Worked examples — Surface area of revolution

Before line one, a reminder of the three tools we lean on, in plain words:

• $ds$ = the true slanted length of a tiny curve piece (see Arc length). It is the length of the "rope" that wraps around.
• $2\pi r$ = the circumference a point traces when spun; $r$ is how far it sits from the pole it spins around.
• substitution (Integration by substitution) = a renaming trick when the stuff outside the square root is (a multiple of) the derivative of the stuff inside.

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The scenario matrix

Every surface-of-revolution problem lives in exactly one of these cells. The examples below are tagged with the cell(s) they hit.

#	Case class	What makes it different	Covered by
A	Rotate $y=f(x)$ about the x-axis	radius $=y$	Ex 1
B	Rotate about the y-axis	radius $=x$, integrate carefully	Ex 2
C	Rotate about a shifted line $y=k$	radius $=|y-k|$ (sign matters!)	Ex 3
D	Parametric curve	$ds$ from $x^{\prime }(t),y^{\prime }(t)$	Ex 4
E	Degenerate: horizontal segment	$dy/dx=0$ → cylinder, a sanity anchor	Ex 5
F	Degenerate: point on the axis (radius $\to 0$)	band shrinks to zero — limiting behaviour	Ex 6
G	Word problem (real object, units)	translate story → curve → integral	Ex 7
H	Exam twist: curve crosses the axis (radius changes sign)	must use $|y|$, split the integral	Ex 8
I	Cross-check with Pappus's theorem	$S=2\pi \bar{r}L$ shortcut	Ex 9

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Example 1 — Rotate about the x-axis (Cell A)

The figure below shows the setup: the magenta curve $y=\sqrt{x}$, a violet arrow giving the radius (the height $y$ to the x-axis), and a short orange segment — one tiny slanted piece $ds$ that wraps around. Notice the orange piece is longer than its horizontal shadow: that is exactly why we use $ds$, not $dx$.

Step 1 — Identify the radius. Every point $(x,y)$ on the curve sits a height $y$ above the x-axis. Spinning it about that axis, it travels a circle of radius $y=\sqrt{x}$ (the violet arrow in the figure). Why this step? The radius is "distance to the axis" — here the axis is the x-axis, so the vertical height $y$ is exactly that distance.

Step 2 — Build $ds$. We need the slope: $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$, so $ds=\sqrt{1+\frac{1}{4x}}dx.$ Why this step? The band wraps along the tilted curve (the orange segment), not its flat shadow — so we need arc length, and $ds$ needs the slope.

Step 3 — Assemble the integral. $S={\int }_0^{4}2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}}dx.$ Why this step? Every band contributes $2\pi \cdot (\text{radius})\cdot ds$; summing (integrating) all bands from $x=0$ to $x=4$ gives the whole skin.

Step 4 — Simplify the integrand. Pull $\sqrt{x}$ inside: $\sqrt{x}\sqrt{1+\frac{1}{4x}}=\sqrt{x+\frac{1}{4}}.$ Why this step? $x\cdot (1+\frac{1}{4x})=x+\frac{1}{4}$ — the awkward $1/x$ vanishes, leaving a plain finite root we can integrate all the way down to $x=0$.

Step 5 — Integrate. With $u=x+\frac{1}{4}$, $du=dx$: $S=2\pi {\int }_0^4\sqrt{x+\frac{1}{4}}dx=2\pi \cdot \frac{2}{3}{\left[(x+\frac{1}{4}{)}^{3/2}\right]}_0^4=\frac{4\pi }{3}\left[{\left(\frac{17}{4}\right)}^{3/2}-{\left(\frac{1}{4}\right)}^{3/2}\right].$ Numerically $S\approx 36.18$.

Verify: ${\left(\frac{17}{4}\right)}^{3/2}=8.755$, ${\left(\frac{1}{4}\right)}^{3/2}=0.125$; difference $8.630$; $\times \frac{4\pi }{3}\approx 36.18$. Positive and finite — good, since after simplification the integrand is bounded on all of $[0,4]$. ✅

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Example 2 — Rotate about the y-axis (Cell B)

Step 1 — Radius is now $x$. Distance from $(x,y)$ to the y-axis is the horizontal distance $x$. Why this step? "Distance to the axis" changed axes; measure horizontally now.

Step 2 — Keep integrating in $x$. We can still use $x$ as our variable; $ds=\sqrt{1+(2x{)}^2}dx$ since $dy/dx=2x$. Why this step? $ds$ is a length — it does not care which axis we spin about; only the radius factor changes.

Step 3 — Integral. $S={\int }_0^{1}2\pi x\sqrt{1+4x^2}dx.$ Why this step? Each band is $2\pi \cdot (\text{radius }x)\cdot ds$; integrating over $x\in [0,1]$ stacks all the bands into the full surface.

