## Volume of an Ellipsoid

The volume of a solid can be found using the formula
$V= \int \int \int_V x \: dy \: dz$
. The equation of the surface of a surface of a ellipsoid of axes
$a, \: b, \: c$
is
$\frac{x^2}{a^2} +\frac{y^2}{b^2} +\frac{z^2}{c^2}=1$
.
Rearranging this gives
$x=a \sqrt{1- \frac{y^2}{b^2}- \frac{z^2}{c^2}}$

The volume in the first quadrant,
$\: 0 \leq x, \: y, \: z \leq a$
is one eighth the volume of the ellipsoid. The integral becomes
\begin{aligned} V &= 8 \int^c_0 \int^{ b\sqrt{1- -z^2 / c^2}}_0 a \sqrt{1- \frac{y^2}{b^2}- \frac{z^2}{c^2}} dy \: dz \\ &= 8 \int^c_0 \int^{ b\sqrt{1- -z^2 / c^2}}_0 a \sqrt{(1- \frac{z^2}{c^2})- \frac{y^2}{b^2}} dy \: dz \end{aligned}

Substitute
$y=b \sqrt{1- \frac{z^2}{c^2}} sin \theta \rightarrow dy =b \sqrt{1- \frac{z^2}{c^2}} cos \theta d \theta$

The limits become
$0, \:\frac{\pi}{2}$

\begin{aligned}V &= 8 \int^c_0 \int^{\pi / 2}_0 a \sqrt{1- \frac{z^2}{c^2}) - (1- \frac{z^2}{c^2}) sin^2 \theta} b \sqrt{1- \frac{z^2}{c^2}} cos \theta d \theta \: dz \\ &= 8ab \int^c_0 \int^{\pi / 2}_0 (1- \frac{z^2}{c^2} ) cos^2 \theta d \theta \: dz \\ &= 8ab \int^c_0 \int^{\pi / 2}_0 (1- \frac{z^2}{c^2} ) (\frac{1}{2} + cos 2 \theta) d \theta \: dz \\ &= 8ab \int^c_0 (1- \frac{z^2}{c^2} ) [ \frac{\theta}{2} + \frac{sin 2 \theta}{2}]^{\pi / 2}_0 dz \\ &= 8ab\int^c_0 (1- \frac{z^2}{c^2}) \frac{\pi}{4} dz \\ &= 8 ab\frac{\pi}{4} [ z - \frac{z^3}{3c^2}]^a_0 \\ &= 2ab \pi (c - \frac{c^3}{3c^2} ) \\ &= \frac{4 \pi abc}{3} \end{aligned}