Surface Area of a Solid of Revolution
Spin a curve around an axis and its skin has area \(S=\displaystyle\int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}\,dx\) — each tiny slanted piece of curve, of length \(ds\), sweeps out a thin frustum band of area \(2\pi r\,ds\), and the integral adds them all up.
By the end you'll be able to set up and evaluate \(\int 2\pi(\text{radius})\,ds\) for a surface of revolution, and pick the correct radius — \(f(x)\) or \(x\) — for the axis you revolve about.
Predict: a band out where the radius is large but the curve is nearly flat, versus a band where the radius is small but the curve is steep — which contributes more surface area per step in x? Slide x to check.
This is \(y=\sqrt{x}\) revolved about the x-axis. The slider positions one thin band: its radius r = f(x) and its slant factor \(\sqrt{1+[f'(x)]^2}\) multiply into the band's area rate \(2\pi r\cdot ds/dx\), plotted in the lower panel and accumulated (shaded) into the running surface area. Hover or focus the band — or hover the lower curve — for exact readouts.
r = f(x) = 1.000 · slant factor √(1+(f′)²) = 1.118 · band area rate 2πr·ds/dx = 7.025 · surface area so far S(0→x) = 5.330
The area of a surface of revolution is \(\int 2\pi(\text{radius})\sqrt{1+(f')^2}\,dx\) — the lateral area \(2\pi r\,ds\) of thin frustum bands, integrated along the curve.
Chop the curve into tiny slanted pieces of length \(ds=\sqrt{1+[f'(x)]^2}\,dx\) — the same arc-length element you built in the previous lesson. Revolved about the x-axis, each piece sweeps a thin band: unroll it and you get a strip of length \(2\pi r\) (the circumference it traces, with \(r=f(x)\)) and width \(ds\) (its slant). Band area \(=2\pi r\,ds\); sum them all: \[S=\int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}\;dx.\] That's exactly what the demo above plots — the lower curve IS \(2\pi r\cdot ds/dx\), and the shaded area under it IS the surface area.
Every formula of this type reads \(\int 2\pi(\text{radius})\,ds\), where the radius is the distance from the curve to the axis of revolution. About the x-axis the radius is \(f(x)\); about the y-axis it's \(x\) itself: \[S=\int_a^b 2\pi x\sqrt{1+[f'(x)]^2}\;dx.\] Engineers use this to price the material in tank shells, nose cones, and lampshades — anything made by spinning a profile. Two pitfalls do most of the damage: using \(ds\) alone (that's arc length, a length not an area) and using \(\pi r^2\) (that's the disk cross-section from volume by slicing).
Revolve \(y=\sqrt{x}\) on \([1,4]\) about the x-axis. Here \(f'(x)=\dfrac{1}{2\sqrt x}\), so \(1+[f'(x)]^2 = 1+\dfrac{1}{4x} = \dfrac{4x+1}{4x}\). Then \[S=\int_1^4 2\pi\sqrt{x}\cdot\frac{\sqrt{4x+1}}{2\sqrt{x}}\,dx =\pi\int_1^4\sqrt{4x+1}\,dx.\] Substitute \(u=4x+1\), \(du=4\,dx\); the limits become \(u=5\) to \(u=17\): \[S=\frac{\pi}{4}\cdot\frac{2}{3}\Big[u^{3/2}\Big]_5^{17} =\frac{\pi}{6}\big(17^{3/2}-5^{3/2}\big)\approx \frac{\pi}{6}(70.09-11.18)\approx\textbf{30.85}.\]
Same curve, a different window: revolve \(y=\sqrt{x}\) on \([2,6]\) about the x-axis. The setup collapses the same way: \(S=\pi\displaystyle\int_2^6\sqrt{4x+1}\,dx\). With \(u=4x+1\) the limits become \(u=9\) to \(u=25\), so \(S=\dfrac{\pi}{6}\big(25^{3/2}-9^{3/2}\big)\). Now finish it: \(25^{3/2}=125\) and \(9^{3/2}=\) ____, so \(S=\) ____.
Reveal the answer
\(9^{3/2}=(\sqrt9)^3=27\), so \(S=\dfrac{\pi}{6}(125-27)=\dfrac{98\pi}{6} =\dfrac{49\pi}{3}\approx 51.3\). Notice it's bigger than the \([1,4]\) answer even though the interval is the same length — out at larger x the radius \(\sqrt{x}\) is bigger, so each band carries more circumference. Set the slider above past x = 2 and watch the band area rate confirm it.
More info — why the slant ds, and not the width dx?
Wrap a strip of paper around the solid where the profile is steep: the strip must lie along the slanted surface, so its width is the slant length \(ds\), which is longer than the horizontal step \(dx\) whenever \(f'\neq 0\). Using \(2\pi r\,dx\) instead of \(2\pi r\,ds\) systematically undercounts — badly so where the curve is steep (in the demo, near \(x=0\), the slant factor blows up even as the radius shrinks). This is the same "don't drop the \(1+\)" trap as in the arc-length element from the previous lesson, because it's literally the same \(ds\). Paul's Online Math Notes (Dive deeper below) runs the frustum-band derivation in full detail.
Check your understanding
Set up (don't evaluate) the integral for the area of the surface generated by revolving \(y = x^3\) on \([0,1]\) about the x-axis.
In the surface-area formula \(S = \int 2\pi r\,ds\), what does the factor \(2\pi r\) represent geometrically?
You revolve \(y = f(x)\), \(a \le x \le b\), about the <em>y</em>-axis instead of the x-axis. In \(S = \int_a^b 2\pi(\text{radius})\,ds\), what is the radius?
The segment \(y = 3x\) on \([0,1]\) has arc length \(L = \sqrt{10}\) (since \(ds = \sqrt{1+3^2}\,dx = \sqrt{10}\,dx\)). Revolving it about the x-axis makes a cone. What is the cone's lateral surface area?
Recap
- About the x-axis: \(S=\int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}\,dx\) — circumference \(2\pi r\) times slant width \(ds\), summed along the curve.
- The radius is the distance to the axis of revolution: \(f(x)\) for the x-axis, \(x\) for the y-axis. Every version reads \(\int 2\pi(\text{radius})\,ds\).
- \(ds=\sqrt{1+[f'(x)]^2}\,dx\) is the same arc-length element as before — surface area is arc length's \(ds\) weighted by the circumference it sweeps.
- Don't confuse \(2\pi r\,ds\) (lateral band area, surface) with \(\pi r^2\,dx\) (disk cross-section, volume) — and never drop the "\(1+\)" inside the root.
Dive deeper
- Paul's Online Math Notes — Surface Area See the frustum-band derivation with additional worked integrals.
Sources
- Surface Area of a Solid of Revolution