Reversing Radial and Vertical Integration Order

Why Reverse the Order?

In cylindrical coordinates, the same solid can often be described by integrating vertically at each radius or radially at each height. One description may be easier to evaluate, or may make the geometry of a problem clearer. Reversing the order does not change the solid: it changes how we list the points in it.

For a solid bounded by an upward-opening paraboloid and a horizontal plane, the key is to draw or reason through a cross-section in the \(r\)-\(z\) plane. At a fixed radius, height runs from the paraboloid up to the plane. At a fixed height, radius runs from the axis out to the paraboloid. These are two descriptions of the same region.

Key idea: Do not reverse bounds by merely swapping their written endpoints. Describe the same \(r\)-\(z\) region from the other direction, and solve the boundary equation for the variable that now varies inside.

The Paraboloid Cap and Its Two Descriptions

Consider the solid between the paraboloid \(z=z_0+a r^2\) and the horizontal plane \(z=H\), where \(a>0\) and \(H>z_0\). Here \(r\) is the distance from the vertical axis. The paraboloid meets the plane where \(z_0+a r^2=H\), so the largest radius is \(R=\sqrt{(H-z_0)/a}\).

At a fixed radius \(r\), the lower height is \(z_0+a r^2\) and the upper height is \(H\). That gives the radial-first description: \(r\) runs from the axis to \(R\), and, at each \(r\), \(z\) runs between the surfaces.

To describe the same region at a fixed height \(z\), first note that the height ranges from the paraboloid's vertex \(z_0\) to the plane \(H\). The inequality \(z_0+a r^2\le z\) then gives \(r^2\le (z-z_0)/a\). Since radius is nonnegative, \(r\) runs from zero to \(\sqrt{(z-z_0)/a}\).

\begin{aligned} \text{Radial first:}\quad &0\le r\le \sqrt{\frac{H-z_0}{a}}, &\qquad z_0+a r^2\le z\le H,\\[4pt] \text{Vertical first:}\quad &z_0\le z\le H, &\qquad 0\le r\le \sqrt{\frac{z-z_0}{a}}. \end{aligned}

The angular coordinate \(\theta\) ranges from \(0\) to \(2\pi\) for a full rotationally symmetric cap. In a cylindrical triple integral, the volume element is \(r\,dz\,dr\,d\theta\), or the same factors ordered according to the chosen integration order. The factor \(r\) remains present when the order changes.

Why the Reversed Bounds Are Correct

The bounds can be justified directly from the inequalities defining the solid. The radial-height region is the set of points satisfying \(0\le r\), \(z_0+a r^2\le z\), and \(z\le H\). When describing this set by height first, its lowest height is \(z_0\), because \(r^2\ge 0\). Its highest height is \(H\). For a fixed \(z\) in that range, rearranging the paraboloid inequality gives the upper radial limit.

Derivation: Since \(a>0\), the inequality \(z_0+a r^2\le z\) is equivalent to \(r^2\le (z-z_0)/a\). With \(r\ge 0\), this is equivalent to \(0\le r\le \sqrt{(z-z_0)/a}\). The height range is \(z_0\le z\le H\), so these bounds describe precisely the same radial-height region as the radial-first bounds.

For any continuous integrand \(f(r,\theta,z)\) on this bounded solid, the two iterated integrals therefore describe the same integral. In one order, each radius is paired with all heights above the paraboloid and below the plane. In the other, each height is paired with all radii between the axis and the paraboloid. The angular range and integrand do not change.

\int_0^{2\pi}\int_0^R\int_{z_0+a r^2}^{H} f(r,\theta,z)\,r\,dz\,dr\,d\theta = \int_{z_0}^{H}\int_0^{2\pi}\int_0^{\sqrt{(z-z_0)/a}} f(r,\theta,z)\,r\,dr\,d\theta\,dz, \qquad R=\sqrt{\frac{H-z_0}{a}}.

Worked Example: Reverse the Bounds for a Paraboloid Cap

Worked Example: Find the Volume in Both Orders

Find the volume of the solid above \(z=r^2\) and below \(z=4\). First find the intersection radius by setting the boundary heights equal: \(r^2=4\), so \(r=2\) (taking the nonnegative radius). Thus the radial-first bounds are \(0\le r\le2\) and \(r^2\le z\le4\). Solving \(r^2\le z\) for \(r\) gives the reversed bounds \(0\le z\le4\) and \(0\le r\le\sqrt z\).

For volume, the integrand is \(1\), and the cylindrical volume element supplies the factor \(r\). Using radial-first bounds gives:

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

Now integrate in the reversed order. The inner integral is with respect to \(r\):

\begin{aligned} V &=\int_0^4\int_0^{2\pi}\int_0^{\sqrt z} r\,dr\,d\theta\,dz\\ &=2\pi\int_0^4\left[\frac{r^2}{2}\right]_0^{\sqrt z}\,dz\\ &=2\pi\int_0^4\frac{z}{2}\,dz\\ &=\pi\left[\frac{z^2}{2}\right]_0^4 =8\pi. \end{aligned}

The two calculations agree. The reversed bounds also match the geometry: at height \(z\), the cross-section is a disk of radius \(\sqrt z\), which expands from radius zero at the vertex to radius two at the plane.

Worked Example: A Shifted, Scaled Paraboloid

Worked Example: Integrate a Radial Weight

Consider the solid above \(z=1+2r^2\) and below \(z=9\). Find \(\iiint r^2\,dV\) in both orders. The boundaries meet when \(1+2r^2=9\), so \(2r^2=8\), \(r^2=4\), and \(R=2\). Solving the lower-surface inequality \(1+2r^2\le z\) for \(r\) gives \(r\le\sqrt{(z-1)/2}\); the height range is \(1\le z\le9\).

