Why Convert to Cylindrical Coordinates?
Cylindrical coordinates describe a point by its distance from the \(z\)-axis, its direction around that axis, and its height. They are especially useful when a point or surface has circular symmetry about the \(z\)-axis, or when a Cartesian equation repeatedly contains \(x^2+y^2\).
A conversion is more than a change of symbols. For a point, the angle must agree with the point’s quadrant. For an equation, the algebra must preserve the entire locus, including any cases lost by division. Finally, the resulting equation should be interpreted geometrically.
Coordinate Formulas and Angle Choice
When \(r>0\), the angle is determined by the direction of \((x,y)\), not just by the ratio \(y/x\). A ratio can give an angle with the right tangent but the wrong quadrant. The signs of \(x\) and \(y\) identify the quadrant; then choose the angle in the stated range. For example, if \(x<0\) and \(y>0\), the angle lies in the second quadrant. If \(r=0\), the point is on the \(z\)-axis, and \(\theta\) is not uniquely determined.
The conversion formulas follow from the right triangle formed by the point’s projection onto the \(xy\)-plane: its hypotenuse has length \(r\), and its horizontal and vertical legs in that plane are \(x\) and \(y\). Thus \(\cos\theta=x/r\) and \(\sin\theta=y/r\), giving \(x=r\cos\theta\) and \(y=r\sin\theta\). Also, \(r^2=x^2+y^2\), so the nonnegative distance is \(r=\sqrt{x^2+y^2}\).
Worked Examples: Converting Points
Worked Example: A Point in the Second Quadrant
Convert the Cartesian point \((-3,3\sqrt{3},4)\) to cylindrical coordinates.
First calculate the distance from the \(z\)-axis:
Here \(x=-3<0\) and \(y=3\sqrt{3}>0\), so the angle is in the second quadrant. Since \(\cos\theta=x/r=-1/2\) and \(\sin\theta=y/r=\sqrt{3}/2\), the angle is \(\theta=2\pi/3\). The height does not change, so \(z=4\). The cylindrical coordinates are \((6,2\pi/3,4)\).
Check by converting back: \(6\cos(2\pi/3)=-3\), \(6\sin(2\pi/3)=3\sqrt{3}\), and \(z=4\). All three Cartesian coordinates match.
Worked Example: A Point in the Third Quadrant
Convert \((-2,-2\sqrt{3},5)\) to cylindrical coordinates.
The radial coordinate is
Both horizontal coordinates are negative, so the angle must be in the third quadrant. Its reference angle is \(\pi/3\), because \(|x|/r=1/2\) and \(|y|/r=\sqrt{3}/2\). Therefore \(\theta=\pi+\pi/3=4\pi/3\), and the height remains \(5\). The cylindrical coordinates are \((4,4\pi/3,5)\).
Check: \(4\cos(4\pi/3)=-2\), \(4\sin(4\pi/3)=-2\sqrt{3}\), and \(z=5\), as required.
Converting Equations: The Main Substitutions
To convert a Cartesian equation, substitute \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\). The last identity often simplifies the result. For example, a Cartesian equation containing \(x^2+y^2\) can become an equation involving only \(r\) and \(z\). The equation \(r=c\), for a positive constant \(c\), describes a circular cylinder of radius \(c\) about the \(z\)-axis.
The identities preserve the locus because every Cartesian point determines cylindrical coordinates, and converting those coordinates back reproduces the same \(x\), \(y\), and \(z\). In the other direction, every allowed cylindrical triple produces a Cartesian point by the defining formulas. Thus equivalent equations describe the same points, provided the coordinate restrictions and any exceptional cases are retained.
Worked Examples: Circles and Cylinders
Worked Example: A Circular Cylinder from an Offset Circle
Convert \(x^2+y^2=6y\) to cylindrical coordinates and describe its surface.
Substitute \(x^2+y^2=r^2\) and \(y=r\sin\theta\):
Factor without dividing by \(r\): \(r(r-6\sin\theta)=0\). For \(r>0\), this gives \(r=6\sin\theta\); the axis case \(r=0\) is also included in the original equation. In Cartesian coordinates, completing the square gives
Each horizontal cross-section is therefore a circle of radius \(3\), centered at \((0,3)\) in the \(xy\)-plane. Since the equation does not involve \(z\), that circle extends unchanged at every height: the surface is a vertical circular cylinder. Check the cylindrical equation by substituting \(r=6\sin\theta\) into \(r^2=6r\sin\theta\); both sides become \(36\sin^2\theta\). The axis point also satisfies the original equation.
