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Integration bounds in a continuous charge distribution of a semicircle

  • Thread starter vantz
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  • #1
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Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
 
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  • #2
gneill
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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.
 
  • #3
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what's the proof? I can't understand why this is the case

Thank you
 
  • #4
gneill
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20,793
2,773
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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