Paraboloid Caps Cut by Horizontal Planes

Introduction

A horizontal plane can cut across an upward-opening paraboloid, leaving a finite solid between the plane and the curved surface. The edge of this solid is a circle: the plane and paraboloid meet at one height and at one distance from the vertical axis.

The key steps are to find that intersection radius, describe the solid with cylindrical bounds, and integrate the vertical distance between the plane and the paraboloid. The radius must come from the equations of both surfaces; it is not an independent bound.

The Paraboloid and the Cutting Plane

Consider the circular paraboloid \(z=z_0+kr^2\), where \(r^2=x^2+y^2\), \(z_0\) is the height of its vertex, and \(k>0\). Its axis is the \(z\)-axis, and it opens upward. Let the horizontal plane be \(z=H\).

A positive-height cap occurs when \(H>z_0\): the plane lies above the vertex and intersects the paraboloid in a circle. At the intersection, the two surfaces have the same \(z\)-coordinate, so \(z_0+kr^2=H\). Solving for the nonnegative radius gives \(R=\sqrt{(H-z_0)/k}\). Here \(R\) is the radius of the circle and of the disk onto which the cap projects.

Definition: For \(k>0\) and \(H>z_0\), the paraboloid cap is the solid above \(z=z_0+kr^2\) and below the plane \(z=H\), over the disk \(0\le r\le R\), where \(R=\sqrt{(H-z_0)/k}\).

At \(r=0\), the lower surface is at the vertex height \(z_0\). At \(r=R\), the lower and upper surfaces meet. For every radius strictly between \(0\) and \(R\), the paraboloid is below the plane. These facts determine the bounds in cylindrical coordinates.

0\le\theta\le 2\pi,\qquad 0\le r\le R,\qquad z_0+kr^2\le z\le H.

The cylindrical volume element is \(r\,dz\,dr\,d\theta\). Integrating \(1\) over these bounds adds up the volume of the cap. The factor \(r\) is part of the volume element; it accounts for the change from Cartesian to cylindrical coordinates.

A Formula for the Cap’s Volume

Let \(d=H-z_0\) denote the cap’s vertical height. Since \(R^2=d/k\), the volume can be expressed using either the height and coefficient or the height and intersection radius.

Theorem: If \(k>0\) and \(H>z_0\), the volume \(V\) of the cap between \(z=z_0+kr^2\) and \(z=H\) is

V=\frac{\pi(H-z_0)^2}{2k} =\frac{\pi R^2(H-z_0)}{2}, \qquad R=\sqrt{\frac{H-z_0}{k}}.

To justify the formula, integrate the vertical thickness \(H-(z_0+kr^2)\) over each circle in the projected disk. The cylindrical bounds include all points of the cap exactly once, so this triple integral computes its volume.

Write the volume integral.
The angular coordinate covers the full circle, the radius ranges from the axis to the intersection circle, and \(z\) runs from the paraboloid to the plane.

Integrate in the vertical direction.
For each fixed \(r\) and \(\theta\), the vertical length is \(H-z_0-kr^2\).

Use the intersection equation.
Substituting \(R^2=(H-z_0)/k\) simplifies the resulting expression to the formula above.

In detail, carrying out those steps gives

\begin{aligned} V &=\int_0^{2\pi}\int_0^R\int_{z_0+kr^2}^{H} r\,dz\,dr\,d\theta\\ &=2\pi\int_0^R (H-z_0-kr^2)r\,dr\\ &=2\pi\left[\frac{(H-z_0)r^2}{2}-\frac{kr^4}{4}\right]_0^R\\ &=\pi(H-z_0)R^2-\frac{\pi kR^4}{2}. \end{aligned}

Set \(d=H-z_0\), so \(R^2=d/k\). Then the first term is \(\pi d^2/k\), while the second is \(\pi k(d/k)^2/2=\pi d^2/(2k)\). Subtracting gives \(V=\pi d^2/(2k)\). Replacing \(d/k\) by \(R^2\) also gives \(V=\pi R^2d/2\), as claimed.

This result says that the cap has half the volume of a cylinder with the same radius \(R\) and height \(d\). This comparison follows from the calculation; it is not an assumption needed to set up the integral.

Worked Examples

Worked Example: A Plane Cutting \(z=2r^2\)

Find the volume between the paraboloid \(z=2r^2\) and the plane \(z=8\). Take all coordinates in consistent length units.

Here \(z_0=0\), \(k=2\), and \(H=8\). At the intersection, \(2R^2=8\), so \(R^2=4\) and \(R=2\). Thus \(0\le\theta\le2\pi\), \(0\le r\le2\), and \(2r^2\le z\le8\).

\begin{aligned} V &=\int_0^{2\pi}\int_0^2\int_{2r^2}^{8}r\,dz\,dr\,d\theta\\ &=2\pi\int_0^2(8-2r^2)r\,dr\\ &=2\pi\left[4r^2-\frac{r^4}{2}\right]_0^2\\ &=2\pi(16-8)=16\pi. \end{aligned}

The antiderivative checks because \(\frac{d}{dr}(4r^2-r^4/2)=8r-2r^3=(8-2r^2)r\). The surface heights also agree at the edge: \(2(2)^2=8\). Finally, the formula gives \(\pi(8-0)^2/(2\cdot2)=16\pi\), matching the integral.

