Unit 8.8 – Volumes with Cross Sections: Triangles and Semicircles

AP® Calculus AB & BC | Advanced Cross-Sectional Volumes

Why This Matters: Building on squares and rectangles, we now tackle triangular and semicircular cross sections! These shapes require specific area formulas but use the same integration principle. Triangles come in several varieties (equilateral, isosceles right, general), each with its own area formula. Semicircles involve π and circular geometry. Master these and you've completed the full range of cross-sectional volume problems tested on AP® exams!

📐 Essential Area Formulas

Area Formulas You MUST Know

Triangle Formulas:

1. General Triangle:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

2. Equilateral Triangle (all sides equal):

\[ A = \frac{s^2\sqrt{3}}{4} \]

where \(s\) = side length

3. Isosceles Right Triangle (45-45-90):

\[ A = \frac{1}{2}b^2 \]

where \(b\) = length of each leg (both legs equal)

Semicircle Formula:

\[ A = \frac{1}{2}\pi r^2 \]

where \(r\) = radius (half the diameter)

📝 Critical: For cross sections, the dimension from the base region (distance between curves) typically becomes the base, diameter, or side of the cross section!

△ Triangular Cross Sections

Triangles Perpendicular to an Axis

General Principle:

The base of each triangular cross section is typically the distance between the curves at that point: \(b(x) = f(x) - g(x)\)

Type 1: Equilateral Triangles

If base = distance between curves:

\[ s = f(x) - g(x) \]

\[ A(x) = \frac{[f(x)-g(x)]^2\sqrt{3}}{4} \]

\[ V = \int_a^b \frac{[f(x)-g(x)]^2\sqrt{3}}{4} \, dx \]

Type 2: Isosceles Right Triangles

If base = hypotenuse = distance between curves:

\[ \text{Each leg} = \frac{f(x)-g(x)}{\sqrt{2}} \]

\[ A(x) = \frac{1}{2}\left(\frac{f(x)-g(x)}{\sqrt{2}}\right)^2 = \frac{[f(x)-g(x)]^2}{4} \]

\[ V = \int_a^b \frac{[f(x)-g(x)]^2}{4} \, dx \]

Type 3: Right Triangles (base and height given)

If base = distance between curves and height specified:

\[ A(x) = \frac{1}{2} \times [f(x)-g(x)] \times h(x) \]

◗ Semicircular Cross Sections

Semicircles Perpendicular to an Axis

Standard Setup:

If diameter = distance between curves:

\[ d = f(x) - g(x) \]

\[ r = \frac{f(x)-g(x)}{2} \]

\[ A(x) = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi \left(\frac{f(x)-g(x)}{2}\right)^2 = \frac{\pi[f(x)-g(x)]^2}{8} \]

\[ V = \int_a^b \frac{\pi[f(x)-g(x)]^2}{8} \, dx \]

⚠️ Watch out: The distance between curves is the DIAMETER, not the radius! Don't forget to divide by 2 before squaring!

📖 Comprehensive Worked Examples

Example 1: Equilateral Triangles

Problem: The base is the region bounded by \(y = x^2\) and \(y = 4\). Cross sections perpendicular to the x-axis are equilateral triangles. Find the volume.

Solution:

Step 1: Find base region bounds

\[ x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]

Region from \(x = -2\) to \(x = 2\)

Step 2: Find side length of equilateral triangle

Base of triangle = vertical distance:

\[ s = 4 - x^2 \]

Step 3: Area of equilateral triangle

\[ A(x) = \frac{s^2\sqrt{3}}{4} = \frac{(4-x^2)^2\sqrt{3}}{4} \]

Step 4: Set up and evaluate integral

\[ V = \int_{-2}^2 \frac{\sqrt{3}}{4}(4-x^2)^2 \, dx \]

\[ = \frac{\sqrt{3}}{4} \int_{-2}^2 (16 - 8x^2 + x^4) \, dx \]

\[ = \frac{\sqrt{3}}{4} \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^2 \]

\[ = \frac{\sqrt{3}}{4} \cdot 2\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{256\sqrt{3}}{15} \]

ANSWER: \(V = \frac{256\sqrt{3}}{15}\) cubic units

Example 2: Semicircles

Problem: Base bounded by \(y = \sin x\) and \(y = 0\) from \(x = 0\) to \(x = \pi\). Cross sections perpendicular to x-axis are semicircles with diameters in the base. Find volume.

