Volume of frustum — derivation and problem solving approach

Question

What is a frustum, how do we derive its volume formula, and how do we solve problems involving frustums?

Solution — Step by Step

A frustum is the portion of a cone that remains when the top part is cut off by a plane parallel to the base. Think of it as a bucket shape — wider at the bottom, narrower at the top, with flat circular faces on both ends.

Parameters: $R$ = radius of larger base, $r$ = radius of smaller base (top), $h$ = height of the frustum.

Think of the frustum as: big cone minus the small cone cut off.

Let the height of the original big cone be $H$ and the height of the removed small cone be $H - h$ .

By similar triangles (the two cones are similar):

\frac{R}{H} = \frac{r}{H - h}

Solving: $H = \frac{Rh}{R - r}$

Volume of big cone: $\frac{1}{3}\pi R^2 H$

Volume of small cone: $\frac{1}{3}\pi r^2 (H - h)$

After substitution and simplification:

V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)

Curved Surface Area:

\text{CSA} = \pi (R + r) l

where $l$ = slant height = $\sqrt{h^2 + (R-r)^2}$

Total Surface Area:

\text{TSA} = \pi (R + r) l + \pi R^2 + \pi r^2

Example: A bucket is 24 cm deep. Its top and bottom radii are 15 cm and 5 cm. Find its capacity.

Given: $h = 24$ cm, $R = 15$ cm, $r = 5$ cm

V = \frac{1}{3}\pi (24)(15^2 + 5^2 + 15 \times 5)

V = \frac{1}{3}\pi (24)(225 + 25 + 75) = \frac{1}{3}\pi (24)(325)

V = 2600\pi \approx 8168.14 \text{ cm}^3 \approx 8.168 \text{ litres}

flowchart TD
    A["Frustum Problem"] --> B["Identify R, r, h from the problem"]
    B --> C{"What to find?"}
    C -->|"Volume"| D["V = 1/3 pi h times R2 + r2 + Rr"]
    C -->|"Slant height"| E["l = sqrt of h2 + R minus r squared"]
    C -->|"CSA"| F["CSA = pi times R + r times l"]
    C -->|"TSA"| G["TSA = CSA + pi R2 + pi r2"]
    A --> H["If not directly given, use similar triangles"]

Why This Works

The volume formula $\frac{1}{3}\pi h(R^2 + r^2 + Rr)$ comes from subtracting two similar cones. The $Rr$ cross-term appears naturally when we expand $(R^2 H - r^2(H-h))$ and substitute the similar triangle relationship. The formula is elegant because it reduces to the volume of a cylinder ( $\pi R^2 h$ ) when $R = r$ , and to the volume of a cone ( $\frac{1}{3}\pi R^2 h$ ) when $r = 0$ .

Alternative Method

If you forget the frustum formula in an exam, derive it on the spot using the subtraction method. Set up the similar triangle ratio, find $H$ in terms of $R$ , $r$ , and $h$ , then subtract the small cone volume from the big cone volume. This takes about 2 extra minutes but guarantees the correct answer.

Common Mistake

Students often confuse $h$ (perpendicular height of the frustum) with $l$ (slant height). The volume formula uses $h$ , while the CSA formula uses $l$ . If the problem gives the slant height, first find $h = \sqrt{l^2 - (R-r)^2}$ before using the volume formula. Using $l$ directly in the volume formula gives a wrong answer. CBSE boards almost always give slant height in at least one frustum problem.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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