Find the surface area of the part of the cone x²+y²=z²
Find the surface area of the part of the cone $x^2+y^2=z^2$ ($z>0$) cut out by the cylinder $(x^2+y^2)^2=2xy$.
Explanation of Method
Like every cone $z=\sqrt{x^2+y^2}$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}$ is the constant $\sqrt2$, so the area is simply $\sqrt2$ times the base region's area, which is a two-lobed lemniscate $r^2=\sin2\theta$ in polar coordinates.
- Compute the (constant) surface-area integrand.
- Convert the cylinder to polar form and identify the two lobes.
- Compute the base area and multiply by $\sqrt2$.
Step 1 · Surface-area integrand
With $z=\sqrt{x^2+y^2}$: $1+z_x^2+z_y^2=2$, so $dS=\sqrt2\,dx\,dy$, and $S=\sqrt2\cdot A(D)$.
Step 2 · The cylinder in polar form
$(x^2+y^2)^2=2xy\iff r^4=r^2\sin2\theta\iff r^2=\sin2\theta$ (for $r\ne0$), real for $\theta\in\left[0,\frac\pi2\right]\cup\left[\pi,\frac{3\pi}2\right]$ — two symmetric lobes.
Step 3 · Base area and result
For one lobe:
$$ A_{\text{one lobe}} = \frac12\int_0^{\pi/2}\sin2\theta\,d\theta = \frac12\left[-\frac{\cos2\theta}2\right]_0^{\pi/2} = \frac12. $$
Both lobes: $A(D)=2\cdot\frac12=1$. Therefore:
$$ S = \sqrt2\cdot1 = \sqrt2. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct polar computation of the base area gives exactly $1$, confirming $S=\sqrt2$.
Result
The surface area is: