Deriving Surface Area From Frustum Bands

From Profile Length to Surface Area

A surface of revolution is formed by rotating a profile curve around an axis. To find its area, it helps to imagine the curve cut into many short pieces. When each piece is straight, rotating it produces a frustum band. The area of each band is determined by its slant length and the radii at its two ends.

Making the pieces shorter improves the approximation of the original curved surface. This tutorial derives the resulting surface-area element and explains what the integral counts if the profile or generated surface overlaps itself.

The Area of One Frustum Band

Use coordinates in a plane containing the axis of rotation: let $x$ measure distance parallel to the axis, and let $r$ be the nonnegative distance from the profile point to the axis. Consider a straight profile segment whose endpoint radii are $r_1$ and $r_2$, and whose length is $\ell$. Rotating that segment produces a frustum band.

The lateral area of a frustum with endpoint radii $r_1,r_2$ and slant height $\ell$ is

A_{\mathrm{band}}=\pi(r_1+r_2)\ell.

One way to see the formula is to imagine cutting the frustum along a generator and flattening its curved side. The flattened band is a sector of an annulus. Its area is the average of the two circumference lengths, $2\pi r_1$ and $2\pi r_2$, multiplied by the slant height. Thus its area is $\frac{2\pi r_1+2\pi r_2}{2}\ell=\pi(r_1+r_2)\ell$.

This formula applies to the lateral band, not the circular end faces. In a surface built from adjacent bands, those end faces are not included in the surface area being approximated.

Worked Example: Area of One Frustum Band

A straight segment joins a point of radius $1$ to a point of radius $4$. The axial separation is $4$. Find the lateral area generated by rotating the segment.

The change in radius is $4-1=3$. The segment’s slant length follows from the right triangle formed by the axial and radial changes:

\ell=\sqrt{4^2+3^2}=\sqrt{16+9}=5.

Substitute the endpoint radii and slant length into the frustum formula:

A_{\mathrm{band}} =\pi(1+4)(5) =25\pi.

Check: the mean circumference is $\frac{2\pi(1)+2\pi(4)}{2}=5\pi$. Multiplying it by the slant height gives $5\pi\cdot5=25\pi$, as required.

Short Bands Approximate a Smooth Profile

Now let a smooth profile be parametrized by $(x(t),r(t))$, for $a\le t\le b$, with $r(t)\ge 0$. Here $x(t)$ is the coordinate parallel to the axis, and $r(t)$ is the distance to it. Divide the parameter interval into pieces $a=t_0<t_1<\cdots<t_n=b$. On the $i$-th piece, replace the curved profile by the straight chord joining its endpoints.

Write $r_i=r(t_i)$, and let $\Delta s_i$ be the chord length between $(x(t_i),r_i)$ and $(x(t_{i+1}),r_{i+1})$. Its value is

\Delta s_i = \sqrt{(x(t_{i+1})-x(t_i))^2+(r_{i+1}-r_i)^2}.

Rotating this chord gives a frustum band. Its lateral area is exactly $\pi(r_i+r_{i+1})\Delta s_i$, or equivalently $2\pi$ times the mean endpoint radius times the chord length. Adding the bands gives the polygonal approximation

A_n = \sum_{i=0}^{n-1} 2\pi\left(\frac{r_i+r_{i+1}}{2}\right)\Delta s_i.

As the partition is refined, the chord lengths approximate the corresponding arc lengths, and the endpoint radii become close to the radius along each short piece. The limiting sum therefore weights each small arc length by its local circumference $2\pi r$.

Surface-area element: For a smooth profile with arc-length element $ds$ and distance $r$ from the axis of rotation, the surface-area element is $$ dA=2\pi r\,ds. $$

Derivation of the Integral

For a differentiable parametrization, the arc-length element is $ds=\sqrt{x'(t)^2+r'(t)^2}\,dt$. Substituting this expression into the surface-area element gives a parameter-based formula. The following argument explains why the frustum sums converge to it.

Write each band’s area.
For the chord from parameter $t_i$ to $t_{i+1}$, the area is $2\pi\bar r_i\Delta s_i$, where $\bar r_i=(r_i+r_{i+1})/2$.

Compare chord length with arc length.
As the parameter intervals shrink, the chord direction and length approach those of the curve on each interval. For a continuously differentiable profile, the sums of these chord lengths, weighted by a continuous radius, converge to the corresponding arc-length integral.

Replace the mean endpoint radius by the local radius.
Continuity of $r$ ensures that the difference between $\bar r_i$ and any radius on the short interval tends uniformly to zero as the largest interval shrinks. The total weighted-sum error therefore tends to zero.

Take the limit.
The limiting weighted arc-length sum is the integral of $2\pi r$ with respect to arc length.

In parameter form, the result is

A = \int_a^b 2\pi r(t)\sqrt{x'(t)^2+r'(t)^2}\,dt.

This derivation assumes a piecewise smooth profile so that the arc length and the limiting sums are well defined. The formula measures the area generated by traversing the profile. If the traversal generates the same surface portion more than once, the integral counts that portion repeatedly. To obtain ordinary geometric area, the generated surface must not retrace or overlap portions of positive area; otherwise, the overlap must be accounted for separately.

