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vantz
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Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
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Integration bounds in a continuous charge distribution of a semicircle
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SUMMARY
The integration bounds for a continuous charge distribution of a semicircle are defined from -π/2 to π/2 due to the angle being measured with respect to the positive y-axis. This range effectively captures the entire semicircular arc, ensuring that all relevant contributions to the charge distribution are included. Using bounds from 0 to π would exclude half of the semicircle, leading to an incomplete analysis. The proof lies in understanding the geometric representation of the semicircle and the orientation of the angle θ.
PREREQUISITES- Understanding of polar coordinates and their application in physics
- Familiarity with continuous charge distributions
- Basic knowledge of integration techniques in calculus
- Ability to interpret geometric diagrams related to semicircles
- Study the concept of polar coordinates in depth
- Explore continuous charge distributions and their mathematical representations
- Review integration techniques, specifically in the context of physics problems
- Analyze geometric interpretations of angles in polar coordinates
Physics students, mathematicians, and engineers who are working with charge distributions and need to understand the implications of integration bounds in semicircular geometries.
Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
- #2
gneill
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vantz said:
Because the angle is with respect to the positive y-axis.
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vantz
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what's the proof? I can't understand why this is the case
Thank you
Thank you
- #4
gneill
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vantz said:
Look at the diagram. Note where the angle θ is. If θ was 2π, where would that put the radius vector?
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