Monday, February 9, 2026

Volume of an ellipsoid using spherical coordinates

 Let's use an example to calculate the volume of an ellipsoid using spherical coordinates.

Example


Solution







Let's use the change of variables that corresponds to an ellipsoid. We still change the variables from rectangular coordinates to spherical coordinates. 




The volume V of the ellipsoid is given by:






Let's change dV = dxdydz in spherical coordinates:

Let's apply the change-of-variables formula

dx dy dz=∣∂(x,y,z)∂(ρ,θ,φ)∣  dρ dθ dφ
.



  Note that the vertical bars represent the determinant of the
dx\,dy\,dz=\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right|\;d\rho\,d\theta\,d\varphi.

So, the task is to compute the Jacobian:

J(ρ,θ,φ)=∂(x,y,z)∂(ρ,θ,φ).J(\rho,\theta,\varphi)=\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}.


Compute the partial derivatives

Differentiate x,y,z with respect to ρ,θ,φ\rho,\theta,\varphi.

With respect to ρ\rho:

∂(x,y,z)∂ρ=(acos⁡φsin⁡θ,  bsin⁡φsin⁡θ,  ccos⁡θ).\frac{\partial(x,y,z)}{\partial\rho} =\Big(a\cos\varphi\sin\theta,\; b\sin\varphi\sin\theta,\; c\cos\theta\Big).

With respect to θ\theta:

∂(x,y,z)∂θ=(aρcos⁡φcos⁡θ,  bρsin⁡φcos⁡θ,  −cρsin⁡θ).\frac{\partial(x,y,z)}{\partial\theta} =\Big(a\rho\cos\varphi\cos\theta,\; b\rho\sin\varphi\cos\theta,\; -c\rho\sin\theta\Big).

With respect to φ\varphi:

∂(x,y,z)∂φ=(−aρsin⁡φsin⁡θ,  bρcos⁡φsin⁡θ,  0).\frac{\partial(x,y,z)}{\partial\varphi} =\Big(-a\rho\sin\varphi\sin\theta,\; b\rho\cos\varphi\sin\theta,\; 0\Big).


 Form the Jacobian determinant

Put those three vectors as columns (or rows—just be consistent). Using columns:

∣∂(x,y,z)∂(ρ,θ,φ)∣=∣acos⁡φsin⁡θaρcos⁡φcos⁡θ−aρsin⁡φsin⁡θbsin⁡φsin⁡θbρsin⁡φcos⁡θ  bρcos⁡φsin⁡θccos⁡θ−cρsin⁡θ0∣. \left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = \left| \begin{matrix} a\cos\varphi\sin\theta & a\rho\cos\varphi\cos\theta & -a\rho\sin\varphi\sin\theta\\ b\sin\varphi\sin\theta & b\rho\sin\varphi\cos\theta & \ \ b\rho\cos\varphi\sin\theta\\ c\cos\theta & -c\rho\sin\theta & 0 \end{matrix} \right|.

 Factor constants (the key simplification)

  • Factor aa from row 1, bb from row 2, cc from row 3

  • Factor ρ\rho from column 2 and ρ\rho from column 3

So

∣∂(x,y,z)∂(ρ,θ,φ)∣=abc ρ2∣cos⁡φsin⁡θcos⁡φcos⁡θ−sin⁡φsin⁡θsin⁡φsin⁡θsin⁡φcos⁡θ cos⁡φsin⁡θcos⁡θ−sin⁡θ0∣.\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = abc\,\rho^2 \left| \begin{matrix} \cos\varphi\sin\theta & \cos\varphi\cos\theta & -\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \sin\varphi\cos\theta & \ \cos\varphi\sin\theta\\ \cos\theta & -\sin\theta & 0 \end{matrix} \right|. 

Evaluating the determinant we obtain:

∣∂(x,y,z)∂(ρ,θ,φ)∣=abc ρ2sin⁡θ.\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| =abc\,\rho^2\sin\theta. 


Final conversion of dV

dx dy dz=abc ρ2sin⁡θ  dρ dθ dφ.\boxed{dx\,dy\,dz=abc\,\rho^2\sin\theta\;d\rho\,d\theta\,d\varphi.}

The volume of the ellipsoid is calculated as follows:




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