Step 4 — Substitute $u=1+4x^2$, $du=8xdx$: $S=\frac{2\pi }{8}{\int }_1^5\sqrt{u}du=\frac{\pi }{4}\cdot \frac{2}{3}{\left[u^{3/2}\right]}_1^5=\frac{\pi }{6}\left(5^{3/2}-1\right)\approx \mathrm{5.33.}$ Why substitution? The $x$ outside is exactly $\frac{1}{8}du$ — a gift.

Verify: $5^{3/2}=11.180$, minus $1=10.180$, $\times \frac{\pi }{6}\approx 5.33$. Matches parent note Example 3. ✅

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Example 3 — Rotate about a shifted line $y=k$ (Cell C)

The figure shows the magenta segment $y=1$, the dashed navy axis $y=3$ above it, and violet arrows measuring the gap between them. Every arrow has the same length $2$ — that constant gap is the radius. Because the strip is flat and the gap never changes, the swept surface is a plain cylinder.

Step 1 — Radius is $|y-k|$. The point sits at height $1$, the axis at height $3$, so radius $=|1-3|=2$ (the violet arrows). Why this step? Distance to a horizontal line at height $k$ is the vertical gap $|y-k|$ — the absolute value keeps it positive even though $y<k$.

Step 2 — $ds$ for a flat line. $dy/dx=0$, so $ds=\sqrt{1+0}dx=dx$. Why this step? A horizontal segment isn't tilted, so its slanted length equals its flat length.

Step 3 — Integral (it's a cylinder!). $S={\int }_0^{3}2\pi \cdot 2\cdot dx=4\pi \cdot 3=12\pi \approx \mathrm{37.70.}$ Why this step? With constant radius $2$ and $ds=dx$, each band is identical $2\pi (2)dx$; summing them from $x=0$ to $3$ just multiplies by the length $3$ — literally unrolling a cylinder.

Verify: A cylinder of radius $2$, length $3$ has lateral area $2\pi rL=2\pi (2)(3)=12\pi$. Exactly what we got. ✅ If we had forgotten the absolute value we might have written radius $=1-3=-2$ and got a negative area — nonsense; the $|\cdot |$ guards us.

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Example 4 — Parametric curve (Cell D)

Step 1 — Radius $=y=t^2$. Rotating about the x-axis, distance to axis is the height $y$. Why this step? Same rule; we just express $y$ through $t$.

Step 2 — Parametric $ds$. $x^{\prime }(t)=3t^2$, $y^{\prime }(t)=2t$, so $ds=\sqrt{(3t^2{)}^2+(2t{)}^2}dt=\sqrt{9t^4+4t^2}dt=t\sqrt{9t^2+4}dt(t\ge 0).$ Why this step? For a parametric curve, arc length is $\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}dt$ — Pythagoras with both coordinates moving. We factor $t^2$ out of the root (safe since $t\ge 0$ makes $\sqrt{t^2}=t$).

Step 3 — Integral. $S={\int }_0^{1}2\pi t^2\cdot t\sqrt{9t^2+4}dt=2\pi {\int }_0^1{t}^3\sqrt{9t^2+4}dt.$ Why this step? Radius $\times ds\times 2\pi$, with everything now written in the single variable $t$ and its limits $0$ to $1$.

Step 4 — Substitute $u=9t^2+4$, $du=18tdt$, and $t^2=\frac{u-4}{9}$: $S=2\pi {\int }_4^{13}\frac{u-4}{9}\sqrt{u}\cdot \frac{du}{18}=\frac{2\pi }{162}{\int }_4^{13}\left(u^{3/2}-4u^{1/2}\right)du.$ Why this step? $t^{3}dt=t^2\cdot tdt$, and $tdt=\frac{1}{18}du$ while $t^2$ becomes a function of $u$ — the whole thing is now polynomial-in-$\sqrt{u}$.

Step 5 — Evaluate. ${\int }_4^{13}\left(u^{3/2}-4u^{1/2}\right)du={\left[\frac{2}{5}{u}^{5/2}-\frac{8}{3}{u}^{3/2}\right]}_4^{13}.$ At $u=13$: $\frac{2}{5}(13^{5/2})-\frac{8}{3}(13^{3/2})=243.86-124.98=118.88$. At $u=4$: $\frac{2}{5}(4^{5/2})-\frac{8}{3}(4^{3/2})=\frac{2}{5}(32)-\frac{8}{3}(8)=12.80-21.33=-8.53$. Bracket $=118.88-(-8.53)=127.41$, so $S=\frac{2\pi }{162}\cdot 127.41\approx \mathrm{4.94.}$

Verify: $\frac{2\pi }{162}\approx 0.03879$; $0.03879\times 127.41\approx 4.94$. Small positive number — sensible for such a tiny curve. ✅

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Example 5 — Degenerate: horizontal segment → cylinder (Cell E)

Step 1 — Radius $=5$ (constant), $ds=dx$ since the slope is $0$. Why this step? No tilt ⇒ arc length equals horizontal length; this is the simplest possible case, our "does the machine still work?" test.

