Find the surface obtained by rotating y=ln x, z=0 about 1^°…

Find the surface obtained by rotating $y=\ln x,\ z=0$ about $1^\circ$ the $x$-axis; $2^\circ$ the $y$-axis.

Explanation of Method

Rotating y = ln x about the x-axis and about the y-axis.

$x$ is the axial coordinate. The radius $|\ln x|$ becomes $\sqrt{y^2+z^2}$: $y^2+z^2=\ln^2x\qquad(x\gt 0)$.

The sections $x=\text{const}$ are circles of radius $|\ln x|$. The surface pinches to a point at $x=1$.

$y$ is the axial coordinate. On the curve $x_0=e^{y}$ is the radius, and it becomes $\sqrt{x^2+z^2}$: $x^2+z^2=e^{2y}$.

The radius $e^y$ tends to $0$ as $y\to-\infty$.

$x$ $x=1$

$1^\circ$: rotating $y=\ln x$ (red) about $Ox$ gives $y^2+z^2=\ln^2x$. The circles have radius $|\ln x|$; the surface pinches at $x=1$.

$y$

$2^\circ$: rotating the same curve about $Oy$ gives $x^2+z^2=e^{2y}$; the radius $e^y$ grows exponentially.

\[\boxed{\,y^2+z^2=\ln^2x\qquad(x>0)\,}\]

\[\boxed{\,x^2+z^2=e^{2y}\,}\]