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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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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.
Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?
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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?
Here is what I tried
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