Volumes Under Paraboloids Over Disks

Introduction

A downward-opening paraboloid can form the upper surface of a bowl-shaped solid. If the solid rests on a circular disk in the \(xy\)-plane, its volume depends on two features: the height of the paraboloid at each radius and how far the disk extends from the vertical axis.

Cylindrical coordinates make both features direct to express. The surface height depends only on the radial coordinate \(r\), and the disk becomes the simple radial range from \(0\) to its radius \(R\). We will use these bounds to set up a triple integral, derive a general volume formula, and check when that setup is valid.

The Solid and Its Conditions

Let \(a>0\) and \(b>0\), and consider the paraboloid \(z=a-br^2\), where \(r^2=x^2+y^2\). Let the base be the disk of radius \(R>0\), centered at the origin in the \(xy\)-plane.

Definition: The solid beneath \(z=a-br^2\) and above the disk of radius \(R\) is \(E=\{(r,\theta,z):0\leq r\leq R,\ 0\leq\theta\leq2\pi,\ 0\leq z\leq a-br^2\}\), provided \(a-br^2\geq0\) throughout the disk.

Because \(b>0\), the height \(a-br^2\) decreases as \(r\) increases. Its smallest value on the disk is therefore at the outer edge, where \(r=R\). The required condition is

a-bR^2\geq 0, \qquad\text{or equivalently}\qquad R\leq\sqrt{\frac{a}{b}}.

If this condition holds, every vertical segment in the solid starts at \(z=0\) and ends at the paraboloid. When equality holds, the surface meets the base at the disk’s edge. If \(R>\sqrt{a/b}\), part of the graph lies below the base plane, so these bounds do not describe a solid beneath the graph and above the entire disk.

Setting Up the Cylindrical Integral

In cylindrical coordinates, the volume element is \(r\,dz\,dr\,d\theta\). The factor \(r\) is essential: it accounts for how cylindrical coordinates scale area as the distance from the axis changes.

For each angle \(0\leq\theta\leq2\pi\), the radius runs from the axis to the disk edge, \(0\leq r\leq R\). At each such point, the vertical coordinate runs from the base \(z=0\) to the surface \(z=a-br^2\). Thus the volume is

V=\int_0^{2\pi}\int_0^R\int_0^{a-br^2} r\,dz\,dr\,d\theta.

Integrating first with respect to \(z\) gives the height of the solid at each radius, multiplied by the cylindrical factor \(r\). The full angular range contributes a factor of \(2\pi\).

\begin{aligned} V &=\int_0^{2\pi}\int_0^R (a-br^2)r\,dr\,d\theta\\ &=2\pi\int_0^R (a-br^2)r\,dr. \end{aligned}

Deriving the Volume Formula

The radial integrand expands as \((a-br^2)r=ar-br^3\). Integrating its two terms separately yields a formula that applies whenever \(a>0\), \(b>0\), and \(0<R\leq\sqrt{a/b}\).

Integrate the height over the disk.
The vertical integral gives \(a-br^2\), so the remaining volume integral is \(2\pi\int_0^R(ar-br^3)\,dr\).

Find an antiderivative.
An antiderivative of \(ar-br^3\) is \(\frac{a}{2}r^2-\frac{b}{4}r^4\), since its derivative is \(ar-br^3\).

Evaluate at the radial limits.
Substitution of \(R\) and \(0\), followed by multiplication by \(2\pi\), gives the volume.

Carrying out the last step gives

\begin{aligned} V &=2\pi\left[\frac{a}{2}r^2-\frac{b}{4}r^4\right]_0^R\\ &=2\pi\left(\frac{a}{2}R^2-\frac{b}{4}R^4\right)\\ &=\pi aR^2-\frac{\pi b}{2}R^4. \end{aligned}

Volume formula: Beneath \(z=a-br^2\) and above a disk of radius \(R\), the volume is \(V=\pi aR^2-\frac{\pi b}{2}R^4\), provided \(a>0\), \(b>0\), and \(R\leq\sqrt{a/b}\).

This derivation also explains why the radial boundary matters. The first term, \(\pi aR^2\), is the volume of a cylinder with base radius \(R\) and height \(a\). The second term subtracts the volume associated with the downward decrease \(br^2\) in the paraboloid’s height. This is an interpretation of the formula; the triple integral supplies its justification.

Worked Examples

Worked Example: A Paraboloid Above a Smaller Disk

Find the volume beneath \(z=12-3r^2\) and above the disk \(0\leq r\leq1\).

