The line x+z=1, y=C rotates about the z -axis. Find the…

The line $x+z=1,\ y=C$ rotates about the $z$-axis. Find the surface of revolution and sketch it.

Explanation of Method

Rotating a line about the z-axis: a one-sheet hyperboloid in general, a cone in the special case.

Step 1: Circles traced by the points

A point of the line is $P(1-t,\ C,\ t)$. It moves on a circle at height $z=t$ whose radius is its distance from $Oz$: $\rho^2=(1-t)^2+C^2$.

So a point $M(x,y,z)$ is on the surface if and only if

\[x^2+y^2-(z-1)^2=C^2\]

Step 2: What the surface is

$C\neq0$: the line is skew to the axis, and the surface is a hyperboloid of one sheet of revolution with center $(0,0,1)$:

\[ \frac{x^2+y^2}{C^2}-\frac{(z-1)^2}{C^2}=1 . \]

Its narrowest circle (the throat) has radius $|C|$, at $z=1$. The rotated line is one of its rulings.

$C=0$: the line meets the axis at $(0,0,1)$, and the surface is the double cone $x^2+y^2=(1-z)^2$. The vertex is $(0,0,1)$ and the half-angle is $45^\circ$.

$x+z=1,\ y=C$ $z$

For $C=1$: the line (red) is skew to $Oz$, and its rotated copies (teal) sweep the one-sheet hyperboloid $x^2+y^2-(z-1)^2=C^2$. The throat circle (blue) has radius $|C|$ at $z=1$. For $C=0$ the throat shrinks to the vertex $(0,0,1)$ and the surface becomes a cone.

Result

\[\boxed{\,x^2+y^2-(z-1)^2=C^2\,}\]

Remark

Correction: the source finds $x^2+y^2=(1-z)^2+C^2$ correctly, but then sets $C=0$ without reason and answers only with the cone. For $C\neq0$, the given line does not meet the axis and the surface is a hyperboloid of one sheet. The cone is only the special case $C=0$.