Area of a Surface of Revolution
Summarize
The lateral area of a frustum of a cone
In this entry we shall see how the area of a surface generated by an arc of a curve rotating about a line is computed by means of the definite integral. As has been repeated several times in the various entries on the subject, integrals have a very close connection with geometry and are not merely analytical tools. Before throwing ourselves into the explanation, though, it is worth taking a small step back and recalling a few notions of elementary geometry.
For this, we shall start from a right circular cone (that is, a cone whose axis is perpendicular to the base) with slant height $\ell$. Recall that the slant height is the length of the sloping segment joining the vertex to a point of the base circle, and it is obtained from the formula:
$$\ell = \sqrt{h^2 + r^2} \tag{1}$$
This expression should be very familiar to you, where $h$ is the height and $r$ is the radius of the base of the cone. It is nothing other than the Pythagorean theorem applied to the cone, where the slant height is the hypotenuse of the right triangle and $h$ and $r$ are its legs. A cone such as the one just described unrolls in the plane into a circular sector of radius $\ell$ whose arc has length $2\pi r,$ equal to the circumference of the base. Since the area of a circular sector is half the product of the radius and the length of the arc, the lateral area of the cone can be written as:
$$S=\pi r\ell \tag{2}$$
Consider now a frustum of a cone with radii $r_1$ and $r_2,$ with $r_1\lt r_2,$ and slant height $\ell.$ We denote by $L_1$ the slant height of the cone of radius $r_1$ and by $L_2$ that of the cone of radius $r_2.$ The two cones are similar because they have the same vertex and parallel bases, while their slant heights satisfy $L_2-L_1=\ell.$ We can therefore write the following relation:
$$\frac{r_1}{L_1}=\frac{r_2}{L_2}$$
From this proportion and from the relation between the three slant heights the following lengths are obtained:
$$L_1=\frac{r_1\ell}{r_2-r_1} \qquad L_2=\frac{r_2\ell}{r_2-r_1}$$
The lateral area of the frustum is given by the difference between the lateral areas of the two cones, and formula $(2)$ can be rewritten as:
$$ \begin{align} S &= \pi r_2L_2-\pi r_1L_1 \\[6pt] &= \frac{\pi\ell r_2^2}{r_2-r_1}-\frac{\pi\ell r_1^2}{r_2-r_1} \\[6pt] &= \frac{\pi\ell(r_2^2-r_1^2)}{r_2-r_1} \\[6pt] &= \pi(r_1+r_2)\ell \end{align} $$
We now introduce the mean radius $\bar r=(r_1+r_2)/2$ so as to write the formula more compactly:
$$S=2\pi\bar r\ell \tag{3}$$
Note that if the two radii coincide the frustum degenerates into a cylinder and the area becomes $2\pi r\ell,$ while if one of the two vanishes the lateral area of the cone is recovered.
From the sum to the integral
After this tedious geometric introduction, we finally come to the part that concerns integrals. Let $f$ be a non-negative function of class $C^1$ (that is, differentiable and with continuous derivative $f'$) on the closed interval $[a,b],$ and let $\Sigma$ be the surface generated by the complete revolution of its graph about the $x$-axis. Consider a partition of the interval:
$$a=x_0 \lt x_1 \lt \dots \lt x_n=b$$
Now, on the subinterval $[x_{k-1},x_k]$ we replace the arc of the curve with the chord joining its endpoints. Rotating the chord generates a frustum of a cone whose radii are the ordinates of the endpoints, while its slant height is the length of the chord itself. Recalling formula $(1)$ we obtain:
$$\ell_k=\sqrt{(\Delta x_k)^2+\big(f(x_k)-f(x_{k-1})\big)^2}$$
By the mean value theorem there exists a point $\xi_k$ interior to the subinterval for which the difference of the ordinates equals $f'(\xi_k)\Delta x_k.$ Factoring out $(\Delta x_k)^2$ under the radical, the formula becomes:
$$\ell_k=\sqrt{1+\big[f'(\xi_k)\big]^2} \ \Delta x_k$$
At this point, adding the contributions of the individual frustums according to formula $(3)$ gives the area of the surface, which is approximately given by the following sum:
$$\Sigma_n=\pi\sum_{k=1}^{n}\big(f(x_{k-1})+f(x_k)\big)\sqrt{1+\big[f'(\xi_k)\big]^2} \ \Delta x_k \tag{4}$$
Careful! Expression $(4)$ is not a Riemann sum, because each summand involves three distinct points of the same subinterval: the ordinates at the two endpoints and the derivative at the point provided by the mean value theorem. Passing to the limit requires a little extra work, which we shall see in a moment.
Let me say straight away that the steps which follow may seem far from immediate at first, but I assure you that on a second reading they will turn out to be considerably clearer. So do not abandon the fight before winning it and hold on a little longer. The moment you feel worn out will be the moment when everything becomes obvious to you. With a bit of abstraction (or rather because somebody has already done it before us), we set:
$$M=\max_{[a,b]}\sqrt{1+[f'(x)]^2}$$
We know that this maximum exists because $f'$ is continuous on a closed and bounded interval. For each index $k$ we now write the sum of the two ordinates, isolating the value at the point $\xi_k$ provided by the mean value theorem:
$$f(x_{k-1})+f(x_k)=2f(\xi_k)+\eta_k$$ $$\eta_k=\big(f(x_{k-1})-f(\xi_k)\big)+\big(f(x_k)-f(\xi_k)\big)$$
Since $f$ is continuous on a compact interval it is also uniformly continuous. Setting $\delta=\max_{1\leq k\leq n}\Delta x_k,$ we denote by $\omega(\delta)$ its modulus of continuity. The points $x_{k-1},$ $x_k$ and $\xi_k$ are at distance at most $\delta$ from one another, so $|\eta_k|\leq2\omega(\delta)$ holds. The difference between $\Sigma_n$ and the sum built with the point $\xi_k$ can be represented by the following relation:
$$\left|\Sigma_n-2\pi\sum_{k=1}^{n}f(\xi_k)\sqrt{1+\big[f'(\xi_k)\big]^2} \ \Delta x_k\right|\leq2\pi M\omega(\delta)(b-a)$$
Letting the mesh $\delta$ of the partition tend to zero, uniform continuity guarantees that $\omega(\delta)$ tends to zero, so the right-hand side of the expression vanishes. The sum that remains is now a Riemann sum of the continuous function $2\pi f\sqrt{1+(f')^2}$ relative to the points $\xi_k,$ and as we know from the Riemann integrability criteria it converges to the integral of that same function. We can therefore write the area of the surface as:
$$S=2\pi\int_a^b f(x)\sqrt{1+\big[f'(x)\big]^2} \ dx \tag{5}$$
For regular surfaces a parametric definition of the area also exists. In the present case it coincides with the limit just obtained, which we take as the definition of the area of $\Sigma.$
Note that if $f$ changes sign, formula $(5)$ remains valid with $f$ replaced by $|f|,$ because the radius of each frustum is the distance between the curve and the axis. A clarification is in order for the disc method, where the radius appears squared and the absolute value is therefore superfluous.
The surface element
Formula $(5)$ is the product of two factors: the length $ds=\sqrt{1+[f'(x)]^2} \ dx$ introduced for the arc length and the length of the circle traced by the point of the curve as it rotates. Denoting by $r$ the distance of the point from the axis of revolution, we define the surface element as:
$$dS=2\pi r \ ds \tag{6}$$
Every variant of the formula is obtained by choosing $r$ and $ds$ according to the configuration of the problem, without redoing the construction from scratch each time, and this is inevitably very useful. For a summary of the various cases, keep the following table in mind:
| Graph of $f$ about the $x$-axis | $$2\pi\int_a^b f(x)\sqrt{1+\big[f'(x)\big]^2} \ dx$$ |
| Graph of $f$ about the $y$-axis, with $a\geq0$ | $$2\pi\int_a^b x\sqrt{1+\big[f'(x)\big]^2} \ dx$$ |
| Graph of $f$ about the line $y=c$ | $$2\pi\int_a^b\lvert f(x)-c\rvert\sqrt{1+\big[f'(x)\big]^2} \ dx$$ |
| Parametric curve about the $x$-axis | $$2\pi\int_{t_0}^{t_1}\lvert y(t)\rvert\sqrt{\big[x'(t)\big]^2+\big[y'(t)\big]^2} \ dt$$ |
In the second row the radius is the abscissa of the point and the condition $a\geq0$ prevents the curve from crossing the axis of revolution. The last row uses the length element in parametric form and covers curves that are not graphs of functions as well. If the rotation covers the same portion of surface more than once, the integral counts its area with the same multiplicity.
Example 1
We compute the area of the sphere of radius $r$ as the surface generated by rotating the semicircle $f(x)=\sqrt{r^2-x^2}$ about the $x$-axis. The function is not of class $C^1$ at the endpoints, so we apply formula $(5)$ on $[-r+\varepsilon,r-\varepsilon]$ and pass to the limit as $\varepsilon\to0.$ For $-r\lt x\lt r$ the derivative of the function equals:
$$f'(x)=-\frac{x}{\sqrt{r^2-x^2}}$$
With a few algebraic substitutions we obtain:
$$1+\big[f'(x)\big]^2=1+\frac{x^2}{r^2-x^2}=\frac{r^2}{r^2-x^2}$$
The factor $\sqrt{r^2-x^2}$ of the ordinate cancels the one in the denominator, and the product appearing in formula $(5)$ reduces to a constant:
$$f(x)\sqrt{1+\big[f'(x)\big]^2}=\sqrt{r^2-x^2}\cdot\frac{r}{\sqrt{r^2-x^2}}=r$$
The product admits a continuous extension equal to $r$ at the endpoints as well, so the improper integral is:
$$S=2\pi\int_{-r}^{r}r \ dx=4\pi r^2$$
The sphere of radius $r$ therefore has area $4\pi r^2,$ as we know from the notions of elementary geometry we carry with us from secondary school. Note how the area of a spherical zone depends only on the distance between the two parallel planes that bound it, and not on their position with respect to the centre. This means that a spherical zone and the portion of the circumscribed cylinder bounded by the same planes have the same area.
Example 2
Consider now the function $f(x)=\sqrt{x}$ on the interval $[0,4]$ and rotate its graph about the $x$-axis, obtaining the paraboloid whose volume is given by the disc method. The derivative and the radicand are:
$$f'(x)=\frac{1}{2\sqrt{x}}$$ $$1+\big[f'(x)\big]^2=1+\frac{1}{4x}=\frac{4x+1}{4x}$$
The factor $\sqrt{x}$ of the ordinate cancels the denominator of the radical, and the integrand reduces to the square root of a linear function:
$$f(x)\sqrt{1+\big[f'(x)\big]^2}=\sqrt{x}\cdot\frac{\sqrt{4x+1}}{2\sqrt{x}}=\sqrt{x+\frac{1}{4}}$$
Evaluating the integral, which is again immediate, we obtain:
$$ \begin{align} S &= 2\pi\int_0^4\sqrt{x+\frac{1}{4}} \ dx \\[6pt] &= 2\pi\cdot\frac{2}{3}\left[\left(x+\frac{1}{4}\right)^{3/2}\right]_0^4 \\[6pt] &= \frac{4\pi}{3}\left(\left(\frac{17}{4}\right)^{3/2}-\left(\frac{1}{4}\right)^{3/2}\right) \\[6pt] &= \frac{4\pi}{3}\cdot\frac{17\sqrt{17}-1}{8} \\[6pt] &= \frac{\pi}{6}\big(17\sqrt{17}-1\big) \end{align} $$
The area of the surface equals $\pi(17\sqrt{17}-1)/6.$ The derivative of $f$ is not bounded in a right neighbourhood of the origin, so $f$ is not of class $C^1$ on the closed interval, but the integrand defined on $(0,4]$ admits a continuous extension to $[0,4]$ because the factor $f(x)$ vanishes at the origin at the same rate at which the radical diverges. The value of the extension at the origin is $1/2.$
Example 3
We now look at an arch of the cycloid, which is described by the following parametric equations, with $t$ varying in $[0,2\pi]$ and $r$ the radius of the circle:
$$x(t)=r(t-\sin t)$$ $$y(t)=r(1-\cos t)$$
When we rotate the arch about the $x$-axis, we reuse the length element already obtained in the computation of the arc length:
$$\sqrt{\big[x'(t)\big]^2+\big[y'(t)\big]^2}=2r\sin\frac{t}{2}$$
The ordinate is rewritten with the half-angle identity as $y(t)=2r\sin^2(t/2),$ and the integrand then contains only powers of the sine of $t/2$:
$$ \begin{align} S &= 2\pi\int_0^{2\pi}2r\sin^2\frac{t}{2}\cdot2r\sin\frac{t}{2} \ dt \\[6pt] &= 8\pi r^2\int_0^{2\pi}\sin^3\frac{t}{2} \ dt \\[6pt] &= 16\pi r^2\int_0^{\pi}\sin^3u \ du \end{align} $$
The last step is the substitution $u=t/2,$ which turns $dt$ into $2 \ du$ and halves the endpoints. Recall that the integral of an odd power of the sine is computed by isolating one factor and using the fundamental identity, and it equals $4/3$:
$$S=16\pi r^2\cdot\frac{4}{3}=\frac{64}{3}\pi r^2$$
The area of a curved surface cannot in general be defined as the limit of the areas of inscribed polyhedra. Even when the diameter of the faces tends to zero, the limit can depend on the way the polyhedra are built, as the Schwarz lantern shows for the cylinder. The approximation by frustums of cones avoids this ambiguity because it ties the inclination of each element to the corresponding chord of the generating curve.
Some closing remarks
In closing we add a few remarks that provide some orientation on the use of what has just been shown.
- Formula $(5)$ was derived with $f$ non-negative and of class $C^1$ on $[a,b],$ a hypothesis that makes the integrand continuous. As the first two examples show, this is a sufficient but not necessary condition. The formula remains valid, for example, if $f$ is continuous on $[a,b],$ of class $C^1$ outside a finite number of points and the improper integrals of the integrand converge.
- In the second row of the table the condition $a\geq0$ allows $x$ to be used as the radius. If the graph crosses the $y$-axis, the radius is $|x|$ and two distinct points generate the same circle only when they have the same ordinate and opposite abscissae. For a parametric curve as well, one has to check that the surface is not traversed more than once.
- The method of parallel cross sections returns the exact volume, but the lateral walls of the approximating cylinders form a stepped surface whose area does not in general converge to the area of $\Sigma.$ The volume of a solid of revolution and the area of its surface require two distinct constructions.
- When the integral in $(5)$ has no elementary antiderivative, the value is determined numerically.
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