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Volumes with Cross Sections

Sarah Miller

Sarah Miller

5 min read

Next Topic - Semicircles as Cross Sections

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Study Guide Overview

This study guide covers calculating volumes of solids with triangular cross-sections using integral calculus. It explains the key concept of integrating the cross-sectional area function A(x). The guide provides the formula for the area of a triangle, a worked example with equilateral triangle cross-sections, practice questions, and exam strategies. Key terms include cross-section, integral, and equilateral triangle.

#Study Notes: Volumes with Cross Sections as Triangles

#Table of Contents

  1. Introduction
  2. Basic Concept
  3. Creating the Cross-Sectional Area Function
  4. Area of a Triangle
  5. Worked Example
  6. Practice Questions
  7. Glossary
  8. Summary and Key Takeaways
  9. Exam Strategy

#Introduction

In this section, we will learn how to find the volume of a solid with a triangular cross-section. This method leverages integral calculus to compute volumes accurately.

#Basic Concept

Key Concept

If the area of the cross-section of a solid is given by A(x)A(x)A(x), and A(x)A(x)A(x) is continuous on [a,b][a, b][a,b], then the volume of the solid from x=ax = ax=a to x=bx = bx=b is given by:

Volume=∫abA(x),dx\mathrm{Volume} = \int_{a}^{b} A(x) , dxVolume=∫ab​A(x),dx

#Creating the Cross-Sectional Area Function

To find the volume, you may need to create the cross-sectional area function A(x)A(x)A(x) based on the information provided in the problem. This function might depend on the values of another function (or functions) given to you in the question.

Exam Tip

Always carefully read the problem to understand how the cross-sectional area is defined and how it relates to the given functions.

#Area of a Triangle

Key Concept

The area of a triangle is calculated as:

Area=12×base×height\mathrm{Area} = \frac{1}{2} \times \mathrm{base} \times \mathrm{height}Area=21​×base×height

#Worked Example

Let's apply these concepts to a specific problem:

**Problem:** Let RRR be the region enclosed by the graph of f(x)=4−xf(x) = \sqrt{4 - x}f(x)=4−x​ and the xxx- and yyy-axes. RRR is the base of a solid. For the solid, at each xxx, the cross-section perpendicular to the xxx-axis is an equilateral triangle. Find the volume of the solid.

Solution:

  1. Determine the cross-sectional area function A(x)A(x)A(x):

    Since the cross-section is an equilateral triangle, we need to find the height of the triangle. For an equilateral triangle with side length sss, the height hhh is given by:

    h=32sh = \frac{\sqrt{3}}{2} sh=23​​s

    In this case, the side length s=f(x)=4−xs = f(x) = \sqrt{4 - x}s=f(x)=4−x​, so:

    h=324−xh = \frac{\sqrt{3}}{2} \sqrt{4 - x}h=23​​4−x​

  2. Calculate the area of the equilateral triangle:

    A(x)=12×base×heightA(x) = \frac{1}{2} \times \text{base} \times \text{height}A(x)=21​×base×height A(x)=12×4−x×324−xA(x) = \frac{1}{2} \times \sqrt{4 - x} \times \frac{\sqrt{3}}{2} \sqrt{4 - x}A(x)=21​×4−x​×23​​4−x​ A(x)=34(4−x)A(x) = \frac{\sqrt{3}}{4} (4 - x)A(x)=43​​(4−x)

  3. Set up the volume integral:

    Volume=∫0434(4−x),dx\mathrm{Volume} = \int_{0}^{4} \frac{\sqrt{3}}{4} (4 - x) , dxVolume=∫04​43​​(4−x),dx

  4. Evaluate the integral:

    Volume=34∫04(4−x),dx\mathrm{Volume} = \frac{\sqrt{3}}{4} \int_{0}^{4} (4 - x) , dxVolume=43​​∫04​(4−x),dx =34[4x−12x2]04= \frac{\sqrt{3}}{4} \left[ 4x - \frac{1}{2} x^2 \right]_{0}^{4}=43​​[4x−21​x2]04​ =34((4⋅4−12⋅42)−0)= \frac{\sqrt{3}}{4} \left( (4 \cdot 4 - \frac{1}{2} \cdot 4^2) - 0 \right)=43​​((4⋅4−21​⋅42)−0) =34(16−8)= \frac{\sqrt{3}}{4} (16 - 8)=43​​(16−8) =34⋅8= \frac{\sqrt{3}}{4} \cdot 8=43​​⋅8 =23≈3.464,units3= 2\sqrt{3} \approx 3.464 , \text{units}^3=23​≈3.464,units3

Thus, the volume of the solid is approximately 3.464 units³ (to 3 decimal places).

#Practice Questions

Practice Question
  1. Let RRR be the region enclosed by the graph of g(x)=9−x2g(x) = \sqrt{9 - x^2}g(x)=9−x2​ and the xxx-axis. RRR is the base of a solid with cross-sections perpendicular to the xxx-axis being equilateral triangles. Find the volume of the solid.

  2. The region bounded by h(x)=x2−4x+4h(x) = x^2 - 4x + 4h(x)=x2−4x+4 and the xxx-axis is the base of a solid. If the cross-sections perpendicular to the xxx-axis are isosceles right triangles with the hypotenuse lying along the base, find the volume of the solid.

#Glossary

  • Cross-section: A surface or shape exposed by making a straight cut through something.
  • Integral: A mathematical concept that represents the area under a curve.
  • Equilateral Triangle: A triangle in which all three sides are equal in length.
  • Continuous Function: A function without breaks, jumps, or holes in its domain.

#Summary and Key Takeaways

  • The volume of a solid with a known cross-sectional area can be calculated using the integral of the area function.
  • For triangular cross-sections, use the formula for the area of a triangle.
  • Always set up the integral correctly by identifying the limits of integration and the area function.

#Exam Strategy

Exam Tip
  • Clearly identify the cross-sectional shape and derive its area function.
  • Double-check the limits of integration.
  • Simplify the integrand before integrating if possible.
  • Practice solving integrals to improve speed and accuracy.

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Question 1 of 10

What is the general formula to find the volume of a solid using cross-sectional areas?

Volume=A(x)∗dxVolume = A(x) * dxVolume=A(x)∗dx

Volume=∫A(x)dxVolume = \int A(x) dxVolume=∫A(x)dx

Volume=∫abA(x),dxVolume = \int_{a}^{b} A(x) , dxVolume=∫ab​A(x),dx

Volume=12∗base∗heightVolume = \frac{1}{2} * base * heightVolume=21​∗base∗height