Converting Points Between Cartesian And Cylindrical Coordinates

From One Coordinate Description to Another

A point in three-dimensional space can be described by its Cartesian coordinates \((x,y,z)\), or by its cylindrical coordinates \((r,\theta,z)\). Cylindrical coordinates describe the same point using its distance from the \(z\)-axis, its direction around that axis, and its height.

Converting between these descriptions is useful only if each coordinate is interpreted correctly. The radius \(r\) and angle \(\theta\) describe the point’s horizontal projection onto the \(xy\)-plane. The coordinate \(z\) is its height and does not change during conversion.

Definition: For a point with Cartesian coordinates \((x,y,z)\), its cylindrical radius is the distance from its horizontal projection \((x,y)\) to the origin of the \(xy\)-plane. Its angle \(\theta\) gives the direction of that projection, measured from the positive \(x\)-direction. Its vertical coordinate remains \(z\).

We will use the usual direction angles between \(0\) and \(2\pi\) when reporting answers. The essential requirement is that the chosen angle points into the same quadrant as the horizontal projection.

The Conversion Formulas

Given cylindrical coordinates, convert to Cartesian coordinates by resolving the horizontal radius into its \(x\)- and \(y\)-components. The vertical coordinate is unchanged:

x=r\cos(\theta),\qquad y=r\sin(\theta),\qquad z=z.

In the other direction, the radius is the distance from \((0,0)\) to \((x,y)\), and the angle points from the origin toward \((x,y)\). Thus,

r=\sqrt{x^2+y^2}.

The angle must be selected using both \(x\) and \(y\), not merely the ratio \(y/x\). When \(x\ne 0\), the ratio satisfies \(\tan(\theta)=y/x\), but tangent alone cannot distinguish directions in opposite quadrants. The signs of \(x\) and \(y\) determine the quadrant:

Signs of \(x\) and \(y\)	Horizontal quadrant	Angle range for \(0\leq\theta<2\pi\)
\(x>0,\ y>0\)	I	\(0<\theta<\frac{\pi}{2}\)
\(x<0,\ y>0\)	II	\(\frac{\pi}{2}<\theta<\pi\)
\(x<0,\ y<0\)	III	\(\pi<\theta<\frac{3\pi}{2}\)
\(x>0,\ y<0\)	IV	\(\frac{3\pi}{2}<\theta<2\pi\)

If one horizontal coordinate is zero, use the corresponding axis direction: \(\theta=0\) on the positive \(x\)-axis, \(\theta=\frac{\pi}{2}\) on the positive \(y\)-axis, \(\theta=\pi\) on the negative \(x\)-axis, and \(\theta=\frac{3\pi}{2}\) on the negative \(y\)-axis.

Important special case: If \(x=y=0\), then \(r=0\) and the point lies on the \(z\)-axis. There is no horizontal direction to measure, so \(\theta\) is undetermined. Any angle gives the same Cartesian point when \(r=0\).

Why the Formulas Work

The conversion formulas follow from the geometry of the horizontal projection. For \(r>0\), the point \((x,y)\) and the origin form a right triangle whose hypotenuse has length \(r\). The angle at the origin is \(\theta\). By the definitions of cosine and sine, the horizontal and vertical legs of this triangle have signed lengths \(r\cos(\theta)\) and \(r\sin(\theta)\). These are precisely the \(x\)- and \(y\)-coordinates.

The reverse radius formula follows from the Pythagorean theorem applied to the same triangle. This also provides a direct check that the two directions of conversion agree.

Derivation: If \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), then

x^2+y^2 =r^2\cos^2(\theta)+r^2\sin^2(\theta) =r^2\bigl(\cos^2(\theta)+\sin^2(\theta)\bigr) =r^2.

Because \(r\) is a distance and therefore nonnegative, taking the nonnegative square root gives \(r=\sqrt{x^2+y^2}\). Conversely, if \(r=\sqrt{x^2+y^2}>0\), choose the angle whose direction agrees with \((x,y)\). Then \(\cos(\theta)=x/r\) and \(\sin(\theta)=y/r\), so \(r\cos(\theta)=x\) and \(r\sin(\theta)=y\). The height remains \(z\), so the conversion recovers the original point.

Worked Conversions

Worked Example: Cartesian Coordinates in Quadrant I

Convert the Cartesian point \((3,3,-2)\) to cylindrical coordinates, using an angle from \(0\) to \(2\pi\).

First calculate the horizontal distance:

r=\sqrt{x^2+y^2} =\sqrt{3^2+3^2} =\sqrt{9+9} =\sqrt{18} =3\sqrt{2}.

Both horizontal coordinates are positive, so the projection is in Quadrant I. Since \(y/x=3/3=1\), the angle in that quadrant is \(\theta=\frac{\pi}{4}\). The height is unchanged, so the cylindrical coordinates are \(\left(3\sqrt{2},\frac{\pi}{4},-2\right)\).

Check by converting back:

x=3\sqrt{2}\cos\left(\frac{\pi}{4}\right) =3\sqrt{2}\cdot\frac{\sqrt{2}}{2}=3, \qquad y=3\sqrt{2}\sin\left(\frac{\pi}{4}\right) =3\sqrt{2}\cdot\frac{\sqrt{2}}{2}=3.

