Volumes with Cross Sections: Triangles and Semicircles - AP Calculus BC Study Guide

The Core Idea: Volumes with Cross Sections: Triangles and Semicircles

This topic introduces a method for calculating the volume of a three-dimensional solid by conceptualizing it as a stack of infinitely thin slices, or cross-sections. We begin with a two-dimensional region in the $xy$-plane, which serves as the base of the solid. We then build the solid up from this base by placing geometric shapes (like triangles or semicircles) on it, with their bases or diameters lying on the region.

The core idea is to find a function, $A(x)$ or $A(y)$, that gives the area of a single cross-section at any point $x$ or $y$. By integrating this area function over the interval that defines the base, we are effectively summing the volumes of all the infinitesimally thin slices, which gives the total volume of the solid. The orientation of the cross-sections (perpendicular to the x-axis or y-axis) determines whether we integrate with respect to $x$ or $y$.

Key Formulas

The calculation of volume using the cross-section method relies on a single principle expressed in two forms, depending on the orientation of the cross-sections.

1. Volume with Cross-Sections Perpendicular to the x-axis:

   If a solid has a known cross-sectional area $A(x)$ for $x$ in the interval $[a,b]$, its volume $V$ is given by the definite integral:

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

2. Volume with Cross-Sections Perpendicular to the y-axis:

   If a solid has a known cross-sectional area $A(y)$ for $y$ in the interval $[c,d]$, its volume $V$ is given by the definite integral:

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

3. Cross-Sectional Area Formulas:

   The function $A(x)$ or $A(y)$ is derived from standard geometric area formulas. For this topic, the key shapes are:

   • Semicircle: The area is $A=\frac{1}{2}\pi r^2$, where $r$ is the radius.

   • Triangle: The area is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height.

The dimensions used in these formulas ($r$, $b$, $h$) must be expressed in terms of $x$ or $y$ before setting up the integral.

Understanding the Integrand $A(x)$ or $A(y)$

The most critical step in solving these problems is correctly constructing the area function, $A(x)$ or $A(y)$. This function is not given directly; it must be derived from the geometry of the solid's base and the shape of its cross-sections.

The process involves finding the length of a representative segment, $s$, that lies across the base region. This length $s$ serves as a key dimension—such as the diameter of a semicircle or the base of a triangle—for the cross-section at that position.

• For cross-sections perpendicular to the x-axis: The length $s$ is a vertical distance. If the base region is bounded above by $y=f(x)$ and below by $y=g(x)$, the length of the segment at $x$ is:

  $$s(x)=f(x)-g(x)(\text{top curve}-\text{bottom curve})$$

• For cross-sections perpendicular to the y-axis: The length $s$ is a horizontal distance. If the base region is bounded on the right by $x=f(y)$ and on the left by $x=g(y)$, the length of the segment at $y$ is:

  $$s(y)=f(y)-g(y)(\text{right curve}-\text{left curve})$$

Once $s$ is determined, it is used to define the area $A$. For example, if the cross-sections are semicircles whose diameters lie on the base region, then the diameter is $s(x)$. The radius is $r(x)=\frac{s(x)}{2}$, and the area function becomes:

$$A(x)=\frac{1}{2}\pi [r(x){]}^2=\frac{1}{2}\pi {\left(\frac{s(x)}{2}\right)}^2=\frac{\pi }{8}[s(x){]}^2$$

If the cross-sections are isosceles right triangles with a leg on the base, then the base $b$ and height $h$ are both equal to $s(y)$. The area function becomes:

$$A(y)=\frac{1}{2}b\cdot h=\frac{1}{2}[s(y)][s(y)]=\frac{1}{2}[s(y){]}^2$$

Core Concepts & Rules

• The volume of a solid can be calculated by integrating the area of its cross-sections.

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

• The integrand, $A(x)$ or $A(y)$, is the area of a single cross-section expressed as a function of its position $x$ or $y$.

• To find the area function $A$, first determine the length of a representative segment $s$ across the base region. This length is typically the difference between two functions ($top-bottom$ or $right-left$).

• The dimensions of the cross-sectional shape (e.g., radius $r$, base $b$, height $h$) must be expressed in terms of this segment length $s$ before being used in the area formula.

