Integration bounds in a continuous charge distribution of a semicircle

AI Thread Summary
The integration bounds for a continuous charge distribution of a semicircle are from -pi/2 to pi/2 because the angle is measured from the positive y-axis. This ensures that the entire semicircle is covered in the integration process. The discussion emphasizes the importance of understanding the geometric representation of the angle θ in relation to the radius vector. A diagram illustrating this relationship can clarify why the bounds are set as they are. Understanding these bounds is crucial for correctly applying integration in this context.
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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vantz said:
Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?

Because the angle is with respect to the positive y-axis.
 
what's the proof? I can't understand why this is the case

Thank you
 
vantz said:
what's the proof? I can't understand why this is the case

Thank you

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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