- #1
vantz
- 13
- 0
Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
Last edited:
Integration bounds in a continuous charge distribution of a semicircle
- Thread starter vantz
- Start date
Click For Summary
Homework Help Overview
The discussion revolves around the integration bounds for a continuous charge distribution modeled as a semicircle. Participants are examining the rationale behind choosing bounds from -π/2 to π/2 instead of 0 to π.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants are questioning the choice of integration bounds and seeking clarification on the geometric interpretation of the angle with respect to the positive y-axis. Some express confusion about the reasoning behind the bounds and request proof or further explanation.
Discussion Status
The discussion is active, with participants reiterating questions about the integration bounds and seeking visual aids or diagrams to support their understanding. There is an indication that some participants are attempting to clarify the geometric setup involved.
Contextual Notes
There is a mention of a diagram that may provide insight into the angle θ and its implications for the radius vector, suggesting that visual representation is a key aspect of the discussion.
Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
- #2
gneill
Mentor
- 20,989
- 2,934
vantz said:
Because the angle is with respect to the positive y-axis.
- #3
vantz
- 13
- 0
what's the proof? I can't understand why this is the case
Thank you
Thank you
- #4
gneill
Mentor
- 20,989
- 2,934
vantz said:
Look at the diagram. Note where the angle θ is. If θ was 2π, where would that put the radius vector?