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

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

The fundamental concept behind this topic is the construction of a three-dimensional solid from a known two-dimensional base region. Imagine a flat shape on the xy-plane (the base). We then build a solid by stacking an infinite number of thin, two-dimensional shapes (called cross-sections) on top of this base. Each cross-section is perpendicular to one of the axes (either the x-axis or the y-axis).

The volume of this solid is found by summing the volumes of all these infinitesimally thin cross-sectional "slices." Calculus allows us to perform this summation using a definite integral. The key is to find a function, $A(x)$ or $A(y)$, that gives the area of a single cross-section at any given point. The definite integral of this area function over the interval of the base region gives the total volume of the solid.

Key Formulas

The calculation of volume with known cross-sections relies on a single master formula and the area formulas for specific geometric shapes.

1. The General Volume Formula

• If the cross-sections are perpendicular to the x-axis, the volume of the solid from $x=a$ to $x=b$ is given by:

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

  where $A(x)$ is the area of a single cross-section at a given x-value.

• If the cross-sections are perpendicular to the y-axis, the volume of the solid from $y=c$ to $y=d$ is given by:

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

  where $A(y)$ is the area of a single cross-section at a given y-value.

2. Cross-Sectional Area Formulas

The function $A(x)$ or $A(y)$ depends on the shape of the cross-section. Let $s$ represent the length of the base of the cross-section within the 2D region.

• For Semicircles: The base length $s$ is the diameter of the semicircle. The radius is $r=\frac{s}{2}$.

  The area is $A=\frac{1}{2}\pi r^2$. Substituting $r=\frac{s}{2}$ gives:

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

• For Triangles: The area is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. The base $b$ is the length $s$. The height $h$ must be expressed in terms of the base $s$. A common case is an isosceles right triangle where a leg is the base. In this case, $h=b=s$, so the area is:

  $$A=\frac{1}{2}{s}^2$$

Understanding the Setup

The most critical part of solving these problems is correctly setting up the definite integral. This involves three key decisions: the variable of integration, the length of the base $s$, and the bounds of integration.

1. Variable of Integration ($dx$ or $dy$):

   • If the problem states that cross-sections are perpendicular to the x-axis, you will integrate with respect to $x$ ($dx$). All functions and bounds must be in terms of $x$.

   • If the problem states that cross-sections are perpendicular to the y-axis, you will integrate with respect to $y$ ($dy$). All functions and bounds must be in terms of $y$.

2. The Base Length ($s$):

   • For a $dx$ integral, the base $s$ of the cross-section is a vertical distance. If the base region is bounded above by $y=f(x)$ and below by $y=g(x)$, the length is:

     $$s=f(x)-g(x)(\text{Top}-\text{Bottom})$$

   • For a $dy$ integral, the base $s$ of the cross-section 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 is:

     $$s=f(y)-g(y)(\text{Right}-\text{Left})$$

3. Bounds of Integration:

   • For a $dx$ integral, the bounds $a$ and $b$ are the minimum and maximum x-values of the base region.

   • For a $dy$ integral, the bounds $c$ and $d$ are the minimum and maximum y-values of the base region.

These three components are combined to build the final integral for the volume.

Core Concepts & Rules

• The volume of a solid with known cross-sections is found by integrating the area of a single cross-section over the appropriate interval.

• The orientation of the cross-sections determines the variable of integration. Perpendicular to the x-axis implies $dx$; perpendicular to the y-axis implies $dy$.

• The area function, $A(x)$ or $A(y)$, must be written entirely in terms of the chosen variable of integration.

• The base of the cross-section, $s$, is the distance between the boundary curves of the 2D region, measured parallel to the cross-section's orientation.

• For semicircular cross-sections, the base $s$ is the diameter. The radius is $r=s/2$. The area is $A=\frac{\pi }{8}{s}^2$.

• For triangular cross-sections, the base is $s$. The height must be determined from the problem description (e.g., for an isosceles right triangle with a leg on the base, height equals base, so $A=\frac{1}{2}{s}^2$).

Step-by-Step Example 1: Semicircles

Problem: Let R be the region in the first quadrant enclosed by the graphs of $y=\sqrt{x}$, $y=0$, and $x=4$. The region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a semicircle. Find the volume of the solid.

Step 1: Determine the Variable of Integration

The problem states the cross-sections are "perpendicular to the x-axis." Therefore, we will use a $dx$ integral.

Step 2: Determine the Base Length, $s$

The base of each cross-section is a vertical segment in the region R. The top curve is $y=\sqrt{x}$ and the bottom curve is $y=0$.

$$s=\text{Top}-\text{Bottom}=\sqrt{x}-0=\sqrt{x}$$

Step 3: Determine the Area Formula, $A(x)$

The cross-sections are semicircles. The base length $s=\sqrt{x}$ is the diameter.

The radius is $r=\frac{s}{2}=\frac{\sqrt{x}}{2}$.

The area of a semicircle is $A=\frac{1}{2}\pi r^2$.

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

Step 4: Determine the Bounds of Integration

The region is bounded by $x=0$ (the y-axis) and $x=4$. So, our bounds are from $a=0$ to $b=4$.

Step 5: Set Up and Evaluate the Integral

$$V={\int }_a^{b}A(x)dx={\int }_0^4\frac{\pi }{8}xdx$$

$$V=\frac{\pi }{8}{\int }_0^{4}xdx=\frac{\pi }{8}{\left[\frac{1}{2}{x}^2\right]}_0^4$$

$$V=\frac{\pi }{16}[x^2{]}_0^4=\frac{\pi }{16}(4^2-0^2)=\frac{\pi }{16}(16)=\pi$$

The volume of the solid is $\pi$.