Step 1: Identify diameter

Diameter = distance between curves:

\[ d = \sin x - 0 = \sin x \]

Radius:

\[ r = \frac{\sin x}{2} \]

Step 2: Area of semicircle

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

Step 3: Evaluate integral

\[ V = \int_0^\pi \frac{\pi \sin^2 x}{8} \, dx = \frac{\pi}{8} \int_0^\pi \sin^2 x \, dx \]

Use identity: \(\sin^2 x = \frac{1 - \cos 2x}{2}\)

\[ V = \frac{\pi}{8} \int_0^\pi \frac{1-\cos 2x}{2} \, dx = \frac{\pi}{16}\left[x - \frac{\sin 2x}{2}\right]_0^\pi = \frac{\pi^2}{16} \]

ANSWER: \(V = \frac{\pi^2}{16}\) cubic units

Example 3: Isosceles Right Triangles

Problem: Base between \(y = x\) and \(y = x^2\). Cross sections perpendicular to x-axis are isosceles right triangles with hypotenuse in the base. Find volume.

Setup:

Intersections: \(x = x^2 \Rightarrow x = 0, 1\)

Hypotenuse = \(x - x^2\)

For isosceles right triangle with hypotenuse \(h\):

\[ A = \frac{h^2}{4} = \frac{(x-x^2)^2}{4} \]

Evaluate:

\[ V = \int_0^1 \frac{(x-x^2)^2}{4} \, dx = \frac{1}{4}\int_0^1 (x^2 - 2x^3 + x^4) \, dx \]

\[ = \frac{1}{4}\left[\frac{x^3}{3} - \frac{x^4}{2} + \frac{x^5}{5}\right]_0^1 = \frac{1}{4}\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{5}\right) = \frac{1}{120} \]

Example 4: Right Triangles with Specified Height

Problem: Base between \(y = \sqrt{x}\), \(y = 0\), from \(x = 0\) to \(x = 4\). Cross sections perpendicular to x-axis are right triangles with base in the xy-plane and height equal to twice the base. Find volume.

Setup:

Base: \(b = \sqrt{x}\)

Height: \(h = 2b = 2\sqrt{x}\)

\[ A(x) = \frac{1}{2}bh = \frac{1}{2} \cdot \sqrt{x} \cdot 2\sqrt{x} = x \]

Evaluate:

\[ V = \int_0^4 x \, dx = \left[\frac{x^2}{2}\right]_0^4 = 8 \]

📊 Complete Formula Reference

Cross Section Volume Formulas
Cross Section Type	Key Dimension	Area Formula
Equilateral Triangle	Side \(s = f(x) - g(x)\)	\(A = \frac{s^2\sqrt{3}}{4}\)
Isosceles Right △	Hypotenuse \(h = f(x) - g(x)\)	\(A = \frac{h^2}{4}\)
Right Triangle	Base \(b\), Height \(h\)	\(A = \frac{1}{2}bh\)
Semicircle	Diameter \(d = f(x) - g(x)\)	\(A = \frac{\pi d^2}{8}\)

💡 Essential Tips & Strategies

✅ Success Strategies:

Memorize area formulas: Equilateral, isosceles right, semicircle
Identify the shape: Read problem carefully for type
Find the key dimension: Usually distance between curves
Watch for diameter vs radius: Semicircles use radius in formula
For equilateral: Remember \(\frac{\sqrt{3}}{4}\) factor
For isosceles right: Remember \(\frac{1}{4}\) factor
Sketch if possible: Visualize the cross section
Check units: Cubic units for volume

🔥 Quick Memory Aids:

Equilateral triangle: "\(\frac{\sqrt{3}}{4}\) times side squared"
Isosceles right triangle: "\(\frac{1}{4}\) times hypotenuse squared"
Semicircle: "Half of \(\pi r^2\)" or "\(\frac{\pi d^2}{8}\)"
Distance between curves → dimension of cross section

