Spin a curve around an axis and it sweeps out a surface: a sphere, a cone, a vase, a satellite dish. How much material does it take to cover it? The answer combines the two previous posts: the arc length \(ds\) of the curve, and the circumference \(2\pi y\) of the circle that each point travels. This post explains why the formula is \(2\pi\int y\,ds\), why the tempting \(2\pi\int y\,dx\) is wrong, and works through the classic shapes, ending with a horn you can fill with paint but never paint.
1. The formula
Rotation about the \(x\)-axis
The area of the surface generated by rotating the arc of the curve \( y = f(x) \ge 0 \), between the points with abscissas \(x = a\) and \(x = b\), about the \(x\)-axis is
\[ P_x = 2\pi\int_a^b y\,ds = 2\pi\int_a^b y\sqrt{1 + y'^2}\,dx. \tag{1} \]
Read it as: (circumference traced by a point) × (length element), added up. A point of the curve at height \(y\) moves on a circle of radius \(y\), of length \(2\pi y\). A small piece \(ds\) of the curve therefore sweeps a thin band of area about \(2\pi y\,ds\).
Computing "the area of a surface" in this way is called complanation (from Latin complanare, to flatten out): conceptually, we cut the surface into thin bands and flatten each of them.
2. Why \(2\pi y\,ds\): thin bands
Replace the curve by an inscribed polygon, as for arc length. Rotating one segment of the polygon gives a frustum of a cone (a cone with the tip cut off). Its lateral area is a classical formula:
\[ (\text{lateral area of a frustum}) = 2\pi\cdot\frac{r_1 + r_2}{2}\cdot l, \]
where \(r_1, r_2\) are the radii of the two rims and \(l\) the slant length. In words: the circumference at the average radius times the slant length. (Cut the frustum along a slant and unroll it: you get a ring sector, whose area works out to exactly this.)
For a segment of the polygon, \( \frac{r_1 + r_2}{2} \approx y \) and \( l = \sqrt{\Delta x^2 + \Delta y^2} \approx ds \). Summing the frustums and passing to the limit gives formula (1).
Figure 1. Rotating the curve \(y = f(x)\) about the \(x\)-axis. A short piece \(ds\) of the curve at height \(y\) sweeps a thin band (green), almost a frustum of a cone, of area \(2\pi y\,ds\).
A common mistake: \(2\pi\int y\,dx\)
For volumes of revolution, slicing into thin cylinders of thickness \(dx\) is fine. For areas it is not: a thin band on a slanted part of the surface is wider than \(dx\); its width is the slant length \(ds = \sqrt{1 + y'^2}\,dx\). Using \(dx\) would, for a cone, give \(\pi rh\) instead of the correct \(\pi rl\), with \(l = \sqrt{r^2 + h^2}\) the slant height. The approximation error does not disappear in the limit, because the slope does not shrink when the bands get thinner.
3. Classic surfaces
Example 1 · The sphere
Rotate \( y = \sqrt{R^2 - x^2} \), \( -R \le x \le R \). Then \( y' = -\frac{x}{y} \), and the root simplifies beautifully:
\[ y\sqrt{1 + y'^2} = y\sqrt{\frac{y^2 + x^2}{y^2}} = \sqrt{x^2 + y^2} = R. \]
\[ P = 2\pi\int_{-R}^RR\,dx = 4\pi R^2. \]
Example 2 · Archimedes' zone theorem
The computation above shows more: the part of the sphere between the planes \(x = x_1\) and \(x = x_2\) (a spherical zone) has area
\[ 2\pi\int_{x_1}^{x_2}R\,dx = 2\pi R\,(x_2 - x_1) = 2\pi Rh. \]
It depends only on the height \(h\) of the zone, not on where it is. A slice of an orange of thickness 1 cm has the same peel area whether it comes from the middle or near the end. This is also why the area of the sphere equals the lateral area of the circumscribed cylinder, \(2\pi R\cdot 2R\), the result Archimedes wanted on his tombstone.
