6.2: Volume by Cross Sections
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 \begin{align*}dV = Area \cdot dx\end{align*}
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 examlike 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
 Student Worksheet: Volume by CrossSection http://www.ck12.org/flexr/chapter/9730, scroll down to the second activity.
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 \begin{align*}dV = \text{Area} \ dx\end{align*}
Part 1 ends with students finding the volume with equilateral triangle cross sections.
Student Solutions

\begin{align*}dx\end{align*}
dx  base times height. The area of a square with side \begin{align*}x\end{align*}
x is \begin{align*}x^2\end{align*}x2 .  \begin{align*}\frac{1}{2} \pi r^2\end{align*}
 base times height. The area of a square with side \begin{align*}x\end{align*}
 \begin{align*}\frac{1}{2}y \frac{\sqrt{3}}{2}y\end{align*}
 \begin{align*}0.433013 \ cm^2\end{align*}
 \begin{align*}\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\end{align*}
If students use \begin{align*}u\end{align*}substitution, \begin{align*}u = 2x^2, du = 4x \ dx\end{align*} and the limits of integration are from 0 to 8.
\begin{align*} \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)\end{align*}
Part 2 – Homework
This section enables students to get a visual of challenging examlike questions. Students should show their work on the first two questions and show their set up on the third question.
Student Solutions
 \begin{align*}\frac{3 \pi}{32} \ units^3\end{align*}
 \begin{align*}2 \ units^3\end{align*}
 \begin{align*}1.57 \ units^3\end{align*}
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