Volumes with Cross Sections: Squares and Rectangles - AP Calculus AB Study Guide

The Core Idea: Volumes with Cross Sections: Squares and Rectangles

The fundamental concept behind this topic is the construction of a three-dimensional solid from a two-dimensional base region. Imagine a flat region defined on the coordinate plane. We can build a solid on top of this base by stacking an infinite number of thin, two-dimensional shapes, called cross sections, perpendicular to an axis. The shape of these cross sections (in this topic, squares or rectangles) is known.

Calculus allows us to find the exact volume of this resulting solid by "summing up" the volumes of all these infinitesimally thin cross-sectional slices. The definite integral is the tool for this summation. We first find a function, $A(x)$ or $A(y)$, that gives the area of a single cross section at any point along the axis. Then, by integrating this area function over the interval of the base region, we accumulate these areas to find the total volume.

Key Formulas

The calculation of volume with known cross sections relies on two primary integral formulas and the geometric area formulas for squares and rectangles.

Volume Formulas:

1. For a solid with a known cross-sectional area $A(x)$ perpendicular to the x-axis, from $x=a$ to $x=b$, the volume $V$ is:

   $$V={\int }_a^{b}A(x)dx$$

2. For a solid with a known cross-sectional area $A(y)$ perpendicular to the y-axis, from $y=c$ to $y=d$, the volume $V$ is:

   $$V={\int }_c^{d}A(y)dy$$

Area Formulas:

1. Square: The area $A$ of a square with side length $s$ is:

   $$A=s^2$$

2. Rectangle: The area $A$ of a rectangle with base $b$ and height $h$ is:

   $$A=b\cdot h$$

Understanding the Orientation of Cross Sections

The most critical nuance in these problems is determining whether to integrate with respect to $x$ or $y$. This choice is dictated entirely by the orientation of the cross sections.

• Cross Sections Perpendicular to the x-axis:

  • The slices are vertical.

  • The side length (or base) of the cross section, let's call it $s(x)$, is a vertical distance. It is typically found by subtracting the bottom function from the top function: $s(x)=y_{top}-y_{bottom}$.

  • The area function, $A(x)$, must be expressed entirely in terms of $x$.

  • The integral will be with respect to $x$ ($dx$), and the bounds of integration, $a$ and $b$, will be x-values.

• Cross Sections Perpendicular to the y-axis:

  • The slices are horizontal.

  • The side length (or base) of the cross section, $s(y)$, is a horizontal distance. It is typically found by subtracting the left function from the right function: $s(y)=x_{right}-x_{left}$.

  • To do this, you must solve the original equations for $x$ in terms of $y$.

  • The area function, $A(y)$, must be expressed entirely in terms of $y$.

  • The integral will be with respect to $y$ ($dy$), and the bounds of integration, $c$ and $d$, will be y-values.

Core Concepts & Rules

• The volume of a solid with a known cross section is found by integrating the area of that cross section.

• The variable of integration is determined by the axis to which the cross sections are perpendicular.

• If cross sections are perpendicular to the x-axis, you must set up an integral with respect to $x$. The area function $A(x)$ and the bounds of integration must be in terms of $x$.

• If cross sections are perpendicular to the y-axis, you must set up an integral with respect to $y$. The area function $A(y)$ and the bounds of integration must be in terms of $y$.

• The side length $s$ of the cross section is the distance across the base region, measured in a direction perpendicular to the axis of integration.

• For square cross sections, the area is $A=s^2$.

• For rectangular cross sections, the area is $A=b\cdot h$. The problem must provide a relationship between the base $b$ and the height $h$ (e.g., "the height is half the length of the base").

Step-by-Step Example 1: Squares Perpendicular to the x-axis

Problem: Find the volume of the solid whose base is the region in the first quadrant bounded by the graph of $y=x^2$, the x-axis, and the line $x=2$. The cross sections of the solid perpendicular to the x-axis are squares.

Step 1: Sketch the Base Region and Identify Bounds

Sketch the parabola $y=x^2$, the x-axis ($y=0$), and the vertical line $x=2$. The region is bounded on the left by $x=0$ and on the right by $x=2$. Therefore, our bounds of integration are $a=0$ and $b=2$.

Step 2: Determine the Side Length of a Cross Section

The cross sections are perpendicular to the x-axis, so their side length, $s(x)$, is a vertical distance. At any given $x$ in the interval $[0,2]$, the top of the region is $y_{top}=x^2$ and the bottom is $y_{bottom}=0$.

$$s(x)=y_{top}-y_{bottom}=x^2-0=x^2$$

Step 3: Write the Area Formula in Terms of $x$

The cross sections are squares, so the area formula is $A=s^2$. Substituting our expression for $s(x)$:

$$A(x)=(s(x){)}^2=(x^2{)}^2=x^4$$

Step 4: Set Up and Evaluate the Definite Integral

Using the volume formula $V={\int }_a^{b}A(x)dx$:

$$V={\int }_0^2{x}^{4}dx$$

Now, find the antiderivative and evaluate using the Fundamental Theorem of Calculus.

$$V={\left[\frac{1}{5}{x}^5\right]}_0^2$$

$$V=\left(\frac{1}{5}(2{)}^5\right)-\left(\frac{1}{5}(0{)}^5\right)$$

$$V=\frac{32}{5}-0=\frac{32}{5}$$

The volume of the solid is $\frac{32}{5}$ cubic units.

