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


Since the ellipsoid centered on the origin is symmetrical around all three coordinate planes, work with integrating half of it and double that result. If you want to work along the x-axis, you'll need to express the elliptical cross-sections parallel to the yz-plane in terms of position along the x-axis. The equation for each ellipse is

$$\frac{y^2}{b'^2} + \frac{z^2}{c'^2} = 1 - \frac{x^2}{a^2}.$$

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.

  • $\begingroup$ I think this is right. I'll try working it out. $\endgroup$ – Neo Apr 19 '13 at 7:42

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