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

dxdydz=(x,y,z)(ρ,θ,φ)  dρdθdφ
.



  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=(acosφsinθ,  bsinφsinθ,  ccosθ).\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=(aρcosφcosθ,  bρsinφcosθ,  cρsinθ).\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=(aρsinφsinθ,  bρcosφsinθ,  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)=acosφsinθaρcosφcosθaρsinφsinθbsinφsinθbρsinφcosθ  bρcosφsinθccosθcρsinθ0\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)=abcρ2cosφsinθcosφcosθsinφsinθsinφsinθsinφcosθ cosφsinθcosθsinθ0.\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)=abcρ2sinθ.\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| =abc\,\rho^2\sin\theta. 


Final conversion of dV

dxdydz=abcρ2sinθ  dρdθdφ.\boxed{dx\,dy\,dz=abc\,\rho^2\sin\theta\;d\rho\,d\theta\,d\varphi.}

The volume of the ellipsoid is calculated as follows: