Find the volume of a right elliptical cone whose base is

Find the volume of a right elliptical cone whose base is the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and whose height is $h$.

We slice the cone into elliptical cross-sections scaled linearly with height, then integrate.

Step 1: Cross-section at height $z$

With the apex at the origin and the base (the given ellipse) at height $h$, similar triangles scale the cross-section at height $z\in[0,h]$ by the factor $z/h$, giving an ellipse with semi-axes $a'=\dfrac{az}{h}$ and $b'=\dfrac{bz}{h}$.

Step 2: Area of the cross-section

\[A(z) = \pi a'b' = \pi\cdot\frac{az}{h}\cdot\frac{bz}{h} = \frac{\pi ab\,z^2}{h^2}\]

Step 3: Integrate over the height

\[V = \int_0^h\frac{\pi ab\,z^2}{h^2}\,dz = \frac{\pi ab}{h^2}\left[\frac{z^3}{3}\right]_0^h = \frac{\pi ab}{h^2}\cdot\frac{h^3}{3} = \frac{\pi abh}{3}\]

\[\boxed{V = \frac{\pi abh}{3} \text{ cubic units}}\]

Step 4: Special case — circular cone

When $a=b=r$ (a circular base): $V=\dfrac{\pi r^2h}{3}$, the familiar cone formula. ✓

Step 5: General principle

Since the base area is $A_{\text{base}}=\pi ab$, the formula is exactly $V=\frac13 A_{\text{base}}\cdot h$ — the same "one-third base times height" rule that holds for a cone over any base shape (not just circles or ellipses), as long as every cross-section is a copy of the base scaled uniformly toward the apex.

Summary

\[\boxed{V = \frac{\pi abh}{3} = \frac13\times(\text{base area})\times(\text{height})}\]

Key insight: The volume of any cone (or pyramid) depends on its base only through its area, not its specific shape — an ellipse of area $\pi ab$ gives exactly the same volume as a circle of the same area, times the same height.