Example 3 · The cone
Rotate \( y = \frac rhx \), \( 0 \le x \le h \): \( \sqrt{1 + y'^2} = \frac{\sqrt{h^2 + r^2}}{h} = \frac lh \).
\[ P = 2\pi\int_0^h\frac rhx\cdot\frac lh\,dx = 2\pi\frac{rl}{h^2}\cdot\frac{h^2}{2} = \pi rl. \]
The school formula for the lateral area of a cone, recovered.
Example 4 · A paraboloid (satellite dish)
Rotate \( y = \sqrt x \), \( 0 \le x \le 1 \): \( y' = \frac{1}{2\sqrt x} \), \( y\sqrt{1 + y'^2} = \sqrt{x + \frac14} \).
\[ P = 2\pi\int_0^1\sqrt{x + \tfrac14}\,dx = \frac{4\pi}{3}\Big[\big(x + \tfrac14\big)^{3/2}\Big]_0^1 = \frac{\pi}{6}\big(5\sqrt5 - 1\big) \approx 5.330. \]
4. Rotation about the \(y\)-axis
Rotation about the \(y\)-axis
If the arc is given as \( x = g(y) \ge 0 \), \( c \le y \le d \), and rotated about the \(y\)-axis:
\[ P_y = 2\pi\int_c^d x\,ds = 2\pi\int_c^d x\sqrt{1 + x'^2}\,dy, \qquad x' = \frac{dx}{dy}. \]
Same idea with the roles of the axes swapped: now the radius of rotation is the distance \(x\) to the \(y\)-axis. In the form \(2\pi\int x\,ds\), the formula can be used with any variable: \(ds\) can just as well be written as \( \sqrt{1 + y'^2}\,dx \), if that is simpler.
Example 5 · The same dish, the other way
Rotate the parabola \( y = x^2 \), \( 0 \le x \le 1 \), about the \(y\)-axis. Using \( ds = \sqrt{1 + 4x^2}\,dx \):
\[ P_y = 2\pi\int_0^1x\sqrt{1 + 4x^2}\,dx = \frac{\pi}{6}\Big[(1 + 4x^2)^{3/2}\Big]_0^1 = \frac{\pi}{6}\big(5\sqrt5 - 1\big). \]
The same number as in Example 4, as it should be: \(y = \sqrt x\) rotated about the \(x\)-axis and \(y = x^2\) rotated about the \(y\)-axis are the same surface, with the roles of the axes swapped.
5. Parametric and polar curves
"If the curve is given in another form, make the corresponding change of variables in formula (1)": keep \( P = 2\pi\int y\,ds \) and express \(y\) and \(ds\) in the curve's own variable.
Curve
Area of the surface (rotation about the \(x\)-axis)
Example 6 · Rotating an arch of the cycloid
\( x = a(t - \sin t) \), \( y = a(1 - \cos t) = 2a\sin^2\frac t2 \), and from the post on arc length, \( ds = 2a\sin\frac t2\,dt \):
\[ P = 2\pi\int_0^{2\pi}2a\sin^2\tfrac t2\cdot 2a\sin\tfrac t2\,dt = 8\pi a^2\int_0^{2\pi}\sin^3\tfrac t2\,dt = 16\pi a^2\int_0^\pi\sin^3u\,du = 16\pi a^2\cdot\frac43 = \frac{64\pi a^2}{3}. \]
Example 7 · Rotating a cardioid
\( \rho = a(1 + \cos\varphi) \) about the polar axis. The upper half (\(0 \le \varphi \le \pi\)) generates the whole surface; \( y = \rho\sin\varphi \), \( ds = 2a\cos\frac\varphi2\,d\varphi \). With \(\varphi = 2u\), \( 1 + \cos\varphi = 2\cos^2u \), \( \sin\varphi = 2\sin u\cos u \):
\[ P = 2\pi\int_0^\pi a(1 + \cos\varphi)\sin\varphi\cdot 2a\cos\tfrac\varphi2\,d\varphi = 32\pi a^2\int_0^{\pi/2}\cos^4u\sin u\,du = \frac{32\pi a^2}{5}. \]
6. Pappus's theorem
Formula (1) can be rewritten in a way that needs almost no computation. The \(y\)-coordinate of the centroid of an arc of length \(L\) is \( \bar y = \frac1L\int y\,ds \). Therefore
Pappus's (Guldin's) first theorem
\[ P = 2\pi\bar y\cdot L: \]
the area of a surface of revolution equals the length of the rotating arc times the distance travelled by its centroid (for an arc that does not cross the axis).
