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

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

The fundamental concept of this topic is to calculate the volume of a three-dimensional solid by considering it as a collection of infinitely thin two-dimensional slices, known as cross sections. We can find the total volume by integrating the area of a representative cross section over a specific interval.

This method involves defining a region in the xy-plane that serves as the base of the solid. We then imagine slicing this solid with planes perpendicular to either the x-axis or the y-axis. Each slice has a specific shape—in this topic, a square or a rectangle—whose area can be expressed as a function, $A(x)$ or $A(y)$. The definite integral of this area function between the boundaries of the solid sums the volumes of these infinite slices, yielding the total volume of the solid.

Key Formulas

The calculation of volume using cross sections relies on a single fundamental principle expressed in two forms, depending on the orientation of the cross sections. The area of the cross section itself depends on the specified geometric shape.

Volume Formulas:

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

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

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

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

Cross-Sectional Area Formulas:

1. Square: If the cross section is a square with side length $s$, its area $A$ is:

   $$A=s^2$$

2. Rectangle: If the cross section is a rectangle with base $b$ and height $h$, its area $A$ is:

   $$A=bh$$

Understanding the Axis of Integration

The most critical step in setting up a volume integral is determining whether to integrate with respect to $x$ or $y$. This choice is dictated by the orientation of the cross sections.

• Perpendicular to the x-axis: If the problem states that the cross sections are perpendicular to the x-axis, you must integrate with respect to $x$. This means:

  • The limits of integration, $a$ and $b$, will be x-values.

  • The dimensions of the cross section (the side $s$ of a square, or the base $b$ and height $h$ of a rectangle) must be expressed as functions of $x$. The base of the cross section typically corresponds to a vertical distance in the xy-plane, calculated as $(topcurve)-(bottomcurve)$.

• Perpendicular to the y-axis: If the problem states that the cross sections are perpendicular to the y-axis, you must integrate with respect to $y$. This means:

  • The limits of integration, $c$ and $d$, will be y-values.

  • The dimensions of the cross section must be expressed as functions of $y$. This often requires solving the original boundary equations for $x$ in terms of $y$. The base of the cross section typically corresponds to a horizontal distance in the xy-plane, calculated as $(rightcurve)-(leftcurve)$.

The choice of $dx$ or $dy$ determines how you define the geometry of the area function $A$.

Core Concepts & Rules

• The volume of a solid can be found by integrating its cross-sectional area function.

• The variable of integration is determined by the axis to which the cross sections are perpendicular. Perpendicular to the x-axis implies integration with respect to $x$; perpendicular to the y-axis implies integration with respect to $y$.

• The area of a square cross section is $A=s^2$, where $s$ is the side length.

• The area of a rectangular cross section is $A=bh$, where $b$ is the base and $h$ is the height.

• The dimensions of the cross section ($s$, $b$, $h$) must be expressed as functions of the chosen variable of integration ($x$ or $y$).

• The length of the base of a cross section is determined by the distance between the boundary curves of the solid's base, measured parallel to the cross-sectional face.

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

Problem: Let R be the region bounded by the graph of $f(x)=2\sqrt{x}$, the x-axis, and the line $x=9$. The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of the solid.

Step 1: Identify the Axis and Variable of Integration

The problem states that the cross sections are perpendicular to the x-axis. Therefore, we will integrate with respect to $x$, and our volume formula is $V={\int }_a^{b}A(x)dx$.

Step 2: Determine the Limits of Integration

The base region is bounded by the x-axis ($y=0$) and $x=9$. The function $f(x)=2\sqrt{x}$ intersects the x-axis at $x=0$. Thus, the solid extends from $x=0$ to $x=9$. Our limits of integration are $a=0$ and $b=9$.

Step 3: Express the Side Length of the Cross Section

For any $x$ in the interval $[0,9]$, the side length $s$ of the square cross section is the vertical distance from the top curve ($f(x)=2\sqrt{x}$) to the bottom curve (the x-axis, $y=0$).

$$s(x)=(\text{top curve})-(\text{bottom curve})=2\sqrt{x}-0=2\sqrt{x}$$

Step 4: Write the Area Function $A(x)$

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

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

Step 5: Set Up and Evaluate the Definite Integral

Now we substitute $A(x)$ and the limits of integration into the volume formula.

$$V={\int }_0^{9}A(x)dx={\int }_0^{9}4xdx$$

Evaluate the integral:

$$V={\left[2x^2\right]}_0^9$$

$$V=(2(9{)}^2)-(2(0{)}^2)=2(81)-0=162$$

The volume of the solid is 162 cubic units.

