How to Calculate an Ellipsoid’s Volume in Cylindrical Coordinates

Cylindrical coordinates are useful for an ellipsoid because its vertical extent can be described as a top surface and a bottom surface above each point in the horizontal plane. If the ellipsoid is centered at the origin and its axes align with the coordinate axes, the triple integral simplifies to the familiar volume formula.

The key detail is that when the two horizontal semiaxes are different, the boundary radius changes with the angle. The derivation below accounts for that variation.

Set up the ellipsoid

Let the ellipsoid be

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,$$

where $a$, $b$, and $c$ are positive semiaxis lengths. In cylindrical coordinates, $x=r\cos\theta$ and $y=r\sin\theta$. Substituting gives

$$\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2}+\frac{z^2}{c^2}=1.$$

Solving for $z$ gives the upper and lower surfaces:

$$z=\pm c\sqrt{1-r^2\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)}.$$

The expression under the square root must be nonnegative. Therefore, for each angle,

$$0\leq r\leq \frac{1}{\sqrt{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}}, \qquad 0\leq\theta\leq 2\pi.$$

The vertical coordinate runs from the lower surface to the upper surface. In cylindrical coordinates, the volume element is $r\,dz\,dr\,d\theta$; the factor $r$ is the Jacobian.

Write and evaluate the triple integral

Integrating first with respect to $z$ uses the ellipsoid’s full vertical height at each $(r,\theta)$:

$$V=\int_0^{2\pi}\int_0^{R(\theta)}\int_{-c\sqrt{1-A(\theta)r^2}}^{c\sqrt{1-A(\theta)r^2}} r\,dz\,dr\,d\theta,$$

where

$$A(\theta)=\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}, \qquad R(\theta)=\frac{1}{\sqrt{A(\theta)}}.$$

After integrating over $z$ and then $r$,

$$\begin{aligned} V&=\int_0^{2\pi}\int_0^{1/\sqrt{A(\theta)}}2cr\sqrt{1-A(\theta)r^2}\,dr\,d\theta\\ &=\frac{2c}{3}\int_0^{2\pi}\frac{1}{A(\theta)}\,d\theta. \end{aligned}$$

The remaining angular integral is

$$\int_0^{2\pi}\frac{d\theta}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}=2\pi ab.$$

For example, symmetry reduces it to four times the integral from $0$ to $\pi/2$. With $t=\tan\theta$, that quarter-circle integral becomes $\int_0^\infty \frac{dt}{1/a^2+t^2/b^2}=\pi ab/2$. Thus

$$V=\frac{2c}{3}(2\pi ab)=\boxed{\frac{4}{3}\pi abc}.$$

Worked example: semiaxes 3, 2, and 4

Consider

$$\frac{x^2}{9}+\frac{y^2}{4}+\frac{z^2}{16}=1.$$

Here $a=3$, $b=2$, and $c=4$. The cylindrical-coordinate radial boundary is

$$R(\theta)=\frac{1}{\sqrt{\cos^2\theta/9+\sin^2\theta/4}}.$$

It varies with $\theta$ because the horizontal cross-section is an ellipse, not a circle. Using the evaluated angular integral,

$$V=\frac{2(4)}{3}\int_0^{2\pi}\frac{d\theta}{\cos^2\theta/9+\sin^2\theta/4} =\frac{8}{3}(2\pi\cdot3\cdot2)=32\pi.$$

The volume is $32\pi$ cubic units, approximately $100.53$ cubic units.

Common mistakes

• Omitting the Jacobian: The cylindrical volume element is $r\,dz\,dr\,d\theta$, not just $dz\,dr\,d\theta$.
• Using a constant radial limit: Unless $a=b$, the radial limit depends on $\theta$. Using $r\leq a$ or $r\leq b$ for every angle does not describe the ellipsoid’s horizontal boundary.
• Integrating only the upper half: The volume integral must cover both surfaces, from negative to positive $z$. Equivalently, you can integrate the upper half and double the result.
• Confusing semiaxes with full axis lengths: In the equation, $a$, $b$, and $c$ are half-lengths. If the full axis lengths are $L_x$, $L_y$, and $L_z$, then the volume is $\pi L_xL_yL_z/6$.

When this setup applies

This derivation assumes the ellipsoid is centered at the origin and aligned with the coordinate axes. A translated but axis-aligned ellipsoid can be handled by shifting coordinates. For a rotated ellipsoid, the equation generally introduces mixed terms, so cylindrical bounds may be less convenient; a rotation of coordinates or an appropriate change of variables is often simpler.