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

Creating the Cross-Sectional Area Function

To find the volume, you may need to create the cross-sectional area function $A(x)$ 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.

Area of a Triangle

Worked Example

Let's apply these concepts to a specific problem:

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

Solution:

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

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

   $$h = \frac{\sqrt{3}}{2} s$$

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

   $$h = \frac{\sqrt{3}}{2} \sqrt{4 - x}$$

2. Calculate the area of the equilateral triangle:

   $$A(x) = \frac{1}{2} \times \text{base} \times \text{height}$$ $$A(x) = \frac{1}{2} \times \sqrt{4 - x} \times \frac{\sqrt{3}}{2} \sqrt{4 - x}$$ $$A(x) = \frac{\sqrt{3}}{4} (4 - x)$$

3. Set up the volume integral:

   $$\mathrm{Volume} = \int_{0}^{4} \frac{\sqrt{3}}{4} (4 - x) , dx$$

4. Evaluate the integral:

   $$\mathrm{Volume} = \frac{\sqrt{3}}{4} \int_{0}^{4} (4 - x) , dx$$ $$= \frac{\sqrt{3}}{4} \left[ 4x - \frac{1}{2} x^2 \right]_{0}^{4}$$ $$= \frac{\sqrt{3}}{4} \left( (4 \cdot 4 - \frac{1}{2} \cdot 4^2) - 0 \right)$$ $$= \frac{\sqrt{3}}{4} (16 - 8)$$ $$= \frac{\sqrt{3}}{4} \cdot 8$$ $$= 2\sqrt{3} \approx 3.464 , \text{units}^3$$

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

Practice Questions

Practice Question

1. Let $R$ be the region enclosed by the graph of $g(x) = \sqrt{9 - x^2}$ and the $x$-axis. $R$ is the base of a solid with cross-sections perpendicular to the $x$-axis being equilateral triangles. Find the volume of the solid.

2. The region bounded by $h(x) = x^2 - 4x + 4$ and the $x$-axis is the base of a solid. If the cross-sections perpendicular to the $x$-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