Step-by-Step Example 1: Semicircular Cross-Sections

Problem: Let $R$ be the region bounded by the graph of $f(x)=4-x^2$ and the x-axis. The region $R$ is the base of a solid. For this solid, the cross-sections perpendicular to the x-axis are semicircles. Find the volume of the solid.

Step 1: Sketch the Base Region and Find Bounds

The base is the region under the parabola $y=4-x^2$ and above the x-axis ($y=0$).

To find the bounds of integration, set the functions equal: $4-x^2=0$ gives $x=\pm 2$.

The interval of integration is $[-2,2]$.

Step 2: Define the Cross-Section Length $s(x)$

The cross-sections are perpendicular to the x-axis, so we need a vertical length $s(x)$.

The top curve is $y=4-x^2$ and the bottom curve is $y=0$.

$$s(x)=(\text{top curve})-(\text{bottom curve})=(4-x^2)-0=4-x^2$$

This length $s(x)$ represents the diameter of the semicircular cross-section at $x$.

Step 3: Write the Area Formula $A(x)$

The cross-sections are semicircles, so the area is $A=\frac{1}{2}\pi r^2$.

The radius $r$ is half the diameter $s(x)$:

$$r(x)=\frac{s(x)}{2}=\frac{4-x^2}{2}$$

Now, substitute this into the area formula:

$$A(x)=\frac{1}{2}\pi {\left(\frac{4-x^2}{2}\right)}^2=\frac{1}{2}\pi \frac{(4-x^2{)}^2}{4}=\frac{\pi }{8}(4-x^2{)}^2$$

Step 4: Set Up and Evaluate the Integral

The volume $V$ is the integral of the area function $A(x)$ from $a=-2$ to $b=2$.

$$V={\int }_{-2}^2\frac{\pi }{8}(4-x^2{)}^{2}dx$$

Expand the integrand:

$$V=\frac{\pi }{8}{\int }_{-2}^2(16-8x^2+x^4)dx$$

Find the antiderivative:

$$V=\frac{\pi }{8}{\left[16x-\frac{8}{3}{x}^3+\frac{1}{5}{x}^5\right]}_{-2}^2$$

Evaluate using the Fundamental Theorem of Calculus:

$$V=\frac{\pi }{8}\left(\left(16(2)-\frac{8}{3}(2{)}^3+\frac{1}{5}(2{)}^5\right)-\left(16(-2)-\frac{8}{3}(-2{)}^3+\frac{1}{5}(-2{)}^5\right)\right)$$

$$V=\frac{\pi }{8}\left(\left(32-\frac{64}{3}+\frac{32}{5}\right)-\left(-32+\frac{64}{3}-\frac{32}{5}\right)\right)$$

$$V=\frac{\pi }{8}\left(64-\frac{128}{3}+\frac{64}{5}\right)=\frac{\pi }{8}\left(\frac{960-640+192}{15}\right)=\frac{\pi }{8}\left(\frac{512}{15}\right)=\frac{64\pi }{15}$$

Step-by-Step Example 2: Triangular Cross-Sections

Problem: Let $R$ be the region in the first quadrant bounded by the graphs of $y=x^3$ and $y=4x$. The region $R$ is the base of a solid. For this solid, each cross-section perpendicular to the y-axis is an equilateral triangle. Find the volume of the solid. (Note: The area of an equilateral triangle with side $s$ is $A=\frac{\sqrt{3}}{4}{s}^2$).

Step 1: Sketch the Base Region and Find Bounds

Find intersection points: $x^3=4x⟹x^3-4x=0⟹x(x^2-4)=0$. Intersections are at $x=0$ and $x=2$ (since we are in the first quadrant). The corresponding y-values are $y=0$ and $y=8$.

The cross-sections are perpendicular to the y-axis, so we will integrate with respect to $y$. The bounds are $y=0$ to $y=8$.

Step 2: Re-express Functions and Define Cross-Section Length $s(y)$

We need the functions in terms of $y$.

$y=x^3⟹x=y^{1/3}$

$y=4x⟹x=\frac{y}{4}$

For a given $y$ between 0 and 8, the right curve is $x=y^{1/3}$ and the left curve is $x=\frac{y}{4}$.

The length of the horizontal segment $s(y)$ is:

$$s(y)=(\text{right curve})-(\text{left curve})=y^{1/3}-\frac{y}{4}$$

This length $s(y)$ is the side length of the equilateral triangle.