Step 2 — Integral. $S={\int }_0^{2}2\pi (5)dx=10\pi \cdot 2=20\pi \approx \mathrm{62.83.}$ Why this step? Every band is the identical $2\pi (5)dx$; integrating just multiplies by the length $2$ — the very definition of unrolling a cylinder into a rectangle of height $2\pi r$ and width $L$.

Verify: Lateral area of a cylinder $=2\pi rL=2\pi (5)(2)=20\pi$. The calculus reduces to grade-school geometry when there's no tilt — exactly as it must. ✅

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Example 6 — Degenerate: radius shrinks to zero (Cell F)

In the figure, the magenta line $y=x$ is the generating curve and the dotted mirror below shows the surface it spins into — a cone. The orange dot at the origin is the tip: it sits right on the axis, so its radius (violet arrow at $x=0.8$) shrinks to zero there. Watch how the arrow gets shorter as you move left toward the tip.

Step 1 — Radius $=y=x$, which $\to 0$ as $x\to 0$. At the tip the circle it traces has radius $0$ — a single point (the orange dot). Why this step? This is the limiting/degenerate check: what happens where the surface pinches to a point?

Step 2 — $ds=\sqrt{1+1}dx=\sqrt{2}dx$ (slope $=1$).

Step 3 — Integral. $S={\int }_0^{1}2\pi x\sqrt{2}dx=2\sqrt{2}\pi {\left[\frac{x^2}{2}\right]}_0^1=\sqrt{2}\pi \approx \mathrm{4.443.}$ Why this step? The integrand $2\pi x\sqrt{2}$ is $0$ at $x=0$ — the pinch point adds nothing, and there's no blow-up. The integral converges cleanly.

Verify: This is a cone of base radius $r=1$, height $h=1$, slant $L=\sqrt{1^2+1^2}=\sqrt{2}$. Formula $\pi rL=\pi (1)\sqrt{2}=\sqrt{2}\pi \approx 4.443$. ✅ The zero-radius tip is harmless.

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Example 7 — Word problem: a lampshade (Cell G)

Step 1 — Radius $=y=1/x$. Distance to x-axis is the height. Why this step? Translate the story: "spin about the horizontal axis" ⇒ radius is vertical height.

Step 2 — Slope and $ds$. $\frac{dy}{dx}=-\frac{1}{x^2}$, so $ds=\sqrt{1+\frac{1}{x^4}}dx.$

Step 3 — Integral. $S={\int }_1^{2}2\pi \cdot \frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx=2\pi {\int }_1^2\frac{\sqrt{x^4+1}}{x^3}dx.$ Why this step? Combine $\frac{1}{x}$ and the root: $\frac{1}{x}\sqrt{1+\frac{1}{x^4}}=\frac{1}{x}\cdot \frac{\sqrt{x^4+1}}{x^2}=\frac{\sqrt{x^4+1}}{x^3}$.

Step 4 — Evaluate numerically with Simpson's rule. The integrand $g(x)=\frac{\sqrt{x^4+1}}{x^3}$ has no clean elementary antiderivative, so we approximate. Split $[1,2]$ into $4$ equal strips ($h=0.25$) and sample:

$g(1)=1.41421$, $g(1.25)=0.83328$, $g(1.5)=0.55372$, $g(1.75)=0.39725$, $g(2)=0.30208$.

Simpson's rule ${\int }_1^{2}gdx\approx \frac{h}{3}[g_0+4g_1+2g_2+4g_3+g_4]$: $\frac{0.25}{3}[1.41421+4(0.83328)+2(0.55372)+4(0.39725)+0.30208]=\frac{0.25}{3}(9.59775)\approx \mathrm{0.7998.}$ Why Simpson's rule? It fits parabolas through the sample points — far more accurate than trapezoids for a smooth curve. Here it lands within about $0.001$ of the true value (a finer grid barely moves the answer), and the story only needs a couple of significant figures of cm².

Step 5 — Multiply by $2\pi$. $S=2\pi (0.7998)\approx 5.03{\text{cm}}^2.$

Verify: Units: radius (cm) × $ds$ (cm) × $2\pi$ (dimensionless) ⇒ cm² ✅. Magnitude: the shade's average radius is about $0.7$ cm over length $\approx 1.13$ cm of slanted curve, giving roughly $2\pi (0.7)(1.13)\approx 4.97$ — same ballpark as $5.03$. ✅

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Example 8 — Exam twist: curve crosses the axis (Cell H)

Step 1 — Radius $=|y|=|x^3|$. A point below the axis is still a positive distance away. Why this step? If we naively used $y=x^3$, the $x<0$ part would give a negative contribution — but area can't cancel like that. We must use $|y|$.

