# How can I calculate the volume of a ellipsoid, using multiple integrals?

In the original, I need to calculate this multiple integral: $$\iiint_V z \ dx \ dy \ dz$$, where $$V$$ is defined by a surfaces: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1, \quad z \ge 0.$$

It's easy to see, that the first inequality is a ellipsoid and the second one means, that we're taking only a half of it, where $$z$$ coordinate is non-negative.

I think, in this case, it's a good idea to use spherical coordinates as a substitute: $$\begin{cases} x = a \cdot r \cdot \cos(\phi) \cos(\psi) \\ y = b \cdot r \cdot \sin(\phi) \cos(\psi) \\ z = c \cdot r \cdot \sin(\psi) \end{cases}$$

The Jacobian is equal to $$a \cdot b \cdot c \cdot r^2 \cdot \cos(\psi)$$, so the integral transforms to $$\iiint_{\Omega} (c \cdot r \cdot \sin(\psi)) \cdot (a \cdot b \cdot c \cdot r^2 \cdot \cos(\psi)) \ dr \ d \phi \ d \psi$$ and here I'm stuck. How can I calculate integration limits? The only thing I can notice, that $$r \leq 1$$, just cause ellipsoid in this coordinates looks like $$r = 1$$.

• How can one intuitively explain that you multiply a, b and c into the standard spherical coordinates, beside the loose intuition, that those factors stretch the uniform sphere into the x, y and z directions? – Ramanujan Nov 12 '18 at 20:03

Hint: except for the factors $a,b,c$, it's the same integration that you would do for the volume of a hemisphere.
• Hmm. Thanks for a tip, so am I right, that $0 \leq \phi \leq 2 \Pi$? – Sergey Potekhin Dec 16 '16 at 1:08
• And $\phi$ should be greater or equal to $0$, because of $z \geq 0$, right? But how can I calculate the top limit? In case of hemisphere it's equal to $\Pi / 2$, which is pretty obvious... – Sergey Potekhin Dec 16 '16 at 1:11
• It's exactly the same as for the hemisphere. The ellipsoid is a sphere that's scaled differently in the $x$, $y$ and $z$ directions. Take a point on the sphere, multiply the $x$, $y$ and $z$ coordinates by $a$, $b$, $c$, and you have a point on the ellipsoid. $r$, $\phi$, $\psi$ are the spherical coordinates for the point on the sphere, not for the point on the ellipsoid. – Robert Israel Dec 16 '16 at 1:38
• @SergeyPotekhin: You can first make the substitution $x=au$, $y=bv$, $z=cw$. Then you get an actual hemisphere in $uvw$-space, and can easily proceed with spherical coordinates as usual. – Hans Lundmark Dec 16 '16 at 9:06