Let us determine the volume of the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1.$$

Suppose $-a \leqq x \leqq a$ When we cut the ellipsoid with a plane parallel to the $yz$ plane, that is, let $x$ be constant, we get the ellipse $$\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1\!-\!\frac{x^2}{a^2},$$ i.e. $$\frac{y^2}{b^2\left(1\!-\!\frac{x^2}{a^2}\right)}+\frac{z^2}{c^2\left(1\!-\!\frac{x^2}{a^2}\right)} = 1,$$ with the semiaxes $$b_1 := b\sqrt{1\!-\!\frac{x^2}{a^2}},\quad c_1 := c\sqrt{1\!-\!\frac{x^2}{a^2}}.$$ The area of this ellipse is $\pi b_1 c_1$ (see area of plane region), and thus we have the function $$A(x) := \pi b c \left(1-\frac{x^2}{a^2}\right)$$ expressing the area cut of the ellipsoid by parallel planes. By the volume formula of the parent entry we can calculate the volume of the ellipsoid as $$V = \int_{-a}^a\!A(x)\,dx = \pi b c \int_{-a}^a\!\left(1\!-\!\frac{x^2}{a^2}\right)\,dx = \pi b c\! \sijoitus{x\,=-a}{\quad a}\left(x-\frac{x^3}{3a^2}\right) = \frac{4}{3}\pi a b c.$$ The special case $a = b = c = r$ , of a sphere is the well-known expression $\frac{4}{3}\pi r^3.$