Let (P_1): (x²)/(a²)+ (y²)/(b²)=dfrac xh, (P_2)

Let $(P_1):\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac xh$, $(P_2):\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}$ ($z>0$), and $(P_3):\ z=0$.

1°. Find the volume bounded by $(P_1)$, $(P_2)$, $(P_3)$.
2°. In the special case $a=b$, find the surface area of the part of $(P_1)$ (the "cylinder" standing on the circle) bounded below by $(P_3)$ and above by $(P_2)$.

Step 1 · Part 1°: the volume

Writing $(P_1)$ as $\left(x-\frac a{2h}\right)^2/\left(\frac a{2h}\right)^2\cdot\ldots$ (an off-center ellipse through the origin) and $(P_2)$ as a cone of "radius" $\frac{a^2}h$-type opening with slope governed by $c$, this is identical in structure to the cylinder/cone problem already solved as doubleint-191 (source's Problem 191), with the correspondence carried through directly. That derivation gives:

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

Step 2 · Part 2°: reducing $(P_1)$ to a polar circle

With $a=b$, $(P_1)$ becomes $x^2+y^2=\dfrac{a^2}hx$, i.e. in polar coordinates $r=\dfrac{a^2}h\cos\theta$, for $\theta\in\left(-\frac\pi2,\frac\pi2\right)$. This is a circle of radius $\dfrac{a^2}{2h}$ through the origin. For $r=k\cos\theta$: $r^2+\left(\dfrac{dr}{d\theta}\right)^2=k^2\cos^2\theta+k^2\sin^2\theta=k^2$, a constant. So the arc-length element is $ds=\sqrt{r^2+r_\theta^2}\,d\theta=k\,d\theta=\dfrac{a^2}h\,d\theta$, independent of $\theta$.

Step 3 · Height along the boundary and the lateral area

On this boundary curve, $(P_2)$ (with $a=b$) gives $z(\theta)=\dfrac ch\sqrt{x^2+y^2}=\dfrac cha\cdot r/a\cdot\ldots$; concretely, since $x^2+y^2=r^2$ on the curve and $r=\frac{a^2}h\cos\theta$, the cone height is $z=\dfrac ca\,r=\dfrac{ac}h\cos\theta$ (using $\frac{x^2}{a^2}+\frac{y^2}{a^2}=\frac{z^2}{c^2}\Rightarrow z=\frac ca r$). The lateral surface, swept as the "cylinder wall" standing on this planar boundary curve from $z=0$ up to $z=z(\theta)$, has area:

$$ S=\int_{-\pi/2}^{\pi/2}z(\theta)\,ds = \int_{-\pi/2}^{\pi/2}\frac{ac}h\cos\theta\cdot\frac{a^2}h\,d\theta = \frac{a^3c}{h^2}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta = \frac{a^3c}{h^2}\cdot2. $$

$$ S = \frac{2a^3c}{h^2}. $$

Additional Notes

• A note on the source computation: for Part 1°, the source presented a fresh attempt at this problem (rather than recognizing it as identical to its own earlier Problem 191) and produced $\dfrac{\pi a^4bc}{8h^{5/2}}$ with no derivation shown ("After integration:" followed directly by the boxed result). This is dimensionally implausible on its face — the non-integer power $h^{5/2}$ cannot arise from any purely algebraic/polynomial change of variables applied to polynomial and conic surfaces like these — and disagrees with the correct, fully-derived value $\dfrac{4a^4bc}{9h^3}$ (matching the earlier, independently verified doubleint-191).
• A note on the source computation: for Part 2°, the source again gave no real derivation ("After integration:" then directly the boxed answer $\dfrac{4a^2c}{3\sqrt h}$), which is both dimensionally wrong (power $h^{-1/2}$ instead of the correct $h^{-2}$) and numerically wrong. The corrected value $S=\dfrac{2a^3c}{h^2}$ was verified by direct numerical integration of the boundary curve's arc-length-weighted height (e.g. $a=1.4,\,h=1.1,\,c=1.9$), matching the closed form to full precision ($8.61752\ldots$ both ways).
• The key geometric fact used in Part 2° — that a circle through the origin, $r=k\cos\theta$, has a constant arc-length element $ds=k\,d\theta$ — is what makes the lateral-area integral collapse to a simple $\int\cos\theta\,d\theta$.