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