The height remains \(-2\), so the result is \((3,3,-2)\), as required.

Worked Example: Cylindrical Coordinates in Quadrant II

Convert \(\left(2,\frac{5\pi}{6},4\right)\) to Cartesian coordinates.

Here \(r=2\), \(\theta=\frac{5\pi}{6}\), and the height is \(4\). Use the conversion formulas and the exact trigonometric values \(\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\) and \(\sin(\frac{5\pi}{6})=\frac{1}{2}\):

x=2\cos\left(\frac{5\pi}{6}\right) =2\left(-\frac{\sqrt{3}}{2}\right) =-\sqrt{3}, \qquad y=2\sin\left(\frac{5\pi}{6}\right) =2\left(\frac{1}{2}\right) =1.

The \(z\)-coordinate is \(4\), so the Cartesian point is \((-\sqrt{3},1,4)\). Its horizontal coordinates have the signs required for Quadrant II.

Check the radius and direction. The radius is \(\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+1}=2\). Also, the direction angle \(\frac{5\pi}{6}\) has cosine \(-\frac{\sqrt{3}}{2}\) and sine \(\frac{1}{2}\), giving \(x=-\sqrt{3}\) and \(y=1\) when multiplied by \(r=2\). The height remains \(4\), so the conversion checks.

Worked Example: A Point with Negative \(y\)-Direction in the Horizontal Plane

Convert the Cartesian point \((0,-3,7)\) to cylindrical coordinates, using an angle from \(0\) to \(2\pi\).

The horizontal projection is \((0,-3)\). It lies along the negative \(y\)-direction, whose angle is \(\frac{3\pi}{2}\). This describes the direction of the horizontal projection; the three-dimensional point itself has height \(7\).

r=\sqrt{0^2+(-3)^2} =\sqrt{9} =3.

Therefore the cylindrical coordinates are \(\left(3,\frac{3\pi}{2},7\right)\). Check by converting back:

x=3\cos\left(\frac{3\pi}{2}\right)=3(0)=0, \qquad y=3\sin\left(\frac{3\pi}{2}\right)=3(-1)=-3.

The height remains \(7\), recovering \((0,-3,7)\).

A Reliable Conversion Procedure

For a Cartesian point, calculate the radius first, then identify the direction from the signs and any zero coordinates of \(x\) and \(y\). Keep the original \(z\)-coordinate. For a cylindrical point, substitute \(r\) and \(\theta\) into the cosine and sine formulas and keep \(z\) unchanged.

Identify the horizontal projection.
Use \((x,y)\); do not treat \(z\) as part of the radius or the direction angle.

Find the radius or components.
For Cartesian input, compute \(r=\sqrt{x^2+y^2}\). For cylindrical input, compute \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\).

Choose or verify the direction.
For Cartesian input, match the angle to the quadrant of \((x,y)\). For cylindrical input, verify that the signs of the resulting components match the angle.

Preserve and check the height.
Copy \(z\) without change, then substitute the converted coordinates back into the formulas.

Common Mistake: Using the Ratio Without Checking the Quadrant

The equation \(\tan(\theta)=y/x\) does not by itself determine the direction. For example, consider \((-1,1,2)\). The ratio is \(y/x=1/(-1)=-1\). Taking the inverse tangent of \(-1\) and reporting \(-\frac{\pi}{4}\) would give a Quadrant IV angle, even though \((-1,1)\) is in Quadrant II.

That wrong angle does not recover the point. With \(r=\sqrt{(-1)^2+1^2}=\sqrt{2}\), it would give

x=\sqrt{2}\cos\left(-\frac{\pi}{4}\right)=1, \qquad y=\sqrt{2}\sin\left(-\frac{\pi}{4}\right)=-1,

which reverses both horizontal signs. The correction is to use the quadrant: \(x<0\) and \(y>0\), so the point is in Quadrant II. Its angle is \(\frac{3\pi}{4}\), and the cylindrical coordinates are \(\left(\sqrt{2},\frac{3\pi}{4},2\right)\). Checking gives \(x=\sqrt{2}(-\frac{\sqrt{2}}{2})=-1\), \(y=\sqrt{2}(\frac{\sqrt{2}}{2})=1\), and \(z=2\).

Takeaway: A conversion is not complete until the angle matches the horizontal direction and the height is copied correctly. Substituting the result back into the conversion formulas is a quick way to catch sign and quadrant errors.

Check Your Understanding

For each question, keep the horizontal projection distinct from the point’s height, and check any conversion by substituting back.

Convert \((0,4,-1)\) to cylindrical coordinates using an angle from \(0\) to \(2\pi\).
Convert \(\left(3,\frac{4\pi}{3},-2\right)\) to Cartesian coordinates.
What is the cylindrical radius of \((-5,-12,6)\), and which quadrant contains its horizontal projection?
Why is an angle undetermined when converting a point \((0,0,z)\) on the \(z\)-axis?
For \((-2,-2,1)\), why is an angle with tangent equal to \(1\) not enough to identify the correct direction?