Surface Area Of Cones And Frustums

From a Straight Profile to a Cone or Frustum

A cone’s lateral surface is generated by revolving a straight line segment from its tip to the rim of a circular base. A frustum’s lateral surface is generated by revolving a straight segment whose endpoints are at two different distances from the axis. In both cases, the segment’s length along its slope—not just its vertical or axial height—determines the surface area.

The key distinction is between vertical height, measured parallel to the cone’s axis, and slant length, measured along the straight profile that is revolved. These lengths are generally different. The slant length is the distance that appears in the lateral-area formulas.

Here we find the lateral surface area only. The circular base or bases are not part of that area unless they are explicitly added.

Definition: The lateral surface area of a cone or frustum is the area of its curved surface, excluding its circular base or bases. The slant length is the length along the straight profile from one end of the lateral surface to the other.

The Surface-Area Element Along a Profile

When a curve is revolved about an axis, a short piece of curve of length $ds$, at distance $\rho$ from the axis, sweeps out a narrow band. Its circumference is $2\pi\rho$, so its area is approximately $2\pi\rho\,ds$. In the limiting integral, the surface area is $2\pi\int \rho\,ds$, provided the profile generates the surface without overlapping itself.

For a straight profile, the distance from the axis changes linearly along the segment. Suppose the endpoint radii are $r$ and $R$, with $0\le r\le R$, and the segment has length $s$. Measure distance $u$ along the segment starting from the endpoint with radius $r$. Then $0\le u\le s$, and the radius at distance $u$ is

\rho(u)=r+\frac{R-r}{s}u.

This linear expression follows because the radius changes by $R-r$ over the full segment length $s$. The surface-area integral now gives the general straight-profile result.

Write the radius along the segment.
At distance $u$, use $\rho(u)=r+\frac{R-r}{s}u$.

Integrate the swept bands.
Use $S=2\pi\int_0^s \rho(u)\,du$.

Simplify.
The result is $\pi(R+r)s$, where $s$ is the slant length.

Derivation of the Frustum Formula

For a frustum, both endpoint radii are positive. Substituting the linear radius into the surface-area integral gives a direct derivation of the formula.

\begin{aligned} S &=2\pi\int_0^s \left(r+\frac{R-r}{s}u\right)\,du\\ &=2\pi\left[ru+\frac{R-r}{2s}u^2\right]_0^s\\ &=2\pi\left(rs+\frac{R-r}{2}s\right)\\ &=2\pi s\left(\frac{R+r}{2}\right)\\ &=\pi(R+r)s. \end{aligned}

The integral uses the average endpoint radius, $(R+r)/2$, multiplied by the segment’s slant length and by $2\pi$. This is why a frustum’s lateral area is the circumference associated with the average radius, multiplied by the slant length.

Theorem: A frustum with larger radius $R$, smaller radius $r$, and slant length $s$ has lateral surface area $$ S=\pi(R+r)s. $$ If its vertical height is $h$, then $$ s=\sqrt{h^2+(R-r)^2}, \qquad S=\pi(R+r)\sqrt{h^2+(R-r)^2}. $$ The right-triangle formula applies because the axial change is $h$ and the radial change is $R-r$; together they are perpendicular components of the slanted segment.

The Cone as a Special Case

A cone is the limiting straight-profile case where one endpoint is the tip on the axis, so its radius is zero. Let the base radius be $r$, the slant length be $l$, and the vertical height be $h$. Substituting $R=r$, $r=0$, and $s=l$ into the straight-profile result gives the cone formula. (Here $r$ denotes the cone’s base radius.)

S=\pi(r+0)l=\pi rl.

The vertical height and base radius form the perpendicular sides of a right triangle, while the slant length is its hypotenuse. Therefore $l=\sqrt{h^2+r^2}$, and an equivalent form of the lateral-area formula is

S=\pi r\sqrt{h^2+r^2}.

Key distinction: In $S=\pi rl$, $l$ is the cone’s slant length, not its vertical height $h$. For a frustum, $s$ is likewise the slanted distance between the two circular rims, not the axial height.

Worked Examples

Worked Example: Lateral Area of a Cone

A cone has base radius $3$ units and vertical height $4$ units. Find its lateral surface area.

First find the slant length from the right triangle formed by the radius, height, and slant:

l=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\text{ units}.

Apply the cone formula using the slant length:

S=\pi rl=\pi(3)(5)=15\pi\text{ square units}.

As a check using the generating segment, let $u$ be distance from the tip, so $0\le u\le5$. The radius increases linearly from $0$ to $3$, hence $\rho(u)=3u/5$. Then

2\pi\int_0^5\frac{3u}{5}\,du =2\pi\left[\frac{3u^2}{10}\right]_0^5 =2\pi\left(\frac{75}{10}\right) =15\pi\text{ square units}.

