Calculate the surface area of the part of the cone
Calculate the surface area of the part of the cone $z^2=x^2+y^2$ cut by the cylinder $x^2+y^2=2x$.
Explanation of Method
For the cone $z=\sqrt{x^2+y^2}$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}$ is the constant $\sqrt2$ — a hallmark of cones, whose surface element is a fixed multiple of the projected area. So the area over any base region is simply $\sqrt2$ times that region's area; here the base region is the disk $(x-1)^2+y^2\le1$ (radius $1$), and both nappes of the cone are included.
- Compute the (constant) surface-area integrand.
- Identify the base region as a disk of radius $1$.
- Evaluate and double for both nappes.
Step 1 · Surface-area integrand
With $z=\sqrt{x^2+y^2}$: $z_x=\dfrac x{\sqrt{x^2+y^2}},\,z_y=\dfrac y{\sqrt{x^2+y^2}}$, so $1+z_x^2+z_y^2=1+1=2$, giving the constant integrand $\sqrt2$.
Step 2 · Base region
$x^2+y^2=2x \iff (x-1)^2+y^2=1$, a disk of radius $1$ centered at $(1,0)$, of area $\pi$.
Step 3 · Evaluate
$$ S_{\text{upper}} = \sqrt2\cdot\pi = \pi\sqrt2. $$
Doubling for both nappes of the cone ($z=\pm\sqrt{x^2+y^2}$):
$$ S = 2\pi\sqrt2. $$
Additional Notes
- This derivation matches the source exactly. The constant integrand $\sqrt2$ reflects the cone's fixed $45^\circ$ half-angle: every unit of projected base area corresponds to $\sqrt2$ units of cone surface area.
- Verified numerically: direct double integration of the constant integrand $\sqrt2$ over the disk gives $\pi\sqrt2$ per nappe, matching $2\pi\sqrt2$ for the total.
Result
The surface area is: