Surface Area: Rotation about y-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
When a curve is rotated about the $y$-axis instead of the $x$-axis, every point on the curve sweeps a circle whose radius is its $x$-coordinate (the horizontal distance from the $y$-axis). The surface area formula is structurally identical to the $x$-axis case, but with $x$ playing the role of the radius instead of $y$.
The two formulas are easy to confuse. The axis of rotation determines the radius: rotate about $x$-axis, radius is $y$; rotate about $y$-axis, radius is $x$. The arc length element $ds$ is the same in both cases.
Quick Reference
Surface area of revolution about the $y$-axis. If $x \geq 0$ on $[a,b]$ and $f'$ is continuous: $$S = 2\pi \int_a^b x\,\sqrt{1 + [f'(x)]^2}\,dx = 2\pi \int_a^b x\,ds.$$
Alternatively, if the curve is given as $x = g(y)$ for $y \in [c,d]$: $$S = 2\pi \int_c^d g(y)\,\sqrt{1 + [g'(y)]^2}\,dy.$$
Key Concepts
1. Why the Radius Changes
In the $x$-axis formula, $S = 2\pi\int y\,ds$, each thin band has radius $y$ (the vertical distance from the axis). When the axis is the $y$-axis, each thin band has radius $x$ (the horizontal distance from the axis). The arc length element $ds$ describes the slant width of the band in both cases.
Summary table:
| Axis of rotation | Radius | Formula |
|---|---|---|
| $x$-axis | $y = f(x)$ | $S = 2\pi\int_a^b f(x)\,\sqrt{1+(f')^2}\,dx$ |
| $y$-axis | $x$ | $S = 2\pi\int_a^b x\,\sqrt{1+(f')^2}\,dx$ |
2. When to Use the $x = g(y)$ Version
If the problem gives $x$ as a function of $y$, use: $$S = 2\pi\int_c^d x\,\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy.$$
This avoids the need to invert the function.
3. Checking Which Axis
Before setting up the integral, identify the axis. “Rotated about the $y$-axis” means the radius is $x$. A common slip is to copy the formula for the $x$-axis (radius = $y$) and substitute the wrong variable.
Worked Example
Find the surface area obtained by rotating $y = x^3$ from $x = 0$ to $x = 1$ about the $y$-axis.
Step 1 -- Identify the radius. Rotation about the $y$-axis, so radius $= x$.
Step 2 -- Differentiate. $f'(x) = 3x^2$.
Step 3 -- Compute $1 + (f')^2$. $1 + 9x^4$.
Step 4 -- Set up the integral. $$S = 2\pi \int_0^1 x\,\sqrt{1+9x^4}\,dx.$$
Step 5 -- Substitution. Let $u = 1 + 9x^4$, $du = 36x^3\,dx$.
This does not simplify the $x\,dx$ factor directly. Rewrite: note that $u = 1 + 9x^4$ does not lead to a clean substitution with just $x\,dx$.
Instead try $u = 1 + 9x^4$: $du = 36x^3\,dx$ -- we have $x\,dx$, not $x^3\,dx$. The substitution does not clear cleanly for this $y = x^3$ example. In Stewart, such integrals are either left in setup form or computed numerically.
Numerical value: $S = 2\pi\int_0^1 x\sqrt{1+9x^4}\,dx \approx 2\pi(0.6063) \approx 3.810$.
Alternatively -- a “nice” version. Rotate $y = \dfrac{x^{3/2}}{3}$ from $x = 0$ to $x = 4$ about the $y$-axis.
$f'(x) = \dfrac{\sqrt{x}}{2}$. $(f')^2 = \dfrac{x}{4}$. $1+(f')^2 = \dfrac{4+x}{4}$.
$S = 2\pi\int_0^4 x \cdot \dfrac{\sqrt{4+x}}{2}\,dx = \pi\int_0^4 x\sqrt{4+x}\,dx$.
Substitution $u = 4+x$, $x = u-4$, $dx = du$. Limits: $u = 4$ to $8$.
$S = \pi\displaystyle\int_4^8 (u-4)\sqrt{u}\,du = \pi\int_4^8 (u^{3/2} - 4u^{1/2})\,du = \pi\left[\frac{2}{5}u^{5/2} - \frac{8}{3}u^{3/2}\right]_4^8$.
$= \pi\left[\left(\frac{2\cdot 128\sqrt{2}}{5} - \frac{8\cdot 16\sqrt{2}}{3}\right) - \left(\frac{2\cdot 32}{5} - \frac{8\cdot 8}{3}\right)\right]$.
$= \pi\left[\sqrt{2}\left(\frac{256}{5}-\frac{128}{3}\right) - \left(\frac{64}{5}-\frac{64}{3}\right)\right] = \pi\left[\sqrt{2}\cdot\frac{128}{15} - \left(-\frac{64}{15}\right)\right] = \dfrac{\pi(128\sqrt{2}+64)}{15}$.
Boxed answer (clean example): $S = \dfrac{64\pi(2\sqrt{2}+1)}{15}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $y$ as the radius for $y$-axis rotation | $S = 2\pi\int f(x)\sqrt{1+(f')^2}\,dx$ when rotating about the $y$-axis | Radius for $y$-axis rotation is $x$: use $S = 2\pi\int x\sqrt{1+(f')^2}\,dx$ |
| Swapping the axis label | Treating “about the $y$-axis” as “about the $x$-axis” | Re-read the problem; the axis of rotation determines which coordinate is the radius |
| Omitting the arc length element | $S = 2\pi\int_a^b x\,dx$ | Surface area always contains $\sqrt{1+(f')^2}$; omitting it gives the projected silhouette area |
Common Misconceptions
when rotating about the $y$-axis, the radius in the surface area formula is $y = f(x)$, the same as for rotation about the $x$-axis.
This is the concept-image-conflicts-definition error. The radius of each circular band is the distance from the point on the curve to the axis of rotation. For the $x$-axis, that distance is $y$; for the $y$-axis, it is $x$, the horizontal coordinate. The two formulas are $S = 2\pi\int y\,ds$ ($x$-axis) and $S = 2\pi\int x\,ds$ ($y$-axis). Using $f(x)$ instead of $x$ when rotating about the $y$-axis produces a different and incorrect integrand; for $y = x^3$ on $[0,1]$, the $y$-axis formula requires $2\pi\int_0^1 x\sqrt{1+9x^4}\,dx$, not $2\pi\int_0^1 x^3\sqrt{1+9x^4}\,dx$.
the arc length element $ds$ changes form depending on the axis of rotation.
This is the concept-image-conflicts-definition error. The arc length element $ds = \sqrt{1+[f'(x)]^2}\,dx$ measures the slant width of each band along the curve and does not depend on which axis the curve is rotated about. Only the radius factor changes: $y$ for the $x$-axis, $x$ for the $y$-axis. Both surface area formulas share the same $ds$ because $ds$ describes the geometry of the curve itself, not the axis of rotation.
Leveled Practice
Level 1 -- Identify the Setup
Problem 1. Set up the surface area integral for $y = 2x^2$ from $x = 0$ to $x = 1$ rotated about the $y$-axis. Do not evaluate.
Show answer
$f'(x) = 4x$. $1+(f')^2 = 1 + 16x^2$.
$S = 2\pi\displaystyle\int_0^1 x\sqrt{1+16x^2}\,dx$.
Level 2 -- Evaluate
Problem 2. Find the surface area of the surface generated by rotating $x = \sqrt{1-y^2}$ from $y = 0$ to $y = 1/2$ about the $y$-axis. (This is part of a sphere.)
Show answer
$g(y) = \sqrt{1-y^2} = (1-y^2)^{1/2}$.
$g'(y) = \dfrac{-y}{\sqrt{1-y^2}}$.
$(g')^2 = \dfrac{y^2}{1-y^2}$.
$1+(g')^2 = \dfrac{1-y^2+y^2}{1-y^2} = \dfrac{1}{1-y^2}$.
$\sqrt{1+(g')^2} = \dfrac{1}{\sqrt{1-y^2}}$.
$S = 2\pi\displaystyle\int_0^{1/2} \sqrt{1-y^2} \cdot \dfrac{1}{\sqrt{1-y^2}}\,dy = 2\pi\int_0^{1/2} 1\,dy = 2\pi \cdot \frac{1}{2} = \pi$.
Boxed answer: $S = \pi$.
(Check: a spherical cap of height $h = 1/2$ on a unit sphere has area $2\pi r h = 2\pi(1)(1/2) = \pi$. Confirmed.)
Mastery Checklist
Mental Model
The two surface-of-revolution formulas share one structure: $S = 2\pi\int (\text{radius})\,ds$. The only thing that changes with the axis of rotation is which coordinate is the radius. Rotation about the $x$-axis: every point sweeps a circle of radius $y$ (its height). Rotation about the $y$-axis: every point sweeps a circle of radius $x$ (its horizontal distance). The arc length element $ds$ is the width of each band and is the same in both cases.
Connections
Looking back
- Surface area: x-axis (Section 8.2): The $x$-axis formula $S = 2\pi\int y\,ds$ is the template; the $y$-axis formula swaps $x$ and $y$ in the radius.
- Arc length element (Section 8.1): Both formulas use $ds = \sqrt{1+(f')^2}\,dx$.
Looking ahead
- Center of mass (Section 8.3): Centroids of flat regions use moment integrals that resemble the surface area element.