Integration bounds in a continuous charge distribution of a semicircle

[vantz]

Why are the integration bounds from -pi/2 to pi/2 and not 0 to pi?

 

[gneill]

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.

 

[vantz]

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

Thank you

 

[gneill]

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?

 

[rwisz]

The reason for choosing integration bounds from -pi/2 to pi/2 for a continuous charge distribution of a semicircle is due to the symmetry of the problem. A semicircle has a symmetrical shape, with its center at the origin and its endpoints at (-1,0) and (1,0). This symmetry allows us to divide the semicircle into two equal halves, each with a width of pi/2. By choosing integration bounds from -pi/2 to pi/2, we are essentially integrating over one half of the semicircle and then multiplying the result by 2 to get the total charge.

On the other hand, choosing integration bounds from 0 to pi would result in integrating over the entire semicircle, which would lead to double counting of the charge. This is because the area under the curve from 0 to pi is equal to the area from pi to 2pi, which is essentially the same area as the first half of the semicircle. Therefore, by choosing integration bounds from -pi/2 to pi/2, we are able to avoid this double counting and obtain an accurate result for the total charge in the semicircle.

In summary, the choice of integration bounds from -pi/2 to pi/2 for a continuous charge distribution of a semicircle is based on the symmetry of the problem, which allows us to accurately calculate the total charge without double counting.