Step-by-Step Example 2: Rectangles Perpendicular to the y-axis

Problem: Let R be the region bounded by the graph of $y=\ln (x)$, the y-axis, and the lines $y=0$ and $y=2$. For the solid with base R, each cross section perpendicular to the y-axis is a rectangle whose height is twice its base. Find the volume of the solid.

Step 1: Sketch the Base Region and Identify Bounds

Sketch the graph of $y=\ln (x)$, the y-axis ($x=0$), and the horizontal lines $y=0$ and $y=2$. The region is bounded below by $y=0$ and above by $y=2$. These are our bounds of integration: $c=0$ and $d=2$.

Step 2: Determine the Base Length of a Cross Section

The cross sections are perpendicular to the y-axis, so their base length, $b(y)$, is a horizontal distance. We need to express the boundary curves in terms of $y$.

The right boundary is $y=\ln (x)⟹e^y=e^{\ln (x)}⟹x=e^y$.

The left boundary is the y-axis, which is $x=0$.

The base length is $b(y)=x_{right}-x_{left}=e^y-0=e^y$.

Step 3: Write the Area Formula in Terms of $y$

The cross sections are rectangles whose height is twice the base.

Base: $b(y)=e^y$

Height: $h(y)=2\cdot b(y)=2e^y$

The area of the rectangle is $A(y)=b(y)\cdot h(y)=(e^y)(2e^y)=2e^{2y}$.

Step 4: Set Up and Evaluate the Definite Integral

Using the volume formula $V={\int }_c^{d}A(y)dy$:

$$V={\int }_0^{2}2e^{2y}dy$$

Find the antiderivative (using u-substitution if necessary, where $u=2y$).

$$V={\left[2\cdot \frac{1}{2}{e}^{2y}\right]}_0^2={\left[e^{2y}\right]}_0^2$$

$$V=(e^{2\cdot 2})-(e^{2\cdot 0})=e^4-e^0=e^4-1$$

The volume of the solid is $e^4-1$ cubic units.

Using Your Calculator

While setting up the integral is a purely analytical skill, a graphing calculator is an essential tool for evaluating the definite integral once it is correctly formulated, especially on calculator-active portions of the AP Exam.

The primary function to use is fnInt (or its equivalent).

To solve a cross-section problem with a calculator:

1. Analyze the Problem: Determine if cross sections are perpendicular to the x-axis or y-axis.

2. Find the Side/Base Length: Determine the expression for $s(x)$ ($top-bottom$) or $s(y)$ ($right-left$). Remember to solve for $x$ in terms of $y$ if necessary.

3. Find the Area Function: Construct $A(x)$ or $A(y)$ using the given shape (e.g., $(s(x){)}^2$ for a square).

4. Use fnInt to Evaluate:

   • Press MATH and select 9: fnInt(.

   • The syntax is fnInt(function, variable, lower_bound, upper_bound).

   • Important: Even if your integral is with respect to $y$, you must use $X$ as the variable in the calculator.

Example: To evaluate $V={\int }_0^{2}2e^{2y}dy$ from Example 2:

• Your calculator entry would be: fnInt(2e^(2X), X, 0, 2)

• The calculator will return a decimal approximation, approximately $53.598$.

The calculator is for evaluating the integral you set up. It cannot determine the bounds, the side length, or the area formula for you.

AP Exam Quick Hit

Common Question Types

• Standard Volume Calculation: You are given two functions that bound a region R (e.g., $f(x)=\sqrt{x}$ and $g(x)=x^2$). You are asked to find the volume of a solid with base R and specified cross sections (e.g., squares, rectangles) perpendicular to the x-axis or y-axis.

• "Set Up, Do Not Evaluate": A multiple-choice or free-response question may ask you to identify the correct integral that represents the volume, but not to calculate the final answer. This purely tests your ability to find the bounds, the side length, and the area function correctly.

• Volume with a Non-Standard Rectangular Relationship: The cross sections are rectangles where the height is defined in terms of the base (e.g., "the height is half the base" or "the height is a constant value of 5"). This requires you to correctly model $A(x)=b(x)\cdot h(x)$.

Common Mistakes

• Integrating Side Length, Not Area: The most common error is to integrate the side length $s(x)$ instead of the area $A(x)=(s(x){)}^2$. Forgetting to square the side length for a square cross section leads to an incorrect answer.

• Incorrect Variable of Integration: Using $dx$ when cross sections are perpendicular to the y-axis, or vice-versa. This often happens when a student finds a horizontal side length $s(y)$ but then integrates with respect to $x$ using x-bounds.

• Incorrect Bounds: Using x-intercepts as bounds for a $dy$ integral, or using y-intercepts as bounds for a $dx$ integral. The bounds must match the variable of integration.

• Parentheses Error: When the side length is a difference of two functions, such as $s(x)=f(x)-g(x)$, students often forget to square the entire quantity. For example, writing $(f(x){)}^2-(g(x){)}^2$ instead of the correct $(f(x)-g(x){)}^2$.

• Rectangle Area Error: For a rectangle where the height is $k$ times the base ($h=kb$), writing the area as $A=k\cdot b$ instead of the correct $A=b\cdot h=b\cdot (kb)=k{b}^2$.