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Unit 8.7 – Volumes with Cross Sections: Squares and Rectangles

AP® Calculus AB & BC | 3D Solids from 2D Regions

Why This Matters: Volumes with cross sections is a powerful technique for finding volumes of 3D solids! Unlike the disk/washer method (rotation), this method works for solids where cross sections perpendicular to an axis have known shapes—squares, rectangles, triangles, semicircles, etc. The idea: slice the solid into thin pieces, find the area of each cross section, and integrate. This topic appears frequently on AP® exams and extends integration to three dimensions!

🎯 The Big Idea: Slicing Method

THE CONCEPT

Imagine slicing a solid perpendicular to an axis (like slicing bread). Each slice has a cross-sectional area. The volume is the sum (integral) of all these thin slices!

The Slicing Principle:

If a solid has cross-sectional area \(A(x)\) at each point \(x\) perpendicular to the x-axis, then:

\[ \text{Volume} = \int_a^b A(x) \, dx \]

📐 The Fundamental Formula

Volume by Cross Sections

THE GENERAL FORMULA:
\[ V = \int_a^b A(x) \, dx \]

Or if slicing perpendicular to y-axis:

\[ V = \int_c^d A(y) \, dy \]
Where:
  • \(A(x)\) = area of cross section at position \(x\)
  • \([a, b]\) = interval over which solid extends
  • Cross sections are perpendicular to axis of integration

⬛ Cross Sections: Squares

Squares Perpendicular to an Axis

For Square Cross Sections:

If the base of the solid is the region between curves \(y = f(x)\) and \(y = g(x)\), and cross sections perpendicular to the x-axis are squares:

\[ \text{Side length} = |f(x) - g(x)| \]
\[ A(x) = [\text{side}]^2 = [f(x) - g(x)]^2 \]
\[ V = \int_a^b [f(x) - g(x)]^2 \, dx \]

📝 Key Insight: The side length of each square equals the height of the base region at that point (distance between curves)!

▭ Cross Sections: Rectangles

Rectangular Cross Sections

For Rectangular Cross Sections:

If rectangles have:

  • One dimension = height of base region: \(h = |f(x) - g(x)|\)
  • Other dimension = some multiple or function of the first

Common cases:

Case 1: Height is k times the base

\[ A(x) = k \cdot [f(x) - g(x)]^2 \]

Case 2: Dimensions specified separately

\[ A(x) = (\text{width}) \times (\text{height}) \]

📋 Step-by-Step Process

Complete Method

The 6-Step Approach:

  1. Identify the base region: What curves bound it? Over what interval?
  2. Determine axis of slicing: Perpendicular to x-axis or y-axis?
  3. Find dimensions of cross section:
    • For squares: side = distance between curves
    • For rectangles: find both dimensions
  4. Write area formula: \(A(x)\) or \(A(y)\)
  5. Set up integral: \(V = \int_a^b A(x)\,dx\)
  6. Evaluate: Find antiderivative and apply FTC

📖 Comprehensive Worked Examples

Example 1: Square Cross Sections

Problem: Find the volume of a solid whose base is the region bounded by \(y = x^2\) and \(y = 4\), where cross sections perpendicular to the x-axis are squares.

Solution:

Step 1: Identify base region

Find intersection points:

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

Base: region between \(y = x^2\) and \(y = 4\) from \(x = -2\) to \(x = 2\)

Step 2: Find side length of square

Side = vertical distance between curves

\[ s(x) = 4 - x^2 \]

Step 3: Find area of cross section

\[ A(x) = [s(x)]^2 = (4 - x^2)^2 \]

Step 4: Set up integral

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

Step 5: Expand and evaluate

\[ V = \int_{-2}^2 (16 - 8x^2 + x^4) \, dx \]
\[ = \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^2 \]
\[ = 2\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{1024}{15} \]

ANSWER: \(V = \frac{1024}{15}\) cubic units

Example 2: Rectangle - Height is 3 Times Base

Problem: The base is bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\). Cross sections perpendicular to the x-axis are rectangles with height 3 times the base. Find the volume.

Step 1: Base region

From \(x = 0\) to \(x = 4\), between \(y = 0\) and \(y = \sqrt{x}\)

Step 2: Rectangle dimensions

Base (width): \(w = \sqrt{x} - 0 = \sqrt{x}\)

Height: \(h = 3w = 3\sqrt{x}\)

Step 3: Area of cross section

\[ A(x) = w \times h = \sqrt{x} \cdot 3\sqrt{x} = 3x \]

Step 4: Set up and evaluate

\[ V = \int_0^4 3x \, dx = \left[\frac{3x^2}{2}\right]_0^4 = 24 \]

ANSWER: \(V = 24\) cubic units

Example 3: Squares with Two Curves

Problem: Base bounded by \(y = x\) and \(y = x^2\). Cross sections perpendicular to x-axis are squares. Find volume.

Find intersections and setup:

\[ x = x^2 \quad \Rightarrow \quad x = 0, 1 \]

Side length: \(s = x - x^2\) (since \(x > x^2\) on \((0,1)\))

\[ A(x) = (x - x^2)^2 \]

Evaluate:

\[ V = \int_0^1 (x - x^2)^2 \, dx = \int_0^1 (x^2 - 2x^3 + x^4) \, dx \]
\[ = \left[\frac{x^3}{3} - \frac{x^4}{2} + \frac{x^5}{5}\right]_0^1 = \frac{1}{3} - \frac{1}{2} + \frac{1}{5} = \frac{1}{30} \]

Example 4: Rectangles - Specified Dimensions

Problem: Base is the region between \(y = \sin x\) and \(y = 0\) from \(x = 0\) to \(x = \pi\). Cross sections perpendicular to x-axis are rectangles with height equal to 2 and width equal to the base dimension. Find volume.