Step 2 — Split at the crossing point $x=0$. The curve crosses the axis exactly at $x=0$, where $|x^3|$ changes its formula. Write $|x^3|=-x^3$ on $[-1,0]$ (since $x^3\le 0$ there) and $|x^3|=x^3$ on $[0,1]$: $S={\int }_{-1}^{0}2\pi (-x^3)\sqrt{1+9x^4}dx+{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Why this step? Wherever $y$ flips sign we must break the integral so that $|y|$ has a single clean formula on each piece — otherwise the antiderivative is wrong.

Step 3 — Use symmetry to combine. With $ds=\sqrt{1+9x^4}dx$, both $|x^3|$ and $\sqrt{1+9x^4}$ are even in $x$, so the two pieces are equal: $S=2{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Why this step? Equal halves let us compute just $[0,1]$ and double — the shortcut that always beats splitting by hand.

Step 4 — Substitute $u=1+9x^4$, $du=36x^{3}dx$: $S=4\pi \cdot \frac{1}{36}{\int }_1^{10}\sqrt{u}du=\frac{\pi }{9}\cdot \frac{2}{3}{\left[u^{3/2}\right]}_1^{10}=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{7.13.}$ Why this step? The factor $x^{3}dx$ outside is exactly $\frac{1}{36}du$ — substitution turns the whole thing into a clean $\sqrt{u}$ integral over $[1,10]$.

Step 5 — State the answer. $S=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx 7.13$.

Verify: $10^{3/2}=31.623$, minus $1=30.623$; $\frac{2\pi }{27}\approx 0.23271$; $0.23271\times 30.623\approx 7.13$. If we had not used $|y|$, the naive signed integral over $[-1,1]$ of $2\pi x^3\sqrt{1+9x^4}$ would be $0$ (odd function) — obviously wrong for a real surface. The $|\cdot |$ fix (and the split at $x=0$) is essential. ✅

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Example 9 — Cross-check with Pappus (Cell I)

Step 1 — Length $L$. From $(0,1)$ to $(3,5)$: $L=\sqrt{3^2+4^2}=5$. Why this step? Pappus needs the arc length of the generating curve; here it's a plain segment.

Step 2 — Centroid height $\bar{r}$. Midpoint of the segment is $\left(\frac{0+3}{2},\frac{1+5}{2}\right)=(1.5,3)$, so $\bar{r}=3$. Why this step? For a uniform straight segment the centroid is the midpoint; distance to the x-axis is its $y$-value.

Step 3 — Pappus. $S=2\pi \bar{r}L=2\pi (3)(5)=30\pi \approx \mathrm{94.25.}$ Why this step? Pappus's shortcut says the swept surface area equals the arc length times the distance its centroid travels in one full turn, $2\pi \bar{r}$ — no integral required.

Step 4 — Confirm by integral. The line is $y=1+\frac{4}{3}x$ on $[0,3]$, slope $\frac{4}{3}$, so $ds=\sqrt{1+\frac{16}{9}}dx=\frac{5}{3}dx$: $S={\int }_0^{3}2\pi \left(1+\frac{4}{3}x\right)\frac{5}{3}dx=\frac{10\pi }{3}{\left[x+\frac{2}{3}{x}^2\right]}_0^3=\frac{10\pi }{3}(3+6)=30\pi .$ Why this step? Two independent methods agreeing is the strongest possible sanity check.

Verify: Both give $30\pi \approx 94.25$. It's a frustum of radii $1$ and $5$, slant $5$: parent's frustum formula $\pi (r_1+r_2)\ell =\pi (1+5)(5)=30\pi$ too — triple agreement. ✅

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Recall Which cell is which? (self-test)

Match each situation to its matrix cell. Rotating $x=g(y)$ about the y-axis ::: Cell B (radius $=x$, but integrate in $y$) A curve dipping below the x-axis, rotated about it ::: Cell H (use $|y|$, split at crossings) Rotating $y=2$ about $y=7$ ::: Cell C (radius $=|2-7|=5$) A curve given as $(x(t),y(t))$ ::: Cell D (parametric $ds$) The point of the generating curve that touches the axis ::: Cell F (radius $\to 0$, contributes nothing)

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Connections

• Surface area of revolution — the parent formula every example applies.
• Arc length — the $ds$ factor, identical here.
• Integration by substitution — used in Ex 1, 2, 4, 8.
• Parametric curves — Ex 4's arc-length form.
• Frustum and cone geometry — sanity anchor for Ex 5, 6, 9.
• Pappus's theorem — the shortcut cross-check in Ex 9.
• Volume of revolution (disk & shell) — same "distance to axis" idea, different power of $r$.