# Using Cavalieri's Principle to find the volume of an ellipsoid

I understand how to use the triple integral + change of variable method to find the volume of an ellipsoid, but given an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$$ whose area is$$A= \pi ab$$

I want to use cavalieri's principle $$v(A) = \int_a^b A(t)dt$$

to show that the volume of ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$$ is $\frac{4}{3}\pi abc$

I think in need to find an area of the ellipse generated from the intersection of the original ellipse and the plane x=t, but im pretty lost from there.

Any help, hints and solutions are appreciated

$$\frac{y^2}{b'^2} + \frac{z^2}{c'^2} = 1 - \frac{x^2}{a^2}.$$
The ellipse at $x = 0$ has semi-axes $b$ and $c$ (and so has area $\pi a b$), while others as you progress along the x-axis will have smaller dimensions. Find the semi-axes $b'$ and $c'$ as a function of $x$, and you'll have the area of each "slice". (Where does the "slice" shrink to zero area?) Integrating these areas will give you half of the ellipsoid's volume.