How to Calculate an Ellipsoid’s Volume with a Triple Integral

A triple integral calculates an ellipsoid’s volume by integrating 1 over the solid. For an ellipsoid centered at the origin with its axes aligned to the coordinate axes, scaled spherical coordinates make the region and limits especially simple.

The result depends on the ellipsoid’s three semiaxes: if their lengths are $a$, $b$, and $c$, then the volume is $\frac{4}{3}\pi abc$. The derivation below shows how this formula comes directly from a triple integral.

Describe the ellipsoid

An axis-aligned ellipsoid centered at the origin has equation

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

where $a$, $b$, and $c$ are positive. They are the distances from the center to the surface along the $x$-, $y$-, and $z$-axes. If a problem gives the full axis lengths, divide each by 2 to get these semiaxes.

The volume is the integral of 1 over the region $E$:

$$V=\iiint_E 1\,dV.$$

Change variables to a unit ball

Use spherical coordinates for a unit ball, then scale each coordinate by the corresponding semiaxis:

$$\begin{aligned} x&=a\rho\sin\phi\cos\theta,\\ y&=b\rho\sin\phi\sin\theta,\\ z&=c\rho\cos\phi. \end{aligned}$$

Here $0\leq\rho\leq1$, $0\leq\phi\leq\pi$, and $0\leq\theta\leq2\pi$. Substituting into the ellipsoid equation gives

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

so the ellipsoid corresponds exactly to the unit ball, $0\leq\rho\leq1$.

A change of variables also changes the volume element. The absolute value of the Jacobian determinant for this transformation is

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}\right|=abc\rho^2\sin\phi.$$

Thus $dV=abc\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. The factor $abc$ accounts for stretching the unit ball by different amounts in the three coordinate directions.

Evaluate the triple integral

Substitute the transformed volume element and the spherical-coordinate limits into the integral:

$$\begin{aligned} V&=\int_0^{2\pi}\int_0^\pi\int_0^1 abc\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=abc\left(\int_0^1\rho^2\,d\rho\right) \left(\int_0^\pi\sin\phi\,d\phi\right) \left(\int_0^{2\pi}d\theta\right)\\ &=abc\left(\frac13\right)(2)(2\pi)\\ &=\frac{4}{3}\pi abc. \end{aligned}$$

This is the volume formula for an ellipsoid. It is the volume of a unit sphere, $4\pi/3$, multiplied by the three scale factors $a$, $b$, and $c$.

Worked example: semiaxes 2, 3, and 4

Suppose the ellipsoid is

$$\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}\leq1.$$

Its semiaxes are $a=2$, $b=3$, and $c=4$. Therefore the triple integral evaluates to

$$V=\frac{4}{3}\pi(2)(3)(4)=32\pi\approx100.53.$$

The volume is $32\pi$ cubic units, or approximately $100.53$ cubic units. The exact value is usually preferable unless a decimal approximation is requested.

Cartesian limits as an alternative

You can also write the volume integral directly in Cartesian coordinates. For the same general ellipsoid, one valid setup is

$$\int_{-a}^{a} \int_{-b\sqrt{1-x^2/a^2}}^{b\sqrt{1-x^2/a^2}} \int_{-c\sqrt{1-x^2/a^2-y^2/b^2}}^{c\sqrt{1-x^2/a^2-y^2/b^2}} 1\,dz\,dy\,dx.$$

These nested limits describe the changing cross-sections of the ellipsoid. This setup is correct, but evaluating it requires more involved limits and integration. Scaling spherical coordinates reduces the region to a unit ball and is usually the cleaner triple-integral method.

Common mistakes

• Using full axis lengths as semiaxes: If the ellipsoid is 6 units wide along an axis, the corresponding semiaxis is 3, not 6.
• Omitting the Jacobian: Coordinate substitutions change the volume element. In this case the required factor is $abc\rho^2\sin\phi$.
• Using spherical limits for the original coordinates without scaling: The ellipsoid becomes a sphere only after dividing the coordinates by their respective semiaxes.
• Mixing up the angles: With the convention used here, $\phi$ runs from 0 to $\pi$ and measures from the positive $z$-axis; $\theta$ runs from 0 to $2\pi$ around that axis.

A translation or rotation of the ellipsoid does not change its volume. For an ellipsoid with different orientation, its three semiaxis lengths still determine the volume through $\frac{4}{3}\pi abc$.