# Volume of ellipsoid via iterated integral

The following is a qual study question:

For positive real numbers $R$ and $r$, let $$E(R,r) = \left\{(x_1, x_2, x_3, x_4) \in \mathbb R^4\ \Big|\ \frac{x_1^2 + x_2^2 + x_3^2}{R^2} + \frac{x_4^2}{r^2} \leq 1\right\}$$ Find the volume of $E(R,r)$ by computing an iterated integral.

I think the most straightforward way to do this is to take a variable, find the bounds in terms of the remaining variables, then project a dimension down. Repeat until you have bounds for all four integrals.

But the quadruple integral that results seems very unwieldy for calculations, especially for a timed qualifying exam. So I think there must be a slick way to do this.

I observed that if you treat $x_4$ as constant, then this becomes the equation for a sphere in 3 dimensions, with radius $R^2(1 - \frac{x_4^2}{r^2})$. My suspicion is that we can change the order of integration so that we have a triple integral w.r.t. $dx_3dx_2dx_1$ that can be replaced with $\frac43 \pi R_0^3$, the volume of a sphere with some radius $R_0$. But I'm having trouble figuring the bounds of integration for $x_4$ in that case. The few things I've tried did not agree with the formula for volume of an $n$-dimensional ellipsoid.

Tranform $x_1,x_2,x_3$ into spherical coordinates. Leave $x_4$ alone:

$x_1 = \rho\cos\theta\sin \phi\\ x_2 = \rho\sin\theta\sin \phi\\ x_3 = \rho\cos \phi\\ x_4 = x_4$

The Jacobian will be the same Jacobian for spherical. $\rho^2\sin\phi$

$\frac {\rho^2}{R^2} + \frac {x_4^2}{r^2} \le 1\\ -r\sqrt {1 - \frac {\rho^2}{R^2}}\le x_4 \le r\sqrt {1 - \frac {\rho^2}{R^2}}$

$\int_0^{2\pi}\int_{0}^{\pi}\int_{0}^R 2r\sqrt{1 - \frac {\rho^2}{R^2}} (\rho^2 \sin\phi) \ d\rho\ d\phi\ d\theta$

Since the limits of $\rho$ don't depend on $\phi,\theta$

$2r\int_0^{2\pi}d\theta\int_{0}^{\pi}\sin\phi\ d\phi\int_{0}^R \rho^2\sqrt{1 - \frac {\rho^2}{R^2}}\ d\rho$

Which we tackle with a basic trig substitution.

$\rho = R\sin u, d\rho = R\cos u\ du$

$(8r\pi)\int_{0}^{\frac {\pi}{2}} R^3 \sin^2 u\cos^2 u \ du\\ (8r\pi)\int_{0}^{\frac {\pi}{2}} R^3\frac 18(1-\cos 4u)\ du\\ (8r\pi R^3 \frac {\pi}{16}) =\frac 12 R^3r\pi^2$

There is a formula that says that the volume of a n-dimensional ball is:

$\frac {\pi^{\frac n2}}{\Gamma(\frac n2 + 1)}R^n$

And then to turn it into its equivalent ellipsoid is just applying the correct compression factor.

• Thank you, this is what I was looking for! Small comment: I think by $\int_0^\phi$, you actually mean $\int_0^\pi$? – CFish May 10 '18 at 17:17
• pi, phi... yes. thanks. – Doug M May 10 '18 at 17:39

The body $E(R,r)\subset{\mathbb R}^4$ is an ellipsoid with semiaxes $R$, $R$, $R$, and $r$. From general principles it then immediately follows that $${\rm vol}\bigl(E(R,r)\bigr)=R^3 r\,\kappa_4\ ,$$ whereby $\kappa_4={\pi^2\over2}$ denotes the volume of the $4$-dimensional unit ball. If we do not have access to $\kappa_4$ we have to compute this volume through an iterated integral.

Fubini's theorem says that $${\rm vol}\bigl(E(R,r)\bigr)=\int_{E(R,r)}1\>{\rm d}({\bf x}',t)=\int_{-r}^r\int_{S_t} 1\>{\rm d}({\bf x}')\>dt$$ whereby the points ${\bf x}'=(x_1,x_2,x_3)\in S_t$ have to satisfy $${x_1^2+x_2^2+x_3^2\over R^2}\leq 1-{t^2\over r^2}\ ,$$ hence fill a $3$-sphere of radius $R\bigl(1-t^2/r^2\bigr)^{1/2}$. It follows that $$\int_{S_t} 1\>{\rm d}({\bf x}')={4\pi R^3\over3}\left(1-{t^2\over r^2}\right)^{3/2}\ ,$$ so that $${\rm vol}\bigl(E(R,r)\bigr)={4\pi R^3\over3}\int_{-r}^r\left(1-{t^2\over r^2}\right)^{3/2}\>dt={\pi^2 R^3 r\over2}\ ,$$ whereby the computation of the last integral requires as much labor as the computation of $\kappa_4$.

• So does $x_4 \in [-r,r]$ just come from the fact that the shape is an ellipsoid, and the $x_4$ term has a denominator of $r^2$? Also, did you have a quick way of evaluating $\int_{-1}^1 (1-t^2)^{3/2} dt$? The way that I know is to use trig substitution followed by integration-by-parts, which seems overly tedious in the context of an analysis qual. – CFish May 11 '18 at 14:43