Why Radial Expressions Matter
Some equations in three variables treat \(x\) and \(y\) as a pair. For example, they may contain \(x^2+y^2\), but not \(x\) and \(y\) separately. This combination has a direct geometric meaning: it is the square of the distance from the point \((x,y,z)\) to the \(z\)-axis.
Cylindrical coordinates package that distance into one coordinate, \(r\). Recognizing the expression \(x^2+y^2\) therefore lets us rewrite equations more compactly and makes their geometry easier to read. The key is to keep track of whether an equation uses the distance \(r\) itself or its square \(r^2\).
The Radial Identity
In cylindrical coordinates, the Cartesian coordinates are related by \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\), with \(r\geq 0\). The coordinate \(r\) measures the distance from the \(z\)-axis; the angle \(\theta\) records direction around that axis.
The identity follows directly from the coordinate formulas. Substituting \(x=r\cos\theta\) and \(y=r\sin\theta\) gives
Here we use \(\cos^2\theta+\sin^2\theta=1\). Since \(r\geq 0\), the distance itself is \(\sqrt{x^2+y^2}=r\). Thus \(x^2+y^2\) becomes \(r^2\), while \(\sqrt{x^2+y^2}\) becomes \(r\). These are related but not interchangeable.
Substituting Carefully
To convert an equation to cylindrical coordinates, replace each occurrence of \(x^2+y^2\) by \(r^2\), then simplify while retaining any restrictions that follow from \(r\geq 0\). If an equation contains other terms, such as \(z\), keep those terms as they are unless the equation also contains \(x\) or \(y\) separately.
Look for \(x^2+y^2\), its square root, or a function of either one.
Use \(x^2+y^2=r^2\) and \(\sqrt{x^2+y^2}=r\).
Remember that \(r\) cannot be negative, and check what the resulting equation describes.
For instance, an expression such as \(g(x^2+y^2)\), where \(g\) is a function, becomes \(g(r^2)\), not \(g(r)\). By contrast, an expression \(h(\sqrt{x^2+y^2})\) becomes \(h(r)\). The input to the function matters.
Worked Example: A Circular Cylinder
Worked Example: A Circular Cylinder
Rewrite the equation \(x^2+y^2=25\) in cylindrical coordinates and identify its basic shape.
Replace \(x^2+y^2\) with \(r^2\):
Taking square roots gives \(r=5\) or \(r=-5\) algebraically. But cylindrical coordinates require \(r\geq 0\), so only \(r=5\) is allowed. The equation is therefore \(r=5\), with no restriction on \(z\) or \(\theta\). It describes all points whose distance from the \(z\)-axis is 5: a circular cylinder of radius 5 around that axis.
Check the conversion at the point \((x,y,z)=(3,4,2)\). In Cartesian coordinates, \(x^2+y^2=3^2+4^2=9+16=25\), so the point satisfies the equation. Its cylindrical radius is \(r=\sqrt{3^2+4^2}=\sqrt{25}=5\), so it also satisfies \(r=5\). The two descriptions agree.
Why Radial Dependence Gives Rotational Symmetry
An equation depends radially on \(x\) and \(y\) when their effect occurs through \(x^2+y^2\), rather than through their separate values. Such an equation cannot distinguish between points with the same distance from the \(z\)-axis. This is why radial expressions are useful for recognizing rotational symmetry.
To justify the claim, consider rotating the \((x,y)\)-coordinates through an angle \(\alpha\). The rotated coordinates are \(x'=x\cos\alpha-y\sin\alpha\) and \(y'=x\sin\alpha+y\cos\alpha\), while \(z\) remains unchanged. Expanding their squared sum gives
The mixed terms cancel, and each pair of squared trigonometric terms sums to 1. Thus a rotation preserves \(x^2+y^2\). It also leaves \(z\) unchanged. If the surface equation uses \(x\) and \(y\) only through \(x^2+y^2\), every term in the equation keeps the same value under the rotation. A point satisfying the equation therefore rotates to another point that satisfies it. This proves the stated symmetry.
Worked Example: A Surface Given as a Graph
Worked Example: A Surface Given as a Graph
Rewrite \(z=\dfrac{1}{1+x^2+y^2}\) in cylindrical coordinates, then check the equation at a point with \(x=1\) and \(y=2\).
