Finding a volume integral in an ellipsoid

Question

I am trying to find the volume integral of $\rho=\rho_{0}\left(\frac{R^{2}-r^{2}}{R^{2}}\right)$ inside an ellipsoid given by $\frac{x^{2}}{(3 R)^{2}}+\frac{y^{2}}{(4 R)^{2}}+\frac{z^{2}}{(5 R)^{2}}=1$

I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships

$x=3 R u$, $y=4 R v$, $z=5 R w$

But the resulting integral is still heavy

$\int\rho_0\left(1-\left(9 u^{2}+16 w^{2}+25 w^{2}\right)\right) 60 dudvdw$

Does anyone have any insight to a more elegant way.

Thank you

Answer 1

The resulting integral in this case is actually very simple due to symmetry: \begin{align} \int_Bu^2\,dV = \int_Bv^2\,dV = \int_Bw^2\,dV=\frac{1}{3}\int_B(u^2+v^2+w^2)\,dV = \frac{1}{3}\int_0^1 r^2\,4\pi r^2\,dr =\frac{4\pi}{15}. \end{align}

This symmetry should be intuitive enough. Otherwise use the change of variables theorem to permute $(u,v,w)$ and the fact that the unit ball remains invariant under this change of coordinates, and that the absolute value of the determinant of this is $1$ if you want a really rigorous proof.

Answer 2

Your last integral is not that heavy.

The constant $1$ yields the volume of the unit sphere. Then using spherical coordinates, the integral of $w^2$ is

$$\int_0^{2\pi}\int_0^\pi\int_0^1 r^2\cos^2\phi\,r^2\sin\phi\,dr\,d\phi\,d\theta=2\pi\frac23\frac15.$$

By symmetry, the integrals of $u^2$ and $v^2$ are equal.