# Polar, Cylindrical, and Spherical Coordinates

Physics I · Multivariable Calculus · https://tryals.app/learn/physics-i/polar-cylindrical-and-spherical-coordinates

## Coordinate Systems

Choosing the right coordinate system simplifies integration dramatically. Match your choice to the region's symmetry.

| Symmetry | System | Formulas |
| --- | --- | --- |
| **Circular (2D)** | **Polar** $(r, \theta)$ | $x=r\cos\theta, y=r\sin\theta, dA=r\,dr\,d\theta$ |
| **Cylindrical** | **Cylindrical** $(r, \theta, z)$ | $x=r\cos\theta, y=r\sin\theta, z=z, dV=r\,dr\,d\theta\,dz$ |
| **Spherical** | **Spherical** $(\rho, \theta, \phi)$ | $x=\rho\sin\phi\cos\theta, y=\rho\sin\phi\sin\theta, z=\rho\cos\phi$ |

Here, $\rho \ge 0$ is the radius, $0 \le \theta \le 2\pi$ is the azimuth, and $0 \le \phi \le \pi$ is the polar angle.

## Integration & Pitfalls

The volume element for **spherical coordinates** is $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.

**Example**: Volume of a sphere of radius $R$:

$$V = \int_0^{2\pi}\int_0^{\pi}\int_0^R \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = \frac{4}{3}\pi R^3$$

> **Common pitfall:** Mathematicians and physicists swap the names of $\theta$ and $\phi$. Always check which angle runs from the pole ($0$ to $\pi$) before using a formula.

> **Physics link:** The Laplacian $\nabla^{2}$ in spherical coordinates uses this geometry to solve Laplace's equations for potentials.

## Practice questions

5 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. Spherical coordinates carry a volume element of $\rho^{2}\sin\varphi$ while Cartesian coordinates carry none. Why does the factor vary with position?

A. Because the coordinate grid stretches away from the origin
B. Because the factor corrects for the curvature of the sphere itself
C. Because angles are measured in radians rather than in units of length
D. Because spherical coordinates are less accurate than Cartesian ones

**Answer:** A. Because the coordinate grid stretches away from the origin

**Why:** A Cartesian grid is uniform and a spherical one is not. The Jacobian records how much real volume each coordinate cell occupies, which is why it is needed at all.

Page: https://tryals.app/practice/physics-i/polar-cylindrical-and-spherical-coordinates/spherical-coordinates-carry-a-volume-element-of-sin-while

### 2. In spherical coordinates ($\rho$, $\theta$, $\varphi$), which ranges cover the full sphere?

A. 0 ≤ $\varphi$ ≤ π
B. 0 ≤ $\rho$ ≤ R
C. 0 ≤ $\varphi$ ≤ 2π
D. 0 ≤ $\theta$ ≤ 2π

**Answer:** A. 0 ≤ $\varphi$ ≤ π; B. 0 ≤ $\rho$ ≤ R; D. 0 ≤ $\theta$ ≤ 2π

**Why:** $\varphi$ ranges from 0 to $\pi$ (pole to pole); $\theta$ goes full circle, 0 to $2\pi$.

Page: https://tryals.app/practice/physics-i/polar-cylindrical-and-spherical-coordinates/in-spherical-coordinates-which-ranges-cover-the-full

### 3. Sort: 2D polar coordinate or 3D coordinate system?

**Answer:**

- 2D polar: (r, $\theta$) with $dA = r$ dr dθ, $x = r \cos \theta, y = r \sin \theta$, Useful for circles and disks in ℝ²
- 3D coordinate system: (r, $\theta$, z) with $dV = r$ dr d$\theta dz$, ($\rho$, $\theta$, $\varphi$) with $dV = \rho^{2} \sin \varphi d\rho d\varphi d\theta$, Useful for spheres and cones in ℝ³

**Why:** Polar coordinates are 2D; cylindrical and spherical are 3D extensions.

Page: https://tryals.app/practice/physics-i/polar-cylindrical-and-spherical-coordinates/sort-2d-polar-coordinate-or-3d-coordinate-system

### 4. Choose the natural coordinate system for integrating over each solid.

**Answer:**

- Cylindrical: A soup can (finite cylinder), A pipe segment (annular cylinder)
- Spherical: A planet (solid ball), An ice-cream cone capped by a sphere chunk
- Cartesian: A shoebox

**Why:** Every boundary that becomes "coordinate = constant" removes one hard limit. Cans and pipes are cylindrical boxes; balls and cones are spherical boxes; crates are Cartesian ones. Geometry chooses the coordinates, not habit.

Page: https://tryals.app/practice/physics-i/polar-cylindrical-and-spherical-coordinates/choose-the-natural-coordinate-system-for-integrating-over-each-solid

### 5. A sphere is awkward in Cartesian coordinates and simple in spherical ones. What does that coordinate change actually buy?

A. Constant limits: the curved boundary becomes $\rho = a$
B. A simpler integrand, since the function becomes constant
C. The removal of the Jacobian factor from the integral
D. Fewer variables, since spherical coordinates need only two

**Answer:** A. Constant limits: the curved boundary becomes $\rho = a$

**Why:** Substitution is chosen for the LIMITS. The Jacobian $\rho^{2}\sin\varphi$ is the price, and it is worth paying because a box is easy and a variable-limit region is not.

Page: https://tryals.app/practice/physics-i/polar-cylindrical-and-spherical-coordinates/a-sphere-is-awkward-in-cartesian-coordinates-and-simple-in-spherical