# Homework Help: Half-plane in R^3

1. Sep 28, 2014

### MathewsMD

1. The problem statement, all variables and given/known data

Describe in words the surface whose equation is given by theta = pi/4

2. The attempt at a solution

It is a fairly simple question, but I'm just trying to understand why this is considered a half-plane that exists for x that is greater than or equal to 0. There are no restriction on r or z. Given this, cannot r be negative? Wouldn't this make a full-plane and let it extend for x less than 0? I just don't quite see why it cannot have the point (-1, pi/4, 0), for example, which would be present for x less than 0. Any feedback is always appreciated!

2. Sep 28, 2014

### LCKurtz

First of all, you haven't told us what coordinate system you are using. $\theta = \frac \pi 4$ makes sense in 2D polar coordinates, 3D cylindrical coordinates, and in spherical coordinates. In spherical coordinates it might be a plane or cone, depending on what convention you use for $\theta$.

But, to answer your question, you seem to understand what is going on perfectly well. Texts are inconsistent about whether or not $r<0$ is used in polar or cylindrical coordinates. If you are using the convention $r\ge 0$ you get a half plane as you say, and if $r$ is allowed to go negative you get the whole plane, as you understand. One problem with disallowing negative values of $r$ is that you don't see all 3 leaves of the rose $r = \sin(3\theta)$ for $0\le\theta\le \pi$.

I wouldn't worry too much about this if I were you since you understand it just fine.