Setup:

Width: \(w = \sin x\)

Height: \(h = 2\)

\[ A(x) = 2\sin x \]

Evaluate:

\[ V = \int_0^\pi 2\sin x \, dx = [-2\cos x]_0^\pi = -2(-1) - (-2)(1) = 4 \]

📊 Key Formulas Summary

Cross Section Volume Formulas
Cross Section Type Area Formula Volume Formula
Square \(A(x) = s^2\) where \(s = f(x) - g(x)\) \(\int_a^b [f(x)-g(x)]^2\,dx\)
Rectangle (h = k·base) \(A(x) = k[f(x)-g(x)]^2\) \(\int_a^b k[f(x)-g(x)]^2\,dx\)
Rectangle (general) \(A(x) = w(x) \times h(x)\) \(\int_a^b w(x)h(x)\,dx\)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Sketch the base region: Visualize the 2D base first
  • Identify the axis: Perpendicular to which axis?
  • Find dimensions carefully: Distance between curves = side/dimension
  • Square the side length: Don't forget to square for squares!
  • Read carefully: Rectangle dimensions often specified differently
  • Units matter: Cubic units for volume
  • Check your bounds: Use intersection points
  • Expand before integrating: Makes integration easier

🔥 Common Setups:

  • "Squares" → Area = (distance)²
  • "Height is twice the base" → Area = 2(base)²
  • "Rectangles with base in region" → Find both dimensions
  • Perpendicular to x-axis → Use \(dx\)
  • Perpendicular to y-axis → Use \(dy\)

❌ Common Mistakes to Avoid

  • Mistake 1: Forgetting to square the side length for squares
  • Mistake 2: Using diameter instead of side for cross section
  • Mistake 3: Wrong distance formula (not subtracting curves correctly)
  • Mistake 4: Misreading rectangle dimensions (which is which?)
  • Mistake 5: Wrong limits of integration
  • Mistake 6: Not expanding before integrating
  • Mistake 7: Sign errors in subtraction
  • Mistake 8: Confusing this with disk/washer method
  • Mistake 9: Integration errors
  • Mistake 10: Wrong units (saying square units instead of cubic)

📝 Practice Problems

Find the volume:

  1. Base: between \(y = x\) and \(y = x^2\). Squares perpendicular to x-axis.
  2. Base: between \(y = 4\) and \(y = x^2\). Squares perpendicular to y-axis.
  3. Base: between \(y = \sqrt{x}\), \(x = 0\), \(x = 4\). Rectangles perpendicular to x-axis with height = 2×base.
  4. Base: triangle with vertices (0,0), (2,0), (0,2). Squares perpendicular to x-axis.

Answers:

  1. \(\frac{1}{30}\) cubic units
  2. \(\frac{256}{15}\) cubic units
  3. 16 cubic units
  4. \(\frac{4}{3}\) cubic units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Show base region: Identify curves and bounds
  • Show dimension calculation: How you got side/dimensions
  • Write area formula: \(A(x) = \ldots\)
  • Set up integral: \(V = \int_a^b A(x)\,dx\) clearly shown
  • Show integration work: Antiderivative before applying bounds
  • Evaluate at bounds: Show substitution
  • Simplify answer: Final numerical value
  • Include units: "cubic units"

💯 Exam Strategy:

  1. Read carefully: Squares or rectangles? What dimensions?
  2. Sketch base region
  3. Find intersection points (limits)
  4. Determine distance between curves (side/dimension)
  5. Write area formula clearly
  6. Set up integral
  7. Expand if needed, then integrate
  8. Evaluate and simplify
  9. State answer with units

⚡ Quick Reference Guide

CROSS SECTIONS ESSENTIALS

General Formula:

\[ V = \int_a^b A(x) \, dx \]

where \(A(x)\) = area of cross section

For Squares:

  • Side = distance between curves
  • Area = (side)²
  • \(V = \int_a^b [f(x)-g(x)]^2\,dx\)

For Rectangles:

  • Find both dimensions
  • Area = width × height
  • \(V = \int_a^b w(x) \cdot h(x)\,dx\)

Remember:

  • Sketch the base region!
  • Distance between curves = dimension
  • Square the side for squares
  • Volume uses cubic units

Master Volumes with Cross Sections! The fundamental formula: \(V = \int_a^b A(x)\,dx\) where \(A(x)\) is the area of the cross section at position \(x\). The slicing method: cut solid perpendicular to an axis, find area of each slice, integrate. For squares: side length = distance between curves, area = (side)² = \([f(x)-g(x)]^2\), so \(V = \int_a^b[f(x)-g(x)]^2\,dx\). For rectangles: find both dimensions (width and height), area = width × height. Common case: "height is k times base" → \(A = k(\text{base})^2\). Process: (1) identify base region and bounds, (2) find dimensions from distance between curves, (3) write area formula \(A(x)\), (4) set up integral, (5) evaluate. Always square for squares! Read rectangle problems carefully for dimensions. This differs from disk/washer (rotation)—these are general cross sections. Units: cubic units (volume). Major AP® exam topic—appears regularly! Practice squares and rectangles until automatic! 🎯✨