How do i plot $(r−2)^2 + z^2 \leq 1$?

Question

How can I plot this equation in 3D? Given in cylindrical coordinates. Have tried in wolfram but couldn't work it out..

$$(r−2)^2 + z^2 \leq1$$

Could you specify your question more clearly? As it's formulated, you have the equation of a circle with Center at (2,0) and radius 1. — Luke, Mar 04 '20 at 15:00
He is working in cylindrical coordinates, not cartesian ones. Plus, it would be a disk rather than a circle. @Luke — nicomezi, Mar 04 '20 at 15:02
Whoops, overlooked that, sorry. Still it nearly solves his problem. — Luke, Mar 04 '20 at 15:05
It depends on what you consider to be near the solution I would say, to me it is the very beginning. ;) @Luke — nicomezi, Mar 04 '20 at 15:07

score 0 · Answer 1 · answered Mar 04 '20 at 15:05

0

You draw a solid disk with radius $1$ at $(x,y,z)=(2,0,0)$ then rotate it a full $2\pi$ radian with respect to $z$ axis.

A solid donut

answered Mar 04 '20 at 15:05

acat3

Mohammad Riazi-Kermani · Answer 2 · 2020-03-04T16:40:15.760

0

Parametrize the surface of $$(r−2)^2 + z^2 \leq1$$ as $$r=2\pm \cos(\theta), z=\pm \sin(\theta)$$

for $$0\le \theta \le 2\pi$$

You get a donut shaped solid with the inner radius of one and outer radius of three where $z$ runs from $-1$ to $1$.

edited Mar 04 '20 at 16:40

answered Mar 04 '20 at 15:26

2 Answers2