First check the height at the disk edge: \(12-3(1)^2=9\), which is positive. The height decreases with \(r\), so it is nonnegative across the disk. The cylindrical integral is

\begin{aligned} V &=\int_0^{2\pi}\int_0^1\int_0^{12-3r^2}r\,dz\,dr\,d\theta\\ &=2\pi\int_0^1(12-3r^2)r\,dr\\ &=2\pi\int_0^1(12r-3r^3)\,dr\\ &=2\pi\left[6r^2-\frac{3}{4}r^4\right]_0^1\\ &=2\pi\left(6-\frac{3}{4}\right)\\ &=\frac{21\pi}{2}. \end{aligned}

For a check, the derivative of \(6r^2-\frac{3}{4}r^4\) is \(12r-3r^3\), the radial integrand. Also, the formula gives \(\pi(12)(1)^2-\frac{\pi(3)}{2}(1)^4=12\pi-\frac{3\pi}{2}=\frac{21\pi}{2}\), agreeing with the integral.

Worked Example: A Disk Reaching the Paraboloid’s Edge

Find the volume beneath \(z=8-2r^2\) and above the disk of radius \(2\).

The height at the edge is \(8-2(2)^2=8-8=0\). It is positive for \(0\leq r<2\), so the whole disk satisfies the required condition. Integrating gives

\begin{aligned} V &=\int_0^{2\pi}\int_0^2\int_0^{8-2r^2}r\,dz\,dr\,d\theta\\ &=2\pi\int_0^2(8r-2r^3)\,dr\\ &=2\pi\left[4r^2-\frac{1}{2}r^4\right]_0^2\\ &=2\pi\left(16-8\right)\\ &=16\pi. \end{aligned}

To check the integration, the derivative of \(4r^2-\frac{1}{2}r^4\) is \(8r-2r^3\), as required. Substituting into the formula gives \(\pi(8)(2)^2-\frac{\pi(2)}{2}(2)^4=32\pi-16\pi=16\pi\). The zero height at the edge does not change the volume integral.

Worked Example: A Disk with an Exact Noninteger Radius

Find the volume beneath \(z=10-2r^2\) and above the disk of radius \(\sqrt{3}\).

At the edge, the height is \(10-2(\sqrt{3})^2=10-6=4\), so it is positive throughout the disk. The integral becomes

\begin{aligned} V &=2\pi\int_0^{\sqrt{3}}(10-2r^2)r\,dr\\ &=2\pi\int_0^{\sqrt{3}}(10r-2r^3)\,dr\\ &=2\pi\left[5r^2-\frac{1}{2}r^4\right]_0^{\sqrt{3}}\\ &=2\pi\left(5(3)-\frac{1}{2}(9)\right)\\ &=2\pi\left(15-\frac{9}{2}\right)\\ &=21\pi. \end{aligned}

The antiderivative checks because \(\frac{d}{dr}(5r^2-\frac{1}{2}r^4)=10r-2r^3\). The general formula gives \(\pi(10)(3)-\frac{\pi(2)}{2}(9)=30\pi-9\pi=21\pi\), confirming the result.

A Common Mistake: Omitting the Factor \(r\)

A frequent error is to integrate the height over \(r\) and \(\theta\) without including the cylindrical volume factor. For the first example, that incorrect setup would be

\int_0^{2\pi}\int_0^1(12-3r^2)\,dr\,d\theta =2\pi\left[12r-r^3\right]_0^1 =22\pi.

This is not the volume: the correct answer is \(\frac{21\pi}{2}\). The incorrect integral treats equal changes in radius as if they always represented equal areas. In fact, a thin cylindrical shell at radius \(r\) has volume proportional to \(r\), which is why the volume element contains \(r\). Restoring that factor gives \(2\pi\int_0^1(12-3r^2)r\,dr\), the integral that evaluates to \(\frac{21\pi}{2}\).

Why the Setup Matters

The formula is useful only after the solid has been described correctly. The radial bound comes from the edge of the base disk, not from the point where the paraboloid reaches the base plane. Those radii coincide only when the disk is chosen to reach the paraboloid’s edge. Likewise, the upper \(z\)-bound must be the actual height \(a-br^2\), not the constant central height \(a\).

Before evaluating, check the solid in this order:

Identify the base disk and its radius \(R\).
Write the surface height as a function of \(r\).
Check that \(a-bR^2\geq0\), so the height is nonnegative across the disk.
Use \(0\leq\theta\leq2\pi\), \(0\leq r\leq R\), and \(0\leq z\leq a-br^2\).
Include the factor \(r\) in the volume element.

These checks separate the geometry of the solid from the integration. Once the bounds match the described region, evaluating the integral is systematic.

Check Your Understanding

Use the bounds and formula developed above to answer these questions.

For \(z=9-r^2\) above a disk of radius \(2\), is the height nonnegative throughout the disk? State the edge height.
Write the cylindrical triple integral for the volume beneath \(z=15-5r^2\) above a disk of radius \(1\).
Derive the volume formula by integrating \(2\pi(ar-br^3)\) from \(0\) to \(R\).
What factor must appear in the cylindrical volume element, and what geometric feature does it account for?
For \(z=6-2r^2\) over a disk of radius \(2\), explain why the stated bounds do or do not describe a solid above the base plane.