Example 8 · The torus
A circle of radius \(a\) whose centre is at distance \(b > a\) from the axis generates a torus (a doughnut). The centroid of the circle is its centre, which travels \(2\pi b\); the length of the circle is \(2\pi a\):
\[ P = 2\pi a\cdot 2\pi b = 4\pi^2ab. \]
(The direct computation with \( x = a\cos t \), \( y = b + a\sin t \) gives \( 2\pi\int_0^{2\pi}(b + a\sin t)\,a\,dt = 4\pi^2ab \) as well.)
7. Gabriel's horn
Finite volume, infinite surface
Rotate \( y = \frac1x \), \( x \ge 1 \), about the \(x\)-axis. The volume is finite:
\[ V = \pi\int_1^\infty\frac{dx}{x^2} = \pi. \]
But the surface area is infinite, since \( \sqrt{1 + y'^2} \ge 1 \):
\[ P = 2\pi\int_1^\infty\frac1x\sqrt{1 + \frac{1}{x^4}}\,dx \ \ge\ 2\pi\int_1^\infty\frac{dx}{x} = \infty. \]
The paradox: you could fill the horn with \(\pi\) litres of paint, yet no amount of paint would cover its inner surface. The resolution is that "a layer of paint" has a thickness, and the horn eventually becomes thinner than any layer.
Problem 2: Rotate the catenary \( y = \cosh x \), \( -1 \le x \le 1 \), about the \(x\)-axis.
\( \sqrt{1 + \sinh^2 x} = \cosh x \), so \( P = 2\pi\int_{-1}^1\cosh^2 x\,dx = 2\pi\Big[\frac x2 + \frac{\sinh 2x}{4}\Big]_{-1}^1 = \pi(2 + \sinh 2) \approx 17.68 \). (This surface, the catenoid, is the shape a soap film takes between two rings.)
Problem 3: Use Pappus to find the lateral area of a cylinder of radius \(r\) and height \(h\).
Rotate a segment of length \(h\) parallel to the axis at distance \(r\): its centroid travels \(2\pi r\), so \( P = 2\pi rh \).
Problem 4: Find the area of the ellipsoid obtained by rotating the ellipse \( \frac{x^2}{4} + y^2 = 1 \) about the \(x\)-axis (major axis).
\( y = \frac12\sqrt{4 - x^2} \), \( y\sqrt{1 + y'^2} = \frac14\sqrt{16 - 3x^2} \). Then \( P = \frac\pi2\int_{-2}^2\sqrt{16 - 3x^2}\,dx \). This is the integral \( \int\sqrt{c^2 - u^2}\,du \) from the post on irrational functions (with \( u = \sqrt3\,x \), \( c = 4 \)), and it gives
\[ P = 2\pi + \frac{8\pi^2}{3\sqrt3} \approx 21.48. \]
For comparison, the sphere of radius 1 has area \(4\pi \approx 12.57\), and the sphere of radius 2 has \(16\pi \approx 50.27\).
Problem 5: Show that a spherical cap of height \(h\) has area \(2\pi Rh\).
A cap is a zone with \( x_1 = R - h \), \( x_2 = R \); Example 2 gives \( 2\pi R\cdot h \). For \(h = R\) (a hemisphere): \(2\pi R^2\).
Problem 6: Why must \(y \ge 0\) in formula (1)?
\(y\) plays the role of a radius. If the arc lies below the axis, the radius is \(|y|\), so use \( 2\pi\int|y|\,ds \); an arc crossing the axis must be split at the crossing points (and parts that are mirror images of each other generate the same surface, so they must not be counted twice).
Final thoughts
The surface-area formula is a product of two earlier ideas: arc length \(ds\) for how long each piece of the curve is, and the circumference \(2\pi y\) for how far it travels. Keep \( P = 2\pi\int(\text{radius})\,ds \) in mind and every variant follows, for either axis and any description of the curve. The classic results are some of the most elegant in calculus: the sphere's \(4\pi R^2\), Archimedes' zones, Pappus's theorem, and a horn with finite volume and infinite surface.