Find the surface areas P_1 and P_2 of the two parts cut
Find the surface areas $P_1$ and $P_2$ of the two parts cut from the ellipsoid $\dfrac{x^2+y^2}6+\dfrac{z^2}2=1$ by the cone $\dfrac{x^2+y^2}2=(z-\sqrt2)^2$.
Explanation of Method
The cone's apex sits exactly at the ellipsoid's north pole $(0,0,\sqrt2)$, so the two surfaces meet at that point and along one circle. Parametrizing the ellipsoid of revolution by the polar angle $\varphi$ (measured from the $z$-axis) reduces both the intersection-finding and the area computation to single-variable calculus: the surface-area element $dS=a\sqrt{a^2\cos^2\varphi+c^2\sin^2\varphi}\,\sin\varphi\,d\varphi\,d\theta$ (with $a^2=6,\,c^2=2$) integrates over $\theta\in[0,2\pi]$ trivially, leaving one radical integral in $\varphi$ (or, after $u=\cos\varphi$, in $u$) of a standard hyperbolic type.
- Find the intersection curve of the ellipsoid and cone.
- Parametrize the ellipsoid and find the surface-area element.
- Locate the $\varphi$-range of the smaller cap $P_1$.
- Evaluate $P_1$, the total ellipsoid area $P$, and $P_2=P-P_1$.
Step 1 · Intersection curve
Substituting $x^2+y^2=2(z-\sqrt2)^2$ into the ellipsoid equation and simplifying gives $5z^2-4\sqrt2\,z-2=0$, with roots $z=\sqrt2$ and $z=-\dfrac{\sqrt2}5$. At $z=\sqrt2$: $x^2+y^2=0$ (the cone's apex, a single point). At $z=-\dfrac{\sqrt2}5$: $x^2+y^2=\dfrac{144}{25}$ — a genuine circle. So the intersection is the apex point plus this circle.
Step 2 · Parametrization and surface element
With $x=\sqrt6\sin\varphi\cos\theta,\,y=\sqrt6\sin\varphi\sin\theta,\,z=\sqrt2\cos\varphi$ (so $\varphi=0$ is the north pole $z=\sqrt2$), the surface-area element for this ellipsoid of revolution is:
$$ dS = \sqrt6\sqrt{6\cos^2\varphi+2\sin^2\varphi}\,\sin\varphi\,d\varphi\,d\theta = 2\sqrt3\sqrt{2\cos^2\varphi+1}\,\sin\varphi\,d\varphi\,d\theta. $$
Step 3 · The cap's angular range
At $z=\sqrt2$: $\varphi=0$. At $z=-\dfrac{\sqrt2}5$: $\cos\varphi=-\dfrac15$, so $\varphi_0=\arccos\!\left(-\dfrac15\right)$, with $\sin\varphi_0=\dfrac{2\sqrt6}5$.
Step 4 · Evaluate $P_1$, $P$, and $P_2$
With $u=\cos\varphi$:
$$ P_1 = 4\pi\sqrt3\int_{-1/5}^1\sqrt{2u^2+1}\,du, \qquad P = 8\pi\sqrt3\int_0^1\sqrt{2u^2+1}\,du. $$
Using $\displaystyle\int\sqrt{2u^2+1}\,du=\frac{u\sqrt{2u^2+1}}2+\frac1{2\sqrt2}\sinh^{-1}(\sqrt2\,u)+C$ and simplifying the resulting logarithms:
$$ P_1 = \frac{168\pi}{25} + \pi\sqrt6\ln\frac{11+4\sqrt6}5, \qquad P = 12\pi+2\pi\sqrt6\ln\big(\sqrt2+\sqrt3\big). $$
Subtracting, using $(\sqrt2+\sqrt3)^2=5+2\sqrt6$:
$$ P_2 = P-P_1 = \frac{132\pi}{25} + \pi\sqrt6\ln\frac{25+10\sqrt6}{11+4\sqrt6}. $$
Additional Notes
- This derivation matches the source exactly, including the surface-area-element formula for an ellipsoid of revolution (verified independently by direct computation of $\|\mathbf r_\varphi\times\mathbf r_\theta\|$), the intersection-curve algebra, and the closed-form logarithmic integrals.
- Verified numerically (with $a^2=6,\,c^2=2$): $P_1\approx32.0805$, total ellipsoid area $P\approx55.3401$, and $P_2=P-P_1\approx23.2595$, all matching the boxed closed forms exactly.
Result
The two surface areas are: