Let (a, b, c > 0). Determine the volume $ V = \iiint\limits_{V} dV$ of the region defined by the inequalities $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1$, $z \geq x$, and $x \geq 0$. When I tried to solve this one I thought about using :

$ x = a r \sin \theta \cos \phi, $

$ y = b r \sin \theta \sin \phi, $

$ z = c r \cos \theta. $

I found that $1 \geq r \geq 0$, $ar$ $\sin \theta \cos \phi \geq 0 $ knowing $\theta$ is bounded: $ \pi \geq \theta \geq 0 $ and so $\phi$ has to be between: $\frac{\pi}{2} \geq \phi \geq \frac{\pi}{2}$. Knowing $ z \geq x $ I found $ \frac{c}{a} \cot(\theta) \geq \cos \phi$ which all together gave me: $ \arccos \frac{a}{c} \geq \theta \geq 0 $, $\arccos(\frac{c}{a} \cot \theta) \geq \phi \geq -\arccos(\frac{c}{a} \cot \theta)$. The problem is that I don't think this integral is solvable with my transformation and I can't figure out what to use.

  • $\begingroup$ You should write down which integral you mean! As far as I can see your approach would work very well. $\endgroup$ Commented May 21 at 22:47
  • $\begingroup$ I suppose it's just V = \iiint\limits_{V} dV $\endgroup$
    – IlllIlII
    Commented May 21 at 23:07
  • $\begingroup$ I would say it is always useful to first put elliptical thing back to circular thing, then apply the techniques you know on circular object (the Jacobian can be computed by the composition of transformation). $\endgroup$
    – Angae MT
    Commented May 22 at 3:33

1 Answer 1


We don't really need to evaluate a triple integral. A suitable scaling transformation $$(u,v,w) = (ax, by, cz)$$ transforms the ellipsoid to a unit sphere, and the plane $z = x$ is transformed to $w/c = u/a$. The plane $x = 0$ becomes $u = 0$. So we have a unit sphere cut into four sections by the planes $u = 0$ and $u = (a/c)w$, and we want the angle $\alpha$ between these two planes such that the region $0 \le u \le (a/c)w$. This angle is simply $\alpha = \arctan \frac{c}{a}$, and because of symmetry, the desired volume in $(u,v,w)$-space is $\frac{\alpha}{2\pi}$ of the total volume of the unit sphere; i.e., $$\frac{4}{3} \pi \cdot \frac{1}{2\pi} \arctan \frac{c}{a} = \frac{2}{3} \arctan \frac{c}{a}.$$ Then in the original $(x,y,z)$ coordinate system, the volume is $abc$ times this, or $$V = \frac{2abc}{3} \arctan \frac{c}{a}.$$


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