The integral agrees with the formula. The vertical height $4$ is not the length of the generating segment; using it in place of $l=5$ would give the wrong area.

Worked Example: Lateral Area of a Frustum

A frustum has larger radius $7$ units, smaller radius $3$ units, and vertical height $3$ units. Find its lateral surface area.

The radial change is $7-3=4$ units. Thus the slant length is the hypotenuse of a right triangle with perpendicular sides $3$ and $4$:

s=\sqrt{3^2+(7-3)^2} =\sqrt{9+16} =5\text{ units}.

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

S=\pi(R+r)s =\pi(7+3)(5) =50\pi\text{ square units}.

Check directly by integrating along the slant. Starting at the smaller rim, the radius is $\rho(u)=3+\frac45u$, for $0\le u\le5$. Therefore

\begin{aligned} S &=2\pi\int_0^5\left(3+\frac45u\right)\,du\\ &=2\pi\left[3u+\frac25u^2\right]_0^5\\ &=2\pi(15+10)\\ &=50\pi\text{ square units}. \end{aligned}

The integral confirms the result. The frustum’s vertical height is $3$ units, but its slant length is $5$ units.

Worked Example: Finding a Frustum Cut from a Cone

A cone has base radius $5$ units and vertical height $12$ units. A plane parallel to the base cuts the cone $3$ units from its tip. Find the lateral area of the remaining frustum.

First, the full cone’s slant length is

l=\sqrt{5^2+12^2}=\sqrt{25+144}=13\text{ units}.

By similarity, the small cone removed at the tip has height-to-full-height ratio $3/12=1/4$. Its radius is therefore $5/4$ units. The frustum has height $12-3=9$ units and radii $R=5$ and $r=5/4$. Its slant length is

s=\sqrt{9^2+\left(5-\frac54\right)^2} =\sqrt{81+\left(\frac{15}{4}\right)^2} =\sqrt{\frac{1296+225}{16}} =\frac{39}{4}\text{ units}.

The frustum formula gives

S=\pi\left(5+\frac54\right)\frac{39}{4} =\pi\left(\frac{25}{4}\right)\frac{39}{4} =\frac{975\pi}{16}\text{ square units}.

As a check, subtract the small cone’s lateral area from the full cone’s. The small cone has radius $5/4$ and slant length $13/4$, so its area is $\pi(5/4)(13/4)=65\pi/16$. The full cone’s area is $\pi(5)(13)=65\pi$. Thus the remaining area is

65\pi-\frac{65\pi}{16} =\frac{1040\pi-65\pi}{16} =\frac{975\pi}{16}\text{ square units}.

This independent calculation agrees with the frustum formula.

A Common Mistake: Using Vertical Height as Slant Length

The most common error is to put the vertical height directly into the lateral-area formula. For the frustum with radii $7$ and $3$ and vertical height $3$, the incorrect calculation would be

S_{\text{wrong}}=\pi(7+3)(3)=30\pi.

This treats the axial height as though it were the distance along the sloping profile. But the radial change is $4$, so the profile length is $s=\sqrt{3^2+4^2}=5$, not $3$. The corrected calculation is

S=\pi(7+3)(5)=50\pi.

A reliable check is to identify the right triangle before using the area formula: its perpendicular sides are the vertical height and the difference in radii, and its hypotenuse is the slant length.

Why the Slant Length Matters

The lateral surface is made of bands swept out by pieces of the profile. Each band’s area depends on the piece’s actual length along the profile, $ds$, as well as its distance from the axis. The axial height alone records only the profile’s vertical displacement; it does not record the full length of the path being revolved. The slant length carries that missing information.

These formulas are useful for finding the curved area of containers, funnels, and other conical shapes. They also provide a geometric check on surface-area integrals: when a straight segment produces a cone or frustum, the integral should simplify to the corresponding formula. Keep the lateral area separate from the area of the circular base or bases, and include those only when the question asks for total surface area.

Check Your Understanding

Answer each question by identifying the radii, axial height, and slant length as needed.

A cone has radius $8$ units and vertical height $15$ units. What is its slant length and lateral surface area?
A frustum has radii $6$ and $2$ units and vertical height $3$ units. Find its slant length and lateral surface area.
Why does the frustum formula use the sum of the endpoint radii?
A cone’s radius and vertical height are both $4$ units. Is its slant length $4$ units? Explain.
Does the lateral surface-area formula include the circular base of a cone? What additional area would be needed for the total area of a closed cone?