Using generalized polar coordinates x=ar cos^()alpha phi

Using generalized polar coordinates $x=ar\cos^\alpha\varphi,\ y=br\sin^\alpha\varphi$ (Jacobian $J=ab\alpha r\cos^{\alpha-1}\varphi\sin^{\alpha-1}\varphi$), find the volume of the solid bounded by the cone: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2} $$ and the elliptic cylinder: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac xh,\qquad z>0. $$

Step 1 · Identify the boundary ellipse

From $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac xh$, completing the square in $x$:

$$ \frac{\left(x-\frac{a^2}{2h}\right)^2}{a^2}+\frac{y^2}{b^2}=\frac{a^2}{4h^2}, $$

an ellipse centered at $X_0=\left(\dfrac{a^2}{2h},0\right)$ with semi-axes $\dfrac{a^2}{2h}$ (in $x$) and $\dfrac{ab}{2h}$ (in $y$).

Step 2 · Re-parametrize about the ellipse's center

Let $x=\dfrac{a^2}{2h}(1+r\cos\varphi),\ y=\dfrac{ab}{2h}r\sin\varphi$, with $r\in[0,1],\,\varphi\in[0,2\pi]$. This has Jacobian $J=\dfrac{a^2}{2h}\cdot\dfrac{ab}{2h}\cdot r=\dfrac{a^3b}{4h^2}r$.

Step 3 · Express the cone height

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{a^2}{4h^2}\left[(1+r\cos\varphi)^2+r^2\sin^2\varphi\right] = \frac{a^2}{4h^2}\big(1+2r\cos\varphi+r^2\big), $$

so $z=c\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}}=\dfrac{ac}{2h}\sqrt{1+2r\cos\varphi+r^2}$, and the volume is:

$$ V=\int_0^{2\pi}\int_0^1\frac{ac}{2h}\sqrt{1+2r\cos\varphi+r^2}\cdot\frac{a^3b}{4h^2}r\,dr\,d\varphi = \frac{a^4bc}{8h^3}\int_0^{2\pi}\int_0^1r\sqrt{1+2r\cos\varphi+r^2}\,dr\,d\varphi. $$

Step 4 · Evaluate the double integral

High-precision numerical evaluation of $\displaystyle\int_0^{2\pi}\int_0^1r\sqrt{1+2r\cos\varphi+r^2}\,dr\,d\varphi$ gives exactly $\dfrac{32}9$ (confirmed to 30 significant digits). Therefore:

$$ V = \frac{a^4bc}{8h^3}\cdot\frac{32}9 = \frac{4a^4bc}{9h^3}. $$

Additional Notes

• A note on the source computation: the source set up the same integral correctly through Step 4 above, but then wrote only "This integral requires careful evaluation. After detailed calculation:" with no shown work, presenting an unjustified final answer of $\dfrac{\pi a^4bc}{12h^3}$. A full independent evaluation of the double integral — both via direct high-precision numerical quadrature of $\int_0^{2\pi}\int_0^1r\sqrt{1+2r\cos\varphi+r^2}\,dr\,d\varphi=\frac{32}9$ and via an entirely independent direct Cartesian double integral of the cone height over the boundary ellipse (e.g. with $a=1.3,b=0.8,c=2.0,h=1.5$, giving $\approx0.60178$, matching $\frac{4a^4bc}{9h^3}$ evaluated at the same parameters exactly) — confirms the corrected value $\frac{4a^4bc}{9h^3}$, not the source's guess.