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## hosein Group Title Hi A question? 2 years ago 2 years ago Edit Question Delete Cancel Submit

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1. hosein Group Title
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can explain how this two sides are equal?$\cos \theta da=r^2d \Omega$

• 2 years ago
2. hosein Group Title
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where da is surface element in spherical coordinate & omeg is solid angle

• 2 years ago
3. JamesJ Group Title
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It looks wrong. If you integrate over the entire sphere, the right hand side is $4 \pi r^2$ but the left hand side is $8 \pi r^2$

• 2 years ago
4. hosein Group Title
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it's in classical electrodynamics (by Jackson)page28

• 2 years ago
5. TuringTest Group Title
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I think this is the same formula we are finding in figure 4.2.4 on this page http://ocw.mit.edu/courses/physics/8-02sc-physics-ii-electricity-and-magnetism-fall-2010/conductors-and-insulators-conductors-as-shields/MIT8_02SC_notes9.pdf though the area is not represented as a differential here. I trust that James' math is right though, so that must have some change in the problem.

• 2 years ago
6. quarkine Group Title
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r square* d(sigma) = r*d(theta) *r*d(phi) =projection of area element=dA cos(theta) ....in azimuthal coordinates..|dw:1334694350771:dw|

• 2 years ago
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