Because \(dV=r\,dz\,dr\,d\theta\), the integrand \(r^2\) produces the total factor \(r^3\). With radial-first bounds:

\begin{aligned} \iiint r^2\,dV &=\int_0^{2\pi}\int_0^2\int_{1+2r^2}^{9}r^3\,dz\,dr\,d\theta\\ &=2\pi\int_0^2 r^3(8-2r^2)\,dr\\ &=2\pi\left[2r^4-\frac{r^6}{3}\right]_0^2\\ &=2\pi\left(32-\frac{64}{3}\right) =\frac{64\pi}{3}. \end{aligned}

For the reversed order, integrate \(r^3\) from zero to \(\sqrt{(z-1)/2}\):

\begin{aligned} \iiint r^2\,dV &=2\pi\int_1^9\int_0^{\sqrt{(z-1)/2}}r^3\,dr\,dz\\ &=2\pi\int_1^9\frac{1}{4}\left(\frac{z-1}{2}\right)^2\,dz\\ &=\frac{\pi}{8}\int_1^9(z-1)^2\,dz\\ &=\frac{\pi}{8}\left[\frac{(z-1)^3}{3}\right]_1^9\\ &=\frac{\pi}{8}\cdot\frac{512}{3} =\frac{64\pi}{3}. \end{aligned}

Both orders give \(64\pi/3\). As a bound check, at \(z=1\) the reversed radial interval has length zero, while at \(z=9\) its upper endpoint is \(\sqrt{8/2}=2\), the intersection radius.

Worked Example: Integrate a Height-Dependent Quantity

Worked Example: Evaluate an Integral with Integrand \(z\)

For the cap above \(z=r^2\) and below \(z=4\), evaluate \(\iiint z\,dV\) in both orders. The bounds remain those from the first example. The integrand is now \(z\), so the cylindrical volume element makes the integrand \(zr\).

In radial-first order, integrate \(z\) between \(r^2\) and \(4\):

\begin{aligned} \iiint z\,dV &=2\pi\int_0^2\int_{r^2}^4 zr\,dz\,dr\\ &=2\pi\int_0^2 r\left[\frac{z^2}{2}\right]_{r^2}^4\,dr\\ &=2\pi\int_0^2\left(8r-\frac{r^5}{2}\right)\,dr\\ &=2\pi\left[4r^2-\frac{r^6}{12}\right]_0^2\\ &=2\pi\left(16-\frac{16}{3}\right) =\frac{64\pi}{3}. \end{aligned}

In vertical-first order, \(z\) is fixed during the radial integration:

\begin{aligned} \iiint z\,dV &=2\pi\int_0^4\int_0^{\sqrt z}zr\,dr\,dz\\ &=2\pi\int_0^4 z\left[\frac{r^2}{2}\right]_0^{\sqrt z}\,dz\\ &=2\pi\int_0^4\frac{z^2}{2}\,dz\\ &=\pi\left[\frac{z^3}{3}\right]_0^4 =\frac{64\pi}{3}. \end{aligned}

The agreement checks both the calculation and the change of bounds. The inner radial integral in the reversed order used \(r=\sqrt z\) as its upper endpoint, obtained from the same paraboloid equation.

A Common Bound-Reversal Mistake

A frequent error is to use the full maximum radius at every height. For the cap \(z=r^2\), \(z=4\), that mistake would assign \(0\le r\le2\) for every \(0\le z\le4\). Those bounds describe a cylinder, not the cap: for example, at \(z=1\) they include \(r=2\), even though the paraboloid condition requires \(r^2\le1\), or \(r\le1\).

The correction is to derive the radial limit at the fixed height by solving the surface inequality: \(r^2\le z\) gives \(r\le\sqrt z\). This limit grows with height, beginning at zero and reaching two at \(z=4\). Checking the endpoints and one intermediate height is a useful way to catch an incorrect reversal.

A Practical Reversal Procedure

Find where the surfaces meet.
Set the paraboloid height equal to the plane height to determine the outer radius and the top of the height interval.

State the original region.
Write the lower and upper heights at a fixed radius, along with the full radial interval.

Fix a height instead.
Use the vertex height and plane height as the vertical limits, then solve the paraboloid inequality for the radial limit.

Check the cross-section.
Confirm that the radial interval starts at zero and reaches the previously computed intersection radius at the plane.

When Reversing the Order Helps

Either order is valid when it describes the same bounded region and the integrand is suitable for iterated integration, as in these examples with continuous integrands. The better order depends on the calculation. A height-first description can simplify an integral when the radial cross-section has a convenient upper radius as a function of height. A radial-first description may be shorter when the vertical thickness is simpler as a function of radius.

The essential discipline is geometric rather than symbolic: identify the region, choose which coordinate varies first, and derive that coordinate's endpoints from the boundary inequalities. For this paraboloid cap, the same cross-section yields both \(z_0+a r^2\le z\le H\) at fixed \(r\) and \(0\le r\le\sqrt{(z-z_0)/a}\) at fixed \(z\).

Check Your Understanding

Use the radial-height cross-section to answer each question.

For the region above \(z=3+r^2\) and below \(z=8\), find the intersection radius and write the bounds in both orders.
For \(z=2+r^2\), \(z=6\), what is the upper radial bound at the fixed height \(z=3\)?
Explain why using the maximum radius at every height gives a larger region than the paraboloid cap.
For the region above \(z=1+3r^2\) and below \(z=7\), state the height range and the radial range at a fixed height.
When reversing the order, which part of the cylindrical volume element must still be included, and why?