Worked Example: A Paraboloid with a Raised Vertex

Find the volume between \(z=1+\frac{r^2}{3}\) and the plane \(z=4\).

The vertex height is \(z_0=1\), the coefficient is \(k=\frac13\), and the cap height is \(d=4-1=3\). At the circular edge, \(1+R^2/3=4\), so \(R^2/3=3\), \(R^2=9\), and \(R=3\). The cylindrical bounds are \(0\le\theta\le2\pi\), \(0\le r\le3\), and \(1+r^2/3\le z\le4\).

\begin{aligned} V &=2\pi\int_0^3\left(4-\left(1+\frac{r^2}{3}\right)\right)r\,dr\\ &=2\pi\int_0^3\left(3-\frac{r^2}{3}\right)r\,dr\\ &=2\pi\left[\frac{3r^2}{2}-\frac{r^4}{12}\right]_0^3\\ &=2\pi\left(\frac{27}{2}-\frac{81}{12}\right)\\ &=2\pi\left(\frac{27}{2}-\frac{27}{4}\right) =2\pi\cdot\frac{27}{4} =\frac{27\pi}{2}. \end{aligned}

Indeed, \(\frac{d}{dr}(3r^2/2-r^4/12)=3r-r^3/3\), the radial integrand. At \(r=3\), the paraboloid height is \(1+9/3=4\). The formula gives \(\pi(3)^2/(2/3)=27\pi/2\), which confirms the result.

Worked Example: A Cap with an Irrational Intersection Radius

Find the volume between \(z=2+\frac{r^2}{2}\) and the plane \(z=7\).

Here \(z_0=2\), \(k=\frac12\), and \(d=7-2=5\). The intersection equation is \(2+R^2/2=7\), so \(R^2/2=5\), \(R^2=10\), and \(R=\sqrt{10}\). The radial bound is therefore \(0\le r\le\sqrt{10}\), not \(r\le10\).

\begin{aligned} V &=2\pi\int_0^{\sqrt{10}}\left(7-\left(2+\frac{r^2}{2}\right)\right)r\,dr\\ &=2\pi\int_0^{\sqrt{10}}\left(5-\frac{r^2}{2}\right)r\,dr\\ &=2\pi\left[\frac{5r^2}{2}-\frac{r^4}{8}\right]_0^{\sqrt{10}}\\ &=2\pi\left(\frac{50}{2}-\frac{100}{8}\right)\\ &=2\pi\left(25-\frac{25}{2}\right)=25\pi. \end{aligned}

The derivative of the antiderivative is \(5r-r^3/2=(5-r^2/2)r\), as required. At the upper radial bound, the lower surface has height \(2+10/2=7\), so the surfaces meet. The formula gives \(\pi(5)^2/(2\cdot\frac12)=25\pi\), independently confirming the integral.

Conditions and Common Mistakes

The assumptions matter. The calculation describes a bounded cap above an upward-opening paraboloid when \(k>0\) and \(H>z_0\). If \(H=z_0\), the intersection radius and volume are both zero. If \(H<z_0\), the plane lies below the vertex, so there is no cap of this specified form. A negative value of \(k\) describes a downward-opening paraboloid and does not satisfy the theorem’s assumptions.

A common error is to divide the height difference by \(k\) and call the result the radius. For example, with \(z=2r^2\) and \(z=18\), the incorrect calculation \(R=18/2=9\) confuses \(R\) with \(R^2\). The correct intersection equation is \(2R^2=18\), hence \(R^2=9\) and \(R=3\). The incorrect bound \(r=9\) cannot be the edge: there the paraboloid has height \(2(9)^2=162\), which is above the plane at height \(18\). The cap ends where the surfaces have equal height, at \(r=3\).

It is also important to distinguish the cap’s height \(H-z_0\) from its radius \(R\). The height is measured vertically from the vertex to the plane; the radius is measured horizontally from the axis to the intersection circle. Their relationship is \(R^2=(H-z_0)/k\), not \(R=H-z_0\).

Why the Method Is Useful

Many rotationally symmetric solids are easiest to describe by their heights at each radius. For a paraboloid cap, the vertical thickness is \(H-z_0-kr^2\), which shrinks from its maximum at the axis to zero at the edge. Cylindrical coordinates encode that changing thickness directly. The intersection equation then supplies the radial limit, ensuring that the integral covers precisely the cap and no region beyond it.

The result also provides a quick check on calculations: after finding \(R\) and the cap height \(d\), the volume must equal \(\pi R^2d/2\). If a direct integral gives a different value, inspect the intersection radius, the vertical bounds, and the cylindrical volume element.

Check Your Understanding

Use the definition and volume formula to answer these questions.

For \(z=3+r^2\) cut by \(z=12\), what are the cap height and intersection radius?
Write the cylindrical bounds for the cap between \(z=2+2r^2\) and \(z=10\).
What condition on \(H\) and \(z_0\) gives a cap with positive volume?
Why does the radius satisfy \(R^2=(H-z_0)/k\), rather than \(R=(H-z_0)/k\)?
What is the volume of a cap with intersection radius \(R\) and vertical height \(d\)?