Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

The base of a solid is the region inside the circle \(x^{2}+y^{2}=4\). Find the volume of the solid if every cross section by a plane perpendicular to the \(x\)-axis is a square. Hint: See Examples 5 and 6 .

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{128}{3}\) cubic units.

Step 1: Identify the Shape of the Base

The base of the solid is the region inside the circle described by the equation \(x^2 + y^2 = 4\). This is a circle centered at the origin with radius 2.

Step 2: Determine the Cross-Section Area Formula

Each cross section of the solid perpendicular to the \(x\)-axis is a square. The side length of each square is determined by the diameter of the circle at the particular \(x\)-coordinate, with \(y = \sqrt{4 - x^2}\). Therefore, the side length of the square is \(2y = 2\sqrt{4 - x^2}\). The area of the square cross section is \((2\sqrt{4 - x^2})^2 = 4(4 - x^2)\).

Step 3: Set Up the Integral for Volume

The volume of the solid can be found by integrating the area of the square cross sections along the \(x\)-axis. The bounds for \(x\) are from \(-2\) to \(2\). The integral to find the volume is:\[ V = \int_{-2}^{2} [4(4 - x^2)] \, dx \].

Step 4: Simplify and Integrate

Simplify the integrand:\[ 4(4 - x^2) = 16 - 4x^2 \]. Now integrate:\[ V = \int_{-2}^{2} (16 - 4x^2) \, dx \]. Split the integral:\[ V = \int_{-2}^{2} 16 \, dx - \int_{-2}^{2} 4x^2 \, dx \].

Step 5: Evaluate the Integrals

First, evaluate \(\int_{-2}^{2} 16 \, dx\):\[ = 16[x]_{-2}^{2} = 16(2 - (-2)) = 16 \times 4 = 64 \]. Now evaluate \(\int_{-2}^{2} 4x^2 \, dx\):\[ = 4[x^3/3]_{-2}^{2} = \frac{4}{3}(8 - (-8)) = \frac{4}{3} \times 16 = \frac{64}{3} \].

Step 6: Calculate the Final Volume

Subtract the results of the integrals to find the volume:\[ V = 64 - \frac{64}{3} = \frac{192}{3} - \frac{64}{3} = \frac{128}{3} \]. The volume of the solid is \(\frac{128}{3}\) cubic units.

Key Concepts

Cross-Sectional Area

In geometry, the cross-sectional area of a solid is crucial for understanding its volume. Specifically, for a solid with a specific base and shape, we look at the slices or sections of the solid perpendicular to a given axis. In this exercise's case, each cross-section is perpendicular to the x-axis and forms a square. The formula for finding the area of these sections comes from the specific geometry of the base.The base of the solid in this exercise is a circle, determined by the equation \(x^2 + y^2 = 4\). This tells us the radius of the circle is 2. At any point along the x-axis within this circle, the maximum y-coordinate or height of the cross-sectional square is \(y = \sqrt{4 - x^2}\). This means the side length of each square is \(2y = 2\sqrt{4 - x^2}\). Consequently, the area of each square cross section is \((2\sqrt{4 - x^2})^2 = 4(4 - x^2)\). By understanding how to determine this cross-sectional area, we set a foundation for calculating the solid's volume.

Circle Geometry

Circle geometry is a cornerstone of understanding this exercise. Here, the base of the solid is represented by a circular region. The circle is given with the equation \(x^2 + y^2 = 4\), which confirms it is centered at the origin \((0,0)\) with a radius of 2.Understanding a circle's properties is essential to solving the problem:

• The radius is constant from the center to any point on the circle.
• The diameter, double the radius, affects the square cross-sections' side length.
• At any x-coordinate, the height or y-coordinate (from the center) is \(y = \sqrt{4-x^2}\).

This geometric insight allows us to translate the circle's properties into a formula for the side length of each square cross section, grounded in its radius and symmetrical properties.

Integral Calculus

Integral calculus is used here to find the volume of the solid by summing up all the infinitesimally small cross-sectional areas along the x-axis. To compute the volume effectively, you need to understand the integration process.Once we've established that each cross-section is a square with an area of \(4(4 - x^2)\), integrating this over the valid range for x, from -2 to 2, allows us to find the total volume. Here’s how it works:- Set up the integral: \(\int_{-2}^{2} 4(4 - x^2) \, dx\).- Simplify to \(\int_{-2}^{2} (16 - 4x^2) \, dx\).
Knowing how to integrate such expressions step by step is much like calculating the area under a curve.This problem demonstrates how the routine of integrating translates cross-sectional geometry into a complete volumetric answer. Evaluating the definite integrals gives us specific numerical results that tell the complete story of the solid's volume.