Section 5.2: Volumes
| Course: MATH162 | Textbook: Stewart Calculus 9th Edition, Section 5.2 |
The Big Picture
How do you measure the volume of an object with no simple formula? The same way you’d figure out how much bread is in a loaf: slice it.
This section develops the fundamental principle that volume equals the integral of cross-sectional area. This single idea—combined with choosing the right slicing direction—handles every volume problem you’ll encounter.
graph TD
subgraph Foundation["Foundation: The Slicing Principle"]
A["Volume by Slicing<br/>V = ∫ A(x) dx"]
end
subgraph Revolution["Solids of Revolution"]
B["Disk Method<br/>A(x) = π[f(x)]²"]
C["Washer Method<br/>A(x) = π(R² - r²)"]
end
subgraph General["General Solids"]
D["Known Cross-Sections<br/>Squares, triangles, etc."]
end
A --> B
A --> C
A --> D
B --> C
style A fill:#d1fae5,stroke:#a565f0,stroke-width:3px
style B fill:#dbeafe,stroke:#3b82f6,stroke-width:2px
style C fill:#dbeafe,stroke:#3b82f6,stroke-width:2px
style D fill:#fef3c7,stroke:#f59e0b,stroke-width:2px
click A "../../skills/ch5-sec2/volume-by-slicing.html"
click B "../../skills/ch5-sec2/disk-method.html"
click C "../../skills/ch5-sec2/washer-method.html"
click D "../../skills/ch5-sec2/volumes-known-cross-sections.html"
Key Equations
| Formula | Name | When to Use |
|---|---|---|
| $V = \displaystyle\int_a^b A(x)\,dx$ | Slicing Formula | Any solid with known cross-sectional area |
| $V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$ | Disk Method | Rotate region about axis it touches |
| $V = \pi\displaystyle\int_a^b \left([R(x)]^2 - [r(x)]^2\right)\,dx$ | Washer Method | Rotate region about axis it doesn’t touch |
Quick Reference: Cross-Section Areas
| Shape | Area Formula | If dimension = $w$ |
|---|---|---|
| Circle (radius $r$) | $\pi r^2$ | $A = \frac{\pi w^2}{4}$ (diameter $w$) |
| Square (side $s$) | $s^2$ | $A = w^2$ |
| Equilateral Triangle (side $s$) | $\frac{\sqrt{3}}{4}s^2$ | $A = \frac{\sqrt{3}}{4}w^2$ |
| Isosceles Right Triangle (hypotenuse $h$) | $\frac{h^2}{4}$ | $A = \frac{w^2}{4}$ |
| Semicircle (diameter $d$) | $\frac{\pi d^2}{8}$ | $A = \frac{\pi w^2}{8}$ |
Learning Path
Phase 1: Master the Foundation
| Skill | What You’ll Learn | Key Takeaway |
|---|---|---|
| Volume by Slicing | The fundamental formula $V = \int A(x)\,dx$ | Volume = sum of thin slices; integral captures continuous variation |
Phase 2: Solids of Revolution
| Skill | What You’ll Learn | Key Takeaway |
|---|---|---|
| Disk Method | When cross-sections are circles: $V = \pi\int [f(x)]^2\,dx$ | Radius = function value; remember to square it |
| Washer Method | When there’s a hole: $V = \pi\int (R^2 - r^2)\,dx$ | Two radii needed; it’s $R^2 - r^2$, not $(R-r)^2$ |
Phase 3: Beyond Rotation
| Skill | What You’ll Learn | Key Takeaway |
|---|---|---|
| Known Cross-Sections | Squares, triangles, semicircles perpendicular to base | Express the cross-section’s dimension in terms of $x$ |
Skills in This Section
| Skill | Description | Difficulty |
|---|---|---|
| Disk Method | Applications of Integration | Intermediate |
| Volume by Slicing | Applications of Integration | Intermediate |
| Volumes with Known Cross-Sections | Applications of Integration | Advanced |
| Washer Method | Applications of Integration | Intermediate |
Exercise Coverage Map
| Exercises | Topic | Skill |
|---|---|---|
| 1–4 | Disk/washer setup and evaluation | Disk Method, Washer Method |
| 5–10 | Set up integrals (no evaluation) | Volume by Slicing |
| 11–28 | Disk and washer computations | Disk Method, Washer Method |
| 29–44 | Rotation about various lines | Washer Method |
| 45–48 | Technology-assisted problems | All skills |
| 49–54 | Interpret integral as volume | Volume by Slicing |
| 55–58 | Applied problems (CAT scan, logs) | Volume by Slicing |
| 59–74 | Known cross-sections | Known Cross-Sections |
| 75–79 | Cavalieri’s Principle, torus | Volume by Slicing |
| 80–87 | Challenge problems | All skills |
Key Mathematical Themes
| Theme | How It Appears | Why It Matters |
|---|---|---|
| Riemann Sum → Integral | Slicing a solid into thin pieces, then taking limit | Connects discrete approximation to exact answer |
| Choosing Variables | Integrate w.r.t. $x$ or $y$ depending on cross-section orientation | Right choice simplifies the area expression |
| The $\frac{1}{3}$ Factor | Pyramids, cones have $V = \frac{1}{3}Bh$ | Linear tapering means integrating $x^2$ gives $\frac{x^3}{3}$ |
| Symmetry | Even integrands let you integrate $0$ to $r$ and double | Simplifies computation and reduces errors |
Self-Assessment Quiz
Test your understanding before moving on:
Q1: What is the volume formula for a solid with cross-sectional area $A(x)$?
