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Volumes with Cross Sections

Sarah Miller

Sarah Miller

5 min read

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Study Guide Overview

This guide covers calculating the volume of solids with semicircular cross sections using integration. It explains the core concept of integrating the cross-sectional area function, A(x), and the formula for the area of a semicircle. A worked example demonstrates finding A(x) from a given base region, setting up the definite integral, and evaluating it to determine the volume. Practice questions and a glossary of terms like radius, diameter, and integral are also included.

Table of Contents

  1. Introduction
  2. Volume of Solids with Semicircular Cross Sections
  3. Worked Example
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

Introduction

In this guide, we will explore how to find the volume of a solid with semicircular cross sections. This involves integrating the area of the cross-section along the axis of the solid.

Volume of Solids with Semicircular Cross Sections

Basic Concept

To find the volume of a solid with a given cross-sectional area:

  • If the area of the cross section of a solid is given by A(x)A(x) and A(x)A(x) is continuous on [a,b][a, b], then the volume of the corresponding solid from x=ax = a to x=bx = b is: Volume=abA(x),dx\mathrm{Volume} = \int_{a}^{b} A(x) , dx

Creating the Cross-Sectional Area Function

You may need to create the cross-sectional area function A(x)A(x) based on the information provided in the problem. Often, A(x)A(x) may depend on the values of another function (or functions) given in the question.

Area of a Semicircle

The area of a circle is π×radius2\pi \times \text{radius}^2. Therefore, the area of a semicircle with radius rr is: Area=12×πr2\mathrm{Area} = \frac{1}{2} \times \pi r^2

Worked Example

Let RR be the triangular region with vertices (0,0)(0, 0), (0,2)(0, 2), and (4,0)(4, 0). RR is the base of a solid. For the solid, at each xx, the cross-section perpendicular to the xx-axis is a semicircle. Find the volume of the solid.

Solution Steps

  1. Find the Equation of the Line:

    • The line passes through points (0,2)(0, 2) and (4,0)(4, 0).
    • Slope (gradient) is: gradient=0240=12\text{gradient} = \frac{0-2}{4-0} = -\frac{1}{2}
    • Equation of the line: y0=12(x4)    y=212xy - 0 = -\frac{1}{2}(x - 4) \implies y = 2 - \frac{1}{2}x
  2. Determine the Radius:

    • The diameter of each semicircle is 2 - \frac{1}{2}x.
    • Therefore, the radius is: radius=212x2=114x\mathrm{radius} = \frac{2 - \frac{1}{2}x}{2} = 1 - \frac{1}{4}x
  3. Find the Cross-Sectional Area A(x)A(x):

    • Area of a semicircle: A(x)=12π(114x)2A(x) = \frac{1}{2} \pi \left(1 - \frac{1}{4}x\right)^2
    • Simplify: A(x)=12π(112x+116x2)=π32(168x+x2)A(x) = \frac{1}{2} \pi \left(1 - \frac{1}{2}x + \frac{1}{16}x^2\right) = \frac{\pi}{32} \left(16 - 8x + x^2\right)
  4. Calculate the Volume:

    • Integrate A(x)A(x) from 0 to 4: Volume=04π32(168x+x2),dx\mathrm{Volume} = \int_{0}^{4} \frac{\pi}{32} \left(16 - 8x + x^2\right) , dx
    • Evaluate the integral: Volume=π32[16x4x2+13x3]04\mathrm{Volume} = \frac{\pi}{32} \left[ 16x - 4x^2 + \frac{1}{3}x^3 \right]_{0}^{4}
    • Substitute and calculate: Volume=π32(6464+643)=π32643=2π3\mathrm{Volume} = \frac{\pi}{32} \left( 64 - 64 + \frac{64}{3} \right) = \frac{\pi}{32} \cdot \frac{64}{3} = \frac{2\pi}{3}
    • Numerical value: Volume2.094,units3\mathrm{Volume} \approx 2.094 , \text{units}^3
The question doesn't specify units, so the units of volume will be units3\text{units}^3.

Practice Questions

Practice Question
  1. Find the volume of a solid with a base region defined by y=xy = \sqrt{x}, from x=0x=0 to x=4x=4, with cross sections perpendicular to the xx-axis that are semicircles.
Practice Question
  1. The base of a solid is a rectangle with length 6 and width 2. Cross sections perpendicular to the length are semicircles. Calculate the volume of the solid.
Practice Question
  1. A solid has a base defined by the region y=x2y = x^2 and y=4y = 4. Cross sections perpendicular to the yy-axis are semicircles. Determine the volume of the solid.

Glossary

  • Cross-Sectional Area (A(x)A(x)): The area of the shape obtained by cutting through a solid perpendicular to a given axis at position xx.
  • Integral: A mathematical operation that calculates the area under a curve.
  • Semicircle: A half of a circle.
  • Radius: The distance from the center of a circle to any point on its circumference.
  • Diameter: The distance across a circle through its center, equal to twice the radius.

Summary and Key Takeaways

Summary

  • To find the volume of a solid with semicircular cross sections, use the integral of the area function A(x)A(x).
  • The area of a semicircle is 12πr2\frac{1}{2} \pi r^2.
  • The integral is evaluated over the range of xx values that define the solid.

Key Takeaways

  • Understand the formula: Volume=abA(x),dx\mathrm{Volume} = \int_{a}^{b} A(x) , dx
  • Identify the radius: Based on the geometry of the problem.
  • Set up the integral: Use the correct limits and area function.
  • Evaluate the integral: Carefully substitute and simplify to find the volume.
Exam Tip

Always check your limits of integration and ensure your area function A(x)A(x) is correctly defined.