Calculate the surface area of the part of the cone
Calculate the surface area of the part of the cone $z^2=x^2+2y^2$ ($z>0$) that lies inside the cylinder $(x^2+y^2)^2=2c^2xy$ ($x>0$).
Explanation of Method
In polar coordinates, the cylinder is $r^2=c^2\sin2\theta$ for $\theta\in\left(0,\frac\pi2\right)$ (a single rose-type loop). Since $z=\sqrt{x^2+2y^2}$ is homogeneous of degree $1$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}$ depends only on the angle $\theta$, not on $r$, so the radial integral is elementary; what remains is a genuinely nontrivial angular integral.
- Compute the (angle-only) surface-area integrand.
- Integrate radially over $r\in\big[0,\,c\sqrt{\sin2\theta}\big]$.
- Reduce the remaining angular integral via $u=\sin^2\theta$ and evaluate it exactly.
Step 1 · Surface-area integrand
With $z=\sqrt{x^2+2y^2}$: $z_x=\dfrac x{z},\,z_y=\dfrac{2y}z$, so $1+z_x^2+z_y^2=1+\dfrac{x^2+4y^2}{x^2+2y^2}=\dfrac{2x^2+6y^2}{x^2+2y^2}$. In polar coordinates ($x=r\cos\theta,\,y=r\sin\theta$), this is angle-only:
$$ \sqrt{1+z_x^2+z_y^2}=\sqrt{\frac{2(1+2\sin^2\theta)}{1+\sin^2\theta}}. $$
Step 2 · Radial integral
$$ S=\int_0^{\pi/2}\sqrt{\frac{2(1+2\sin^2\theta)}{1+\sin^2\theta}}\left[\int_0^{c\sqrt{\sin2\theta}}r\,dr\right]d\theta = \frac{c^2}2\int_0^{\pi/2}\sqrt{\frac{2(1+2\sin^2\theta)}{1+\sin^2\theta}}\,\sin2\theta\,d\theta. $$
Step 3 · Substitute $u=\sin^2\theta$ and evaluate
With $u=\sin^2\theta,\,du=\sin2\theta\,d\theta$:
$$ S = \frac{c^2}2\int_0^1\sqrt{\frac{2(1+2u)}{1+u}}\,du. $$
This integral does not reduce to a clean radical; evaluating it in closed form (e.g. via the substitution $t=\sqrt{(1+2u)/(1+u)}$, which rationalizes it into an inverse-hyperbolic form) gives:
$$ \int_0^1\sqrt{\frac{2(1+2u)}{1+u}}\,du = 2\sqrt3-\sqrt2-\operatorname{acosh}2+\operatorname{acosh}\sqrt2. $$
Using $\operatorname{acosh}2=\ln\big(2+\sqrt3\big)$ and $\operatorname{acosh}\sqrt2=\ln\big(\sqrt2+1\big)$, this simplifies to:
$$ \int_0^1\sqrt{\frac{2(1+2u)}{1+u}}\,du = 2\sqrt3-\sqrt2+\ln\!\left(\frac{\sqrt2+1}{2+\sqrt3}\right). $$
Therefore:
$$ S = \frac{c^2}2\left[2\sqrt3-\sqrt2+\ln\!\left(\frac{\sqrt2+1}{2+\sqrt3}\right)\right]. $$
Additional Notes
- A note on the source computation: the source correctly derived Steps 1–2 above (including the same angle-only integrand and the same substitution $u=\sin^2\theta$), but then claimed the resulting $u$-integral evaluates to the clean radical expression $\dfrac{2c^2\big(2\sqrt6-3\big)}3$. This is incorrect: the integral $\displaystyle\int_0^1\sqrt{2(1+2u)/(1+u)}\,du$ does not simplify to elementary radicals — it genuinely requires an inverse-hyperbolic (equivalently, logarithmic) term, as derived above.
- Verified numerically (e.g. $c=1.3$): direct evaluation of the original polar double integral gives $1.364086662716181$, matching the corrected closed form $\frac{c^2}2\big[2\sqrt3-\sqrt2+\ln\big(\frac{\sqrt2+1}{2+\sqrt3}\big)\big]$ exactly, while the source's claimed value $\frac{2c^2(2\sqrt6-3)}3$ evaluates to $2.139516887071428$ — a large, confirmed discrepancy.
Result
The surface area is: