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Surface Area and Scalar Integrals · Tutorial 858 of 999

Area of Cylinders, Cones, and Surfaces of Revolution

Use parametrizations to derive and calculate the areas of cylindrical, conical, and general surfaces of revolution.

Advanced 10 min read

What You'll Learn

  • Parametrize a surface of revolution using a profile curve and an angular variable
  • Derive the surface-area element from the cross product of tangent vectors
  • Calculate lateral area for cylinders and cones, excluding any end caps
  • Use a graph profile to express the area in terms of height or radius
  • Distinguish a full revolution from a partial angular patch
  • Check parametrization bounds and avoid counting overlapping surface area

From a Profile Curve to a Surface

A cylinder or cone can be described by a curve in a vertical plane, rotated around an axis. The same construction works for many other surfaces. To calculate their area, we first parametrize the surface and then measure how its two parameters stretch a small rectangle into a small piece of the surface.

Take the vertical axis as the axis of rotation. Let the profile curve be given by \((\rho(t),z(t))\), where \(\rho(t)\) is its distance from the axis and \(z(t)\) is its height. We require \(\rho(t)\geq 0\). Rotating the profile through an angle \(\theta\) gives the parametrization

$$ \mathbf{r}(t,\theta) = \bigl(\rho(t)\cos\theta,\ \rho(t)\sin\theta,\ z(t)\bigr). $$

Here \(t\) runs over an interval \([a,b]\), and \(\theta\) runs over an angular interval \([\alpha,\beta]\). If \(\rho(t)>0\) and \(\beta-\alpha=2\pi\), each profile point traces a full circle around the axis. A narrower angular interval describes only an angular patch, not the full surface.

Definition: For a regular parametrized surface \(\mathbf{r}(t,\theta)\), its area element is \(dS=\lVert\mathbf{r}_t\times\mathbf{r}_\theta\rVert\,dt\,d\theta\). The area over a parameter domain \(D\) is the double integral of this factor over \(D\), provided the parametrization is sufficiently smooth and does not count overlapping pieces more than once.

Deriving the Area Element

Differentiate the parametrization with respect to each parameter. The \(t\)-derivative follows the profile as it rotates; the \(\theta\)-derivative moves around a circle of radius \(\rho(t)\).

$$ \mathbf{r}_t = \bigl(\rho'(t)\cos\theta,\ \rho'(t)\sin\theta,\ z'(t)\bigr), \qquad \mathbf{r}_\theta = \bigl(-\rho(t)\sin\theta,\ \rho(t)\cos\theta,\ 0\bigr). $$

The cross product is

$$ \mathbf{r}_t\times\mathbf{r}_\theta = \bigl(-\rho(t)z'(t)\cos\theta,\ -\rho(t)z'(t)\sin\theta,\ \rho(t)\rho'(t)\bigr). $$

Squaring its length and using \(\cos^2\theta+\sin^2\theta=1\) gives

$$ \begin{aligned} \lVert\mathbf{r}_t\times\mathbf{r}_\theta\rVert^2 &=\rho(t)^2z'(t)^2\cos^2\theta +\rho(t)^2z'(t)^2\sin^2\theta +\rho(t)^2\rho'(t)^2\\ &=\rho(t)^2\bigl(z'(t)^2+\rho'(t)^2\bigr). \end{aligned} $$

Since \(\rho(t)\geq0\), taking the square root yields the area factor. This calculation explains why the radius appears as a multiplier: an angular change \(d\theta\) sweeps out a distance \(\rho(t)d\theta\), while a change in \(t\) moves a distance \(\sqrt{\rho'(t)^2+z'(t)^2}\,dt\) along the profile.

Area formula: For a profile \((\rho(t),z(t))\), the area over \(a\leq t\leq b\) and \(\alpha\leq\theta\leq\beta\) is $$ A=\int_\alpha^\beta\int_a^b \rho(t)\sqrt{\rho'(t)^2+z'(t)^2}\,dt\,d\theta. $$ If the profile is traced once without producing overlapping surface pieces, and the angular interval has width exactly \(2\pi\), this gives the area of the full surface of revolution (apart from any pieces on the axis, which have zero area).

Because the area factor does not depend on \(\theta\), the angular integral is simply the interval width \(\beta-\alpha\). Thus

$$ A=(\beta-\alpha)\int_a^b \rho(t)\sqrt{\rho'(t)^2+z'(t)^2}\,dt. $$

When the profile is a graph \(\rho=\rho(z)\), with \(z\) increasing from \(z_0\) to \(z_1\), take \(t=z\). For a full revolution, this becomes

$$ A=2\pi\int_{z_0}^{z_1} \rho(z)\sqrt{1+\left(\frac{d\rho}{dz}\right)^2}\,dz. $$

Alternatively, if the profile is expressed as \(z=z(\rho)\), the same parametrization with \(t=\rho\) gives \(A=2\pi\int \rho\sqrt{1+(dz/d\rho)^2}\,d\rho\). These graph formulas require the profile to be represented by the indicated single-valued function over the interval; the parameter formula applies more generally.

Worked Example: Lateral Area of a Cylinder

Worked Example: Lateral Area of a Cylinder

Find the lateral area of a cylinder of radius \(3\) and height \(4\). The profile is the vertical line \(\rho(t)=3,\ z(t)=t\), for \(0\leq t\leq4\). Use \(0\leq\theta\leq2\pi\) for the full cylinder.

The derivatives are \(\rho'(t)=0\) and \(z'(t)=1\). Therefore the area factor is

$$ \rho(t)\sqrt{\rho'(t)^2+z'(t)^2} =3\sqrt{0^2+1^2}=3. $$

Integrate over the height and the full angular interval:

$$ \begin{aligned} A &=\int_0^{2\pi}\int_0^4 3\,dt\,d\theta\\ &=\int_0^{2\pi}12\,d\theta\\ &=24\pi. \end{aligned} $$

As a check, the circumference of each horizontal circle is \(2\pi(3)=6\pi\), so circumference times height is \(6\pi\cdot4=24\pi\), as obtained. This is the lateral area only; neither circular end cap is included.

Cones and Their Sloping Profiles

A cone has a straight-line profile whose radius changes with height. The area factor includes the slant of that line, not just its vertical height. For a cone with base radius \(R\) and height \(H\), the profile can be written \(\rho(z)=Rz/H\) for \(0\leq z\leq H\). Then \(d\rho/dz=R/H\), so the full lateral area is

$$ A=2\pi\int_0^H\frac{Rz}{H} \sqrt{1+\left(\frac{R}{H}\right)^2}\,dz =\pi R\sqrt{R^2+H^2}. $$

The square root \(\sqrt{R^2+H^2}\) is the slant height. The formula follows directly from the surface-area element; it does not include the base disk.

Worked Example: Lateral Area of a Cone

Consider a cone of height \(4\) and top radius \(2\). Its profile is \(\rho(z)=z/2\) for \(0\leq z\leq4\). Find the full lateral area.

First, \(d\rho/dz=1/2\), so the graph-profile formula gives

$$ \begin{aligned} A &=2\pi\int_0^4 \frac{z}{2}\sqrt{1+\left(\frac12\right)^2}\,dz\\ &=2\pi\int_0^4 \frac{z}{2}\cdot\frac{\sqrt5}{2}\,dz\\ &=2\pi\cdot\frac{\sqrt5}{4} \left[\frac{z^2}{2}\right]_0^4\\ &=2\pi\cdot\frac{\sqrt5}{4}\cdot 8\\ &=4\pi\sqrt5. \end{aligned} $$

For a check, the slant height is \(\sqrt{4^2+2^2}=2\sqrt5\). The cone formula \(\pi R\) times slant height gives \(\pi\cdot2\cdot2\sqrt5=4\pi\sqrt5\), agreeing with the integral. The radius is zero at the tip, where the parametrization degenerates; that single point contributes no area.

A General Profile Need Not Be a Graph

The profile-parameter formula also handles curves that are inconvenient to express as \(\rho(z)\) or \(z(\rho)\). For a regular profile curve, its speed is \(\sqrt{\rho'(t)^2+z'(t)^2}\). The area swept out by a thin profile segment is its length times the circumference at that radius, \(2\pi\rho(t)\). Integrating these contributions gives the full-revolution formula. The cross-product derivation above makes this argument precise.

