Mathwords logoMathwords

Cylindrical Coordinates — Definition, Formula & Examples

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height axis zz, representing each point in space as (r,θ,z)(r, \theta, z) where rr is the distance from the zz-axis, θ\theta is the angle measured from the positive xx-axis, and zz is the vertical height.

A point PP in R3\mathbb{R}^3 is expressed in cylindrical coordinates as the ordered triple (r,θ,z)(r, \theta, z) where r≥0r \geq 0 is the radial distance from the zz-axis, θ∈[0,2π)\theta \in [0, 2\pi) is the azimuthal angle in the xyxy-plane measured counterclockwise from the positive xx-axis, and z∈(−∞,∞)z \in (-\infty, \infty) is the signed distance from the xyxy-plane. The coordinate surfaces are concentric cylinders (r=constr = \text{const}), half-planes through the zz-axis (θ=const\theta = \text{const}), and horizontal planes (z=constz = \text{const}).

Key Formula

x = r\cos\theta, \quad y = r\sin\theta, \quad z = z$$ $$r = \sqrt{x^2 + y^2}, \quad \theta = \arctan\!\left(\frac{y}{x}\right), \quad z = z
Where:
  • rr = Radial distance from the z-axis (r ≥ 0)
  • θ\theta = Azimuthal angle in the xy-plane, measured from the positive x-axis
  • zz = Height above (or below) the xy-plane
  • x,yx, y = Cartesian coordinates in the horizontal plane

How It Works

To convert from cylindrical to Cartesian coordinates, project the radial component onto the xx- and yy-axes using cosine and sine. To go the other direction, compute rr from xx and yy using the Pythagorean theorem and find θ\theta with the inverse tangent. Cylindrical coordinates simplify problems that have rotational symmetry about one axis — cylinders, cones, helices, and many physical systems. When setting up triple integrals in cylindrical coordinates, the volume element becomes r dr dθ dzr\,dr\,d\theta\,dz, which accounts for the fact that small patches farther from the zz-axis sweep out more area.

Worked Example

Problem: Convert the Cartesian point (3,33,5)(3, 3\sqrt{3}, 5) to cylindrical coordinates.
Find r: Compute the radial distance from the z-axis.
r=x2+y2=32+(33)2=9+27=36=6r = \sqrt{x^2 + y^2} = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6
Find θ: Use the inverse tangent. Since both x and y are positive, the point is in the first quadrant.
θ=arctan⁡ ⁣(333)=arctan⁡(3)=π3\theta = \arctan\!\left(\frac{3\sqrt{3}}{3}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}
Identify z: The z-coordinate stays the same in both systems.
z=5z = 5
Answer: The cylindrical coordinates are (r,θ,z)=(6,  π3,  5)(r, \theta, z) = \left(6,\; \dfrac{\pi}{3},\; 5\right).

Another Example

Problem: Evaluate ∭EdV\displaystyle\iiint_E dV where EE is the solid cylinder x2+y2≤4x^2 + y^2 \leq 4, 0≤z≤30 \leq z \leq 3.
Set up bounds: In cylindrical coordinates the cylinder x2+y2≤4x^2 + y^2 \leq 4 becomes 0≤r≤20 \leq r \leq 2, and θ\theta ranges over a full revolution.
0≤r≤2,0≤θ≤2π,0≤z≤30 \leq r \leq 2, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq z \leq 3
Write the integral: Replace dVdV with the cylindrical volume element r dr dθ dzr\,dr\,d\theta\,dz.
V=∫02π∫02∫03r dz dr dθV = \int_0^{2\pi}\int_0^{2}\int_0^{3} r\,dz\,dr\,d\theta
Evaluate: Integrate in order: first zz, then rr, then θ\theta.
V=∫02π∫023r dr dθ=∫02π3⋅r22∣02 dθ=∫02π6 dθ=12πV = \int_0^{2\pi}\int_0^{2} 3r\,dr\,d\theta = \int_0^{2\pi} 3\cdot\frac{r^2}{2}\Big|_0^2\,d\theta = \int_0^{2\pi} 6\,d\theta = 12\pi
Answer: The volume of the cylinder is 12π12\pi cubic units.

Why It Matters

Cylindrical coordinates appear throughout multivariable calculus courses (Calculus III) whenever you evaluate triple integrals over cylinders, cones, or regions with axial symmetry. Engineers use them constantly in electromagnetism — for instance, finding the electric field around a long charged wire — and in fluid dynamics when modeling flow through pipes. Mastering the conversion formulas and the volume element r dr dθ dzr\,dr\,d\theta\,dz saves significant computation time on exams and in professional work.

Common Mistakes

Mistake: Forgetting the extra factor of rr in the volume element
Correction: The cylindrical volume element is r dr dθ dzr\,dr\,d\theta\,dz, not dr dθ dzdr\,d\theta\,dz. The factor rr arises because arc length at radius rr is r dθr\,d\theta. Omitting it gives an incorrect integral.
Mistake: Using arctan⁡(y/x)\arctan(y/x) without adjusting for the correct quadrant
Correction: The basic arctan⁡\arctan function only returns values in (−π/2,π/2)(-\pi/2, \pi/2). When x<0x < 0, you must add π\pi (or use the two-argument atan2(y,x)\text{atan2}(y, x)) to place θ\theta in the correct quadrant.