<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Calculus Teacher's Edition Go to the latest version.

# 6.2: Volume by Cross Sections

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Calculus, Chapter 5, Lesson 2.

ID: 12280

Time Required: 15 minutes

## Activity Overview

In this activity, students will be introduced to the concept of finding the volume of a solid formed by cross sections of a function that form certain shapes. Since volume is the area of the base times the height and $dV = Area \cdot dx$, students review areas of various shapes like squares, semicircles and equilateral triangles. Calculator screenshots are used to help students get a visual of the volume under consideration. Students will practice what they learn with exam-like questions.

Topic: Volume by Cross Sections

• Applications of integration
• Volume by cross sections

Teacher Preparation and Notes

• Part 1 of this activity takes less than 15 minutes. Part 2 contains three exam-like questions that have accompanying visuals that can be used as an extension or homework.
• Students will write their responses on the accompanying handout where space is provided for students to show work when applicable.

Associated Materials

## Part 1 – Setting Up The Problem And Understanding The Concept

In this section students are introduced to the concept of finding the volume of a solid formed by cross sections of a function that form certain shapes. Since volume is the area of the base times the height and $dV = \text{Area} \ dx$, students review areas of various shapes like squares, semicircles and equilateral triangles.

Part 1 ends with students finding the volume with equilateral triangle cross sections.

Student Solutions

1. $dx$
1. base times height. The area of a square with side $x$ is $x^2$.
2. $\frac{1}{2} \pi r^2$
2. $\frac{1}{2}y \frac{\sqrt{3}}{2}y$
3. $0.433013 \ cm^2$
4. $\int \limits_{0}^{2} \frac{1}{2}y \frac{\sqrt{3}}{2} \ y \ dx & = \int \limits_{0}^{2} \frac{1}{2} \left(\sqrt{x} \cdot e^{-x^2}\right) \frac{\sqrt{3}}{2} \left(\sqrt{x} \cdot e^{-x^2} \right)dx \\& = \int \limits_{0}^{2} \frac{\sqrt{3}}{4}x \cdot e^{-2x^2}dx$

If students use $u-$substitution, $u = -2x^2, du = -4x \ dx$ and the limits of integration are from 0 to -8.

$- \frac{\sqrt{3}}{16} \int \limits_{0}^{-8}e^u du = -\frac{\sqrt{3}}{16}(e^{-8} -1) = \frac{\sqrt{3}}{16}\left(1 - \frac{1}{e^8}\right)$

## Part 2 – Homework

This section enables students to get a visual of challenging exam-like questions. Students should show their work on the first two questions and show their set up on the third question.

Student Solutions

1. $\frac{3 \pi}{32} \ units^3$
2. $2 \ units^3$
3. $1.57 \ units^3$

Feb 23, 2012

Nov 04, 2014