Worked Example: A Curved Profile Segment

Rotate the profile \(\rho(t)=2+\cos t,\ z(t)=\sin t\), for \(0\leq t\leq\pi/2\), through a full angle \(0\leq\theta\leq2\pi\). Find the area of the resulting patch.

The radius stays positive because \(2+\cos t\geq2\) on this interval. The derivatives are \(\rho'(t)=-\sin t\) and \(z'(t)=\cos t\), so the profile speed is

$$ \sqrt{\rho'(t)^2+z'(t)^2} =\sqrt{\sin^2t+\cos^2t}=1. $$

Substitute into the full-revolution formula:

$$ \begin{aligned} A &=2\pi\int_0^{\pi/2}(2+\cos t)(1)\,dt\\ &=2\pi\left[2t+\sin t\right]_0^{\pi/2}\\ &=2\pi\left(\pi+1-0\right)\\ &=2\pi(\pi+1). \end{aligned} $$

As a check, the profile speed is exactly \(1\), so the area equals \(2\pi\) times the integral of the radius with respect to profile length. The computed integral of the radius is \(\pi+1\), giving the same area \(2\pi(\pi+1)\). This is a curved surface patch, not a complete closed surface.

Angular Patches, Full Surfaces, and Overlap

The width of the angular interval is part of the geometric description. For a profile with positive radius, an interval of width \(2\pi\) sweeps out a full circle at each profile point. A width smaller than \(2\pi\) leaves out part of those circles, while a width greater than \(2\pi\) may trace parts repeatedly. The formula calculates area over the parameter domain, so repeated coverage would count the overlap more than once.

For example, take \(\rho(t)=1,\ z(t)=t\) on \(0\leq t\leq1\), but use \(0\leq\theta\leq\pi\). The area factor is \(1\), so the area is

$$ \int_0^\pi\int_0^1 1\,dt\,d\theta=\pi. $$

This is half of the unit cylinder’s lateral area, not the full cylinder. For the full lateral surface the angular interval must have width exactly \(2\pi\), giving area \(2\pi\). The profile is traced once in both cases; tracing it once does not compensate for using only half the angular range.

Common pitfall: Do not treat an angular interval of width at most \(2\pi\) as a full revolution. A width below \(2\pi\) describes an angular patch. For a full surface with positive radius, use a width of exactly \(2\pi\), and ensure the profile does not retrace a portion that would create overlapping surface area.

Likewise, a lateral-area calculation does not automatically include caps or bases. Such pieces are separate surfaces and must be added separately if the question asks for the total area of a closed solid’s boundary. At points where \(\rho=0\), the angular parameter collapses to the axis; this can make the parametrization singular there. For cylinders and cones, the collapse occurs at most at isolated endpoints and does not alter the area integral. If a profile retraces itself or the rotated surface intersects itself, the integral counts parametrized area with multiplicity; the geometric area of a union must instead be arranged so each piece is counted once.

Why the Parametrization Method Is Useful

The cylinder and cone formulas are special cases of one construction. The method also applies when a profile is curved, when only a range of heights is wanted, or when only an angular sector is present. The key choices are to describe the profile accurately, use the correct parameter bounds, and include the radius and profile speed in the area factor. Checking whether the angle covers a full revolution and whether any pieces overlap is as important as evaluating the integral.

Check Your Understanding

Use the area element and parameter bounds to answer the following.

  1. For \(\rho(t)=5,\ z(t)=t\), with \(0\leq t\leq3\), what is the area factor, and what is the lateral area for a full revolution?
  2. Why does a cone’s lateral-area integral include the factor \(\sqrt{1+(d\rho/dz)^2}\)?
  3. For a full surface of revolution with positive radius, what angular interval width is required? What does a smaller width describe?
  4. For a profile \(\rho(t),z(t)\), state the full-revolution area formula in terms of \(t\).
  5. Does the lateral-area formula for a cone include its base disk? Explain how you would account for the base if total boundary area were requested.

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