Surface Area: Rotation about x-axis
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 8.2: “Area of a Surface of Revolution” |
Opening Scenario
Imagine peeling the surface off a solid of revolution. If you rotate the curve $y = f(x)$ from $x = a$ to $x = b$ around the $x$-axis, you get a 3D surface shaped like a vase, funnel, or trumpet. The area of that surface is not the same as the area under the curve -- it involves the curve’s length, not just its height.
The key insight: slicing the surface into thin rings (frustums of cones) and summing their lateral areas gives an integral involving both the height $y$ (which sets the radius of each ring) and the arc length element $ds$ (which sets the slant height of each ring).
Quick Reference
Surface area of revolution about the $x$-axis. If $f(x) \geq 0$ and $f'$ is continuous on $[a,b]$: $$S = 2\pi \int_a^b f(x)\,\sqrt{1 + [f'(x)]^2}\,dx = 2\pi \int_a^b y\,ds.$$
Here $ds = \sqrt{1+(f')^2}\,dx$ is the arc length element.
Key Concepts
1. Derivation Using Frustum Slices
Slice the curve into small pieces of arc length $\Delta s_i$. Rotating piece $i$ around the $x$-axis traces out a thin band (a frustum of a cone). The lateral area of a frustum with average radius $r$ and slant height $\ell$ is $2\pi r \ell$.
Here the average radius is approximately $f(x_i^*)$ (the $y$-value at a sample point) and the slant height is $\Delta s_i \approx \sqrt{1+[f'(x_i^*)]^2}\,\Delta x$.
Summing and taking the limit: $$S = \lim_{n\to\infty} \sum_{i=1}^n 2\pi f(x_i^*)\sqrt{1+[f'(x_i^*)]^2}\,\Delta x = 2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx.$$
2. The Formula Requires y >= 0
The formula uses $y = f(x)$ as the radius of rotation. If $f(x) < 0$ on part of $[a,b]$, write $|f(x)|$ as the radius. In problems on MATH162 exams, the curve is typically above the $x$-axis.
3. The Perfect-Square Trick (Again)
Just like arc length, the integrand $\sqrt{1+(f')^2}$ almost never produces an elementary antiderivative unless $1+(f')^2$ is a perfect square. Stewart’s exercises are designed so that this simplification works. The strategy is:
- Compute $f'(x)$ and $(f'(x))^2$.
- Add 1 and check whether the result is a perfect square.
- If yes, take the square root and integrate.
Worked Example
Find the surface area obtained by rotating $y = \sqrt{x}$ from $x = 0$ to $x = 1$ about the $x$-axis.
Step 1 -- Differentiate. $f'(x) = \dfrac{1}{2\sqrt{x}}$.
Step 2 -- Compute $1 + (f')^2$. $\left(\dfrac{1}{2\sqrt{x}}\right)^2 = \dfrac{1}{4x}$. So $1 + (f')^2 = 1 + \dfrac{1}{4x} = \dfrac{4x+1}{4x}$.
Step 3 -- Set up the integral. $$S = 2\pi \int_0^1 \sqrt{x} \cdot \sqrt{\frac{4x+1}{4x}}\,dx = 2\pi\int_0^1 \sqrt{x} \cdot \frac{\sqrt{4x+1}}{2\sqrt{x}}\,dx = \pi\int_0^1 \sqrt{4x+1}\,dx.$$
Step 4 -- Evaluate. Substitution: $u = 4x+1$, $du = 4\,dx$.
When $x=0$: $u=1$. When $x=1$: $u=5$.
$$S = \pi \int_1^5 \sqrt{u}\,\frac{du}{4} = \frac{\pi}{4}\left[\frac{2}{3}u^{3/2}\right]_1^5 = \frac{\pi}{6}\left(5^{3/2} - 1\right) = \frac{\pi(5\sqrt{5}-1)}{6}.$$
Boxed answer: $S = \dfrac{\pi(5\sqrt{5}-1)}{6} \approx 5.33$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $2\pi\int_a^b f(x)\,dx$ (volume disk formula) | Computing the volume instead of surface area | Surface area is $2\pi\int f(x)\sqrt{1+(f')^2}\,dx$; it includes the arc length factor $\sqrt{1+(f')^2}$ |
| Dropping the $\sqrt{1+(f')^2}$ factor | Writing $S = 2\pi\int y\,dx$ | Without the arc length element, the formula integrates the silhouette area, not the surface area |
| Using the wrong radius | Rotating about the $x$-axis but writing $2\pi\int x\,ds$ | Radius for rotation about the $x$-axis is $y = f(x)$; radius for the $y$-axis is $x$ |
Common Misconceptions
the surface area obtained by rotating $y = f(x)$ about the $x$-axis is $2\pi\int_a^b f(x)\,dx$, the same integral used for volume by the disk method (without the squaring).
