---
title: "Surface of revolution: equation, points, area"
subject: "Computer Aided Design"
chapter: 4
key: "semester_4/computer_aided_design:surface-of-revolution-numerical"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/surface-of-revolution-numerical
site: BCT Engineering
---

# Surface of revolution: equation, points, area

Computer Aided Design · Chapter 4: Representation of Surfaces and Solids

- Asked in: year not given, 6 marks
- Video: [easy concept of #Surface of #revolution.....](https://www.youtube.com/watch?v=xQ0X7T77xSI) — Sky Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/surface-of-revolution-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-4

## Question

Describe the surface of revolution. A straight line from (2, 0, 0) to (4, 0, 6) in the XZ plane is revolved about the Z axis. Write the parametric equation of the surface, find the points for (s, phi) = (0.5, 90 deg), (1, 180 deg) and (0.25, 45 deg), and calculate the curved surface area.

## Short answer

Rotating the line from (2,0,0) to (4,0,6) about the Z-axis yields P(s,phi)=((2+2s)cos phi, (2+2s)sin phi, 6s). Points: (0,3,3); (-4,0,6); (1.768,1.768,1.5). Area = pi(2+4)*sqrt(40) = 119.2 square units.

## Answer

A **surface of revolution** is formed by rotating a plane curve (the generatrix or profile) about an axis in its plane. Every point of the curve moves on a circle (parallel); each position of the curve is a meridian. For a profile $r=f(s)$, $z=g(s)$ about the Z axis:

$$
\mathbf P(s,\phi)=\big(f(s)\cos\phi,\ f(s)\sin\phi,\ g(s)\big),\qquad 0\le\phi\le2\pi
$$

#### Generatrix
The line is $r=2+2s$, $z=6s$ for $0\le s\le1$.

$$
\mathbf P(s,\phi)=\big((2+2s)\cos\phi,\ (2+2s)\sin\phi,\ 6s\big)
$$

The surface is the side of a frustum of a cone with bottom radius 2, top radius 4 and height 6.

#### Points
| $s$ | $\phi$ | $r=2+2s$ | $\mathbf P$ |
|---|---|---|---|
| 0.5 | 90 deg | 3 | (0, 3, 3) |
| 1 | 180 deg | 4 | (-4, 0, 6) |
| 0.25 | 45 deg | 2.5 | (1.768, 1.768, 1.5) |

#### Curved surface area
Slant length $l=\sqrt{(4-2)^2+6^2}=\sqrt{40}=6.325$.

$$
A=\pi\,(r_1+r_2)\,l=\pi(2+4)(6.325)=119.2\ \text{square units}
$$

**Answer: P(s, phi) = ((2+2s)cos phi, (2+2s)sin phi, 6s); points (0,3,3), (-4,0,6), (1.768,1.768,1.5); area = 119.2 square units.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the generatrix radius r(s)?
   A. 4 + 2s
   B. 2 + 2s
   C. 2s
   D. 6s
   Answer: B. r goes from 2 to 4 linearly. (Key point: Generatrix)

2. What is the point at s = 0.5, φ = 90°?
   A. (3, 0, 3)
   B. (0, 4, 3)
   C. (0, 3, 3)
   D. (0, 3, 6)
   Answer: C. r = 3, so x = 0, y = 3, z = 3. (Key point: Point)

3. What is the frustum curved area?
   A. 75.4
   B. 226
   C. 150
   D. 119.2
   Answer: D. π(2+4)(6.325) = 119.2. (Key point: Area)

4. What is the slant length?
   A. 6.325
   B. 6
   C. 2
   D. 8
   Answer: A. sqrt(2^2 + 6^2) = sqrt(40). (Key point: Slant length)

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