Step 3: Write the Area Formula $A(y)$

The cross-sections are equilateral triangles with side length $s(y)$. The area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}(\text{side}{)}^2$.

$$A(y)=\frac{\sqrt{3}}{4}[s(y){]}^2=\frac{\sqrt{3}}{4}{\left(y^{1/3}-\frac{y}{4}\right)}^2$$

Step 4: Set Up the Integral

The volume $V$ is the integral of the area function $A(y)$ from $c=0$ to $d=8$.

$$V={\int }_0^8\frac{\sqrt{3}}{4}{\left(y^{1/3}-\frac{y}{4}\right)}^{2}dy$$

On a free-response question where a calculator is permitted, this setup would be a complete answer. To evaluate by hand, we would expand the integrand:

$$V=\frac{\sqrt{3}}{4}{\int }_0^8\left((y^{1/3}{)}^2-2(y^{1/3})\left(\frac{y}{4}\right)+{\left(\frac{y}{4}\right)}^2\right)dy$$

$$V=\frac{\sqrt{3}}{4}{\int }_0^8\left(y^{2/3}-\frac{1}{2}{y}^{4/3}+\frac{1}{16}{y}^2\right)dy$$

This integral can then be evaluated using the power rule for antiderivatives.

Using Your Calculator

While the setup of the integral is an analytical task, a graphing calculator is an efficient tool for evaluating the final definite integral, especially with complex integrands.

To calculate the volume from Example 1, $V={\int }_{-2}^2\frac{\pi }{8}(4-x^2{)}^{2}dx$:

1. Define the Area Function: In the Y= editor, enter the area function $A(x)$.

   • Y1 = (π/8)*(4 - X^2)^2

2. Use the Numerical Integration Function: From the main screen, access the numerical integration function (e.g., fnInt on a TI-84, typically found in the MATH menu).

   • The syntax is fnInt(expression, variable, lower bound, upper bound).

   • Enter: fnInt(Y1, X, -2, 2)

3. Execute: Press ENTER. The calculator will return the numerical value of the integral, which is approximately $13.404$. You can verify that $64\pi /15$ is approximately $13.404$.

The primary skill is deriving the correct expression for $A(x)$ or $A(y)$. The calculator is a tool for the final computation.

AP Exam Quick Hit

Common Question Types

• Functions Defining a Base: You are given two or more functions, e.g., $f(x)=\sqrt{x}$ and $g(x)=x^2$, that define the base of a solid. You are asked to find the volume of the solid with specified cross-sections (e.g., isosceles right triangles with a leg on the base) perpendicular to an axis.

• Graph or Table Defining a Base: The base of the solid is described by a region $R$ shown in a graph. You must determine the "top minus bottom" or "right minus left" relationship and the bounds of integration directly from the visual information.

• Setup, Do Not Evaluate: A common free-response prompt asks you to write an integral expression for the volume but explicitly tells you not to evaluate it. This tests your ability to correctly identify the bounds, the variable of integration, and the formula for the area of the cross-section, $A(x)$ or $A(y)$.

Common Mistakes

• Forgetting Constants in Area Formulas: A frequent error is omitting the $1/2$ for a triangle's area or the $\pi /2$ for a semicircle's area.

• Incorrectly Deriving $A(x)$ from $s(x)$: For semicircles, a very common mistake is forgetting that $s(x)$ is the diameter, not the radius. This leads to using $A(x)=\frac{1}{2}\pi [s(x){]}^2$ instead of the correct $A(x)=\frac{1}{2}\pi {\left(\frac{s(x)}{2}\right)}^2=\frac{\pi }{8}[s(x){]}^2$.

• Mixing $dx$ and $dy$ Setups: Choosing the wrong variable of integration. If cross-sections are perpendicular to the y-axis, you must express all functions in terms of $y$ and integrate with respect to $y$. Using $x$ functions with $dy$ is a fundamental error.

• Algebraic Errors in the Integrand: When expanding the expression for $A(x)$ or $A(y)$, such as $(f(x)-g(x){)}^2$, students often make algebraic mistakes, for example, writing $f(x{)}^2-g(x{)}^2$ instead of $f(x{)}^2-2f(x)g(x)+g(x{)}^2$.