where $a$ and $b$ are where the solid begins and ends along the $x$-axis.
Q2: When do you use the washer method instead of the disk method?
Use the washer method when the region being rotated does NOT touch the axis of rotation. This creates a solid with a hole in it, so you need both an outer radius $R$ and an inner radius $r$.
Key formula: $A(x) = \pi R^2 - \pi r^2$ (not $\pi(R-r)^2$!)
Q3: A solid has a circular base of radius 2. Cross-sections perpendicular to the $x$-axis are squares. What is $A(x)$?
The circle $x^2 + y^2 = 4$ gives $y = \pm\sqrt{4-x^2}$.
The width of the base at position $x$ is $2\sqrt{4-x^2}$.
Since cross-sections are squares with side equal to this width: \(A(x) = (2\sqrt{4-x^2})^2 = 4(4-x^2) = 16 - 4x^2\)
Q4: Why does a cone have volume $\frac{1}{3}\pi r^2 h$ while a cylinder has $\pi r^2 h$?
A cylinder has constant cross-sectional area $\pi r^2$ at every height.
A cone’s radius grows linearly from 0 to $r$, so its cross-sectional area grows as the square of the height. When you integrate $x^2$, you get $\frac{x^3}{3}$—the cubic growth with the $\frac{1}{3}$ factor.
The average cross-sectional area of the cone is $\frac{1}{3}$ of the cylinder’s constant area.
Q5: You want to find the volume of a solid of revolution. The region is bounded by $y = \sqrt{x}$ and $y = x$. What are the limits of integration?
Find where the curves intersect: $\sqrt{x} = x$ means $x = x^2$, so $x(x-1) = 0$.
The curves intersect at $x = 0$ and $x = 1$.
The limits of integration are 0 to 1.
Deep Connections
Why “Slicing” Is a Universal Strategy
The technique in this section—breaking a complex object into simple pieces—appears throughout mathematics and science:
| Domain | Application |
|---|---|
| Medical Imaging | CT scans measure cross-sectional areas; computers integrate to estimate organ volumes |
| Engineering | Structural analysis computes stress by integrating over cross-sections |
| Physics | Moment of inertia is $\int r^2\,dm$—integrating over “slices” of mass |
| Probability | Joint distributions are integrated over “slices” (conditional densities) |
Connection to Arc Length and Surface Area
The same “slice and integrate” pattern continues:
| Quantity | What You Slice | What You Sum |
|---|---|---|
| Volume | 3D solid into 2D slices | Cross-sectional areas |
| Arc Length | Curve into line segments | $\sqrt{1 + [f’(x)]^2}\,dx$ |
| Surface Area | Surface into circular bands | $2\pi r\sqrt{1 + [f’(x)]^2}\,dx$ |
Common Mistakes to Avoid
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Forgetting to square the radius | $V = \pi\int r\,dx$ | $V = \pi\int r^2\,dx$ |
| Writing $(R-r)^2$ for washers | This is the square of thickness, not area | Use $R^2 - r^2$ |
| Wrong limits of integration | Limits should be where solid starts/ends | Find intersection points carefully |
| Using $x$ when should use $y$ | Rotation about $y$-axis needs $x = g(y)$ | Match variable to axis of rotation |
| Forgetting the $\pi$ | Disk/washer areas include $\pi$ | Cross-section area = $\pi r^2$ |
Practice Problems
Practice problems are available on each skill page:
- Volume by Slicing Practice (Levels 1–5)
- Disk Method Practice (Levels 1–5)
- Washer Method Practice (Levels 1–5)
- Known Cross-Sections Practice (Levels 1–5)
| Previous | Up | Next |
|---|---|---|
| Section 5.1 | Chapter 5 | Section 5.3 |