❌ Common Mistakes to Avoid

Mistake 1: Forgetting \(\sqrt{3}\) in equilateral triangle formula
Mistake 2: Using diameter instead of radius for semicircles
Mistake 3: Wrong factor for isosceles right triangles (using \(\frac{1}{2}\) instead of \(\frac{1}{4}\))
Mistake 4: Not squaring the dimension before applying formula
Mistake 5: Confusing which dimension is base, side, or diameter
Mistake 6: Arithmetic errors with fractions (\(\frac{\sqrt{3}}{4}\), \(\frac{\pi}{8}\), etc.)
Mistake 7: Wrong limits of integration
Mistake 8: Not simplifying \(\sin^2 x\) in semicircle problems
Mistake 9: Integration errors
Mistake 10: Forgetting to include constant factors in final answer

📝 Practice Problems

Find the volume:

Base between \(y = x^2\) and \(y = 4\). Semicircular cross sections perpendicular to x-axis.
Base between \(y = \sqrt{x}\) and \(y = 0\), \(x \in [0, 4]\). Equilateral triangle cross sections perpendicular to x-axis.
Base between \(y = x\) and \(y = x^2\). Semicircular cross sections perpendicular to y-axis.
Base: triangle with vertices (0,0), (3,0), (0,3). Isosceles right triangle cross sections perpendicular to x-axis.

Answers:

\(\frac{32\pi}{15}\) cubic units
\(\frac{16\sqrt{3}}{3}\) cubic units
\(\frac{\pi}{60}\) cubic units
\(\frac{9}{8}\) cubic units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Identify cross section type: State equilateral, semicircle, etc.
Show dimension calculation: How you got side, diameter, etc.
Write area formula: \(A(x) = \ldots\) with correct formula
Set up integral correctly: Right formula, right bounds
Show integration work: Especially for trig integrals
Simplify before integrating: Expand, use identities
Evaluate at bounds: Show substitution
Simplify final answer: Include \(\pi\), \(\sqrt{3}\), etc.
Include units: "cubic units"

💯 Exam Strategy:

Read carefully: What type of cross section?
Write down the appropriate area formula
Find the dimension (usually distance between curves)
Substitute into area formula
Set up integral: \(V = \int_a^b A(x)\,dx\)
For semicircles: Remember to use \(\sin^2 x\) identity if needed
Show all work
Simplify answer completely
Check: Does answer make sense? Right form?

⚡ Quick Reference Guide

TRIANGLES & SEMICIRCLES ESSENTIALS

Equilateral Triangle:

\[ A = \frac{s^2\sqrt{3}}{4} \quad \text{where } s = \text{side} \]

Isosceles Right Triangle:

\[ A = \frac{h^2}{4} \quad \text{where } h = \text{hypotenuse} \]

Semicircle:

\[ A = \frac{\pi d^2}{8} = \frac{\pi r^2}{2} \quad \text{where } d = \text{diameter}, r = \text{radius} \]

Remember:

Distance between curves = key dimension
Memorize the area formulas!
Volume = \(\int_a^b A(x)\,dx\)
Include \(\sqrt{3}\), \(\pi\) in answers

Master Triangular and Semicircular Cross Sections! The fundamental approach: \(V = \int_a^b A(x)\,dx\) where \(A(x)\) uses the specific area formula. For equilateral triangles: side = distance between curves, area = \(\frac{s^2\sqrt{3}}{4}\), giving \(V = \int_a^b \frac{\sqrt{3}}{4}[f(x)-g(x)]^2\,dx\). For isosceles right triangles: hypotenuse = distance, area = \(\frac{h^2}{4}\). For semicircles: diameter = distance, radius = \(\frac{d}{2}\), area = \(\frac{1}{2}\pi r^2 = \frac{\pi d^2}{8}\). Critical: memorize these area formulas—they're not given on the exam! Distance between curves provides the key dimension (side, diameter, base). For semicircles with trig functions, use \(\sin^2 x = \frac{1-\cos 2x}{2}\) identity. Common errors: forgetting \(\sqrt{3}\), using diameter instead of radius, wrong factors. Process: identify shape, find dimension, apply area formula, integrate. This completes cross-sectional volumes—practice all types! 🎯✨