---
title: "Cartesian, cylindrical, spherical conversion"
subject: "Computer Aided Design"
chapter: 2
key: "semester_4/computer_aided_design:coordinate-systems-conversion"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/coordinate-systems-conversion
site: BCT Engineering
---

# Cartesian, cylindrical, spherical conversion

Computer Aided Design · Chapter 2: Geometric Modeling Fundamentals

- Asked in: year not given, 6 marks
- Video: [Cylindrical and Spherical Conversions](https://www.youtube.com/watch?v=zdIB_HASOmM) — Firefly Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/coordinate-systems-conversion
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-2

## Question

Explain the Cartesian, cylindrical and spherical coordinate systems used in geometric modelling. (a) Convert the Cartesian point (3, 4, 5) into cylindrical and spherical coordinates. (b) Convert the spherical point r = 10, theta = 60 deg (measured from the +Z axis), phi = 45 deg (azimuth from +X in the XY plane) into Cartesian coordinates.

## Short answer

Cartesian uses x,y,z; cylindrical uses rho, phi, z; spherical uses r, theta from +Z, and phi from +X. Part (a): (3,4,5) becomes cylindrical (5, 53.13 deg, 5) and spherical (7.071, 45 deg, 53.13 deg). Part (b): r=10, theta=60 deg, phi=45 deg gives Cartesian (6.124, 6.124, 5.000).

## Answer

#### Coordinate systems
- **Cartesian** $(x, y, z)$: three mutually perpendicular axes; simplest for prismatic shapes and used internally by CAD.
- **Cylindrical** $(\rho, \phi, z)$: radial distance in the XY plane, azimuth angle from +X, and height; suited to shafts, holes and revolved parts.
- **Spherical** $(r, \theta, \phi)$: distance from origin, polar angle from +Z, azimuth from +X; suited to spheres and radial scanning.

Relations:

$$
\rho=\sqrt{x^2+y^2},\quad \phi=\tan^{-1}\frac{y}{x},\quad r=\sqrt{x^2+y^2+z^2},\quad \theta=\cos^{-1}\frac{z}{r}
$$

$$
x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta
$$

#### (a) Point (3, 4, 5)
$$
\begin{aligned}
\rho &= \sqrt{3^2+4^2} = 5 \\
\phi &= \tan^{-1}(4/3) = 53.13^\circ \\
r &= \sqrt{9+16+25} = 7.071 \\
\theta &= \cos^{-1}(5/7.071) = 45^\circ
\end{aligned}
$$

Cylindrical: $(5,\ 53.13^\circ,\ 5)$. Spherical: $(7.071,\ 45^\circ,\ 53.13^\circ)$.

#### (b) Spherical to Cartesian
$$
\begin{aligned}
x &= 10\sin 60^\circ\cos 45^\circ = 10(0.8660)(0.7071) = 6.124 \\
y &= 10\sin 60^\circ\sin 45^\circ = 6.124 \\
z &= 10\cos 60^\circ = 5.000
\end{aligned}
$$

**Answer: (a) cylindrical (5, 53.13 deg, 5); spherical (7.071, 45 deg, 53.13 deg). (b) Cartesian (6.124, 6.124, 5.000) in the same length units.**

## Check yourself

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

1. In the spherical system (r, theta, phi) used here, what is theta?
   A. Azimuth from +X
   B. Height above XY plane
   C. Angle from +Y
   D. Polar angle from +Z
   Answer: D. theta is measured from the +Z axis and phi is the azimuth from +X. (Key point: Spherical angles)

2. What is the cylindrical form of (3, 4, 5)?
   A. (5, 53.13°, 5)
   B. (7.071, 53.13°, 5)
   C. (5, 36.87°, 5)
   D. (5, 45°, 7.071)
   Answer: A. rho = sqrt(9+16) = 5, phi = atan(4/3) = 53.13°, z stays 5. (Key point: Cylindrical conversion)

3. For (3, 4, 5), what is theta?
   A. 53.13°
   B. 45°
   C. 36.87°
   D. 90°
   Answer: B. r = 7.071 and theta = acos(5/7.071) = 45°. (Key point: Spherical conversion)

4. What is z for r = 10, theta = 60°?
   A. 8.660
   B. 6.124
   C. 5.000
   D. 7.071
   Answer: C. z = r cos(theta) = 10 cos 60° = 5.000. (Key point: Spherical to Cartesian)

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