14.1 Cylindrical coordinate system and transformation

Cylindrical Coordinate System

Defining Cylindrical Coordinates

Cylindrical coordinates describe points in 3D space using three values: $(r, \theta, z)$. Think of it as polar coordinates in the xy-plane with a height coordinate tacked on.

• $r$ is the radial distance from the z-axis to the point's projection in the xy-plane.
  • Always non-negative: $r \geq 0$
  • Computed as $r = \sqrt{x^2 + y^2}$
• $\theta$ is the angle in the xy-plane, measured counterclockwise from the positive x-axis.
  • Range: $0 \leq \theta < 2\pi$
  • Computed as $\theta = \arctan\!\left(\frac{y}{x}\right)$, with the usual quadrant adjustments (use $\text{atan2}(y, x)$ to get the correct quadrant automatically)
• $z$ is the height along the z-axis, exactly the same as in Cartesian coordinates. It can be positive, negative, or zero.

This system is ideal whenever the region or integrand has circular symmetry about the z-axis (cylinders, cones, paraboloids, etc.), because the geometry lines up naturally with $r$ and $\theta$.

Relationship to Cartesian Coordinates

Cylindrical → Cartesian:

$x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z$

Cartesian → Cylindrical:

$r = \sqrt{x^2 + y^2}, \qquad \theta = \arctan\!\left(\frac{y}{x}\right), \qquad z = z$

The $z$-coordinate is unchanged in both directions. The first two equations are just the polar-coordinate conversions you already know from Calculus II/III, applied to the xy-plane.

Quick example: Convert the Cartesian point $(1, \sqrt{3}, 5)$ to cylindrical.

• $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$
• $\theta = \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
• $z = 5$

So the cylindrical form is $(2,\, \frac{\pi}{3},\, 5)$.

Coordinate Transformation

Cartesian to Cylindrical Transformation

When you convert a triple integral from Cartesian to cylindrical coordinates, you need to do three things:

1. Substitute the coordinate expressions into the integrand:

   • Replace every $x$ with $r\cos\theta$
   • Replace every $y$ with $r\sin\theta$
   • Leave $z$ alone

2. Replace the volume element $dx\,dy\,dz$ with $r\,dr\,d\theta\,dz$ (the extra factor of $r$ comes from the Jacobian; see below).

3. Transform the limits of integration to describe the same region in $(r, \theta, z)$ variables. This is often the trickiest step: sketch the region and express its boundaries in terms of $r$, $\theta$, and $z$.

Example: Consider the integral over the upper half of a unit cylinder:

$\int_{0}^{1}\!\int_{0}^{\sqrt{1-x^2}}\!\int_{0}^{1} dz\,dy\,dx$

The xy-limits describe the upper half of the unit disk ($x$ from 0 to 1, $y$ from 0 to $\sqrt{1-x^2}$). In cylindrical coordinates that region is $0 \leq r \leq 1$, $0 \leq \theta \leq \frac{\pi}{2}$. The $z$-limits stay the same. So the integral becomes:

$\int_{0}^{\pi/2}\!\int_{0}^{1}\!\int_{0}^{1} r\,dz\,dr\,d\theta$

Notice the limits changed to $\theta \in [0, \pi/2]$ (first quadrant only), not $[0, 2\pi]$, because the original Cartesian region only covers the first quadrant.

Jacobian Determinant and Volume Element

Why does that extra $r$ appear? It comes from the Jacobian determinant of the coordinate transformation. The Jacobian matrix is:

$J = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \$4pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \$4pt] \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{pmatrix} = \begin{pmatrix} \cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Computing the determinant gives:

$|J| = \cos\theta \cdot (r\cos\theta) - (-r\sin\theta)\cdot(\sin\theta) = r\cos^2\theta + r\sin^2\theta = r$

So $|J| = r$, and the volume element transforms as:

$dV = dx\,dy\,dz = r\,dr\,d\theta\,dz$

Geometrically, this makes sense: a small "box" in cylindrical coordinates is a wedge-shaped piece. Its radial side has length $dr$, its height is $dz$, but its arc-length side is $r\,d\theta$ (not just $d\theta$). The product of these three gives $r\,dr\,d\theta\,dz$.

The general transformation formula for a triple integral is:

$\iiint_D f(x,y,z)\,dx\,dy\,dz = \iiint_{D'} f(r\cos\theta,\, r\sin\theta,\, z)\; r\,dr\,d\theta\,dz$

where $D'$ is the region $D$ described in cylindrical coordinates. Don't forget the $r$ factor; dropping it is one of the most common mistakes on exams.