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

Problem: Let R be the region in the first quadrant bounded by the graph of $y=\frac{1}{4}{x}^2$, the y-axis, and the line $y=4$. The region R is the base of a solid. For this solid, 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: Identify the Axis and Variable of Integration

The problem states that the cross sections are perpendicular to the y-axis. Therefore, we will integrate with respect to $y$, and our volume formula is $V={\int }_c^{d}A(y)dy$.

Step 2: Determine the Limits of Integration

The base region is bounded by the y-axis and the line $y=4$. The region starts at $y=0$ (the vertex of the parabola) and extends to $y=4$. Our limits of integration are $c=0$ and $d=4$.

Step 3: Express the Dimensions of the Cross Section in terms of $y$

The base $b$ of the rectangular cross section is the horizontal distance within the region R. We must express the boundary curve in terms of $y$.

$$y=\frac{1}{4}{x}^2⟹4y=x^2⟹x=\sqrt{4y}=2\sqrt{y}$$

The base $b$ is the distance from the right curve ($x=2\sqrt{y}$) to the left curve (the y-axis, $x=0$).

$$b(y)=(\text{right curve})-(\text{left curve})=2\sqrt{y}-0=2\sqrt{y}$$

The problem states the height $h$ is twice the base:

$$h(y)=2\cdot b(y)=2(2\sqrt{y})=4\sqrt{y}$$

Step 4: Write the Area Function $A(y)$

The cross sections are rectangles, so the area is $A=bh$. Substituting our expressions for $b(y)$ and $h(y)$:

$$A(y)=b(y)\cdot h(y)=(2\sqrt{y})(4\sqrt{y})=8y$$

Step 5: Set Up and Evaluate the Definite Integral

Substitute $A(y)$ and the limits into the volume formula.

$$V={\int }_0^{4}A(y)dy={\int }_0^{4}8ydy$$

Evaluate the integral:

$$V={\left[4y^2\right]}_0^4$$

$$V=(4(4{)}^2)-(4(0{)}^2)=4(16)-0=64$$

The volume of the solid is 64 cubic units.

Using Your Calculator

While setting up the integral is an analytical skill, a graphing calculator is an efficient tool for evaluating the final definite integral, especially on calculator-active portions of the AP Exam.

To calculate the volume from Example 1, $V={\int }_0^{9}4xdx$:

1. Determine the Area Function: First, you must analytically determine that $A(x)=4x$. This step cannot be done by the calculator.

2. Access the Integral Function: On a TI-84 style calculator, press the $[math]$ key and select 9: fnInt(.

3. Enter the Arguments: The syntax is $fnInt(integrand,variable,lowerlimit,upperlimit)$.

   • For the integrand, enter the area function: 4X

   • For the variable, enter: $X$

   • For the lower limit, enter: $0$

   • For the upper limit, enter: $9$

4. Execute: Your screen should show fnInt(4X, X, 0, 9). Press `[ENTER]to get the result, which will be $162.

The calculator is used to perform the final computation after you have correctly modeled the problem by finding the area function $A(x)$ or $A(y)$ and the limits of integration.

AP Exam Quick Hit

Common Question Types

• Standard Calculation: Given a region R bounded by one or more functions and lines, find the volume of a solid with base R and specified cross sections (e.g., squares) perpendicular to an axis. This is the most direct application, similar to Example 1.

• Setup Only: A multiple-choice question might provide a description of a solid's base and cross sections and ask you to choose the correct definite integral that represents its volume, without requiring you to evaluate it. This tests your ability to correctly identify the integrand $A(x)$ and the limits of integration.

• Variable Relationships: The dimensions of the cross section are defined in relation to each other. For example, "cross sections perpendicular to the y-axis are rectangles whose height is three times the length of the base." This requires an extra step to define the area function, as seen in Example 2.

Common Mistakes

• Integrating Side Length, Not Area: For square cross sections, a common error is to integrate the side length $s(x)$ instead of the area $s(x{)}^2$. Remember, volume is the integral of area.

  • Incorrect: $V={\int }_a^{b}s(x)dx$

  • Correct: $V={\int }_a^b[s(x){]}^{2}dx$

• Mixing Up Variables: Using $dx$ when cross sections are perpendicular to the y-axis, or vice-versa. This leads to an incorrect expression for the side/base length and incorrect limits of integration. Always check the perpendicular axis first.

• Incorrect Base/Side Length: For a region between two curves, $f(x)$ and $g(x)$, incorrectly defining the side length as just $f(x)$ instead of the correct $f(x)-g(x)$ (assuming $f(x)$ is the top curve). The same error occurs for $dy$ integrals with $(rightcurve)-(leftcurve)$.

• Forgetting to Solve for the Correct Variable: When integrating with respect to $y$, forgetting to solve the boundary equations for $x$ in terms of $y$ before finding the length of the cross section's base. For example, using $y=x^2$ directly instead of $x=\sqrt{y}$.