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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 \begin{align*}dV = Area \cdot dx\end{align*}dV=Areadx, 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 \begin{align*}dV = \text{Area} \ dx\end{align*}dV=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. \begin{align*}dx\end{align*}dx
    1. base times height. The area of a square with side \begin{align*}x\end{align*}x is \begin{align*}x^2\end{align*}x2.
    2. \begin{align*}\frac{1}{2} \pi r^2\end{align*}
  2. \begin{align*}\frac{1}{2}y \frac{\sqrt{3}}{2}y\end{align*}
  3. \begin{align*}0.433013 \ cm^2\end{align*}
  4. \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 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. \begin{align*}\frac{3 \pi}{32} \ units^3\end{align*}
  2. \begin{align*}2 \ units^3\end{align*}
  3. \begin{align*}1.57 \ units^3\end{align*}

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