). Solving for \(s\), we get \(s = \frac{2\sqrt{x}}{\sqrt{2}} = \sqrt{2x}\). The area \(A(x)\) of the square is \(s^2 = (\sqrt{2x})^2 = 2x\).

Step 2d: Area formula for equilateral triangles

For equilateral triangles with bases in the \(xy\)-plane, the base \(b\) is the diameter \(2\sqrt{x}\). The area \(A\) of an equilateral triangle is given by \(A = \frac{\sqrt{3}}{4}b^2\). Thus, \(A(x) = \frac{\sqrt{3}}{4}(2\sqrt{x})^2 = \sqrt{3}x\).

Key Concepts

Circular Disks

The concept of circular disks in this context revolves around using the distance between two curves to form the diameter of a circle. Here, the curves are defined by the parabolas: \( y=-\sqrt{x} \) and \( y=\sqrt{x} \). This means the distance between the parabolas acts as the diameter of the disks.
To find the area of these circular cross-sections, we first determine the diameter, which is the vertical distance: \( 2\sqrt{x} \). From this, we get the radius \( r \) as half of the diameter or \( \sqrt{x} \). With the radius in hand, the area \( A(x) \) of a circular disk is calculated using the formula for the area of a circle:

• Radius \( r = \sqrt{x} \)
• Area \( A(x) = \pi r^2 = \pi (\sqrt{x})^2 = \pi x \)

This result tells us how large the cross-section is at any point \( x \) along the specified range of the solid.

Squares with Bases

Squares with bases in the \( xy \)-plane utilize the same distance between the parabolas \( y=-\sqrt{x} \) and \( y=\sqrt{x} \) to define their sides. Here, the side of each square is equal to the diameter of the circular disk, which is \( 2\sqrt{x} \).
This makes it straightforward to calculate the area \( A(x) \) of such squares, as follows:

• Side length \( s = 2\sqrt{x} \)
• Area \( A(x) = (2\sqrt{x})^2 = 4x \)

By squaring the side length, we get the formula for the area of the square, which grows linearly with \( x \). This area calculation shows how the section size evolves as you move along the \( x \)-axis of the solid.

Equilateral Triangles

In utilizing equilateral triangles, the base is taken as the same diameter \( 2\sqrt{x} \) of our cross-sections. This creates a unique shape for the cross-section because all sides of an equilateral triangle are equal, and its height relates to its base through basic geometry.
The area \( A(x) \) of these triangles is determined using the formula for equilateral triangles:

• Base \( b = 2\sqrt{x} \)
• Area \( A(x) = \frac{\sqrt{3}}{4}b^2 = \frac{\sqrt{3}}{4}(2\sqrt{x})^2 = \sqrt{3}x \)

This formula accounts for the properties of equilateral triangles and illustrates how the area increases with \( x \). The equilateral triangle’s symmetry is a critical feature that makes this calculation straightforward, as the simple relationship between its side and height leads directly to the area formula.

Parabolas

Parabolas play a pivotal role as they outline the boundaries of the cross-sections. Defined by the functions \( y=-\sqrt{x} \) and \( y=\sqrt{x} \), they create a symmetric opening about the \( y \)-axis.
The distance between these parabolas is essential as it determines the linear size of different shapes whether it be a diameter, side, or base. This distance \( 2\sqrt{x} \) directly influences all area calculations for the cross-sections above:

• Diameter for Circular Disks: \( 2\sqrt{x} \)
• Side of Squares with Bases: \( 2\sqrt{x} \)
• Base of Equilateral Triangles: \( 2\sqrt{x} \)

Understanding and employing the geometry of parabolas helps in accurately constructing and calculating the necessary cross-sections. This knowledge is crucial as it ensures that each shape aligns correctly within the solid’s defined space on the \( xy \)-plane.