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Distinguishing Radial Bounds From Cartesian Bounds

Learn to convert planar and solid boundaries into radial limits by using the geometry of \(r\), rather than treating it as a Cartesian coordinate.

Intermediate 9 min read

What You'll Learn

  • Distinguish a radial distance from a signed Cartesian coordinate
  • Convert disk and annulus inequalities into radial intervals
  • Recognize when a Cartesian inequality produces an angle-dependent radial bound
  • Derive radial limits from a cylinder-paraboloid intersection
  • Detect and correct radial bounds that are too large

Why Radial Bounds Need Their Own Geometry

In Cartesian coordinates, \(x\) and \(y\) are signed coordinates: either can be positive, negative, or zero. In cylindrical coordinates, \(r\) is different. It is the distance from the \(z\)-axis, so it cannot be negative. This distinction affects how a region’s boundaries translate into limits of integration.

A frequent error is to replace a Cartesian bound on \(x\) with the same-looking bound on \(r\). But a condition such as \(x\leq 3\) does not generally say \(r\leq 3\), and a disk described using \(x^2+y^2\) must be converted by using \(r^2=x^2+y^2\). The right limits follow from the region’s geometry, not from matching variable names.

Definition: In cylindrical coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r\geq 0\). The radial coordinate is the distance \(\sqrt{x^2+y^2}\) from the \(z\)-axis. Thus \(x^2+y^2=r^2\).

From a Disk to a Radial Interval

A disk of radius \(R\), centered at the origin, consists of points satisfying \(x^2+y^2\leq R^2\). Substituting \(x^2+y^2=r^2\) gives \(r^2\leq R^2\). Since \(r\geq 0\), the equivalent radial bound is \(0\leq r\leq R\). The nonnegativity condition matters: without it, taking square roots of the inequality would not specify the radial interval correctly.

More generally, an annulus with inner radius \(a\) and outer radius \(b\), where \(0\leq a\leq b\), satisfies \(a^2\leq x^2+y^2\leq b^2\). Its radial limits are \(a\leq r\leq b\). A radial interval measures distance outward from the axis; it does not describe a left-to-right coordinate range.

Takeaway: For a region centered on the \(z\)-axis, translate bounds on \(x^2+y^2\) into bounds on \(r^2\), then use \(r\geq 0\) to write the radial limits.

Worked Example: A Disk Described by a Cartesian Inequality

Worked Example: A Disk Described by a Cartesian Inequality

Find cylindrical-coordinate limits for the disk \(x^2+y^2\leq 9\), and use them to find the volume of the vertical cylinder of height \(1\) above the disk.

First use \(x^2+y^2=r^2\). The disk condition becomes \(r^2\leq 9\). Together with \(r\geq 0\), this gives \(0\leq r\leq 3\). The full disk has \(0\leq\theta\leq 2\pi\).

The vertical height is \(1\). The cylindrical volume element is \(r\,dz\,dr\,d\theta\), so the volume is

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^3\int_0^1 r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^3 r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^3d\theta\\ &=\int_0^{2\pi}\frac{9}{2}\,d\theta\\ &=\frac{9}{2}(2\pi)=9\pi. \end{aligned} $$

As a check, the base disk has area \(\pi(3)^2=9\pi\), and a cylinder of height \(1\) has volume equal to its base area times its height: \(9\pi\cdot1=9\pi\). The integral and geometric calculation agree.

When a Cartesian Boundary Is Not a Radial Boundary

Not every Cartesian inequality becomes a constant upper or lower limit on \(r\). For example, \(x\leq c\) becomes \(r\cos\theta\leq c\). If \(\cos\theta>0\), this implies \(r\leq c/\cos\theta\); if \(\cos\theta<0\), dividing by \(\cos\theta\) reverses the inequality. When \(\cos\theta=0\), the inequality imposes no restriction on \(r\) if \(c\geq0\). In general, the resulting radial condition depends on the angle and must be interpreted together with the other boundaries.

This is why it is unsafe to look at a Cartesian coordinate limit and simply reuse its endpoint for \(r\). A rectangle, for example, is not generally a disk or an annulus. Its radial extent can vary with \(\theta\), and it may be necessary to split the region into angular pieces to describe it with one consistent set of limits.

Worked Example: An Annulus

Worked Example: An Annulus

Find the radial limits for \(1\leq x^2+y^2\leq 9\), and calculate the volume of the solid of height \(2\) above this annulus.

Replace \(x^2+y^2\) by \(r^2\). The inequalities become \(1\leq r^2\leq9\). Because \(r\geq0\), taking square roots gives \(1\leq r\leq3\), not \(-3\leq r\leq3\). The region goes all the way around the axis, so \(0\leq\theta\leq2\pi\). The height bounds are \(0\leq z\leq2\).

Integrating with the cylindrical volume element gives

$$ \begin{aligned} V &=\int_0^{2\pi}\int_1^3\int_0^2 r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_1^3 2r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[r^2\right]_1^3d\theta\\ &=\int_0^{2\pi}(9-1)\,d\theta\\ &=8(2\pi)=16\pi. \end{aligned} $$

For a check, the annulus area is \(\pi(3^2)-\pi(1^2)=9\pi-\pi=8\pi\). Multiplying by height \(2\) gives \(16\pi\), as obtained by integration.

Deriving Limits from a Cylinder-Paraboloid Intersection

Radial limits for a solid can come from more than a disk stated in the \(xy\)-plane. They can also arise by examining where bounding surfaces meet. For a standard upward-opening circular paraboloid \(z=r^2\), a horizontal plane \(z=H\), with \(H\geq0\), meets it where \(r^2=H\). Since radius is nonnegative, the intersection circle has radius \(\sqrt H\).