Substitute \(r^2\) for \(x^2+y^2\):
For \(x=1\) and \(y=2\), the squared radius is \(r^2=1^2+2^2=1+4=5\), so \(r=\sqrt{5}\). The cylindrical equation gives \(z=\dfrac{1}{1+5}=\dfrac{1}{6}\). Check in the original Cartesian equation: \(z=\dfrac{1}{1+1^2+2^2}=\dfrac{1}{1+1+4}=\dfrac{1}{6}\). The values agree.
This surface equation uses the horizontal coordinates only through their distance from the \(z\)-axis. For example, replacing \((1,2)\) with \((2,1)\) still gives \(x^2+y^2=5\), and hence the same \(z\)-value. The rotational symmetry is visible in the form \(z=\dfrac{1}{1+r^2}\).
Worked Example: A More Complicated Radial Equation
Worked Example: A More Complicated Radial Equation
Convert \((x^2+y^2-4)^2+z^2=1\) to cylindrical coordinates and verify that \((2,0,1)\) lies on the resulting surface.
First replace the full occurrence of \(x^2+y^2\) with \(r^2\). The subtraction and square remain part of the expression:
At \((x,y,z)=(2,0,1)\), the radius is \(r=\sqrt{2^2+0^2}=\sqrt{4}=2\). Substituting into the cylindrical equation gives
Checking directly in Cartesian coordinates gives \((2^2+0^2-4)^2+1^2=(4-4)^2+1=1\). Thus the point satisfies both equations. The cylindrical form also shows that the equation depends on horizontal position through \(r\) alone, so rotating a point about the \(z\)-axis preserves the equation.
A Common Mistake: Dropping the Square
A frequent error is to replace \(x^2+y^2\) with \(r\). That confuses a squared distance with the distance itself. The correct substitution is \(x^2+y^2=r^2\).
Worked Example: Correcting a Radial Substitution
Consider \(x^2+y^2=9\). An incorrect conversion is \(r=9\), obtained by replacing the left side with \(r\). The correction is:
where the last step uses \(r\geq 0\). To check the difference, take the point \((3,0,0)\). Its squared radial distance is \(3^2+0^2=9\), and its radius is \(r=3\). It satisfies the original equation and the correct equation \(r=3\), but not the mistaken equation \(r=9\). The incorrect substitution changes the geometric set.
Another error is to take a square root without considering the coordinate restriction. From \(r^2=25\), the equation \(r=\pm5\) is a valid algebraic intermediate result, but \(r=-5\) is not a cylindrical coordinate. The equivalent cylindrical equation is \(r=5\).
When This Recognition Is Useful
Radial substitutions can simplify an equation, make symmetry evident, and reduce a three-variable expression to one involving \(r\) and \(z\). This is especially helpful when comparing points at different angles: if their \(r\) and \(z\) coordinates agree, an equation depending only on \(r\) and \(z\) treats them identically.
The substitution does not remove every variable from every equation. For example, an equation containing \(x\) by itself cannot be converted merely by replacing \(x^2+y^2\) with \(r^2\); the separate \(x\)-dependence remains and must be expressed using the angular coordinate as well. The radial shortcut applies to the combination \(x^2+y^2\), or to expressions built from that combination, not to arbitrary appearances of \(x\) or \(y\).
When interpreting an equation, also check whether it has any solutions. Since \(r^2\geq 0\), an equation such as \(x^2+y^2=-1\) has no real points: its cylindrical form would require \(r^2=-1\), which is impossible for real \(r\). More generally, taking square roots or solving for \(r\) must respect both the nonnegativity of \(r\) and any restrictions imposed by the rest of the equation.
Check Your Understanding
Try each question by identifying the radial quantity before simplifying.
- Rewrite \(x^2+y^2=16\) in cylindrical coordinates. Why is the negative square root not included?
- Convert \(z=3+\sqrt{x^2+y^2}\) to cylindrical coordinates. Why does the answer use \(r\), not \(r^2\)?
- Rewrite \(z=\dfrac{2}{1+x^2+y^2}\) in cylindrical coordinates.
- Explain why rotating a point about the \(z\)-axis preserves the value of \(x^2+y^2\).
- Does \(x^2+y^2=-4\) have any real solutions? Explain using \(r\).