Checking the Formula on a Straight Profile

Worked Example: A Line Segment with Changing Radius

Let $x(t)=3t$ and $r(t)=2+3t$ for $0\le t\le1$. The profile is a straight segment, so its revolution is one frustum band. Compute its area in both ways.

The endpoint radii are $r(0)=2$ and $r(1)=5$. The axial change is $x(1)-x(0)=3$, and the radial change is $5-2=3$. Hence the slant length is

\ell=\sqrt{3^2+3^2}=3\sqrt{2}.

The frustum formula gives

A_{\mathrm{band}} =\pi(2+5)(3\sqrt{2}) =21\sqrt{2}\pi.

For the integral, $x'(t)=3$ and $r'(t)=3$, so $ds=\sqrt{3^2+3^2}\,dt=3\sqrt{2}\,dt$. Therefore

\begin{aligned} A &=\int_0^1 2\pi(2+3t)(3\sqrt{2})\,dt\\ &=6\pi\sqrt{2}\int_0^1(2+3t)\,dt\\ &=6\pi\sqrt{2}\left[2t+\frac{3t^2}{2}\right]_0^1\\ &=6\pi\sqrt{2}\left(2+\frac{3}{2}\right) =21\sqrt{2}\pi. \end{aligned}

Check: the integral agrees with the exact frustum calculation. The profile has increasing $x$, so it does not retrace a meridian segment, and the generated bands do not overlap in area.

Checking the Formula on a Curved Profile

Worked Example: A Curved Profile with an Exact Integral

Consider $x(t)=t^2$, $r(t)=t$, for $0\le t\le1$. The curve is regular because $x'(t)=2t$, $r'(t)=1$, and $x'(t)^2+r'(t)^2=4t^2+1>0$ throughout the interval. Also, $x(t)$ increases strictly, so the profile does not retrace itself.

The arc-length factor is

\frac{ds}{dt} = \sqrt{(2t)^2+1^2} = \sqrt{1+4t^2}.

Applying the derived formula gives

\begin{aligned} A &=\int_0^1 2\pi t\sqrt{1+4t^2}\,dt. \end{aligned}

Set $u=1+4t^2$, so $du=8t\,dt$, or $t\,dt=du/8$. When $t=0$, $u=1$; when $t=1$, $u=5$. Thus

\begin{aligned} A &=2\pi\cdot\frac18\int_1^5u^{1/2}\,du\\ &=\frac{\pi}{4}\left[\frac23u^{3/2}\right]_1^5\\ &=\frac{\pi}{6}\left(5\sqrt{5}-1\right). \end{aligned}

Check: differentiating the antiderivative $\frac{\pi}{6}(1+4t^2)^{3/2}$ gives $2\pi t\sqrt{1+4t^2}$, the integrand. Evaluating it at $1$ and $0$ produces the stated area.

Multiplicity and a Common Pitfall

A regular parametrization has nonzero velocity, but that alone does not mean it traces the profile just once. If it goes around the same profile twice, the frustum sums include the same bands twice. The integral then gives area counted with multiplicity, not the ordinary area of the union of the generated surface.

Worked Example: A Regular Profile Traversed Twice

Take $x(t)=\cos t$, $r(t)=2+\sin t$, for $0\le t\le4\pi$. Its speed is

\sqrt{x'(t)^2+r'(t)^2} = \sqrt{\sin^2t+\cos^2t} =1.

The radius is positive because $r(t)\ge1$. But both coordinates repeat after $2\pi$, so the profile is traced twice on the given interval. The surface-area integral over $0\le t\le4\pi$ is therefore twice the integral over $0\le t\le2\pi$. For the once-traced surface, restrict the parameter interval to $0\le t\le2\pi$. Regularity and positive radius do not remove this repetition.

A related mistake is replacing the slant length by the axial change alone. For the segment in the first example, using $4$ instead of the correct slant length $5$ would give $\pi(1+4)(4)=20\pi$, not $25\pi$. The correction is to use the length along the profile: for endpoints separated by changes $\Delta x$ and $\Delta r$, it is $\sqrt{(\Delta x)^2+(\Delta r)^2}$.

Why the Element Matters

The expression $2\pi r\,ds$ separates the geometry of a thin band into two parts: $2\pi r$ is the circumference at its local radius, and $ds$ is the small distance traveled along the profile. It is the limiting form of the frustum calculation, not a rule that assumes the profile is a straight line.

When using the formula, identify the radius as the distance to the axis, use arc length along the profile rather than just its axial change, and check whether the parametrization or generated surface repeats. These checks determine whether the integral represents the desired geometric area or counts some parts more than once.

Check Your Understanding

Use the frustum-band derivation to answer the following questions.

Why does a straight profile segment produce a frustum band when revolved?
What quantities determine the slant length of a chord joining two profile points?
For a parametrized profile $(x(t),r(t))$, what is the arc-length factor $ds/dt$?
Why does the surface-area integral count a surface patch twice when the profile traces it twice?
What error results from using axial change instead of slant length for the first worked example?