Step-by-Step Example 2: Exam-Style Application

Problem: Let R be the region enclosed by the graphs of $x=y^2$ and $x=y+2$. The region R is the base of a solid. For this solid, each cross-section perpendicular to the y-axis is an isosceles right triangle with one leg in the region R. Find the volume of the solid.

Step 1: Determine the Variable of Integration

The problem states the cross-sections are "perpendicular to the y-axis." Therefore, we will use a $dy$ integral.

Step 2: Determine the Base Length, $s$

The base of each cross-section is a horizontal segment. We need to identify the right and left boundary curves. The line $x=y+2$ is to the right of the parabola $x=y^2$ in the enclosed region.

$$s=\text{Right}-\text{Left}=(y+2)-y^2$$

Step 3: Determine the Area Formula, $A(y)$

The cross-sections are isosceles right triangles with a leg on the base. This means the height $h$ is equal to the base $b$. Here, the base is $s$.

So, $b=s$ and $h=s$.

The area of a triangle is $A=\frac{1}{2}bh$.

$$A(y)=\frac{1}{2}{s}^2=\frac{1}{2}(y+2-y^2{)}^2$$

Step 4: Determine the Bounds of Integration

We need the y-values where the curves intersect.

$$y^2=y+2$$

$$y^2-y-2=0$$

$$(y-2)(y+1)=0$$

The intersection points are at $y=-1$ and $y=2$. These are our bounds, $c=-1$ and $d=2$.

Step 5: Set Up the Integral

$$V={\int }_c^{d}A(y)dy={\int }_{-1}^2\frac{1}{2}(y+2-y^2{)}^{2}dy$$

This is a calculator-active problem. Evaluating this by hand is tedious and not typically required. The setup is the key part. For completeness, the value is $\frac{81}{20}$.

Using Your Calculator

Most volume-by-cross-section problems on the AP exam appear on the calculator-active section. Your calculator is used to find bounds and evaluate the final definite integral.

Problem: Find the volume of the solid whose base is the region bounded by $f(x)=e^x$ and $g(x)=4-x^2$, and whose cross-sections perpendicular to the x-axis are semicircles.

Step 1: Graph and Find Bounds

• Enter `Y_1 = e^x$ and $Y_2=4-x^2$.

• Graph the functions. You will see two intersection points.

• Use the calculator's intersection feature (2nd -> TRACE -> 5: intersect) to find the bounds.

  • Left intersection (lower bound): $x\approx -1.9646$ (Store this as $A$).

  • Right intersection (upper bound): $x\approx 1.0580$ (Store this as $B$).

Step 2: Set Up the Integral on Paper

• Variable: Perpendicular to x-axis, so $dx$.

• Base $s$: The top curve is $Y_2$ and the bottom is $Y_1$. So, $s=(4-x^2)-e^x$.

• Area $A(x)$: Semicircles. $A(x)=\frac{\pi }{8}{s}^2=\frac{\pi }{8}((4-x^2)-e^x{)}^2$.

• Integral:$V={\int }_A^B\frac{\pi }{8}((4-x^2)-e^x{)}^{2}dx$

Step 3: Evaluate Using fnInt

• On the home screen, access the numerical integration function (MATH -> 9: fnInt).

• Enter the integral using your stored bounds and $Y_1,Y_2$ variables.

• Syntax: fnInt( (π/8)*(Y₂ - Y₁)^2 , X, A, B )

  • Note: To get $Y_1$ and $Y_2$, press VARS -> Y-VARS -> 1: Function....

• The calculator will return the volume, which is approximately $4.515$.

AP Exam Quick Hit

Common Question Types

• Set Up, Do Not Evaluate: You are given a region and a cross-section shape and asked only to write the definite integral that gives the volume. This tests your ability to choose the correct variable, bounds, and area formula without requiring calculation.

  • Example: "Let R be the region bounded by $y=\ln (x)$, $x=e$, and the x-axis. Write, but do not evaluate, an integral expression for the volume of the solid with base R and cross-sections perpendicular to the x-axis that are squares."

• Calculator-Active Full Problem: You are given a region bounded by functions that are difficult to work with by hand (e.g., trig, exponential, logs) and asked to find the volume. This tests the entire process, including using the calculator to find bounds and evaluate the final integral.

  • Example: "The region R is bounded by $y=\sin (\pi x)$ and $y=x^3-4x$. Find the volume of the solid with base R and cross-sections perpendicular to the x-axis that are semicircles."

• Perpendicular to the y-axis: A common variation is to require a $dy$ integral, forcing you to rewrite equations in the form $x=f(y)$ and use a $Right-Left$ setup for the base length.

Common Mistakes

• Radius vs. Diameter Error: For semicircles, a very common mistake is to forget to divide the base length $s$ by 2 before squaring. Students use $A=\frac{1}{2}\pi s^2$ instead of the correct $A=\frac{1}{2}\pi (\frac{s}{2}{)}^2=\frac{\pi }{8}{s}^2$.

• Forgetting Constants: Leaving out the $\frac{1}{2}$ for a triangle's area or the $\frac{\pi }{8}$ (or $\frac{\pi }{2}$) for a semicircle's area.

• Squaring Error: Incorrectly squaring the base expression, for example, writing $(f(x)-g(x){)}^2$ as $f(x{)}^2-g(x{)}^2$.

• Incorrect Bounds: Using x-value intersection points as bounds for a $dy$ integral, or y-value intersection points for a $dx$ integral.

• Mixing Variables: Setting up a $dx$ integral but having the area function $A(y)$ in terms of $y$, or vice-versa. The variable in the area function must match the differential ($dx$ or $dy$).