This is the concept-image-conflicts-definition error. The integral $2\pi\int_a^b f(x)\,dx$ gives the lateral area of a cylinder of radius $f(x)$ and width $dx$ at each cross-section, not the surface of revolution. The surface area formula is $S = 2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx$, because each thin band is a frustum whose slant height is the arc length element $ds$, not the horizontal width $dx$. For $y = \sqrt{x}$ on $[0,1]$, using $2\pi\int_0^1\sqrt{x}\,dx = \tfrac{4\pi}{3}$ instead of the correct $\tfrac{\pi(5\sqrt{5}-1)}{6}$ underestimates the surface area by ignoring the curve’s slope.
the product $f(x)\sqrt{1+(f')^2}$ in the surface area integrand can be split as the product of two separate integrals.
This is the multiplicative-not-additive error. The integrand is a product of the radius $y = f(x)$ and the arc length element $\sqrt{1+(f')^2}$, and these two factors must be integrated together. There is no rule that allows $\int f(x)\sqrt{1+(f')^2}\,dx$ to be written as $\bigl(\int f(x)\,dx\bigr)\bigl(\int\sqrt{1+(f')^2}\,dx\bigr)$; such a split would be numerically incorrect in every nontrivial example.
Leveled Practice
Level 1 -- Setup
Problem 1. Set up (do not evaluate) the surface area integral for $y = x^3$ rotated about the $x$-axis from $x = 0$ to $x = 1$.
Show answer
$f'(x) = 3x^2$. $1+(f')^2 = 1+9x^4$.
$S = 2\pi\displaystyle\int_0^1 x^3\sqrt{1+9x^4}\,dx$.
Level 2 -- Evaluation
Problem 2. Find the surface area obtained by rotating $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 2$ about the $x$-axis.
Show answer
$f'(x) = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$.
$(f')^2 = \dfrac{x^4}{4} - \dfrac{1}{2} + \dfrac{1}{4x^4}$.
$1+(f')^2 = \dfrac{x^4}{4} + \dfrac{1}{2} + \dfrac{1}{4x^4} = \left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^2$.
$\sqrt{1+(f')^2} = \dfrac{x^2}{2} + \dfrac{1}{2x^2}$.
$S = 2\pi\displaystyle\int_1^2 \left(\dfrac{x^3}{6}+\dfrac{1}{2x}\right)\left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)dx$.
Multiply out: $\dfrac{x^5}{12} + \dfrac{x}{12} + \dfrac{x}{4} + \dfrac{1}{4x^3} = \dfrac{x^5}{12} + \dfrac{x}{3} + \dfrac{1}{4x^3}$.
$\displaystyle\int_1^2\left(\dfrac{x^5}{12}+\dfrac{x}{3}+\dfrac{1}{4x^3}\right)dx = \left[\dfrac{x^6}{72}+\dfrac{x^2}{6}-\dfrac{1}{8x^2}\right]_1^2$
$= \left(\dfrac{64}{72}+\dfrac{4}{6}-\dfrac{1}{32}\right)-\left(\dfrac{1}{72}+\dfrac{1}{6}-\dfrac{1}{8}\right) = \dfrac{63}{72}+\dfrac{3}{6}+\dfrac{3}{32} = \dfrac{7}{8}+\dfrac{1}{2}+\dfrac{3}{32} = \dfrac{28+16+3}{32} = \dfrac{47}{32}$.
$S = 2\pi \cdot \dfrac{47}{32} = \dfrac{47\pi}{16}$.
Boxed answer: $S = \dfrac{47\pi}{16}$.
Mastery Checklist
Mental Model
A surface of revolution is built from thin rings. Each ring has circumference $2\pi r$ (where $r = y$ for rotation about the $x$-axis) and width $ds$ (the arc length of the generating curve over the ring). Its area is $2\pi r\,ds$. Summing: $S = 2\pi\int y\,ds$. This is the simplest way to remember the formula -- it says area equals circumference times arc length, integrated along the curve.
Connections
Looking back
- Arc length formula (Section 8.1): The arc length element $ds = \sqrt{1+(f')^2}\,dx$ is the key piece.
- Volumes by disks/shells (Section 6.2-6.3): Volumes also arise from rotating curves, but the formula is $\pi y^2\,dx$ (disk) or $2\pi x y\,dx$ (shell), not $2\pi y\,ds$.
Looking ahead
- Surface area: y-axis (Section 8.2): When the axis of rotation is the $y$-axis, the radius becomes $x$ and the formula changes to $S = 2\pi\int x\,ds$.