If a vertical cylinder \(r\leq R\) is also present, then the solid must satisfy both the cylinder bound and any radial restriction required for the vertical bounds to be in the correct order. The admissible upper radial limit is whichever condition is more restrictive. In particular, the interval between \(z=r^2\) and \(z=H\) exists only where \(r^2\leq H\).

Derivation: For points between the paraboloid \(z=r^2\) and the plane \(z=H\), the vertical coordinate must satisfy \(r^2\leq z\leq H\). Such a \(z\) exists only if \(r^2\leq H\). As \(r\geq0\), this requires \(r\leq\sqrt H\). If the cylinder also requires \(r\leq R\), the radial interval is \(0\leq r\leq\min(R,\sqrt H)\).

This conclusion follows directly from the vertical inequalities. It is not enough for a point to lie inside the cylinder: if the paraboloid is already above the proposed top plane, there is no vertical interval between those two surfaces at that radius.

Worked Example: A Cylinder, a Paraboloid, and a Plane

Worked Example: A Cylinder, a Paraboloid, and a Plane

Find the cylindrical limits and volume of the solid inside the cylinder \(x^2+y^2\leq4\), above the paraboloid \(z=x^2+y^2\), and below the plane \(z=4\).

The cylinder condition becomes \(r^2\leq4\), so \(0\leq r\leq2\). The lower surface is \(z=r^2\), and the upper surface is \(z=4\). These surfaces are in the correct order for every \(r\) in that interval, since \(r^2\leq4\). At the outer edge \(r=2\), they meet: \(z=r^2=4\). Thus the cylinder, paraboloid, and plane meet consistently at the boundary circle.

The angular interval covers a full revolution, \(0\leq\theta\leq2\pi\). Therefore the volume is

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^2\int_{r^2}^4 r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 (4-r^2)r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 (4r-r^3)\,dr\,d\theta\\ &=\int_0^{2\pi}\left[2r^2-\frac{r^4}{4}\right]_0^2d\theta\\ &=\int_0^{2\pi}\left(8-4\right)d\theta\\ &=4(2\pi)=8\pi. \end{aligned} $$

For a direct check of the radial endpoint, substituting \(r=2\) into the paraboloid gives \(z=2^2=4\), exactly the plane’s height. At \(r=0\), the lower surface is \(z=0\) and the upper surface is \(z=4\), so the vertical height is \(4\). Throughout \(0\leq r\leq2\), the height \(4-r^2\) is nonnegative. These checks confirm that the limits describe the solid with the stated surface order.

A Common Mistake: Using a Disk’s Squared Radius as Its Radius

Suppose the condition is \(x^2+y^2\leq9\). Replacing the left side by \(r^2\) gives \(r^2\leq9\), not \(r\leq9\). The correct step is to use \(r\geq0\), giving \(r\leq3\).

For example, if someone mistakenly uses \(0\leq r\leq9\) for the height-one cylinder above this disk, the resulting integral is

$$ \int_0^{2\pi}\int_0^9 r\,dr\,d\theta = 2\pi\left[\frac{r^2}{2}\right]_0^9 = 81\pi. $$

This is wrong: it describes a cylinder with radius \(9\), not radius \(3\). Its base area is \(81\pi\), whereas the original disk’s area is \(9\pi\). Correcting the radial limit to \(3\) gives

$$ \int_0^{2\pi}\int_0^3 r\,dr\,d\theta = 2\pi\left[\frac{r^2}{2}\right]_0^3 = 9\pi, $$

which agrees with the disk’s area. The mistake is not a failure of integration; it is a failure to translate the boundary correctly.

A Reliable Boundary Check

Before evaluating an integral, test the proposed radial endpoints against the original geometry. For a disk or annulus, substitute \(r\) into \(x^2+y^2=r^2\) and check that the boundary values match the stated squared radii. For a solid between surfaces, substitute the radial endpoint into the relevant surface equations and check where the surfaces meet. Also verify that the lower \(z\)-bound does not exceed the upper \(z\)-bound anywhere in the proposed radial interval.

1
Translate the boundary.
Use \(x^2+y^2=r^2\) for circular boundaries, or substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) for other Cartesian boundaries.
2
Respect nonnegative radius.
Use \(r\geq0\) when converting squared radial inequalities into intervals.
3
Check all restrictions together.
For a solid, combine the cylinder bound with the condition that the proposed lower vertical surface lies below the upper surface.
4
Test the endpoints.
Substitute them into the original boundaries or surface equations to confirm that they represent the intended edges.

These checks are useful because the radial bound determines which points the integral includes. A correct integrand cannot compensate for an incorrect region. Distinguishing radial distance from Cartesian position is therefore essential whenever a cylindrical integral is set up from a geometric description.

Check Your Understanding

Use the geometry of the boundary to answer each question.

  1. What are the radial limits for \(x^2+y^2\leq16\), and why is the upper limit not \(16\)?
  2. Write the radial interval for the annulus \(4\leq x^2+y^2\leq25\).
  3. When \(x\leq2\) is written in cylindrical coordinates, why is \(r\leq2\) not generally an equivalent condition?
  4. For the region between \(z=r^2\) and \(z=9\), what restriction does the vertical ordering place on \(r\)?
  5. If both \(r\leq5\) and \(z=r^2\leq z\leq4\) apply, what is the largest admissible value of \(r\), and which condition makes it smaller?

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