Worked Example: A Cylinder Tangent to the \(z\)-Axis
Convert \((x-2)^2+y^2=4\) and describe the resulting surface.
Expand the Cartesian equation first:
Now substitute \(x^2+y^2=r^2\) and \(x=r\cos\theta\):
For \(r>0\), the cylindrical equation is \(r=4\cos\theta\). The origin, where \(r=0\), also belongs to the original surface. In the \(xy\)-plane the original equation is a circle centered at \((2,0)\) with radius \(2\), since it is already in the form \((x-2)^2+y^2=2^2\). Extending that circle vertically gives a cylinder tangent to the \(z\)-axis. Check the algebra in reverse: \(r=4\cos\theta\) implies \(r^2=4r\cos\theta\), hence \(x^2+y^2=4x\), which is the expanded Cartesian equation. The separately retained origin satisfies it as well.
Rotational Surfaces
A surface whose equation depends on \(x\) and \(y\) only through \(x^2+y^2\) is unchanged by rotations around the \(z\)-axis. In cylindrical coordinates, that dependence becomes dependence on \(r\) alone. This makes the geometry visible: an equation relating \(r\) and \(z\), but not \(\theta\), specifies the same radial-height profile at every angle. The resulting surface is rotational about the \(z\)-axis.
Worked Example: A Paraboloid
Convert \(z=x^2+y^2\) and identify the surface.
Using \(x^2+y^2=r^2\) gives \(z=r^2\). Since \(r\ge0\), this is equivalently \(r=\sqrt{z}\) for \(z\ge0\). At each fixed height \(z>0\), the cross-section is a circle of radius \(\sqrt{z}\); at \(z=0\), it is the point on the axis. The surface is an upward-opening circular paraboloid with vertex at the origin.
Check by substituting \(r^2=z\) and \(x^2+y^2=r^2\): the Cartesian equation becomes \(z=x^2+y^2\), exactly the starting equation. The restriction \(z\ge0\) follows from \(r^2\ge0\), so the converted equation introduces no points below the \(xy\)-plane.
Worked Example: A Double Cone
Convert \(x^2+y^2=4z^2\) and describe the surface.
Substitution gives \(r^2=4z^2\). Equivalently, \(r=2|z|\), because \(r\ge0\). The absolute value matters: \(z\) may be positive or negative. At height \(z\), the cross-section has radius \(2|z|\), and the radius is zero when \(z=0\). The surface is a double circular cone with vertex at the origin and axis along the \(z\)-axis.
Check by squaring \(r=2|z|\): \(r^2=4z^2\). Since \(r^2=x^2+y^2\), this recovers the original equation. Conversely, \(r^2=4z^2\) and \(r\ge0\) imply \(r=2|z|\), so neither nappe is lost.
A Common Angle Mistake
Using \(\tan\theta=y/x\) without checking the quadrant can produce a wrong point. For \((x,y)=(-1,-\sqrt{3})\), the ratio is \(y/x=\sqrt{3}\). The angle \(\pi/3\) has that tangent, but it is in the first quadrant, so it cannot represent a point with both coordinates negative. The correct angle is \(4\pi/3\), in the third quadrant. Indeed, with \(r=2\), the correct formulas give \(x=2\cos(4\pi/3)=-1\) and \(y=2\sin(4\pi/3)=-\sqrt{3}\).
Another important check is whether a conversion step divided by \(r\). At \(r=0\), that division is not valid. Factoring first, as in the cylinder examples, makes the exceptional axis case visible and prevents accidental loss of points.
How to Check a Conversion
Calculate \(\sqrt{x^2+y^2}\), keeping \(z\) unchanged.
Use the signs of \(x\) and \(y\) to select the quadrant, then verify both sine and cosine.
Use \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\).
Identify the geometry and substitute the converted equation back into the Cartesian relation, checking any cases excluded by division.
These checks connect algebra to geometry. An equation independent of \(z\) often describes a vertical extension of a planar curve; an equation independent of \(\theta\) often describes a rotational surface. The earlier material on checking surface equations with traces can provide additional geometric confirmation, but the conversion itself should already preserve the locus algebraically.
Check Your Understanding
For each question, give a reason for your answer and check any conversion by substituting back.
- Convert \((0,-5,2)\) to cylindrical coordinates using the stated angle range.
- Convert \(x^2+y^2=8x\) to cylindrical form and identify the circle that generates the vertical cylinder.
- Why can the ratio \(y/x\) alone fail to determine the correct angle?
- Convert \(z=4(x^2+y^2)\) to cylindrical coordinates and describe its horizontal cross-sections.
- When converting \(r^2=5r\sin\theta\), what case must be considered before dividing by \(r\)?